The Flat Earth Society

Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: cikljamas on October 15, 2014, 04:39:43 AM

Title: "Equator" problem
Post by: cikljamas on October 15, 2014, 04:39:43 AM
Hi guys! Since this is my first post at this forum, maybe the best possible way to present myself to you is this: http://energeticforum.com/renewable-energy/17050-north-south.html (http://energeticforum.com/renewable-energy/17050-north-south.html)

As you can see from above link i am one of the most convinced flat earthers in the world. However, i had become a flat earther very recently, that is why i think no one should be too much surprised with the fact that i (as one of the most convinced FE in the world) am opening this thread with such (elementary) following question:

The measurement of the diameter of the earth, and the measurement for circumference of the equator do not agree. Why?

Sorry for my broken english!
Title: Re: "Equator" problem
Post by: inquisitive on October 15, 2014, 05:38:02 AM
Hi guys! Since this is my first post at this forum, maybe the best possible way to present myself to you is this: http://energeticforum.com/renewable-energy/17050-north-south.html (http://energeticforum.com/renewable-energy/17050-north-south.html)

As you can see from above link i am one of the most convinced flat earther in the world. However, i had become a flat earther very recently, that is why i think no one should be too much surprised with the fact that i (as one of the most convinced FE in the world) am opening this thread with such (elementary) following question:

The measurement of the diameter of the earth, and the measurement for circumference of the equator do not agree. Why?

Sorry for my broken english!
Numbers please.
Title: Re: "Equator" problem
Post by: cikljamas on October 15, 2014, 06:08:11 AM
Numbers?

I can demonstrate it with a following example:

Paris (France) - Accra (Ghana) distance = 2986 miles (4777 km)
Accra - Equator distance = 381 miles (609 km)
Paris - North Pole distance = 2846 miles (4553 km)

2986 + 381 + 2846 = 6213 miles (9940 km)

So, North Pole - Equator distance = 6213 miles

6213 * 2 * 3,14 = 39017 miles (62428 km) This number should represent circumference of the Equator on Flat Earth...

But as we all know (and i doubt that Round Earthers could easily falsify that number) the Equator's circumference is equal to 25 000 miles (40 000 km)...

I believe that this "Equator" problem could have been the main reason for strange kind of FE shaping as it is presented in this map : http://www.thehistoryblog.com/archives/11651: (http://www.thehistoryblog.com/archives/11651:) http://www.thehistoryblog.com/archives/11651 (http://www.thehistoryblog.com/archives/11651)
Title: Re: "Equator" problem
Post by: markjo on October 15, 2014, 06:27:17 AM
Numbers?

I can demonstrate it with a following example:

Paris (France) - Accra (Ghana) distance = 2986 miles (4777 km)
Accra - Equator distance = 381 miles (609 km)
Paris - North Pole distance = 2846 miles (4553 km)
Who measured those distances?
Title: Re: "Equator" problem
Post by: BJ1234 on October 15, 2014, 06:45:13 AM
Numbers?

I can demonstrate it with a following example:

Paris (France) - Accra (Ghana) distance = 2986 miles (4777 km)
Accra - Equator distance = 381 miles (609 km)
Paris - North Pole distance = 2846 miles (4553 km)

2986 + 381 + 2846 = 6213 miles (9940 km)

So, North Pole - Equator distance = 6213 miles

6213 * 2 * 3,14 = 39017 miles (62428 km) This number should represent circumference of the Equator on Flat Earth...

But as we all know (and i doubt that Round Earthers could easily falsify that number) the Equator's circumference is equal to 25 000 miles (40 000 km)...

I believe that this "Equator" problem could have been the main reason for strange kind of FE shaping as it is presented in this map : http://www.thehistoryblog.com/archives/11651: (http://www.thehistoryblog.com/archives/11651:) http://www.thehistoryblog.com/archives/11651 (http://www.thehistoryblog.com/archives/11651)
North pole through Paris through Accra to the equator would be 1/4th of the circumference.  So if you take your 6213 miles and quadrupled it, you get roughly 24,400 miles.  Which is pretty darn close to the round earth circumference of 24,900 miles.

I guess you are right.  There is an issue with the circumference of the equator.  Unfortunately, it supports round earth and not flat earth as you are trying to portray.
Title: Re: "Equator" problem
Post by: Rama Set on October 15, 2014, 06:47:34 AM
Numbers?

I can demonstrate it with a following example:

Paris (France) - Accra (Ghana) distance = 2986 miles (4777 km)
Accra - Equator distance = 381 miles (609 km)
Paris - North Pole distance = 2846 miles (4553 km)

2986 + 381 + 2846 = 6213 miles (9940 km)

So, North Pole - Equator distance = 6213 miles

6213 * 2 * 3,14 = 39017 miles (62428 km) This number should represent circumference of the Equator on Flat Earth...

But as we all know (and i doubt that Round Earthers could easily falsify that number) the Equator's circumference is equal to 25 000 miles (40 000 km)...

I believe that this "Equator" problem could have been the main reason for strange kind of FE shaping as it is presented in this map : http://www.thehistoryblog.com/archives/11651: (http://www.thehistoryblog.com/archives/11651:) http://www.thehistoryblog.com/archives/11651 (http://www.thehistoryblog.com/archives/11651)

The circumference measured around the North-South circumference is less that the circumference around the Equator. 
Title: Re: "Equator" problem
Post by: cikljamas on October 15, 2014, 06:59:01 AM
Who measured those distances?
[/quote]

Notwithstanding the "accuracy with which all the operations had been conducted," the skill and ingenuity and perfection of the instruments employed were such that after measuring base lines far apart and triangulating from summit to summit of the hills, between the stations the actually measured and the mathematically calculated results "did not differ more than one inch." Such exactitude was never scarcely contemplated, and certainly could not be surpassed, if at all equalled, by the ordnance officers or practical surveyors of any other country in the world; and yet they failed to corroborate the assumption of polar depression or diminution in the axial radius of the earth. "For instead of the degrees increasing as we proceed from north to south, they appear to decrease, as if the earth were an oblong instead of an oblate spheroid."

http://www.sacred-texts.com/earth/za/za40.htm (http://www.sacred-texts.com/earth/za/za40.htm)

Even Rowbotham talks (about this kind of possibility) that it seems as if the earth were an oblong (instead of an oblate) spheroid...And i didn't notice any kind of mistrust (in correctness of the measurments, quite contrary) in Rowbotham' s words...Why is it so, what do you think?

Does your question imply suspicion in the validity of these numbers? And if so, why?

I am flat earther and i don't see any threat or danger (to my already very firmly established conviction in trueness of FET) of dealing with this (or any other) problem, only we have to be reasonable while we walk along entire path of our investigation which is a search for truth and nothing but the truth!!!
Title: Re: "Equator" problem
Post by: ausGeoff on October 15, 2014, 10:16:34 AM
I am flat earther and i don't see any threat or danger (to my already very firmly established conviction in trueness of FET) of dealing with this (or any other) problem, only we have to be reasonable while we walk along entire path of our investigation which is a search for truth and nothing but the truth!


I'm sorry, but you seem to be struggling under the misapprehension that there's actually a science-based argument occurring about the geometry of the planet.  But there isn't.  The 21st-century scientific status quo is that it's an oblate spheroid, and all practical means of testing and proving this confirm it to be so.

Currently there is no available empirical evidence to suggest any other shape; in fact all the evidence proves that the earth cannot be and is not a flat plane.

Your comments (on the linked site) seem to be a farrago of biblical myth, pseudoscience, guesswork, and a vivid imagination.

And it's the task of the flat earthers to prove their claim, rather than for the round earthers to disprove it.  If i claim I can fly like a bird, I can't expect you to "prove" I'm lying;  I have to prove my claim by jumping off the roof.  And if I don't, then you have every right to call me a liar.  Similar scenario.
Title: Re: "Equator" problem
Post by: shannonator96 on October 15, 2014, 11:52:29 AM
Well since the diameter of a sphere is not equal to it's circumference, the values of the earth's circumference and diameter will also not be equal.
-Shannon
Title: Re: "Equator" problem
Post by: inquisitive on October 15, 2014, 11:58:39 AM
Hi guys! Since this is my first post at this forum, maybe the best possible way to present myself to you is this: http://energeticforum.com/renewable-energy/17050-north-south.html (http://energeticforum.com/renewable-energy/17050-north-south.html)

As you can see from above link i am one of the most convinced flat earther in the world. However, i had become a flat earther very recently, that is why i think no one should be too much surprised with the fact that i (as one of the most convinced FE in the world) am opening this thread with such (elementary) following question:

The measurement of the diameter of the earth, and the measurement for circumference of the equator do not agree. Why?

Sorry for my broken english!
Numbers please.
Title: Re: "Equator" problem
Post by: cikljamas on October 15, 2014, 12:00:32 PM
I am flat earther and i don't see any threat or danger (to my already very firmly established conviction in trueness of FET) of dealing with this (or any other) problem, only we have to be reasonable while we walk along entire path of our investigation which is a search for truth and nothing but the truth!


I'm sorry, but you seem to be struggling under the misapprehension that there's actually a science-based argument occurring about the geometry of the planet.  But there isn't.  The 21st-century scientific status quo is that it's an oblate spheroid, and all practical means of testing and proving this confirm it to be so.

Currently there is no available empirical evidence to suggest any other shape; in fact all the evidence proves that the earth cannot be and is not a flat plane.

Your comments (on the linked site) seem to be a farrago of biblical myth, pseudoscience, guesswork, and a vivid imagination.

And it's the task of the flat earthers to prove their claim, rather than for the round earthers to disprove it.  If i claim I can fly like a bird, I can't expect you to "prove" I'm lying;  I have to prove my claim by jumping off the roof.  And if I don't, then you have every right to call me a liar.  Similar scenario.

I have selected these few (NORTH-SOUTH) arguments just for you:

"Polar Night" argument : http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)
"Daily rate of annual motion of the stars - a constant" argument : http://www.energeticforum.com/265053-post497.html (http://www.energeticforum.com/265053-post497.html)
"Lighthouses" argument : http://www.energeticforum.com/264766-post457.html (http://www.energeticforum.com/264766-post457.html)
"No rotation - No revolution" argument : http://www.energeticforum.com/264738-post450.html (http://www.energeticforum.com/264738-post450.html)
"Midnight Sun - Polaris - Southern Cross" argument : http://www.energeticforum.com/264204-post367.html (http://www.energeticforum.com/264204-post367.html)
"Antarctica nothing alike Arctic" argument : http://www.energeticforum.com/264212-post368.html (http://www.energeticforum.com/264212-post368.html)
"Water level experiments" argument : http://www.energeticforum.com/264258-post373.html (http://www.energeticforum.com/264258-post373.html)
"Southern Hemisphere" argument : http://www.energeticforum.com/263904-post281.html (http://www.energeticforum.com/263904-post281.html)
"Moon" argument: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)
"Sextant" argument : http://www.energeticforum.com/264972-post486.html (http://www.energeticforum.com/264972-post486.html)
"Triangulation" argument : http://www.energeticforum.com/264976-post488.html (http://www.energeticforum.com/264976-post488.html)

Regarding last (triangulation) one of above arguments one addition:

With the eye at water level at one angle and the sun at water level at the other, the line joining them is the base of the triangle — a straight line, of which we have already heard so much. But if water be convex, when the height of eye is deducted and the observation reduced to the datum line— the sea, then the eye and the sun are both at the surface of the convex water, consequently the base of the triangle is the arc of the circle between the two points, and another allowance must be made to reduce this arc of a circle to a straight line, in order to determine the true angle of the plane triangle. That this is not only never done, but that no work on Navigation ever published makes the slightest reference to the need for such a correction, and that all triangulation in Navigation is plane, proves incontestably that the surface of the ocean is a plane surface.

The fact that water is flat like a sheet of paper (when undisturbed by wind and tide) is my " working anchor," and the powerful " ground tackle " of all those who reject the delusions of modern theoretical astronomy. Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.

Cheers!
Title: Re: "Equator" problem
Post by: cikljamas on October 15, 2014, 12:05:55 PM
Hi guys! Since this is my first post at this forum, maybe the best possible way to present myself to you is this: http://energeticforum.com/renewable-energy/17050-north-south.html (http://energeticforum.com/renewable-energy/17050-north-south.html)

As you can see from above link i am one of the most convinced flat earther in the world. However, i had become a flat earther very recently, that is why i think no one should be too much surprised with the fact that i (as one of the most convinced FE in the world) am opening this thread with such (elementary) following question:

The measurement of the diameter of the earth, and the measurement for circumference of the equator do not agree. Why?

Sorry for my broken english!
Numbers please.

I already gave you your numbers, specify what you are asking for so that you can get what you want!!!
Title: Re: "Equator" problem
Post by: sokarul on October 15, 2014, 01:36:29 PM


I have selected these few (NORTH-SOUTH) arguments just for you:

"Polar Night" argument : http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)
Incorrect claim and incorrect usage of math.
Quote
"Daily rate of annual motion of the stars - a constant" argument : http://www.energeticforum.com/265053-post497.html (http://www.energeticforum.com/265053-post497.html)
Solar time vs sidereal time, learn it.

Quote
"Lighthouses" argument : http://www.energeticforum.com/264766-post457.html (http://www.energeticforum.com/264766-post457.html)
Mirage, learn it.

Quote
"No rotation - No revolution" argument : http://www.energeticforum.com/264738-post450.html (http://www.energeticforum.com/264738-post450.html)
Foucault Pendulum, learn it.

Quote
"Midnight Sun - Polaris - Southern Cross" argument : http://www.energeticforum.com/264204-post367.html (http://www.energeticforum.com/264204-post367.html)
Shows the Earth to be round, thanks?

Quote
"Antarctica nothing alike Arctic" argument : http://www.energeticforum.com/264212-post368.html (http://www.energeticforum.com/264212-post368.html)
Who would have guess two points on opposite sides of the globe didn't behave the same?

Should I continue?
Title: Re: "Equator" problem
Post by: cikljamas on October 15, 2014, 02:39:07 PM
Should I continue?

You shouldn't have started this shameless parade of stupidity in the first place, because you haven't slightest idea what you are talking about...I have thought that this forum doesn't tolerate trolls like you...I was obviously wrong...And if you dared to contradict me claiming that you are not a troll, then you should bare in mind that you will have to prove correctness of your claims by elaborating validity of your undoubtedly false claims with concrete (counter) arguments. Arrogantly parading with your false assertions won't make your (non-existant) "round" arguments any stronger, and won't make you look any smarter than you really are also!
Title: Re: "Equator" problem
Post by: inquisitive on October 15, 2014, 02:56:01 PM
Should I continue?

You shouldn't have started this shameless parade of stupidity in the first place, because you haven't slightest idea what you are talking about...I have thought that this forum doesn't tolerate trolls like you...I was obviously wrong...And if you dared to contradict me claiming that you are not a troll, then you should bare in mind that you will have to prove correctness of your claims by elaborating validity of your undoubtedly false claims with concrete (counter) arguments. Arrogantly parading with your false assertions won't make your (non-existant) "round" arguments any stronger, and won't make you look any smarter than you really are also!
Please explain sunrise and sunset times proving a round earth.
Title: Re: "Equator" problem
Post by: sokarul on October 15, 2014, 03:50:26 PM
Should I continue?

You shouldn't have started this shameless parade of stupidity in the first place, because you haven't slightest idea what you are talking about...I have thought that this forum doesn't tolerate trolls like you...I was obviously wrong...And if you dared to contradict me claiming that you are not a troll, then you should bare in mind that you will have to prove correctness of your claims by elaborating validity of your undoubtedly false claims with concrete (counter) arguments. Arrogantly parading with your false assertions won't make your (non-existant) "round" arguments any stronger, and won't make you look any smarter than you really are also!
Your post is cute, I'm going to stick it up on my fridge.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 15, 2014, 05:30:53 PM
I have thought that this forum doesn't tolerate trolls like you...

That's the most patently ridiculous thing I've ever seen posted here, and there have been some doozies!!

Now what could possibly have made you think that? "sokarul" is a complete, rank, and total amateur compared to many here. I know you're new, but have you read any threads on this site???
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 15, 2014, 07:08:21 PM
Should I continue?

You shouldn't have started this shameless parade of stupidity in the first place, because you haven't slightest idea what you are talking about...I have thought that this forum doesn't tolerate trolls like you...I was obviously wrong...And if you dared to contradict me claiming that you are not a troll, then you should bare in mind that you will have to prove correctness of your claims by elaborating validity of your undoubtedly false claims with concrete (counter) arguments. Arrogantly parading with your false assertions won't make your (non-existant) "round" arguments any stronger, and won't make you look any smarter than you really are also!
Emphasis added.

*Ahem* Instead of just slagging each other, lets look at one "proof" claimed in one of your links. I picked the "Southern Cross (http://www.energeticforum.com/264204-post367.html)" example because I think Crux is a cool constellation. The refutation for the visibility of Acrux at -63.16 declination from 28 North latitude is incorrect, but that's not what I want to point out. We can talk about that later if you want.

Consider the next "proof", as summarized in this illustration from the same link:

(http://www.zaslike.com/files/gf6lodg7stz7nmwacqp2.jpg)

Refer to the link for the full text of the explanation offered, but here's the problem with the argument, starting about 2/3 of the way through the explanation:

"Persons living further north than this [Tropic of Cancer] have always to look in a southerly direction for the sun at noon ; and it ought therefore never to be seen to the north of them at any time, so we must place the sun in the diagram somewhere on the line P F G. Let it be placed at any point P. Now it is manifest that for an observer at M, near the latitude of Haparanda, to see the sun at midnight at P, over the tropic at Cancer, he would have to to look downwards and be able to see right through the "Globe" for about five or six thousand miles along the dotted line MR!!"

Two fatal errors are immediately obvious.

One of which is that "Point P" can't be "any" point along the line FG. If P were immediately to the left of F instead of its presented position, the elevation angle to P from R would become lower; if P were moved further away, the elevation angle from R would increase. In fact, the elevation angle to the Sun, represented by P in the drawing, from R in the situation described, is fixed by nature, so the assertion that P can be anywhere along line through FG is false.

Even worse, if the Sun is directly above the Tropic of Cancer, it would have to be along the line through line EF (also KEF), not the line through line FG. Call this point P' (P prime). Add to this that P' has to be effectively at infinity because the Sun is vastly farther away than the radius of the circle, EM, and badda-bing, badda-boom(!) the rays P'M (shown as the line through Point 2 to Point M in Diagram 1) are parallel to ray P'F(E(K)). Thus, you see Point P' (the Sun) from Point M without looking through the Earth.

QED.

OK, since we're here, let's look at Acrux anyway. The author of the quote just brushes off "altitude" and "refraction" as factors in being able to see stars that "should" be one degree below the horizon.

As in the overly blustery (but not unusual when you're trying to bluff):
Quote
No "altitude" or/and "refraction" excuse will suffice to cover up the only possible solution that we can use to convincingly explain this phenomena, and that one and only convincing explanation is of course FLAT EARTH theory and nothing but the FLAT EARTH obvious fact
Let's actually look at the data instead of just dismissing it, shall we?

From an altitude of 2300m above sea level, a sea-level horizon will be about 170 km away and a little more than 1.5 degrees below horizontal. Further, while refraction can add about 1/2 degree of apparent elevation to an object on the horizon when viewed from sea level, if you let that ray continue through the relatively thick atmosphere for another 170 km, it will continue to look even higher. Not a lot higher, mind you, but it ain't gonna be less than 1/2 degree. So now, in this situation, a star that is geometrically 1 degree below horizontal will appear about 1 degree above your horizon. These things matter.

This is pretty poor stuff. Is there any reason to think the rest of the links contain anything better?

[Edit] Minor correction indicated by strikethroughs.
Title: Re: "Equator" problem
Post by: cikljamas on October 16, 2014, 01:23:26 AM
@Alpha2Omega, here we go:

The North Star can even be seen slightly south of the Equator [COLOR="Red"]because of atmospheric refraction[/COLOR][/B]) (wiki quote)

(http://zaslike.com/files/8kx71vx5q0yvayyskbj.jpg)

Line of sight CD being A TANGENT to the Earth beyond the equator E must diverge from the axis N and could not by any known possibility cause the star P to be visible to an observer at C. NO MATTER HOW DISTANT THE STAR P, the line CD being divergent from the direction NP COULD NEVER COME IN CONTACT WITH IT!!!

If "Tamatea" can look up the London Times of May 13th, 1962, in the Naval and Military Inteligence, he may read as follows: -"On the 19th of April, in latitude 23,53, longitude 35.46, Captain Wilkins reports that the Southern Cross and the Polar Star were both DISTINCTLY visible at midnight."- If the event of this being considered as an error of some kind, I may state that Captain Edward Gillett states that he has observed the same thing between the 12th and 13th degree of South Latitude. While, if there were any curvature such as we read so much about, the whole thing would be an impossibility instead of a well authenticated fact. 

Here is another nut for "Tamatea" to crack, and when he has cracked this I can give him some more: -On the last trip of the R.M.S. Kaikoura, Captain W. C. Crutchley R.N.R. sighted Mount Peel at a distance of 118 miles. Take off for elevation of observer 7 miles, which leaves 111 miles; the curvature in that distance according to science, is 8,214 feet; take the height of Mount Peel 5,500 feet from this, and it leaves 2,714 feet. The top of Mount Peel should have been below the horizon, and could not be seen at the distance named, if the world were a globe.-

http://www.zaslike.com/files/zqumw2ph80a6jojd60y.jpg (http://www.zaslike.com/files/zqumw2ph80a6jojd60y.jpg)

Pay attention to two things:

1. Meaning of the word TANGENT
2. Destroying function of the TANGENT to all your refraction excuses

A) Apply TANGENT destroying function to the Midnight Sun example.

B) Don't apply your refraction excuse to "Polar Night" argument because it would become even much worse case for you, because my argument against Round Earther's dreams would become much stronger!!! Why? Because if refraction is bending objects upwards then you should even see the Sun directly (it wouldn't be just twilight) at latitude 66 S in NOON position!!! http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)

C) Don't apply your refraction excuse in your explanations of the Eclipses of the Moon because refraction is bending objects up, but in the same time refraction is bending shadows of the objects DOWN, and that presents quite fatal refraction-excuse-flaw in your (RET) funny tries of explaining that phenomena.

Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.

And if you can't prove it then there is still no reason for concern  ;D, i can provide you with countless proofs in favor of perfect flatness of all the waters on the Earth!  And i will do it on your demand, just say a word and your wishes will come true!!!
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 16, 2014, 06:30:00 AM
@Alpha2Omega, here we go:

The North Star can even be seen slightly south of the Equator because of atmospheric refraction) (wiki quote)

(http://zaslike.com/files/8kx71vx5q0yvayyskbj.jpg)

Line of sight CD being A TANGENT to the Earth beyond the equator E must diverge from the axis N and could not by any known possibility cause the star P to be visible to an observer at C. NO MATTER HOW DISTANT THE STAR P, the line CD being divergent from the direction NP COULD NEVER COME IN CONTACT WITH IT!!!
The North Star isn't exactly at the pole - it's about 3/4 degree from it. Because of this alone, you should be able to see it at times from anywhere north of about 3/4 degrees south of the Equator. Refraction, as noted before, makes objects on the horizon appear about 1/2 degree higher, so with the addition of refraction, I'd expect it to be barely visible from about 1.25 or so degrees south.

Fig. 20 presumes the observer is at the Tropic of Capricorn - 23.5 degrees south. That ain't exactly "slightly south of the Equator"; it's more than a quarter of the way to the pole.
Quote
If "Tamatea" can look up the London Times of May 13th, 1962, in the Naval and Military Inteligence, he may read as follows: -"On the 19th of April, in latitude 23,53, longitude 35.46, Captain Wilkins reports that the Southern Cross and the Polar Star were both DISTINCTLY visible at midnight."-
Where is the problem with this? Acrux is north of -64 degrees declination, so it should be easily visible from 24 degrees north. Obviously, Polaris is visible from 24 N.
Quote
If the event of this being considered as an error of some kind, I may state that Captain Edward Gillett states that he has observed the same thing between the 12th and 13th degree of South Latitude. While, if there were any curvature such as we read so much about, the whole thing would be an impossibility instead of a well authenticated fact. 
Can you vouch for Captain Edward Gillett's statement? Why isn't this sort of thing reported all the time? It appears to be an error to me, not a well-authenticated fact. Do you believe everything you read in the newspaper?

Quote
Here is another nut for "Tamatea" to crack, and when he has cracked this I can give him some more: -On the last trip of the R.M.S. Kaikoura, Captain W. C. Crutchley R.N.R. sighted Mount Peel at a distance of 118 miles. Take off for elevation of observer 7 miles, which leaves 111 miles; the curvature in that distance according to science, is 8,214 feet; take the height of Mount Peel 5,500 feet from this, and it leaves 2,714 feet. The top of Mount Peel should have been below the horizon, and could not be seen at the distance named, if the world were a globe.-

http://www.zaslike.com/files/zqumw2ph80a6jojd60y.jpg (http://www.zaslike.com/files/zqumw2ph80a6jojd60y.jpg)

Pay attention to two things:

1. Meaning of the word TANGENT
2. Destroying function of the TANGENT to all your refraction excuses

A) Apply TANGENT destroying function to the Midnight Sun example.

B) Don't apply your refraction excuse to "Polar Night" argument because it would become even much worse case for you, because my argument against Round Earther's dreams would become much stronger!!! Why? Because if refraction is bending objects upwards then you should even see the Sun directly (it wouldn't be just twilight) at latitude 66 S in NOON position!!! http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)

C) Don't apply your refraction excuse in your explanations of the Eclipses of the Moon because refraction is bending objects up, but in the same time refraction is bending shadows of the objects DOWN, and that presents quite fatal refraction-excuse-flaw in your (RET) funny tries of explaining that phenomena.

I've got to pack for a trip, so don't have time to look at the Mount Peel exercise; it will take some calculations. I'll look at it when I get back unless someone else addresses it in the meantime. Remind me if I haven't done that by the end of next week.

In general, however:

Atmospheric refraction bends light rays down (toward the ground). This makes the object where the ray originated appear higher than it actually is. Atmospheric refraction affects light that travels all the way through the atmosphere most, thus, it affects light from objects near your horizon most since it passes through way more air than rays coming from higher, but diminishes rapidly as the objects elevation above the horizon increases, being negligible for most purposes by 20 degrees or so. It has zero effect for objects at the zenith.

Quote
Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.
The next time you take a trans-oceanic jet trip, request a window seat. Look out the window. (This works best if the flight is during daylight)

Quote
And if you can't prove it then there is still no reason for concern  ;D , i can provide you with countless proofs in favor of perfect flatness of all the waters on the Earth!  And i will do it on your demand, just say a word and your wishes will come true!!!
No worries here. Why don't you start with just one? I will be away for a few days, so no hurry.
Title: Re: "Equator" problem
Post by: markjo on October 16, 2014, 06:50:16 AM
@Alpha2Omega, here we go:

The North Star can even be seen slightly south of the Equator because of atmospheric refraction) (wiki quote)

(http://zaslike.com/files/8kx71vx5q0yvayyskbj.jpg)

Line of sight CD being A TANGENT to the Earth beyond the equator E must diverge from the axis N and could not by any known possibility cause the star P to be visible to an observer at C. NO MATTER HOW DISTANT THE STAR P, the line CD being divergent from the direction NP COULD NEVER COME IN CONTACT WITH IT!!!
Would you consider point C to be "slightly" beyond the equator?  If so, then I see your first problem.
Title: Re: "Equator" problem
Post by: cikljamas on October 16, 2014, 09:59:15 AM
I just have noticed that this "Equator" issue has been already raised at least a dozen times at this forum, and i wonder how much time have passed since the first time this problem had appeared at this forum?

Many years have passed by since then, but all of these years not even one Flat Earther has come up with at least one single reasonable idea attempting to solve this problem?

If this is so, then what could be the real reason for further keeping this forum up?

I am absolutely amazed with the disclosure of this remarkable fact.  ??? ??? ???
Title: Re: "Equator" problem
Post by: liimr on October 16, 2014, 10:48:11 AM
I just have noticed that this "Equator" issue has been already raised at least a dozen times at this forum, and i wonder how much time have passed since the first time this problem had appeared at this forum?

Many years have passed by since then, but all of these years not even one Flat Earther has come up with at least one single reasonable idea attempting to solve this problem?

If this is so, then what could be the real reason for further keeping this forum up?

I am absolutely amazed with the disclosure of this remarkable fact.  ??? ??? ???

I totally agree with you. Something's odd with their stance. You would expect that they would conduct experiments, publish results, check/verify things, love true science, seek the truth, improve their knowledge etc. but no they just sit around and idly monitor debates/threads. Almost like AI bots...
Title: Re: "Equator" problem
Post by: cikljamas on October 16, 2014, 12:14:08 PM
I totally agree with you. Something's odd with their stance. You would expect that they would conduct experiments, publish results, check/verify things, love true science, seek the truth, improve their knowledge etc. but no they just sit around and idly monitor debates/threads. Almost like AI bots...

You are right, totaly mind-boggling....But wait, there is something even much more mind-boggling: Samuel Rowbotham, Thomas Winship, D.W. Scott, Voliva, and other great zetetic names had knew very well how to measure Sun's distance from the Earth; according to their explanations (which methods i consider to be accurate btw.) distance of the Sun (when he's above the Equator) is equal with the distance from the Equator to the latitude at 45 degree (North or South - doesn't matter).

That distance is 2700 nautical miles. 2 * 2700 miles = 5400 nautical miles.

Given that all those great zetetic names knew these numbers very well, it is absolutely stunning that no one of them had ever refered to that issue although knowing that the circumference of the equator (according to above numbers) on the Flat Earth must be 33912 nautical miles instead of 25000 statute miles.

This is absolutely shocking, so shocking that i have just run out of words...or am i just missing something all along?

Title: Re: "Equator" problem
Post by: ausGeoff on October 16, 2014, 02:18:24 PM
You shouldn't have started this shameless parade of stupidity in the first place, because you haven't slightest idea what you are talking about...I have thought that this forum doesn't tolerate trolls like you...I was obviously wrong...And if you dared to contradict me claiming that you are not a troll, then you should bare in mind that you will have to prove correctness of your claims by elaborating validity of your undoubtedly false claims with concrete (counter) arguments. Arrogantly parading with your false assertions won't make your (non-existant) "round" arguments any stronger, and won't make you look any smarter than you really are also!


Thanks for the major LULZ.....  This is one of the absolutely funniest rants I've read on these forums for many a day.  10/10 cikljamas.    ;D

And all from a guy who accepts 150-year-old pseudo-science as legitimate in the 21st century!
Title: Re: "Equator" problem
Post by: 29silhouette on October 16, 2014, 05:54:25 PM
This is absolutely shocking, so shocking that i have just run out of words...or am i just missing something all along?
Yes, there's a likely reason some of the FE numbers don't add up.

24,901 miles (approx) circumference of the equator seems to be the agreed upon distance according to FE'rs and RE'rs alike.  This requires a radius and surface measurement from equator to pole of 3,963 miles, and does fit with the RE distance of the axis to equator.  All distances I can find however show 6,210 surface miles from the pole to the equator. 

According to the FE mono-pole model, the Tropic of Capricorn in the southern hemisphere should have a circumference of 49,178 miles (based off known distance from the north pole), which would mean the sun needs to move at a speed of 2,049 mph to complete one day during the summer season in the south.  The known distance to the Tropic of Cancer is 4,602 miles, which would result in a circumference of 28,915 miles.  The sun would have to move 1,204mph during summer in the north.

So if there's some evidence out there of the surface distance from the north pole to the equator being 3,963 miles (preferably backed by measured distances between cities, etc, in conjunction with lines of latitude, and/or some evidence of the sun traveling overhead almost twice as fast in December than it does in June, we'd love to see it.
Title: Re: "Equator" problem
Post by: 29silhouette on October 16, 2014, 06:01:41 PM
Prove water to be convex,
Images of objects visibly sinking as they're blocked from view by the curvature?
Title: Re: "Equator" problem
Post by: ausGeoff on October 16, 2014, 06:56:10 PM
Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.

Too easy.  If I travel by ship from New York to London, I travel in a curved line—known as a "great circle"—rather than a straight line.

(http://euler.slu.edu/escher/upload/thumb/b/b3/Great-circle-NY-London.svg/404px-Great-circle-NY-London.svg.png)


This is to conserve fuel (and save money) simply because a straight line isn't the shortest distance between two points on the surface of a sphere.  IF the earth were a flat plane, then obviously the ship would travel in a dead straight line.

Your knowledge of non-Euclidean geometry is obviously lacking.

Title: Re: "Equator" problem
Post by: markjo on October 16, 2014, 08:22:43 PM
Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.
(http://i.telegraph.co.uk/multimedia/archive/02153/ant-water-drop_2153832i.jpg)
Title: Re: "Equator" problem
Post by: ausGeoff on October 16, 2014, 11:31:36 PM
Also a convex meniscus in a pipette...



(http://sydney.edu.au/science/chemistry/~mjtj/EDUP1004/images/Q15.jpg)


EDIT:  [img] coding corrected.



Title: Re: "Equator" problem
Post by: inquisitive on October 17, 2014, 12:05:46 AM
Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.

Too easy.  If I travel by ship from New York to London, I travel in a curved line—known as a "great circle"—rather than a straight line.

(http://euler.slu.edu/escher/upload/thumb/b/b3/Great-circle-NY-London.svg/404px-Great-circle-NY-London.svg.png)


This is to conserve fuel (and save money) simply because a straight line isn't the shortest distance between two points on the surface of a sphere.  IF the earth were a flat plane, then obviously the ship would travel in a dead straight line.

Your knowledge of non-Euclidean geometry is obviously lacking.
It is a straight line with no change of direction left or right, just following the curvature of the earth.
Title: Re: "Equator" problem
Post by: Saros on October 17, 2014, 12:31:46 AM
Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.
(http://i.telegraph.co.uk/multimedia/archive/02153/ant-water-drop_2153832i.jpg)

This is not water around a solid object, but a water droplet. There is a difference. Show me an example of water encircling a solid object/sphere and actually sticking to it without spilling. Otherwise, nice try.
Title: Re: "Equator" problem
Post by: Saros on October 17, 2014, 12:34:02 AM
Prove water to be convex, and we will at once and forever recant and grant you anything you like to demand.

Too easy.  If I travel by ship from New York to London, I travel in a curved line—known as a "great circle"—rather than a straight line.

(http://euler.slu.edu/escher/upload/thumb/b/b3/Great-circle-NY-London.svg/404px-Great-circle-NY-London.svg.png)


This is to conserve fuel (and save money) simply because a straight line isn't the shortest distance between two points on the surface of a sphere.  IF the earth were a flat plane, then obviously the ship would travel in a dead straight line.

Your knowledge of non-Euclidean geometry is obviously lacking.
It is a straight line with no change of direction left or right, just following the curvature of the earth.

How do you know if a line is straight or curved? What is your reference point?
Title: Re: "Equator" problem
Post by: cikljamas on October 17, 2014, 01:44:16 AM
Hi Saros, nice to see you here also!

Shall we fight with them like adults?

@ AUSGeoff, it's nice to hear you laughing, especially when i begin to feel desperate like Charlie in this scene:  (http://) Also, do you remember captain Bligh (Trevor Howard) how funny he looked like while attempting to dance tahitian dance?

Let's get serious now:

@ 29silhouette, calculating the differences of the different speeds of the Sun (when the Sun is above different latitudes) caused me to stumble upon "Equator" problem...

Few days ago (when there was sunny here) i have tried to measure the speed of the Sun (with a welding glass and a stopwatch). I noticed that the Sun moves one of his diameter in two minutes, that is roughly in accordance with zetetic reckoning of the diameter of the Sun AND with the RET & FET agreement over Equator circumference.

30 * 2 (minutes) = 1 hour
1666 / 30 = 55,5 km = 32 nautical miles = 34 statute miles = diameter of the Sun

So far, so good, but yes, you are right, the differences between the different speeds of the Sun (when the Sun is above different latitudes) would be too great to be acceptable if we calculated those differences assuming applicability of simple mathematical rules in the case of the Sun although we don't really know how Sun works at all...

That is why we should maybe reconsider these words of Sandokhan more carefully:

Quote
Therefore, statements such as: On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles.  and  If a navigator neglects to apply the sun's radius to his observation at sea, he is 16 nautical miles (nearly) out in calculating the position his ship is in. A minute of arc on the sextant represents a nautical mile, and if the radius of the sun is 16 miles, the diameter is of course 32 miles. And as measured by the sextant, the sun's diameter is 32 minutes of arc, that is 32 nautical miles in diameter. cannot be true given the effect of the many layers of aether (of various densities) upon the light emitted by the Sun. Also, measuring the angle of the sun from a latitude of 30 degrees or 60 degrees will give different results.

And there are further issues to be dealt with, if we use simple triangulation to obtain possible figures for the Earth-Sun distance:

Read more, post #12 on this page: http://www.theflatearthsociety.org/forum/index.php?topic=39728.msg994892#msg994892 (http://www.theflatearthsociety.org/forum/index.php?topic=39728.msg994892#msg994892)

Plane Sailing 33 000 miles: http://www.energeticforum.com/265962-post590.html (http://www.energeticforum.com/265962-post590.html)
Proving water to be convex: http://www.energeticforum.com/255875-post14.html (http://www.energeticforum.com/255875-post14.html)
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 17, 2014, 02:17:08 AM
This is not water around a solid object, but a water droplet. There is a difference. Show me an example of water encircling a solid object/sphere and actually sticking to it without spilling.
(http://cdn.theatlantic.com/static/mt/assets/science/cool-space-picture-5.jpg)
Title: Re: "Equator" problem
Post by: ausGeoff on October 17, 2014, 04:02:16 AM
How do you know if a line is straight or curved? What is your reference point?


The red line on my map follows a vertical curve along the circumference of the planet.  Hence my reference to Euclid.  In plan view, the line is straight.

The globe as shown is an (approximate) isometric projection.

Title: Re: "Equator" problem
Post by: cikljamas on October 17, 2014, 06:07:14 AM
No worries here. Why don't you start with just one? I will be away for a few days, so no hurry.

Here is just one: http://www.energeticforum.com/265962-post590.html (http://www.energeticforum.com/265962-post590.html)

Here is another one: http://www.energeticforum.com/265994-post593.html (http://www.energeticforum.com/265994-post593.html)

Many more to come....

edit: Here you can take a look how had ended up one of previous FET debate on this very subject:
http://theflatearthsociety.org/forum/index.php?topic=58467.30;wap2 (http://theflatearthsociety.org/forum/index.php?topic=58467.30;wap2)
Title: Re: "Equator" problem
Post by: markjo on October 17, 2014, 07:13:26 AM
This is not water around a solid object, but a water droplet. There is a difference.
You asked for an example of water being convex and I provided you one.  If you wanted an example of water sticking to an object, then you should have asked for one.  Quit moving the goal post.

Show me an example of water encircling a solid object/sphere and actually sticking to it without spilling. Otherwise, nice try.
Better?
(http://www.wired.com/wp-content/uploads/blogs/insights/wp-content/uploads/2013/03/snowball.jpg)
Title: Re: "Equator" problem
Post by: sceptimatic on October 17, 2014, 07:17:58 AM
I think he meant water not ice, markjo.
Title: Re: "Equator" problem
Post by: BJ1234 on October 17, 2014, 07:58:17 AM
I think he meant water not ice, markjo.
Ice is water  ::)
Title: Re: "Equator" problem
Post by: Son of Orospu on October 17, 2014, 08:09:01 AM
I think he meant water not ice, markjo.
Ice is water  ::)

Incorrect.  Just because something is made out of something else, that does not make them the same thing. 
Title: Re: "Equator" problem
Post by: sceptimatic on October 17, 2014, 08:40:05 AM
I think he meant water not ice, markjo.
Ice is water  ::)
Ice is frozen water. He meant water and you know it. You can't answer him because you have no clue how in the hell oceans stick to a supposed massive ball, except to mention gravity as a lame excuse.
Title: Re: "Equator" problem
Post by: markjo on October 17, 2014, 09:21:12 AM
I think he meant water not ice, markjo.
Shhh... Don't tell anyone, but I'm just messing with him because I know that he knows that he's asking for an impossible demonstration within earth's gravitational influence.
Title: Re: "Equator" problem
Post by: cikljamas on October 17, 2014, 09:38:13 AM
@Markjo, You bet that he is asking for an impossible demonstration, let alone asking for this kind of demonstration:

http://www.energeticforum.com/261779-post206.html (http://www.energeticforum.com/261779-post206.html)
Title: Re: "Equator" problem
Post by: markjo on October 17, 2014, 02:05:39 PM
@Markjo, You bet that he is asking for an impossible demonstration, let alone asking for this kind of demonstration:

http://www.energeticforum.com/261779-post206.html (http://www.energeticforum.com/261779-post206.html)
Actually, the one in the link would be a lot easier because you could simulate the zero gee with magnetic levitation.
Title: Re: "Equator" problem
Post by: sokarul on October 17, 2014, 02:54:58 PM
@Markjo, You bet that he is asking for an impossible demonstration, let alone asking for this kind of demonstration:

http://www.energeticforum.com/261779-post206.html (http://www.energeticforum.com/261779-post206.html)
The Earth's spin is slowing down and the axis of rotation isn't fixed. Please note this.
Title: Re: "Equator" problem
Post by: The Ellimist on October 17, 2014, 07:10:59 PM
I think he meant water not ice, markjo.
Ice is water  ::)
Did you really expect him to know that? This is sceptimatic we're talking about here.
Title: Re: "Equator" problem
Post by: BJ1234 on October 17, 2014, 11:54:41 PM
I think he meant water not ice, markjo.
Ice is water  ::)

Incorrect.  Just because something is made out of something else, that does not make them the same thing.
Ice is water just in a solid form.  Just as steam is water just is gaseous form.  Now if he had asked for an example of liquid water in a sphere, then by all means.  But he didn't now did he?
Title: Re: "Equator" problem
Post by: cikljamas on October 18, 2014, 02:03:41 AM
While you are searching for an example that Saros has asked you for, maybe next two arguments (one of which arguments has been named by the name of my friend Saros, because it consists from his own words) can be of some help in speeding up your search and make it (your futile and absurd search) more productive...

http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D
Title: Re: "Equator" problem
Post by: sceptimatic on October 18, 2014, 03:02:55 AM
While you are searching for an example that Saros has asked you for, maybe next two arguments (one of which arguments has been named by the name of my friend Saros, because it consists from his own words) can be of some help in speeding up your search and make it (your futile and absurd search) more productive...

http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D
Keep up the good work. It's nice to see more people thinking for themselves. I had a topic similar to what you are talking about on the water running to the level of the sea.
Stick around as I'm enjoying your input as well as a few otehrs, like Legion, saros, etc. It makes a refreshing change to the tefal heads and their ready made scientific page turning MS answers.
Title: Re: "Equator" problem
Post by: ausGeoff on October 18, 2014, 05:53:05 AM
This is not water around a solid object, but a water droplet. There is a difference. Show me an example of water encircling a solid object/sphere and actually sticking to it without spilling.
Too easy.....

UK Daily Mail Australia: April 2011:

(http://i.dailymail.co.uk/i/pix/2011/04/01/article-1371416-0B620F7C00000578-935_964x697.jpg)



This image by photographer Adam Gormley, from Noosaville, QLD, Australia
shows a dead ant trapped in a tiny perfect sphere of water, totally unable
to escape, after being caught in a sudden heavy downpour.



It never ceases to amaze me how pedantic flat earthers are, or how often they play with semantics whenever they're backed into a corner.  In this case, surface tension is the answer.



Title: Re: "Equator" problem
Post by: BJ1234 on October 18, 2014, 07:08:05 AM


http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D

Just want to point out that you are either misunderstanding or misrepresenting the "no up or down".  On a globe there IS a down.  It is towards the center of the Earth.
Title: Re: "Equator" problem
Post by: sokarul on October 18, 2014, 09:38:18 AM
While you are searching for an example that Saros has asked you for, maybe next two arguments (one of which arguments has been named by the name of my friend Saros, because it consists from his own words) can be of some help in speeding up your search and make it (your futile and absurd search) more productive...

http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D
No, rivers on a globe would not have to run uphill. Please note this.
Title: Re: "Equator" problem
Post by: cikljamas on October 18, 2014, 10:10:37 AM
While you are searching for an example that Saros has asked you for, maybe next two arguments (one of which arguments has been named by the name of my friend Saros, because it consists from his own words) can be of some help in speeding up your search and make it (your futile and absurd search) more productive...

http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D
Keep up the good work. It's nice to see more people thinking for themselves. I had a topic similar to what you are talking about on the water running to the level of the sea.
Stick around as I'm enjoying your input as well as a few otehrs, like Legion, saros, etc. It makes a refreshing change to the tefal heads and their ready made scientific page turning MS answers.

Thanks for your words of encouragement, it's a pity that today's people are so brainwashed, so that thoughts of every trully free thinking man in today's world are precious as if they have been made of pure gold!!!

@BJ1234, according to you, rivers flow towards the centre of the Earth? Don't be ridiculous!

@sokarul, rivers on a globe would not have to run uphill, that is true, rivers on a globe would have to run downhill in both cases as it is described in my "Rivers" argument, only every day let's say at midnight downhill would be uphill, and at noon downhill would be again downhill as we know it from our everyday experience. And if you think that uphill would be impossibility because of inclinations of the river beds then i suggest you to read this for clarification:

Quote
" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles."  Read more: http://www.energeticforum.com/265601-post587.html (http://www.energeticforum.com/265601-post587.html)

In fact, rivers on a globe wouldn't flow at all, they would be spilled out into space together with all other water on such imagined globe. Don't you see how insulting for sanity of every man on the Earth is that outrageously idiotic fairytale about a imagined globular planet which our beautiful-true world supposed to be???
Title: Re: "Equator" problem
Post by: BJ1234 on October 18, 2014, 10:24:04 AM
While you are searching for an example that Saros has asked you for, maybe next two arguments (one of which arguments has been named by the name of my friend Saros, because it consists from his own words) can be of some help in speeding up your search and make it (your futile and absurd search) more productive...

http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D
Keep up the good work. It's nice to see more people thinking for themselves. I had a topic similar to what you are talking about on the water running to the level of the sea.
Stick around as I'm enjoying your input as well as a few otehrs, like Legion, saros, etc. It makes a refreshing change to the tefal heads and their ready made scientific page turning MS answers.

Thanks for your words of encouragement, it's a pity that today's people are so brainwashed, so that thoughts of every trully free thinking man in today's world are precious as if they have been made of pure gold!!!

@BJ1234, according to you, rivers flow towards the centre of the Earth? Don't be ridiculous!

@sokarul, rivers on a globe would not have to run uphill, that is true, rivers on a globe would have to run downhill in both cases as it is described in my "Rivers" argument, only every day let's say at midnight downhill would be uphill, and at noon downhill would be again downhill as we know it from our everyday experience. And if you think that uphill would be impossibility because of inclinations of the river beds then i suggest you to read this for clarification:

Quote
" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles."  Read more: http://www.energeticforum.com/265601-post587.html (http://www.energeticforum.com/265601-post587.html)

In fact, rivers on a globe wouldn't flow at all, they would be spilled out into space together with all other water on such imagined globe. Don't you see how insulting for sanity of every man on the Earth is that outrageously idiotic fairytale about a imagined globular planet which our beautiful-true world supposed to be???
You are ignoring one huge factor.  The thing called gravity!!!!
And yes, rivers flow towards the center of the earth.  They never quite make it there though, you know with all the rock and ground in the way.  But yes, they flow from a higher elevation to a lower elevation.  Elevation is based on the distance from the center of the earth.  The river bed might only get closer to the center of the earth by a few feet over a few miles, but the water will flow that way.
Title: Re: "Equator" problem
Post by: 29silhouette on October 18, 2014, 10:31:59 AM
@ 29silhouette, calculating the differences of the different speeds of the Sun (when the Sun is above different latitudes) caused me to stumble upon "Equator" problem...

Few days ago (when there was sunny here) i have tried to measure the speed of the Sun (with a welding glass and a stopwatch). I noticed that the Sun moves one of his diameter in two minutes, that is roughly in accordance with zetetic reckoning of the diameter of the Sun AND with the RET & FET agreement over Equator circumference.
I would suggest someone along the Tropic of Cancer in June using a stick to measure the speed of the shadow as the sun travels directly overhead, and someone along the Tropic of Capricorn in December, again using stick of the same height, to measure the shadow speed.

According to the mono-pole model, which seems to be the most popular, the sun would have to move almost twice as fast in December.  If someone can come up with a way to explain how the sun could travel the same speed along two vastly different distances, yet complete each trip in the same amount of time, and visibly appear to move the same speed, we'd all love to hear it.

Quote
30 * 2 (minutes) = 1 hour
1666 / 30 = 55,5 km = 32 nautical miles = 34 statute miles = diameter of the Sun

So far, so good, but yes, you are right, the differences between the different speeds of the Sun (when the Sun is above different latitudes) would be too great to be acceptable if we calculated those differences assuming applicability of simple mathematical rules in the case of the Sun although we don't really know how Sun works at all...

That is why we should maybe reconsider these words of Sandokhan more carefully:
Indeed, calculating the FE diameter of the sun.  I was talking about how fast it travels though.
Title: Re: "Equator" problem
Post by: cikljamas on October 18, 2014, 10:55:24 AM
@ 29silhouette, calculating the differences of the different speeds of the Sun (when the Sun is above different latitudes) caused me to stumble upon "Equator" problem...

Few days ago (when there was sunny here) i have tried to measure the speed of the Sun (with a welding glass and a stopwatch). I noticed that the Sun moves one of his diameter in two minutes, that is roughly in accordance with zetetic reckoning of the diameter of the Sun AND with the RET & FET agreement over Equator circumference.
I would suggest someone along the Tropic of Cancer in June using a stick to measure the speed of the shadow as the sun travels directly overhead, and someone along the Tropic of Capricorn in December, again using stick of the same height, to measure the shadow speed.

According to the mono-pole model, which seems to be the most popular, the sun would have to move almost twice as fast in December.  If someone can come up with a way to explain how the sun could travel the same speed along two vastly different distances, yet complete each trip in the same amount of time, and visibly appear to move the same speed, we'd all love to hear it.

Quote
30 * 2 (minutes) = 1 hour
1666 / 30 = 55,5 km = 32 nautical miles = 34 statute miles = diameter of the Sun

So far, so good, but yes, you are right, the differences between the different speeds of the Sun (when the Sun is above different latitudes) would be too great to be acceptable if we calculated those differences assuming applicability of simple mathematical rules in the case of the Sun although we don't really know how Sun works at all...

That is why we should maybe reconsider these words of Sandokhan more carefully:
Indeed, calculating the FE diameter of the sun.  I was talking about how fast it travels though.

That was my first measurement of that kind, next measurments i will make sometime between 4th and 10th november, and then many more measurments to come...We have to make measurments by ourself, because this is the hoax-matrix reality in which we live, and if you want to figure out how our reality really looks like, you have to do everything by yourself, there is no help from anyone....Even Flat Earth Society is not competent (or willing)  to give you concrete answer to such elementary question as it is one that i have posed in the first post of this thread...
Title: Re: "Equator" problem
Post by: 29silhouette on October 18, 2014, 11:30:42 AM
http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)
LMAO... The concept of a globe with a force pulling matter against it's surface sure is a tough one.  ::)

Would an MS-paint diagram help?
Title: Re: "Equator" problem
Post by: 29silhouette on October 18, 2014, 12:10:25 PM
Keep up the good work.
Thanks for your words of encouragement,

Your new friend (sceptimatic) said he conducted an elaborate experiment with a high-powered laser on a frozen lake that proves Earth is flat.  Perhaps you can talk him in to providing the photos and evidence of it to you.  :)
Title: Re: "Equator" problem
Post by: sokarul on October 18, 2014, 01:43:35 PM

@sokarul, rivers on a globe would not have to run uphill, that is true, rivers on a globe would have to run downhill in both cases as it is described in my "Rivers" argument, only every day let's say at midnight downhill would be uphill, and at noon downhill would be again downhill as we know it from our everyday experience. And if you think that uphill would be impossibility because of inclinations of the river beds then i suggest you to read this for clarification:

Quote
" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles."  Read more: http://www.energeticforum.com/265601-post587.html (http://www.energeticforum.com/265601-post587.html)

In fact, rivers on a globe wouldn't flow at all, they would be spilled out into space together with all other water on such imagined globe. Don't you see how insulting for sanity of every man on the Earth is that outrageously idiotic fairytale about a imagined globular planet which our beautiful-true world supposed to be???
Gravitation, learn it.
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 01:23:52 AM
Gravitation, learn it.

http://www.energeticforum.com/257792-post142.html (http://www.energeticforum.com/257792-post142.html) Learn it!!!

In addition:

(http://i.imgur.com/4cQVajW.jpg)
(http://i.imgur.com/dgYAXWc.jpg)
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 19, 2014, 02:34:33 AM
Are you serious?  :P
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 02:58:25 AM
Regarding water level experiments:

http://www.energeticforum.com/264162-post361.html (http://www.energeticforum.com/264162-post361.html)

In addition:

(http://i.imgur.com/6LdPzpT.jpg)

Accompanying video: # (http://#)

One question for JimmyTheCrab: Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Title: Re: "Equator" problem
Post by: Macpie on October 19, 2014, 04:25:54 AM
(http://i.imgur.com/4cQVajW.jpg)
(http://i.imgur.com/dgYAXWc.jpg)

Now this is moronic. You know why usually trails look blurry and jagged when the camera is moved? Because it shakes like hell while being moved. Earth is rotating as smoothly as possible - there is literally NOTHING to shake it while the long exposure is taken.
Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 05:24:23 AM
Regarding water level experiments:

http://www.energeticforum.com/264162-post361.html (http://www.energeticforum.com/264162-post361.html)

In addition:

(http://i.imgur.com/6LdPzpT.jpg)

Accompanying video: # (http://#)

One question for JimmyTheCrab: Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Ship is clearly not on or past the horizon.
Title: Re: "Equator" problem
Post by: sceptimatic on October 19, 2014, 05:27:22 AM
Regarding water level experiments:

http://www.energeticforum.com/264162-post361.html (http://www.energeticforum.com/264162-post361.html)

In addition:

(http://i.imgur.com/6LdPzpT.jpg)

Accompanying video: # (http://#)

One question for JimmyTheCrab: Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Ship is clearly not on or past the horizon.
Take notice of the zoom in and out. the horizon is simply down to your vision over distance.
Do you think the horizon is a drawn line or something.  ::)
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 06:21:52 AM
Quote
Now this is moronic. You know why usually trails look blurry and jagged when the camera is moved? Because it shakes like hell while being moved. Earth is rotating as smoothly as possible - there is literally NOTHING to shake it while the long exposure is taken.

NOTHING to shake it, of course, because the Earth is immovable and doesn't rotate, and doesn't revolve, that is why there is literally NOTHING to shake it, couldn't you figure it out without my help?

The Earth is rotating as smoothly as possible only in RET dizzy-rotund heads, the Earth is revolving as smoothly as possible only in RET drunk-round heads, Earth is traveling around galactic centre 250 km/sec only in deluded globular RET heads!!!

Sorry, but that is ruthless truth!!!

Quote
Ship is clearly not on or past the horizon.

Stop the video at 17 sec. and pay attention to the object on the right side of the screen, has this much more distant ship passed the horizon? Do you think that there is any chance that you could see any part of this (more distant) ship with naked eyes?

So, this first ship should have been used to illustrate the principle of how objects passing the line of horizon disappear out of our perspective. If you forgot how it works read again Rowbotham's explanation of a TRUE law of perspective!!!

http://www.energeticforum.com/255927-post20.html (http://www.energeticforum.com/255927-post20.html)

The main point of this conversation is that majority of members of this forum are very skilled in skipping capital questions like it is a question posed in the opening post of this thread, and like it is a question posed in my last post, i repeat it:

Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 06:52:39 AM
Regarding water level experiments:

http://www.energeticforum.com/264162-post361.html (http://www.energeticforum.com/264162-post361.html)

In addition:

(http://i.imgur.com/6LdPzpT.jpg)

Accompanying video: # (http://#)

One question for JimmyTheCrab: Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Ship is clearly not on or past the horizon.
Take notice of the zoom in and out. the horizon is simply down to your vision over distance.
Do you think the horizon is a drawn line or something.  ::)
And as you notice, the ship in the first picture is in front of the horizon.  Therefore, it is not restoring a ship that has gone past the horizon.
Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 06:58:10 AM
Stop the video at 17 sec. and pay attention to the object on the right side of the screen, has this much more distant ship passed the horizon? Do you think that there is any chance that you could see any part of this (more distant) ship with naked eyes?
  Then maybe the guy recording the video should have zoomed in on the more distant object?  But from my experience, zooming in on an object past the horizon do not restore more of the object.  Just make what is visible larger.
Quote
So, this first ship should have been used to illustrate the principle of how objects passing the line of horizon disappear out of our perspective. If you forgot how it works read again Rowbotham's explanation of a TRUE law of perspective!!!
No it doesn't.  First, the ship is not past the horizon.  Second, it shows the limitations of the camera's digital sensors to discern details until the optical zoom is employed.
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 07:45:09 AM
@BJ1234, There are countless recorded (and minutely described and explained on the basis of a true law of perspective) testimonies (experiments)  of a different people about their successful visual restorations of hulls of a sailing away ships by applying telescopes or binoculars or strong optical lenses of cameras. Would you like to assert that all minutely described experiences of that kind are just false testimonies?

(http://i.imgur.com/a679K9y.jpg)


Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 07:58:16 AM
@BJ1234, There are countless recorded (and minutely described and explained on the basis of a true law of perspective) testimonies (experiments)  of a different people about their successful visual restorations of hulls of a sailing away ships by applying telescopes or binoculars or strong optical lenses of cameras. Would you like to assert that all minutely described experiences of that kind are just false testimonies?

(http://i.imgur.com/a679K9y.jpg)


Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?


Then show me a video where the hull of a ship has been restored AFTER it has gone past the horizon.  So fare, every video posted has been of ships in front of the horizon.
Title: Re: "Equator" problem
Post by: markjo on October 19, 2014, 08:29:22 AM
Regarding water level experiments:
What do water level experiments have to do with the equator?  ???
Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 08:41:46 AM
Also tell me why, if the disappearance is due to perspective, the hulls disappear first, which are much larger than masts or stacks on a ship?  Logically, these should disappear first as they are smaller.
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 08:55:38 AM
Then show me a video where the hull of a ship has been restored AFTER it has gone past the horizon.  So fare, every video posted has been of ships in front of the horizon.

(http://i.imgur.com/QKLhI41.jpg)
(http://i.imgur.com/LCOcwIQ.jpg)
Flat Earth Response Video To Everyone its so flat__________________________ (http://#) (Starts at 1min. 30 sec.)
Amazing video : #t=47 (http://#t=47)
Time lapse photographs relating video: Flat motionless earth vs globe challenge (http://#)




Quote
What do water level experiments have to do with the equator?

I am still waiting for that answer!

Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 09:16:36 AM
Then show me a video where the hull of a ship has been restored AFTER it has gone past the horizon.  So fare, every video posted has been of ships in front of the horizon.

(http://i.imgur.com/QKLhI41.jpg)
(http://i.imgur.com/LCOcwIQ.jpg)
Flat Earth Response Video To Everyone its so flat__________________________ (http://#) (Starts at 1min. 30 sec.)
Amazing video : #t=47 (http://#t=47)
Time lapse photographs relating video: Flat motionless earth vs globe challenge (http://#)



Those pictures are not showing what I am asking.  They show that raising elevation allows you to see further.  They do not show a hull being restored after it has gone past the horizon.  And I assumed I didn't need to add by zooming in on the ship since the previous video showed a camera zooming.

Don't have time to watch the video provided right now.  Will watch it later.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 19, 2014, 09:32:32 AM
Regarding water level experiments:

http://www.energeticforum.com/264162-post361.html (http://www.energeticforum.com/264162-post361.html)

In addition:

(http://i.imgur.com/6LdPzpT.jpg)


The ship is not on the horizon.   ::)
Title: Re: "Equator" problem
Post by: 29silhouette on October 19, 2014, 10:54:33 AM
Ships and hillside sinking below curvature of Earth as point of view is lowered.
*original thread- http://theflatearthsociety.org/forum/index.php?topic=50707.0#.VEP3m2fvgSk (http://theflatearthsociety.org/forum/index.php?topic=50707.0#.VEP3m2fvgSk)

Ship 1, 8x mag. 12m alt.
(http://farm7.static.flickr.com/6155/6155738378_71f20689ea_o.jpg) (http://www.flickr.com/photos/pitdroidtech/6155738378/)
Ships Below the Horizon (http://www.flickr.com/photos/pitdroidtech/6155738378/#) by max_wedge (http://www.flickr.com/people/pitdroidtech/), on Flickr

Ship 1, 8x mag. 0m alt.
(http://farm7.static.flickr.com/6068/6155193283_8a080a6256_o.jpg) (http://www.flickr.com/photos/pitdroidtech/6155193283/)
Ships Below the Horizon (http://www.flickr.com/photos/pitdroidtech/6155193283/#) by max_wedge (http://www.flickr.com/people/pitdroidtech/), on Flickr



Ship 2, 8x mag. 0m alt.
(http://farm7.static.flickr.com/6207/6155192949_68fa85794e_o.jpg) (http://www.flickr.com/photos/pitdroidtech/6155192949/)
Ships Below the Horizon (http://www.flickr.com/photos/pitdroidtech/6155192949/#) by max_wedge (http://www.flickr.com/people/pitdroidtech/), on Flickr

Ship 2, 8x mag. 12m alt.
(http://farm7.static.flickr.com/6164/6155192737_9c088a5df7_o.jpg) (http://www.flickr.com/photos/pitdroidtech/6155192737/)
Ships Below the Horizon (http://www.flickr.com/photos/pitdroidtech/6155192737/#) by max_wedge (http://www.flickr.com/people/pitdroidtech/), on Flickr


No restoration here.

Ship 3, 8x mag. 0m alt.
(http://farm7.static.flickr.com/6179/6155737584_cfc0fb517a_o.jpg) (http://www.flickr.com/photos/pitdroidtech/6155737584/)
Ships Below the Horizon (http://www.flickr.com/photos/pitdroidtech/6155737584/#) by max_wedge (http://www.flickr.com/people/pitdroidtech/), on Flickr

Ship 3, 28x mag. 0m alt. 
(http://farm7.static.flickr.com/6086/6155192491_a10315b3ca_b.jpg) (http://www.flickr.com/photos/pitdroidtech/6155192491/)
Ships Below the Horizon (http://www.flickr.com/photos/pitdroidtech/6155192491/#) by max_wedge (http://www.flickr.com/people/pitdroidtech/), on Flickr
Title: Re: "Equator" problem
Post by: 29silhouette on October 19, 2014, 11:08:17 AM
In addition:

http://i.imgur.com/4cQVajW.jpg (http://i.imgur.com/4cQVajW.jpg)
http://i.imgur.com/dgYAXWc.jpg (http://i.imgur.com/dgYAXWc.jpg)
The camera is stationary in relation to the surface and moves right along with Earth's rotation, thus the stars become streaks across the sky and the ground sharply focused. 

Learn some photography and then get back to us.
Title: Re: "Equator" problem
Post by: 29silhouette on October 19, 2014, 11:47:39 AM
Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Because (assuming calm water) any one spot on the surface of a given body of water is level (not tilted).   
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 11:54:03 AM
Answer to post #76 : http://www.sacred-texts.com/earth/za/za34.htm (http://www.sacred-texts.com/earth/za/za34.htm)

Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Because (assuming calm water) any one spot on the surface of a given body of water is level (not tilted).

Not tilted? Like the Earth? Hahahahaha...
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 12:24:08 PM
This is the video that you were asking for: (http://)
Title: Re: "Equator" problem
Post by: cikljamas on October 19, 2014, 01:21:07 PM
1st hd camera at the Edge of Space:

(http://i.imgur.com/4hQgcN4.jpg)

accompanying video: 1st HD Camera At The Edge Of Space (http://#ws)

Flat Earth/Terra Plana : Flat Earth/Terra Plana (http://#ws)

Now that we are 100 % sure that the Earth is flat, you can freely spell out your suggestions about the most logical possible solutions for resolving Equator problem, don't be afraid, you got nothing to lose, you already lost this game long time ago, since the creation of the world...

Title: Re: "Equator" problem
Post by: BJ1234 on October 19, 2014, 06:50:34 PM
This is the video that you were asking for: (http://)
As you can see in that video, no more of the hull is appearing.  The portions of the ship are staying the same relative size.  The hull isn't gaining in proportional size.
Title: Re: "Equator" problem
Post by: ausGeoff on October 19, 2014, 08:25:18 PM
1st hd camera at the Edge of Space:

(http://i.imgur.com/4hQgcN4.jpg)


I'm afraid selecting a single frame from the video that happens to show a horizontal cloud band (rather than the earth's horizon) does nothing for your cause—except weaken it.  Sorry.

At least the two videos you've inadvertently (?) linked to show the actual curved horizon at various stages.
Title: Re: "Equator" problem
Post by: 29silhouette on October 19, 2014, 08:35:28 PM
http://i.imgur.com/QKLhI41.jpg (http://i.imgur.com/QKLhI41.jpg)
Nice picture showing curvature there.

Quote
Flat Earth Response Video To Everyone its so flat__________________________ (http://#) (Starts at 1min. 30 sec.)
Not sure what going on with the music drowning out the narration.  Anyway, at 1:30 we see a nice example of resolution getting a boost from magnification, with additional help from infrared.  Not seeing any example of 'restoration' though.

Quote
Amazing video : #t=47 (http://#t=47)
A ship traveling through large waves.  Neat.

Quote
Time lapse photographs relating video: Flat motionless earth vs globe challenge (http://#) 
LMAO... this guy can't even get the speed of rotation correct ("1000mph faster than the speed of sound"?), nor can he comprehend one revolution per day.  Spin a merry-go-round at that rate and get back to us.
Title: Re: "Equator" problem
Post by: cikljamas on October 20, 2014, 03:34:16 AM
IT has already been proved that the astronomers of the Copernican school merely assumed the rotundity of the earth as a doctrine which enabled them to explain certain well-known phenomena. "What other explanation can be imagined except the sphericity of the earth?" is the language of Professor de Morgan, and it expresses the state of mind of all who hold that the earth is a globe. There is on their part an almost amusing innocence of the fact, than in seeking to explain phenomena by the assumption of rotundity, another assumption is necessarily involved, viz., that nothing else will explain the phenomena in question but the foregone and gratuitous conclusion to which they have committed themselves. To argue, for instance, that because the lower part of an outward-bound vessel disappears before the mast-head, the water must be round, is to assume that a round surface only can produce such an effect. But if it can be shown that a simple law of perspective in connection with a plane surface necessarily produces this appearance, the assumption of rotundity is not required, and all the misleading fallacies and confusion involved in or mixed up with it may be avoided.

Before explaining the influence of perspective in causing-the hull of a ship to disappear first when outward bound, it is necessary to remove an error in its application, which artists and teachers have generally committed, and which if persisted in will not only prevent their giving, as it has hitherto done, absolutely correct representations of natural things, but also deprive them of the power to understand the cause of the lower part of any receding object disappearing to the eye before any higher portion--even though the surface on which it moves is admittedly and provably horizontal.

"The range of the eye, or diameter of the field of vision, is 110°; consequently this is the largest angle under which an object can be seen. The range of vision is from 110° to 1°. . . . The smallest angle under which an object can be seen is upon an average, for different sights, the sixtieth part of a degree, or one minute in space; so that when an object is removed from the eye 3000 times its own diameter, it will only just be distinguishable; consequently the greatest distance at which we can behold an object like a shilling of an inch in diameter, is 3000 inches or 250 feet." 1

The above may be called the law of perspective. It may be given in more formal language, as the following:. when any object or any part thereof is so far removed that its greatest diameter subtends at the eye of the observer, an angle of one minute or less of a degree, it is no longer visible.

From the above it follows:--

1.--That the larger the object the further will it require to go from the observer before it becomes invisible.

2.--The further any two bodies, or any two parts of the same body, are asunder, the further must they recede before they appear to converge to the same point.

3.--Any distinctive part of a receding body will be-come invisible before the whole or any larger part of the same body.
Read more: http://www.sacred-texts.com/earth/za/za32.htm (http://www.sacred-texts.com/earth/za/za32.htm)

Classical example:

(http://i.imgur.com/iHng4Us.jpg)

Accompanying video: (http://)

In addition: Ship :

(http://)
(http://)

Now you can proceed with your wordplay (obscuring the meanings of the words)....

Title: Re: "Equator" problem
Post by: BJ1234 on October 20, 2014, 05:44:58 AM
Yes, Iunderstand how you claim it works.  However, none of these videos actually are supporting your claims of ships being restored from beyond the horizon.
Title: Re: "Equator" problem
Post by: cikljamas on October 20, 2014, 06:12:36 AM
Yes, Iunderstand how you claim it works.  However, none of these videos actually are supporting your claims of ships being restored from beyond the horizon.

Or maybe being restored from beyond vanishing point, shouldn't we use that phrase ("vanishing point") instead of "horizon", so that you cannot play with words any more?  ;)
Title: Re: "Equator" problem
Post by: Goth on October 20, 2014, 06:31:16 AM
The Earth is Flat, Rory Cooper says so! Part II: The Horizon Problem (http://#ws)


The Earth is Flat, Rory Cooper says so!
Title: Re: "Equator" problem
Post by: BJ1234 on October 20, 2014, 06:32:37 AM
Yes, Iunderstand how you claim it works.  However, none of these videos actually are supporting your claims of ships being restored from beyond the horizon.

Or maybe being restored from beyond vanishing point, shouldn't we use that phrase ("vanishing point") instead of "horizon", so that you cannot play with words any more?  ;)
Now who's vanishing point? Yours or the camera's?  The digital camera's sensor might not be adequate enough to see the boat out on the ocean while a human eye can? 
Title: Re: "Equator" problem
Post by: cikljamas on October 20, 2014, 07:18:43 AM
Now who's vanishing point? Yours or the camera's?  The digital camera's sensor might not be adequate enough to see the boat out on the ocean while a human eye can?

What exactly you don't understand?

Do these (following) words can help you to finally comprehend the true law of perspective:

From the several cases now advanced, which are selected from a great number of instances involving the same law, the third proposition (on page 203) that "any distinctive part of a body will become invisible before the whole or any larger part of the same body," is sufficiently demonstrated. It will therefore be readily seen that the hull of a receding ship obeying the same law must disappear on a plane surface, before the mast head. If it is put in the form of a syllogism the conclusion is inevitable:--

Any distinctive part of a receding object becomes invisible before the whole or any larger part of the same object.

The hull is a distinctive part of a ship.

Ergo, the hull of a receding or outward bound ship must disappear before the whole, inclusive of the mast head.

To give the argument a more practical and nautical character it may be stated as follows:

That part of any receding body which is nearest to the surface upon which it moves, contracts, and becomes in-visible before the parts which are further away from such surface--as shown in figs. 63, 64, 65, 66, 67, 68, 69, and 70.

The hull of a ship is nearer to the water--the surface on which it moves--than the mast head.

Ergo, the hull of an outward bound ship must be the first to disappear.

This will be seen mathematically in the following diagram, fig. 83.
Title: Re: "Equator" problem
Post by: BJ1234 on October 20, 2014, 07:41:26 AM
Now who's vanishing point? Yours or the camera's?  The digital camera's sensor might not be adequate enough to see the boat out on the ocean while a human eye can?

What exactly you don't understand?

Do these (following) words can help you to finally comprehend the true law of perspective:

From the several cases now advanced, which are selected from a great number of instances involving the same law, the third proposition (on page 203) that "any distinctive part of a body will become invisible before the whole or any larger part of the same body," is sufficiently demonstrated. It will therefore be readily seen that the hull of a receding ship obeying the same law must disappear on a plane surface, before the mast head. If it is put in the form of a syllogism the conclusion is inevitable:--

Any distinctive part of a receding object becomes invisible before the whole or any larger part of the same object.

The hull is a distinctive part of a ship.

Ergo, the hull of a receding or outward bound ship must disappear before the whole, inclusive of the mast head.

To give the argument a more practical and nautical character it may be stated as follows:

That part of any receding body which is nearest to the surface upon which it moves, contracts, and becomes in-visible before the parts which are further away from such surface--as shown in figs. 63, 64, 65, 66, 67, 68, 69, and 70.

The hull of a ship is nearer to the water--the surface on which it moves--than the mast head.

Ergo, the hull of an outward bound ship must be the first to disappear.

This will be seen mathematically in the following diagram, fig. 83.
And this addresses my question how?  My question is if the digital sensor inside the camera has the ability to discern the ship?  Or possibly the video encoding that took place muddied the distinction between the hull and the water line.

How large do you think a hull is versus how large a mast is?  If a hull is 30 feet, according to the law of perspective you quoted, the vanishing point is 3000 times that so 90,000 feet.  Which is between 15 and 20 miles away.  How would that disappear before a mast of say 2 feet in diameter which should vanish from sight in 6000 feet?
Title: Re: "Equator" problem
Post by: cikljamas on October 20, 2014, 09:23:06 AM
And this addresses my question how?  My question is if the digital sensor inside the camera has the ability to discern the ship?  Or possibly the video encoding that took place muddied the distinction between the hull and the water line.

How large do you think a hull is versus how large a mast is?  If a hull is 30 feet, according to the law of perspective you quoted, the vanishing point is 3000 times that so 90,000 feet.  Which is between 15 and 20 miles away.  How would that disappear before a mast of say 2 feet in diameter which should vanish from sight in 6000 feet?

Your question is somehow misleading, the truth is that the vanishing point is at the distance of 3000 diameters of an receding object, the truth is also that RET and FET are in agreement concerning the fact that a hull of a ship disappears from our perspective before a mast, only the question is why and how it happens. A hull of a ship has larger diameter than a mast, but obviously in this case it is not the primary factor which decidedly determines order of sequences of ship's disappearance on the horizon, the primary factor is rather a true law of perspective. All i could do in my last attempt to help you out comprehending the real cause of this phenomena is to emphasise (once more) two excerpts from already linked  Rowbotham's explanatory chapter that deals with this phenomena:

(http://i.imgur.com/0wz0fzM.jpg)
(http://i.imgur.com/KGIrjIx.jpg)


Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 20, 2014, 10:41:36 AM

Any distinctive part of a receding object becomes invisible before the whole or any larger part of the same object.
Nonsense.

Quote
The hull is a distinctive part of a ship.

Why does the bottom of the hull disappear first?
Title: Re: "Equator" problem
Post by: Rama Set on October 20, 2014, 12:26:50 PM

Any distinctive part of a receding object becomes invisible before the whole or any larger part of the same object.
Nonsense.

Quote
The hull is a distinctive part of a ship.

Why does the bottom of the hull disappear first?

Naturally because the bottom of the hull is a distinctive part of the ship.  Duh...
Title: Re: "Equator" problem
Post by: sokarul on October 20, 2014, 03:45:50 PM
Gravitation, learn it.

http://www.energeticforum.com/257792-post142.html (http://www.energeticforum.com/257792-post142.html) Learn it!!!
Unfounded opinions don't really prove anything.
Title: Re: "Equator" problem
Post by: 29silhouette on October 20, 2014, 09:05:30 PM
Answer to post #76 : http://www.sacred-texts.com/earth/za/za34.htm (http://www.sacred-texts.com/earth/za/za34.htm)
And when waves or swells are minimal, objects still sink below the horizon.  Also, if one's line of sight is above the average wave height, then there's nothing to block the line of sight.

Quote
Why do we put word "level" in a phrase "water level" if the surface of all waters is not leveled but curved?
Because (assuming calm water) any one spot on the surface of a given body of water is level (not tilted).

Not tilted? Like the Earth? Hahahahaha...
Yeah, don't worry about it.  Some people can understand round objects, some can't.

Any distinctive part of a receding object becomes invisible before the whole or any larger part of the same object.

The hull is a distinctive part of a ship.
So what about those little 'distinctive' parts of a ship, such as people and windows in your "classical example" in post #85, that all disappear before the 'larger' parts, such as the hull?
Title: Re: "Equator" problem
Post by: cikljamas on October 21, 2014, 01:42:20 AM
If the Earth is round then you should be able to plausibly explain this:

1: Lighthouses: http://www.energeticforum.com/264766-post457.html (http://www.energeticforum.com/264766-post457.html)

2: Experiments on lake Michigan: http://www.sacred-texts.com/earth/cc/cc21.htm (http://www.sacred-texts.com/earth/cc/cc21.htm)

3: Exact formula for counting visibility range for two heights and striking examples which render validity of FET beyond dispute: http://www.energeticforum.com/258148-post188.html (http://www.energeticforum.com/258148-post188.html)

4: At sea the datum line is always a horizontal line - spherical trigonometry is never used - That all triangulation used at sea is PLANE, PROVES THAT THE SEA IS A PLANE : http://www.energeticforum.com/265962-post590.html (http://www.energeticforum.com/265962-post590.html)

5: Railways: In 1862, the Houses of Lords and Commons issued an Order that all Railways were to be constructed on a Datum Horizontal line without allowing one inch for curvature.

6: Canals:

Example A: " The German Emperor performed the ceremony of opening the Gates of the Baltic and North Sea Canal, in the spring of 1891. The canal starts at Hollenau, on the south side of Kiel Hay, and Joins the Elbe 15 miles above its mouth, It is 61 miles long, 200 feet wide at the surface and 85 feet at bottom, the depth being 28 feet. No locks are required, as the surface of the two seas is level."

Example B: Let those who believe it is the practice for surveyors to make allowance for "curvature" ponder over the following from the Manchester Ship Canal Company, — (Earth Review, October, 1893), " It is customary in Railway and Canal constructions for all levels to be referred to a datum which is nominally horizontal, and is so shown on all sections. It is not the practice in laying out Public Works to make allowance for the curvature of the earth." — Manchester Ship Canal Co., Engineer's Office, 19th February, 1892!

Example C:
A surveyor, Mr. T. Westwood, writes to the Earth Review for January, 1896, as follows :

" In leveling, I work from Ordnance marks, or canal levels, to get the height above sea level I work sometimes from what is known as the Wolverhampton level, this is said to be 473.19 feet above sea level ; sometimes I work from the Birmingham level, this is said to be 453.04 feet above sea level. Sometimes I work from the Walsal level, this is said to be 407.89 feet above sea level. The puzzle to me used to be, that, though each extends several miles, each level was and is treated throughout its whole length as the same level from end to end ; not the least allowance being made for curvature, although if the earth were a globe, 112 feet ought to be allowed... One of the civil engineers in this district, after some amount of argument on each side as to the reason why no allowance for curvature was made, said he did not believe anybody would know the shape of the earth in this life."

7: London to Moscow proof : In " Chambers' Information for the People," section on Physical Geography, page 513, the following occurs:

"In North America, the basin or drainage of the Mississippi is estimated at 1.300.000 square miles, and that of the St. Lawrened at 600,000; while northward of the 50th parallel, extends an inhospitable FLAT of perhaps greater dimensions. . . . Next in order of importance is that section of Europe extending from the German Sea, through Prussia. Poland, and Russia, towards the Ural Mountains, presenting indifferently tracts of heath, sand and open pasture, and regarded by geographers as ONE VAST PLANE. So flat is the general profile of the region, that It has been remarked, IT IS POSSIBLE TO DRAW A LINE FROM LONDON TO MOSCOW, WHICH WOULD NOT PERCEPTIBLY VARY FROM A DEAD LEVEL."

The foregoing is a London-to-Moscow proof that the surface of the world is not globular.

8: Extraordinary flatness of the Earth's crust: These extracts clearly prove that the surface of the earth is a level surface, and that, therefore, the world is not a globe. And when we come to consider the surface of the world under the sea, we shall find the same unformity of evidence against the popular view. In " Nature and Man," by Professor W. B. Carpenter, article " The Deep Sea and its Contents," pages 320 and 321, the writer says :

"Nothing seems to have struck the "Challenger" surveyors more than the extraordinary FLATNESS (except in the neighbourhood of land) of that depressed portion of the earth's crust which forms the FLOOR OF THE GREAT OCEANIC AREA. . . . If the bottom of mid-ocean were laid dry, an observer standing on any spot of it would Jind himself surrounded BY A PLAIN, only comparable to that of the North American prairies or the South American pampas The form of the depressed area which lodges the water of the deep ocean is rather, indeed, to be likened to that of a FLAT WAITER or TEA TRAY, surrounded bj- an elevated and deeply -sloping rim, than to that of the basin with which it is commonly compared."

9: RIVERS:
From the "Atlas of Physical Geography," by the Rev. T.Milner, M.A., I extract the following:

" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles."

"The plains of Venezuela and New Granada, in South America, chiefiy on the left of the Orinoco, are termed llanos, or level fields. Often in the space of 270 square miles THE SURFACE DOES NOT VARY A SINGLE FOOT."

"The Amazon only falls 12 feet in the last 700 miles of its course; the La Plata has only a descent of one thirty-third of an inch a mile,"

NOT THAT YOU CANNOT PLAUSIBLY EXPLAIN ABOVE FET UNDENIABLE PROOFS, YOU CANNOT EVEN START TO ARGUE AGAINST THE OBVIOUSNESS OF ABOVE ARGUMENTS!!!

One additional undeniable proof no. 10: EYES LEVELED HORIZON NO MATTER HOW HIGH YOU LIFT YOUR 'EYES' UP!!!

So, there is no doubt about the general shape of the surface of the Earth, the question is: How to resolve "Equator" problem, KNOWING VERY WELL that the surface of the Earth is Flat???

Be of some use after so long time (spent on this forum)!
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 21, 2014, 02:20:42 AM
@cikljamas, what's with your 19th century obsession?
Title: Re: "Equator" problem
Post by: cikljamas on October 21, 2014, 03:03:22 AM
@cikljamas, what's with your 19th century obsession?

Let me show you one picture from 19th century:

(http://i.imgur.com/vw594Y7.jpg)

Accompanying video: Flat Earth/Terra Plana (http://#ws)

It is interesting how round earthers always evade concrete challenges by use of ridiculous "19th century "argument"" , but right after usage of that idiotic "argument" they aint got any problems with bringing forth "Eratostens" or "Pitagora" or any other 2 500 years old "argument".... i mean, i mean, i mean...#t=2m54s.... (http://#t=2m54s....)
Title: Re: "Equator" problem
Post by: Rama Set on October 21, 2014, 04:12:03 AM
It's amazing reading over your so called proof's, how little you understand of physics, mathematics and engineering.
Title: Re: "Equator" problem
Post by: cikljamas on October 21, 2014, 05:08:34 AM
It's amazing reading over your so called proof's, how little you understand of physics, mathematics and engineering.

Let me show you how you little understand of anything:

Heliocentricity is not a logically plausible (let alone irrefutable) theory that is based on scientific data but is actually, purely based on a series of assumptions that were built-up over the last 200 years. For example many (but not all) of the assertions regarding astronomical distances between celestial bodies are based on the necessary assumption that the earth must be revolving around the sun.
 
But at the same time, these assumed distances have another function whereby they are deployed as some sort of supportive argument for the "trueness" of the heliocentric hypothesis. For example we are told that sun is too big to revolve around the earth, despite the fact that the sun's size was determined in the first place by assuming how big it must have to be in order to allow a heliocentric premise! Go figure. Other needed assumptions include:
 
■ the bendover earth (alleged 'tilt' of the earth's axis - a desperately needed heliocentric variable that has no basis in the physical world where the sun simply spirals from the Tropic of Cancer to the Tropic of Capricorn annually. Both of these tropic latitude lines are not tilted - they are at a 0° angle (= parallel) to the equator. The word "tropic" itself comes from the Greek term tropos, meaning turn, referring to the fact that the sun "turns back" at these lines that aren't tilted in any way,
 
■ the earth supposedly jittering around the sun at various speed levels (it orbits at a faster speed at one time, and then it goes relatively slower at another - then back faster again) but somehow, all this alleged speed-change remains unnoticeable),
 
■ the moon also being dragged along exactly at those same speed levels (100% complete synchronization with the wobbly earth despite being hundreds of thousands of miles away from it(!) Now how about that?,
 
■ even atmospheric gas (the air) being attached to the earth's surface (again completely synchronized but somehow (simultaneously) free-flowing enough to blow in every direction). These are just samples of the never shown, never detected, never scientifically observed absurdities that are required to save the appearances of the heliocentric model.

 Unproven Heliocentric Assumptions
When Heliocentrists failed to disprove the geocentric nature that we live in, they resorted to inventing assumptions, many of which are so absurd that the inventors themselves admit that they are unfalsifiable (by implication unscientific) thought-experiments. Some of these assumptions include:

    -    the alleged tilt of the earth's axis,
     
    -    the so called Copernican principle,
     
    -    positive stellar parallax,
     
    -    uniformitiy of the speed of light,
     
    -    lengh contraction
     
    -    time dilation
     
    -    denial of inertia (only accepting an imaginary and isolated "chosen" inertial frame of reference)
     
    -    the earth supposedly moving at a various speeds (in order to account for the observed eclipses)

These and many other assumptions are presented as evidence to each other. In other words one assumption is used in order to prove another assumption. In fact these assumptions are so fundamentally dependent on each other that one becomes meaningless without the other, which shows that heliocentrists don't refrain from applying deceit (circular reasoning in this case) in order to make their assertions believable.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 21, 2014, 05:44:16 AM
Yeah, but what's with your 19th century obsession?
Title: Re: "Equator" problem
Post by: cikljamas on October 21, 2014, 06:09:58 AM
Yeah, but what's with your 19th century obsession?

Can' believe how fast you forget things, let me refresh your memory:

Quote
@cikljamas, what's with your 19th century obsession?

Let me show you one picture from 19th century:

(http://i.imgur.com/vw594Y7.jpg)

Accompanying video: Flat Earth/Terra Plana (http://#ws)

It is interesting how round earthers always evade concrete challenges by use of ridiculous "19th century "argument"" , but right after usage of that idiotic "argument" they aint got any problems with bringing forth "Eratostens" or "Pitagora" or any other 2 500 years old "argument".... i mean, i mean, i mean...#t=2m54s.... (http://#t=2m54s....)

Now, when are you going to answer to these irrefutable arguments:

If the Earth is round then you should be able to plausibly explain this:

1: Lighthouses: http://www.energeticforum.com/264766-post457.html (http://www.energeticforum.com/264766-post457.html)

2: Experiments on lake Michigan: http://www.sacred-texts.com/earth/cc/cc21.htm (http://www.sacred-texts.com/earth/cc/cc21.htm)

3: Exact formula for counting visibility range for two heights and striking examples which render validity of FET beyond dispute: http://www.energeticforum.com/258148-post188.html (http://www.energeticforum.com/258148-post188.html)

4: At sea the datum line is always a horizontal line - spherical trigonometry is never used - That all triangulation used at sea is PLANE, PROVES THAT THE SEA IS A PLANE : http://www.energeticforum.com/265962-post590.html (http://www.energeticforum.com/265962-post590.html)

5: Railways: In 1862, the Houses of Lords and Commons issued an Order that all Railways were to be constructed on a Datum Horizontal line without allowing one inch for curvature.

6: Canals:

Example A: " The German Emperor performed the ceremony of opening the Gates of the Baltic and North Sea Canal, in the spring of 1891. The canal starts at Hollenau, on the south side of Kiel Hay, and Joins the Elbe 15 miles above its mouth, It is 61 miles long, 200 feet wide at the surface and 85 feet at bottom, the depth being 28 feet. No locks are required, as the surface of the two seas is level."

Example B: Let those who believe it is the practice for surveyors to make allowance for "curvature" ponder over the following from the Manchester Ship Canal Company, — (Earth Review, October, 1893), " It is customary in Railway and Canal constructions for all levels to be referred to a datum which is nominally horizontal, and is so shown on all sections. It is not the practice in laying out Public Works to make allowance for the curvature of the earth." — Manchester Ship Canal Co., Engineer's Office, 19th February, 1892!

Example C:
A surveyor, Mr. T. Westwood, writes to the Earth Review for January, 1896, as follows :

" In leveling, I work from Ordnance marks, or canal levels, to get the height above sea level I work sometimes from what is known as the Wolverhampton level, this is said to be 473.19 feet above sea level ; sometimes I work from the Birmingham level, this is said to be 453.04 feet above sea level. Sometimes I work from the Walsal level, this is said to be 407.89 feet above sea level. The puzzle to me used to be, that, though each extends several miles, each level was and is treated throughout its whole length as the same level from end to end ; not the least allowance being made for curvature, although if the earth were a globe, 112 feet ought to be allowed... One of the civil engineers in this district, after some amount of argument on each side as to the reason why no allowance for curvature was made, said he did not believe anybody would know the shape of the earth in this life."

7: London to Moscow proof : In " Chambers' Information for the People," section on Physical Geography, page 513, the following occurs:

"In North America, the basin or drainage of the Mississippi is estimated at 1.300.000 square miles, and that of the St. Lawrened at 600,000; while northward of the 50th parallel, extends an inhospitable FLAT of perhaps greater dimensions. . . . Next in order of importance is that section of Europe extending from the German Sea, through Prussia. Poland, and Russia, towards the Ural Mountains, presenting indifferently tracts of heath, sand and open pasture, and regarded by geographers as ONE VAST PLANE. So flat is the general profile of the region, that It has been remarked, IT IS POSSIBLE TO DRAW A LINE FROM LONDON TO MOSCOW, WHICH WOULD NOT PERCEPTIBLY VARY FROM A DEAD LEVEL."

The foregoing is a London-to-Moscow proof that the surface of the world is not globular.

8: Extraordinary flatness of the Earth's crust: These extracts clearly prove that the surface of the earth is a level surface, and that, therefore, the world is not a globe. And when we come to consider the surface of the world under the sea, we shall find the same unformity of evidence against the popular view. In " Nature and Man," by Professor W. B. Carpenter, article " The Deep Sea and its Contents," pages 320 and 321, the writer says :

"Nothing seems to have struck the "Challenger" surveyors more than the extraordinary FLATNESS (except in the neighbourhood of land) of that depressed portion of the earth's crust which forms the FLOOR OF THE GREAT OCEANIC AREA. . . . If the bottom of mid-ocean were laid dry, an observer standing on any spot of it would Jind himself surrounded BY A PLAIN, only comparable to that of the North American prairies or the South American pampas The form of the depressed area which lodges the water of the deep ocean is rather, indeed, to be likened to that of a FLAT WAITER or TEA TRAY, surrounded bj- an elevated and deeply -sloping rim, than to that of the basin with which it is commonly compared."

9: RIVERS:
From the "Atlas of Physical Geography," by the Rev. T.Milner, M.A., I extract the following:

" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles."

"The plains of Venezuela and New Granada, in South America, chiefiy on the left of the Orinoco, are termed llanos, or level fields. Often in the space of 270 square miles THE SURFACE DOES NOT VARY A SINGLE FOOT."

"The Amazon only falls 12 feet in the last 700 miles of its course; the La Plata has only a descent of one thirty-third of an inch a mile,"

NOT THAT YOU CANNOT PLAUSIBLY EXPLAIN ABOVE FET UNDENIABLE PROOFS, YOU CANNOT EVEN START TO ARGUE AGAINST THE OBVIOUSNESS OF ABOVE ARGUMENTS!!!

One additional undeniable proof no. 10: EYES LEVELED HORIZON NO MATTER HOW HIGH YOU LIFT YOUR 'EYES' UP!!!

So, there is no doubt about the general shape of the surface of the Earth, the question is: How to resolve "Equator" problem, KNOWING VERY WELL that the surface of the Earth is Flat???

Be of some use after so long time (spent on this forum)!

Title: Re: "Equator" problem
Post by: Rama Set on October 21, 2014, 07:35:17 AM

If the Earth is round then you should be able to plausibly explain this:

1: Lighthouses: http://www.energeticforum.com/264766-post457.html (http://www.energeticforum.com/264766-post457.html)

The elevation of the Italian coast in the closest area to the light house you mention is in the 45m-92m range.  Couple that with the 90m-100m height of the lighthouse and refraction, and 70 miles does not sound too improbable.  Also, one should not believe every story from an old navigator.  They often have syphilis.

Quote
2: Experiments on lake Michigan: http://www.sacred-texts.com/earth/cc/cc21.htm (http://www.sacred-texts.com/earth/cc/cc21.htm)

Your own source shows that when the observers lowered their elevation the ships disappeared.  It sounds like another case of refraction not being accounted for.

Quote
3: Exact formula for counting visibility range for two heights and striking examples which render validity of FET beyond dispute: http://www.energeticforum.com/258148-post188.html (http://www.energeticforum.com/258148-post188.html)

Probably refraction again.

Quote
4: At sea the datum line is always a horizontal line - spherical trigonometry is never used - That all triangulation used at sea is PLANE, PROVES THAT THE SEA IS A PLANE : http://www.energeticforum.com/265962-post590.html (http://www.energeticforum.com/265962-post590.html)

Datum lines use mean height above sea level, which is curved, but can be normalized to appear as a straight line.

Quote
5: Railways: In 1862, the Houses of Lords and Commons issued an Order that all Railways were to be constructed on a Datum Horizontal line without allowing one inch for curvature.

Railroads frequently exhibit much more flexibility that that required by the curvature of the Earth.  Why should a special allowance be needed?

Quote
6: Canals:

Example A: " The German Emperor performed the ceremony of opening the Gates of the Baltic and North Sea Canal, in the spring of 1891. The canal starts at Hollenau, on the south side of Kiel Hay, and Joins the Elbe 15 miles above its mouth, It is 61 miles long, 200 feet wide at the surface and 85 feet at bottom, the depth being 28 feet. No locks are required, as the surface of the two seas is level."

Example B: Let those who believe it is the practice for surveyors to make allowance for "curvature" ponder over the following from the Manchester Ship Canal Company, — (Earth Review, October, 1893), " It is customary in Railway and Canal constructions for all levels to be referred to a datum which is nominally horizontal, and is so shown on all sections. It is not the practice in laying out Public Works to make allowance for the curvature of the earth." — Manchester Ship Canal Co., Engineer's Office, 19th February, 1892!

Example C:
A surveyor, Mr. T. Westwood, writes to the Earth Review for January, 1896, as follows :

" In leveling, I work from Ordnance marks, or canal levels, to get the height above sea level I work sometimes from what is known as the Wolverhampton level, this is said to be 473.19 feet above sea level ; sometimes I work from the Birmingham level, this is said to be 453.04 feet above sea level. Sometimes I work from the Walsal level, this is said to be 407.89 feet above sea level. The puzzle to me used to be, that, though each extends several miles, each level was and is treated throughout its whole length as the same level from end to end ; not the least allowance being made for curvature, although if the earth were a globe, 112 feet ought to be allowed... One of the civil engineers in this district, after some amount of argument on each side as to the reason why no allowance for curvature was made, said he did not believe anybody would know the shape of the earth in this life."

7: London to Moscow proof : In " Chambers' Information for the People," section on Physical Geography, page 513, the following occurs:

"In North America, the basin or drainage of the Mississippi is estimated at 1.300.000 square miles, and that of the St. Lawrened at 600,000; while northward of the 50th parallel, extends an inhospitable FLAT of perhaps greater dimensions. . . . Next in order of importance is that section of Europe extending from the German Sea, through Prussia. Poland, and Russia, towards the Ural Mountains, presenting indifferently tracts of heath, sand and open pasture, and regarded by geographers as ONE VAST PLANE. So flat is the general profile of the region, that It has been remarked, IT IS POSSIBLE TO DRAW A LINE FROM LONDON TO MOSCOW, WHICH WOULD NOT PERCEPTIBLY VARY FROM A DEAD LEVEL."

The foregoing is a London-to-Moscow proof that the surface of the world is not globular.

Please note the word "perceptibly".  Important when considering cute anecdotes.  Nowhere does it say in fact that it is dead level.

Quote
8: Extraordinary flatness of the Earth's crust: These extracts clearly prove that the surface of the earth is a level surface, and that, therefore, the world is not a globe. And when we come to consider the surface of the world under the sea, we shall find the same unformity of evidence against the popular view. In " Nature and Man," by Professor W. B. Carpenter, article " The Deep Sea and its Contents," pages 320 and 321, the writer says :

"Nothing seems to have struck the "Challenger" surveyors more than the extraordinary FLATNESS (except in the neighbourhood of land) of that depressed portion of the earth's crust which forms the FLOOR OF THE GREAT OCEANIC AREA. . . . If the bottom of mid-ocean were laid dry, an observer standing on any spot of it would Jind himself surrounded BY A PLAIN, only comparable to that of the North American prairies or the South American pampas The form of the depressed area which lodges the water of the deep ocean is rather, indeed, to be likened to that of a FLAT WAITER or TEA TRAY, surrounded bj- an elevated and deeply -sloping rim, than to that of the basin with which it is commonly compared."

Interesting observation, but considering the ocean floor was almost completely unmapped topographically at that timeI am not sure how seriously we should take this.

Quote
9: RIVERS:
From the "Atlas of Physical Geography," by the Rev. T.Milner, M.A., I extract the following:

" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles." 

"The plains of Venezuela and New Granada, in South America, chiefiy on the left of the Orinoco, are termed llanos, or level fields. Often in the space of 270 square miles THE SURFACE DOES NOT VARY A SINGLE FOOT."

"The Amazon only falls 12 feet in the last 700 miles of its course; the La Plata has only a descent of one thirty-third of an inch a mile,"

NOT THAT YOU CANNOT PLAUSIBLY EXPLAIN ABOVE FET UNDENIABLE PROOFS, YOU CANNOT EVEN START TO ARGUE AGAINST THE OBVIOUSNESS OF ABOVE ARGUMENTS!!!

Why can't this phenomenon exist on RE?  Not all portions of the Earth slope like the circumference of a sphere.

Quote
One additional undeniable proof no. 10: EYES LEVELED HORIZON NO MATTER HOW HIGH YOU LIFT YOUR 'EYES' UP!!!

If the Earth were flat, the horizon would be higher than is observed.

Quote
So, there is no doubt about the general shape of the surface of the Earth, the question is: How to resolve "Equator" problem, KNOWING VERY WELL that the surface of the Earth is Flat???

Be of some use after so long time (spent on this forum)!

Cough.
Title: Re: "Equator" problem
Post by: cikljamas on October 21, 2014, 08:13:42 AM
Datum lines use mean height above sea level, which is curved, but can be normalized to appear as a straight line.

Define sea LEVEL!!!

If the Earth were flat, the horizon would be higher than is observed.

Prove it.

A horizon observations (eye leveled-from the edge of space) as we know it, are due to the fact that the Earth is flat, otherwise a curvature of the sphere would be very noticeably observable, let alone that we would notice curvature of the Earth (if the Earth were a sphere) even from the sea level. Even from the sea level we would notice curved horizon (horizon line that stretches from left to right wouldn't be flat but curved). Both (left and right) ends of that line would be inclined (no matter how slightly) downwards, and when you climbed up to the mountains (especially to the HIGHest mountains) these inclinations (at both ends of that horizontal-horizon line) would be of course much more noticeable. If the Earth were a sphere a horizon horizontal line wouldn't be horizontal any more, so i wonder if  we would (in that case) still call that horizontal line "a horizon"???
Title: Re: "Equator" problem
Post by: Rama Set on October 21, 2014, 08:22:08 AM
Datum lines use mean height above sea level, which is curved, but can be normalized to appear as a straight line.

Define sea LEVEL!!!

Google not working for you?

http://en.wikipedia.org/wiki/Sea_level (http://en.wikipedia.org/wiki/Sea_level)

Quote
If the Earth were flat, the horizon would be higher than is observed.

Prove it.

A horizon observations (eye leveled-from the edge of space) as we know it, are due to the fact that the Earth is flat, otherwise a curvature of the sphere would be very noticeably observable, let alone that we would notice curvature of the Earth (if the Earth were a sphere) even from the sea level. Even from the sea level we would notice curved horizon (horizon line that stretches from left to right wouldn't be flat but curved). Both (left and right) ends of that line would be inclined (no matter how slightly) downwards, and when you climbed up to the mountains (especially to the HIGHest mountains) these inclinations (at both ends of that horizontal-horizon line) would be of course much more noticeable. If the Earth were a sphere a horizon horizontal line wouldn't be horizontal any more, so i wonder if  we would (in that case) still call that horizontal line "a horizon"???
[/quote]

No, you prove it.

Anyway:

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/ (http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/)
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 21, 2014, 09:52:50 AM
This shotgun copy-pasta style of debate will get us nowhere.

I'm intrigued by this one, however

Quote
5: Railways: In 1862, the Houses of Lords and Commons issued an Order that all Railways were to be constructed on a Datum Horizontal line without allowing one inch for curvature.
Have you an actual link to the "order"?  Or is this something you just keep copy-pastering without actually finding a source?


Do you realise that surveyors use calculations that account for the curvature of the earth, every single day?  They only use these corrections, as they work.  If the earth isn't round, then why do they work?

They generally use this to correct for curvature and refraction:

(http://upload.wikimedia.org/math/b/4/f/b4f3665435a1aa088220d78015ecf231.png)
Title: Re: "Equator" problem
Post by: ausGeoff on October 21, 2014, 10:34:28 AM
@cikljamas, what's with your 19th century obsession?

My guess?  He (and a lot of his peers) quote "science" that's 150 years out of date because it's the only science that supports their flat earth proposition.  Any science from the second half of the 20th century proves unequivocally that the planet is an oblate spheroid.

During the historical period that cikljamas seems enthralled with, there were numerous other loony tunes notions out there as well...

Iridology;  each part of the eye's iris is connected to specific bodily organ.
Spiritualism;   séances, spirit photography, mediums, ectoplasm.
Phrenology;  determining a person's character by feeling the bumps on their scalp.  [see image below]
Homeopathy and Isopathy;  like cures like.
Atlantis;  an ancient lost continent of advanced intellectuals.
Martian Canals;  proof of intelligent life on the planet.
Mesmerism;  an invisible natural force exerted by animals.
Anthropometry;   classifying potential criminals by facial characteristics.
Ear Candling;  used to cleanse and harmonise a person’s energy fields or auras.

All of these were considered as legitimate scientific notions in the mid-nineteenth century.  At the same time Mr Samuel Birley Rowbotham was writing about his flat earth proposition.  What an interesting (but unsurprising) coincidence.



(http://en.citizendium.org/images/9/9b/Phrenologychart.png)


Popular in the middle of the 19th century, phrenology held that
mental faculties are localised to different parts of the brain, that
they develop differently in different individuals, and that these
differences are reflected in measurable differences in the
external form of the cranium.





Title: Re: "Equator" problem
Post by: sokarul on October 21, 2014, 02:53:58 PM
cikljamas, you need to realize on round surface with a curved gravitation field would act the same as a flat surface. I don't think you posted anything that isn't possible on a round earth. You also need to look up what canals actually are.
Title: Re: "Equator" problem
Post by: cikljamas on October 22, 2014, 03:57:45 AM
cikljamas, you need to realize on round surface with a curved gravitation field would act the same as a flat surface. I don't think you posted anything that isn't possible on a round earth. You also need to look up what canals actually are.

It is very obvious, that if the Earth were experiencing a daily rotation; the air flow at the surface of the Earth would have variable velocity (not the thermal), variable pressure (not the static), and variable density (not the normal). The science of aerodynamic is clear in this issue; an air flow (dynamic) should be generated from the region of M = 1.33 to the region of M = 0; from the region of high density (at the equator) to the region of low density (at the poles). Such air flow and such air pressure regimes do not exist. The localities of these regimes would terminate the life of all species on Earth. The fact that, the static sea level pressure is equal to one atmosphere (= 1013.25 mbar (millibar) = 0.101325 x 106 Pascal) at standard conditions concludes the absence of rotating Earth. You can confirm this fact by applying the Pitot static-tube along the Earth’s latitudes. In a non windy day, the Pitot static-tube must indicate a total pressure of 1.0 atm which is equal to the static pressure; otherwise, it must be calibrated. This verifies that the speed of the air flow has no other component than the thermal static speed. If you hold the Pitot static tube so that the portion-B faces the air flow and the portion-A faces the air loft; the tube will read 1.0 atm in a standard day, in all directions. It is true and fact. Read more: http://www.energeticforum.com/256388-post62.html (http://www.energeticforum.com/256388-post62.html)

Now you should consider these assertions as a proven facts:

1. There is no Earth's revolution around anything!
2. There is no Earth's rotation on it's axis!
3. There is no axial tilt of the Earth!
4. Whole universe is centred to the Earth! (Celestial equator is aligned with the Earth's equator)
5. There is no gravitation as such. Newton invented it with one and only purpose (to prove that the Earth revolves around the Sun)


Now, we should consider the shape of the Earth:

What is gravity, what is the mechanics behind it? Nobody knows, and nobody will ever find out, because there is no such thing as gravity. See this: http://www.universetoday.com/74015/what-causes-gravity/ (http://www.universetoday.com/74015/what-causes-gravity/)

Let say that the best description of the "gravity" has been found in these words:

Take the Earth, for instance. Classical physics sees the force of gravity as some type of almost magical attractive force between stars and planets. Ether theory has a totally different view. The reason we fall back to the Earth when we jump up is not this mystical force of gravity, but rather it is because the Earth is constantly absorbing a tremendous amount of ether to keep all of its elementary particles spinning. We are just in the way of this influx. This view explains what gravity is, and also explains Tesla’s seemingly odd statement that the sun is absorbing more energy than it is radiating. The more you think about it, the more this seemingly nutty idea makes perfect sense. The sun requires a gargantuan amount of etheric energy to keep its integrity.

Once it is realised that electrons spin at speeds in excess of the speed of light, a new paradigm is born. The idea simply is that the elementary particles, by their nature, are absorbing ether all the time. This influx is what gravity is. As ether is absorbed two things happen. (1) The process enables the elementary particles to maintain their spin, and (2) Simultaneously, this etheric energy, probably stemming from what some physicists call the zero point energy realm, which is a vast reservoir of untapped energy, is transformed into electromagnetic energy. That is Grand Unification, Einstein’s dream of how to combine gravity with electromagnetism. Read more: http://www.energeticforum.com/253761-post225.html (http://www.energeticforum.com/253761-post225.html)


If there is no gravitation as such, there must be something else that keeps us from floating off into space. What is that? Let's call it pressure instead of "Ether influx"...

Now, if the Earth were a globe, there should be two different pressures in action on the same spot (on the surface of the Earth), 1: one which would keep us from floatting off into space (this kind of pressure shouldn't be too strong, otherwise we would be squashed instantly), and 2: one which would keep oceans from being spilled out into space (this kind of pressure should be much, much, much stronger, otherwise oceans would be spilled out into space)!

Now, someone should be able to explain parallel existence of these two completely different pressures on the same spot (on the surface of the ROUND Earth)!

Now, read this:

Quote
The latitude that corresponds to the Mach number M =1 is 45 degree north and south. The third principal region is designated for the subsonic and incompressible flow, M = 0.3. Consider the mean temperature of the air to be 10 degree C (= 283 degree K). (The mean temperature drops from the equatorial value as you move toward north and south poles.) In this case, the speed of sound is equal to 336 m/s, since V = 0.3 a, hence the spinning speed is equal to 100.8 m/s (363 km/hr). The latitude that corresponds to the Mach number M = 0.3 is 77 degree north and south.

Now, try to figure out "the mystery" of a blue region on the next picture:

(http://i.imgur.com/SZtO8nI.jpg)
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 22, 2014, 06:14:57 AM
Quote
the air flow at the surface of the Earth would have variable velocity
It's called wind - look it up.

as to the rest: tldr
Title: Re: "Equator" problem
Post by: cikljamas on October 22, 2014, 08:10:09 AM
Datum lines use mean height above sea level, which is curved, but can be normalized to appear as a straight line.

Define sea LEVEL!!!

Google not working for you?

http://en.wikipedia.org/wiki/Sea_level (http://en.wikipedia.org/wiki/Sea_level)

Is the sea the same level all over the world ? What do you think? If you don't know i will tell you, don't worry! In fact i can tell right now! The answer is "Yes, it is!"...Now, let's see some interesting excerpts from Rowbotham's "Earth not a globe":

FACT 12. The earth has a tremulous motion more or less at all times.

PROOF. If a delicate spirit-level be firmly fixed on a rock or on the most solid foundation it is possible to construct, and far away from the influence of any railway, or blasting or mining operations, the curious phenomenon will be observed of continual but irregular change in the position of the air bubble. However carefully the level may be adjusted, and the instrument protected from the atmosphere, the "bubble" will not maintain its position long together.

A similar effect is noticed in the most favourably situated astronomical observatories, where instruments of the very best construction, and placed in the most approved positions, cannot always be relied upon. without occasional and systematic readjustment.

The following quotation affords a good illustration of the tremour above described:---

"MARCH 12TH, 1822, in Adventure Bay, Island of South Georgia, we anchored in seven fathoms water, latitude 54° 2´ 48″ S., longitude 38° 8´ 4″ W. The head of this Bay being surrounded with mountains, I ascended the top of one of them for the purpose of taking the altitude of the sun when at some distance from the meridian; but after planting my artificial horizon, I was surprised to find that although there was not a breath of wind, and everything around perfectly still, yet the mercury had so tremulous a motion that I could not get an observation."


FACT 13. Tides in the extreme south are very small, and in some parts are scarcely perceptible.

PROOF. "The rise and fall of tide in Christmas Harbour, latitude 48° 41´ S, longitude 69° 3´ 35″ E., is remarkably small; not on any occasion amounting to more than 30 inches and the usual spring tides are generally less than two feet. The neap tide varies from four to twelve inches, and the diurnal inequality is, comparatively, very considerable."

FACT 16.--If, at any hour of the night, a telescope is firmly fixed, securely lashed to any solid object, and turned to the pole-star, it will be found on continuing the observation for some hours that the star "Polaris" does not maintain its position, but seems to slowly rise and fall in the field of view of the telescope. The line-of-sight will be sometimes above it; in about twelve hours it will be below it; and in another twelve hours it will again be above the star.

Many more facts could be added to the foregoing collection, but already the number is sufficient to enable us to form a definite conclusion as to what is the real cause of the tides.

The facts 1 to 7 fully enable us to establish syllogistically the groundwork of the reply. All bodies floating in an incompressible medium, and exposed to atmospheric pressure, fluctuate, or rise and fall in that medium.

The earth is a vast irregular structure, stretched out upon and standing or floating in the incompressible waters of the "great deep."

Ergo--The earth has, of necessity, a motion of fluctuation.

Hence, when by the pressure of the atmosphere, the earth is depressed or forced slowly down into the "great deep," the waters immediately close in upon the receding bays and headlands, and produce the flood tide; and when, by reaction, the earth slowly ascends, the waters recede, and the result is the ebb tide. Read more: http://www.sacred-texts.com/earth/za/za30.htm (http://www.sacred-texts.com/earth/za/za30.htm)

Now, since the sea level is the same all over the world how come that tides in the extreme south are very small, and in some parts scarcely perceptible? How does this agree with the idea of supposedly globular Earth? Why is the air pressure lesser in extreme south? See the map in my previous post above! What would be the consequences of these irregularities on a globular Earth?
Title: Re: "Equator" problem
Post by: Rama Set on October 22, 2014, 08:35:30 AM
What does sea level have to do with the strength of tides?  From the maps I looked at, tides are stronger towards the eliptic of the Earth's orbit, and less so near the poles.  Seems to be exactly what you would expect with the current model.
Title: Re: "Equator" problem
Post by: 29silhouette on October 22, 2014, 08:54:46 AM
Now, if the Earth were a globe, there should be two different pressures in action on the same spot (on the surface of the Earth), 1: one which would keep us from floatting off into space (this kind of pressure shouldn't be too strong, otherwise we would be squashed instantly), and 2: one which would keep oceans from being spilled out into space (this kind of pressure should be much, much, much stronger, otherwise oceans would be spilled out into space)!

Now, someone should be able to explain parallel existence of these two completely different pressures on the same spot (on the surface of the ROUND Earth)!
Yet another subject in which FE'rs always seem to struggle.  What is your mass compared to that of the oceans?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 22, 2014, 01:50:11 PM
FACT 16.--If, at any hour of the night, a telescope is firmly fixed, securely lashed to any solid object, and turned to the pole-star, it will be found on continuing the observation for some hours that the star "Polaris" does not maintain its position, but seems to slowly rise and fall in the field of view of the telescope. The line-of-sight will be sometimes above it; in about twelve hours it will be below it; and in another twelve hours it will again be above the star.

This one is easy. Polaris isn't located exactly at the pole, so this is exactly what you would expect to see. It's currently about 3/4 degree from the true celestial pole and getting closer; in 1900 it was more than a degree from the pole, and in 1822, almost two degrees.

What is the source of these numbered "Facts"? Are the rest like these - either incorrect, or obvious and not inconsistent with current models?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 22, 2014, 03:08:29 PM
@Alpha2Omega, here we go:

Quote
Here is another nut for "Tamatea" to crack, and when he has cracked this I can give him some more: -On the last trip of the R.M.S. Kaikoura, Captain W. C. Crutchley R.N.R. sighted Mount Peel at a distance of 118 miles. Take off for elevation of observer 7 miles, which leaves 111 miles; the curvature in that distance according to science, is 8,214 feet; take the height of Mount Peel 5,500 feet from this, and it leaves 2,714 feet. The top of Mount Peel should have been below the horizon, and could not be seen at the distance named, if the world were a globe.-

Getting back to this, I find that the top of Mt. Peel at 5,500 ft. should be geometrically about 1/4 degree below the horizon from a distance of 111 miles.

There are several possibilities that explain this report:

1) It's due to ordinary atmospheric refraction, and no big deal.

2) It's an unusually large amount of refraction (thus it's a news item).

3) Distance and/or height are wrong.

4) The whole report is completely bogus.

Do you disbelieve everything you read in newspapers except this?

Without actually running the numbers, I lean toward 2). Even if this is somewhat more refraction than you'd expect under normal conditions (I'm not sure if it is or not), it hardly seems excessive; light from objects outside the atmosphere normally has been refracted twice that amount when it appears to be on the horizon, and most of that occurs in the lower parts of the atmosphere where the air is densest. The exact amount of refraction for two points within the atmosphere is difficult to calculate, even knowing the exact density profile of the air (which we never do). Approximations using standard models for the atmosphere are possible, but sometimes local conditions will make the real value less or more than 'normal'.

This begs the larger question whether this has ever been observed since? If true, but only seen once or very rarely, I fail to see how this could be anything other than unusual atmospheric conditions. If it's common, then it's less unusual or even normal atmospheric conditions.

If unusual, I also fail to see how this particularly supports a flat-earth model. What I've seen here requires "bendy light" (not refracted like the above, but bent upward by some unknown mechanism) that mimics the sharp horizon inherent in the spherical-earth model. If this is "impossible" with a spherical earth, then why is it more plausible in a model that requires bent light; what caused the temporary local aberration?

Since you responded to the last paragraph in the earlier reply and none of the others, I presume you are satisfied with the explanations given. I'll take a look at the links you replied with later.

Title: Re: "Equator" problem
Post by: sokarul on October 22, 2014, 04:45:44 PM


It is very obvious, that if the Earth were experiencing a daily rotation; the air flow at the surface of the Earth would have variable velocity (not the thermal), variable pressure (not the static), and variable density (not the normal).
Why? The atmosphere moves with the earth, no need for it to be variable.

Quote
The science of aerodynamic is clear in this issue; an air flow (dynamic) should be generated from the region of M = 1.33 to the region of M = 0; from the region of high density (at the equator) to the region of low density (at the poles).
You are missing units on those numbers. But anyways, yes there are areas of high and low pressure. Watch the weather channel.

Quote
Such air flow and such air pressure regimes do not exist.
They do exist,watch the weather channel to find them.

Quote
The localities of these regimes would terminate the life of all species on Earth. The fact that, the static sea level pressure is equal to one atmosphere (= 1013.25 mbar (millibar) = 0.101325 x 106 Pascal) at standard conditions concludes the absence of rotating Earth.

Pressure would never be zero. so no, life wouldn't die.

Quote
You can confirm this fact by applying the Pitot static-tube along the Earth’s latitudes. In a non windy day, the Pitot static-tube must indicate a total pressure of 1.0 atm which is equal to the static pressure; otherwise, it must be calibrated. This verifies that the speed of the air flow has no other component than the thermal static speed. If you hold the Pitot static tube so that the portion-B faces the air flow and the portion-A faces the air loft; the tube will read 1.0 atm in a standard day, in all directions. It is true and fact. Read more: http://www.energeticforum.com/256388-post62.html (http://www.energeticforum.com/256388-post62.html)
I don't know what they are but it would be hard for them to read one atm here at 5,280 feet. So no, an instrument on a plane does not prove the earth is flat,

Quote
Now you should consider these assertions as a proven facts:

1. There is no Earth's revolution around anything!
That is not what astronomy says.
Quote
2. There is no Earth's rotation on it's axis!
Foucault pendulums, still.

Quote
3. There is no axial tilt of the Earth!
Seasons.

Quote
4. Whole universe is centred to the Earth! (Celestial equator is aligned with the Earth's equator)
Astronomy again.

Quote
5. There is no gravitation as such. Newton invented it with one and only purpose (to prove that the Earth revolves around the Sun)
Or to explain falling objects. Today we know he was not 100% correct.

Quote
Now, we should consider the shape of the Earth:

What is gravity, what is the mechanics behind it? Nobody knows, and nobody will ever find out, because there is no such thing as gravity. See this: http://www.universetoday.com/74015/what-causes-gravity/ (http://www.universetoday.com/74015/what-causes-gravity/)

I don't have to read those to know that mass, energy, and momentum cause gravitation.

Quote
Let say that the best description of the "gravity" has been found in these words:

Take the Earth, for instance. Classical physics sees the force of gravity as some type of almost magical attractive force between stars and planets. Ether theory has a totally different view. The reason we fall back to the Earth when we jump up is not this mystical force of gravity, but rather it is because the Earth is constantly absorbing a tremendous amount of ether to keep all of its elementary particles spinning. We are just in the way of this influx. This view explains what gravity is, and also explains Tesla’s seemingly odd statement that the sun is absorbing more energy than it is radiating. The more you think about it, the more this seemingly nutty idea makes perfect sense. The sun requires a gargantuan amount of etheric energy to keep its integrity.
Any type of ether or aether has never been found or detected. Why does gravity not drop off in mines as the mass above would be absorbing the ether so there would be less underground?
Quote
Once it is realised that electrons spin at speeds in excess of the speed of light, a new paradigm is born.
Electrons do no spin at faster than the speed of light. We know the mass of electrons, therefore they cannot have a velocity faster than the speed of light.

Quote
The idea simply is that the elementary particles, by their nature, are absorbing ether all the time.
This influx is what gravity is. As ether is absorbed two things happen. (1) The process enables the elementary particles to maintain their spin,
They don't need anything to keep their spin. Your idea would allow their spin to be any number rather than a quantized number.

Quote
and (2) Simultaneously, this etheric energy, probably stemming from what some physicists call the zero point energy realm, which is a vast reservoir of untapped energy, is transformed into electromagnetic energy. That is Grand Unification, Einstein’s dream of how to combine gravity with electromagnetism. Read more: http://www.energeticforum.com/253761-post225.html (http://www.energeticforum.com/253761-post225.html)

We can detect electromagnetic energies. What you described is not detected anywhere. Your ether is not delectable but everything absorbs it?

Quote
If there is no gravitation as such, there must be something else that keeps us from floating off into space. What is that? Let's call it pressure instead of "Ether influx"...

Now, if the Earth were a globe, there should be two different pressures in action on the same spot (on the surface of the Earth), 1: one which would keep us from floatting off into space (this kind of pressure shouldn't be too strong, otherwise we would be squashed instantly), and 2: one which would keep oceans from being spilled out into space (this kind of pressure should be much, much, much stronger, otherwise oceans would be spilled out into space)!

Gravitation and your "ether flux" should be able to have one pressure hold both down.

Quote
Now, read this:

Quote
The latitude that corresponds to the Mach number M =1 is 45 degree north and south. The third principal region is designated for the subsonic and incompressible flow, M = 0.3. Consider the mean temperature of the air to be 10 degree C (= 283 degree K). (The mean temperature drops from the equatorial value as you move toward north and south poles.) In this case, the speed of sound is equal to 336 m/s, since V = 0.3 a, hence the spinning speed is equal to 100.8 m/s (363 km/hr). The latitude that corresponds to the Mach number M = 0.3 is 77 degree north and south.
Mach as in speed of sound? Speed of sound changes from air density. You arn't making sence sine you aren't defining your terms.
Quote
Now, try to figure out "the mystery" of a blue region on the next picture:

(http://i.imgur.com/SZtO8nI.jpg)
Small picture with the label unreadable. What am I looking at?

Now let me ask you a question, what is the difference between gold and silver?
Title: Re: "Equator" problem
Post by: BJ1234 on October 22, 2014, 10:05:29 PM

Now let me ask you a question, what is the difference between gold and silver?
Well, currently it is about 1200 dollars an ounce... Not sure what that has to do with blue regions on a map though ;D
Title: Re: "Equator" problem
Post by: guv on October 22, 2014, 10:38:28 PM
Difference between gold and silver 32 protons.
Title: Re: "Equator" problem
Post by: cikljamas on October 23, 2014, 04:27:22 AM
@Alpha2Omega...

(http://i.imgur.com/9R43xLI.jpg)

...those mountains are 100 miles away (picture has been taken out from one FET debate on this forum). Does anybody notice 2 000 m high hill of water between those mountains and the guy who shot this picture? I don't! Not just that i don't see 2 000 m high hill of water, i don't see no bulge of water whatsoever...   

In addition: (http://)

Gravitation and your "ether flux" should be able to have one pressure hold both down.

You don't get that, do you?

Let's see another aspect of this problem:

The mechanics of tides that has been described in post #112 of this thread is quite imaginable and consistent with what is shown in this picture:

(http://i.imgur.com/Gn3v4sb.jpg)

All continents being submerged in great deep, slightly drift (up and down/in and out) so to create tides. If the surface of the Earth is flat no problems at all with that kind of mechanics of tides. But if the Earth is round, we have to count with the "breathing", better to say "contracting-expanding" Globe (inhale/exhale), which is hardly imaginable.

Can you imagine continents pushing and squeezing waters of the Globe like this from all directions/sides simultaneously towards the centre of the supposedly round Earth:

(http://i.imgur.com/8KfGce2.jpg)

REs even try to persuade us that similar phenomena already is in action on the supposedly round Earth:

The only difference is that globularists claim that the continents are at rest, and only the oceans go up and down due to gravitational influence of the Moon:

(http://i.imgur.com/n15jTKS.jpg)

I believe that if we deeper analyzed this problem we (FEs) could use this argument so to develop one of the best FE proofs so far.

So, what's gonna be?

1. The continents float and by doing so cause the tides, the Earth is at rest!
2. The continents are at rest, the Earth whirls and fly through space and the oceans dance like hell and nobody notices any of these multiple insanely fast motions and crazy dances!
Title: Re: "Equator" problem
Post by: Rama Set on October 23, 2014, 07:12:51 AM
Why do you never include elevation information in these photos.  You act as if it is not crucial to the calculation of the curvature.  Also, you are probably calculating the obscuring curvature starting at the observer instead of at the horizon, which would be the correct method.  Why?  The amount of curvature previous to the horizon obscuring the subject is zero.
Title: Re: "Equator" problem
Post by: QuQu on October 23, 2014, 07:17:12 AM
1. The continents float and by doing so cause the tides, the Earth is at rest!
2. The continents are at rest, the Earth whirls and fly through space and the oceans dance like hell and nobody notices any of these multiple insanely fast motions and crazy dances!

3. The green turtle that supports the flat earth has periodic sexual arousals and this causes tides.
Title: Re: "Equator" problem
Post by: cikljamas on October 23, 2014, 07:25:31 AM
Why do you never include elevation information in these photos.  You act as if it is not crucial to the calculation of the curvature.  Also, you are probably calculating the obscuring curvature starting at the observer instead of at the horizon, which would be the correct method.  Why?  The amount of curvature previous to the horizon obscuring the subject is zero.

You don't tell me...

This is how it is:

http://www.energeticforum.com/258148-post188.html (http://www.energeticforum.com/258148-post188.html)

http://www.energeticforum.com/258162-post190.html (http://www.energeticforum.com/258162-post190.html)

3. The green turtle that supports the flat earth has periodic sexual arousals and this causes tides.

Something for you son:

http://www.energeticforum.com/255859-post9.html (http://www.energeticforum.com/255859-post9.html)
Title: Re: "Equator" problem
Post by: Rama Set on October 23, 2014, 07:45:48 AM
Why do you never include elevation information in these photos.  You act as if it is not crucial to the calculation of the curvature.  Also, you are probably calculating the obscuring curvature starting at the observer instead of at the horizon, which would be the correct method.  Why?  The amount of curvature previous to the horizon obscuring the subject is zero.

You don't tell me...

This is how it is:

http://www.energeticforum.com/258148-post188.html (http://www.energeticforum.com/258148-post188.html)

http://www.energeticforum.com/258162-post190.html (http://www.energeticforum.com/258162-post190.html)

3. The green turtle that supports the flat earth has periodic sexual arousals and this causes tides.

Something for you son:

http://www.energeticforum.com/255859-post9.html (http://www.energeticforum.com/255859-post9.html)


Oh great.  So how did you calculate Db?  Seems to me you have too little information to do so.  You do not have an elevation with which to derive the distance to the horizon and the photo is clearly taken from an elevation as shown by being on a cliff.

Title: Re: "Equator" problem
Post by: cikljamas on October 23, 2014, 08:20:08 AM
Oh great.  So how did you calculate Db?  Seems to me you have too little information to do so.  You do not have an elevation with which to derive the distance to the horizon and the photo is clearly taken from an elevation as shown by being on a cliff.

Informations about an elevations and distances Alex Tomasovich provided for you here (post #55) : http://theflatearthsociety.org/forum/index.php?topic=674.40#.VEkaQ_LLKXV (http://theflatearthsociety.org/forum/index.php?topic=674.40#.VEkaQ_LLKXV)

As for formula, since you don't believe me, Tom Bishop explained it for you here (post #68) : http://www.theflatearthsociety.org/forum/index.php?topic=18114.msg319626#msg319626 (http://www.theflatearthsociety.org/forum/index.php?topic=18114.msg319626#msg319626)
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 23, 2014, 09:05:03 AM
@Alpha2Omega...

(http://i.imgur.com/9R43xLI.jpg)

...those mountains are 100 miles away (picture has been taken out from one FET debate on this forum). Does anybody notice 2 000 m high hill of water between those mountains and the guy who shot this picture? I don't! Not just that i don't see 2 000 m high hill of water, i don't see no bulge of water whatsoever...   
What mountains are those? Where was the picture taken from? In other words, how do you know those mountains are 100 miles away? I'm not disputing this, just wondering how you know it.

How high are they?

In a recent reply you refer to Post #55 in http://theflatearthsociety.org/forum/index.php?topic=674.40#.VEkaQ_LLKXV (http://theflatearthsociety.org/forum/index.php?topic=674.40#.VEkaQ_LLKXV) for the elevations and distances involved, but that seems to refer to a different photograph over Lake Michigan (the link is broken). This ain't a photo over Lake Michigan, so the questions still stand.

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.

(http://i26.photobucket.com/albums/c118/FromVegaButNotVegan/tfesMtnsInDistance3_zpsc828fd5f.png)

Quote
Let's see another aspect of this problem:

The mechanics of tides that has been described in post #112 of this thread is quite imaginable and consistent with what is shown in this picture:

<snip picture of "floating continent">

All continents being submerged in great deep, slightly drift (up and down/in and out) so to create tides. If the surface of the Earth is flat no problems at all with that kind of mechanics of tides. But if the Earth is round, we have to count with the "breathing", better to say "contracting-expanding" Globe (inhale/exhale), which is hardly imaginable.

Can you imagine continents pushing and squeezing waters of the Globe like this from all directions/sides simultaneously towards the centre of the supposedly round Earth:

<snip picture of man in gym>
No, I can't. I don't know anyone who proposes this as a mechanism, either, except perhaps as a strawman.

Quote
REs even try to persuade us that similar phenomena already is in action on the supposedly round Earth:

The only difference is that globularists claim that the continents are at rest, and only the oceans go up and down due to gravitational influence of the Moon:

(http://i.imgur.com/n15jTKS.jpg)
Yes, that's the model. What, specifically, is wrong with it? "I don't like it" or "it looks stupid." Aren't problems with the model. If you claim "it cannot work that way", you need to explain why you think so.

Quote
I believe that if we deeper analyzed this problem we (FEs) could use this argument so to develop one of the best FE proofs so far.
Please do.

Quote
So, what's gonna be?

1. The continents float and by doing so cause the tides, the Earth is at rest!
2. The continents are at rest, the Earth whirls and fly through space and the oceans dance like hell and nobody notices any of these multiple insanely fast motions and crazy dances!
I pick 2. The motions are noticeable in the apparent motion of objects not moving along with us, i.e. celestial objects; you look at the numbers and think "wow - that's fast!" but since everything on earth is traveling along with us, there's no relative motion for us to see. The "crazy dances" you refer to are the tides. Are you really suggesting that no one notices the tides? What have we just been discussing?
Title: Re: "Equator" problem
Post by: 29silhouette on October 23, 2014, 10:29:11 AM
Informations about an elevations and distances Alex Tomasovich provided for you here (post #55) : http://theflatearthsociety.org/forum/index.php?topic=674.40#.VEkaQ_LLKXV (http://theflatearthsociety.org/forum/index.php?topic=674.40#.VEkaQ_LLKXV)
And a few pages later in that same thread Alex Tomasovich and myself provide some diagrams showing how it works with RET (edited to add- ,and why one simply doesn't directly subtract the viewing elevation from the height that hidden by the curvature.  Factoring in the distance from the horizon to the target object is required.)

We're still waiting on jroa's diagram supporting the FE'rs argument.
http://theflatearthsociety.org/forum/index.php?topic=674.80#.VEkxCGfvgSk (http://theflatearthsociety.org/forum/index.php?topic=674.80#.VEkxCGfvgSk)

Quote
As for formula, since you don't believe me, Tom Bishop explained it for you here (post #68) : http://www.theflatearthsociety.org/forum/index.php?topic=18114.msg319626#msg319626 (http://www.theflatearthsociety.org/forum/index.php?topic=18114.msg319626#msg319626)
That's because he wasn't looking 33 miles.  The distance is 22 miles, and he was probably looking the wrong direction at a beach much closer.  Also, he never has provided photographic proof of being able to see that much detail 22 miles away through a telescope that I know of.   http://theflatearthsociety.org/forum/index.php?topic=57282.msg1444217#msg1444217 (http://theflatearthsociety.org/forum/index.php?topic=57282.msg1444217#msg1444217)
Title: Re: "Equator" problem
Post by: 29silhouette on October 23, 2014, 10:56:49 AM
What mountains are those? Where was the picture taken from?
lat 57.296791  lon -135.840717
Sea Lion Cove looking toward a coastal range (Mt. Bertha, Mt. Crillon, etc).  The partial clouds make it difficult to match up in GE, but a lot of the lower elevations do appear cut off.  The picture was originally posted if I recall, in response to TB and Thork saying a person can only see about 30 miles or so at sea level because the air is too thick.
Title: Re: "Equator" problem
Post by: cikljamas on October 23, 2014, 11:00:50 AM

What mountains are those? Where was the picture taken from? In other words, how do you know those mountains are 100 miles away? I'm not disputing this, just wondering how you know it.

How high are they?

http://theflatearthsociety.org/forum/index.php?topic=59073.0#.VEkpfvLLKXU (http://theflatearthsociety.org/forum/index.php?topic=59073.0#.VEkpfvLLKXU)

Please do.

Would it be so difficult to determine if the continents float and drift or not? What's more, i would say that it has been already determined, as i already have shown in my post #112

What would happen if you squeezed northern part of North America and southern part of South America simultaneously? If the Earth were round you would break up Mexico into two parts, if the Earth is flat, you would just cause tide...

That's because he wasn't looking 33 miles.  The distance is 22 miles, and he was probably looking the wrong direction at a beach much closer.  Also, he never has provided photographic proof of being able to see that much detail 22 miles away through a telescope that I know of. 

He provided correct formula for reckoning accurate height of a supposed bulge of water that would stand between observed object and observer on the round Earth (which phenomena nobody ever saw), that was the whole point...
Title: Re: "Equator" problem
Post by: Rama Set on October 23, 2014, 11:56:47 AM
What mountains are those? Where was the picture taken from?
lat 57.296791  lon -135.840717
Sea Lion Cove looking toward a coastal range (Mt. Bertha, Mt. Crillon, etc).  The partial clouds make it difficult to match up in GE, but a lot of the lower elevations do appear cut off.  The picture was originally posted if I recall, in response to TB and Thork saying a person can only see about 30 miles or so at sea level because the air is too thick.

Do you know off-hand what the lower elevations are there?  If not, I might research it myself.
Title: Re: "Equator" problem
Post by: 29silhouette on October 23, 2014, 12:36:19 PM
That's because he wasn't looking 33 miles.  The distance is 22 miles, and he was probably looking the wrong direction at a beach much closer.  Also, he never has provided photographic proof of being able to see that much detail 22 miles away through a telescope that I know of. 

He provided correct formula for reckoning accurate height of a supposed bulge of water that would stand between observed object and observer on the round Earth (which phenomena nobody ever saw), that was the whole point...
He provided a formula for calculating the drop, 600' over 30 miles, sure.  The amount of "bulge" will be far less.  Anyway, there are doubts regarding his account of seeing people throwing frisbies and such from 22 miles away.
Title: Re: "Equator" problem
Post by: 29silhouette on October 23, 2014, 12:48:36 PM
Do you know off-hand what the lower elevations are there?  If not, I might research it myself.
The last set of hills visible before the snow covered range is 2600ft iirc.  There are some islands a couple hundred feet high just off that point that aren't visible in the picture. 
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 23, 2014, 12:51:05 PM

What mountains are those? Where was the picture taken from? In other words, how do you know those mountains are 100 miles away? I'm not disputing this, just wondering how you know it.

How high are they?

http://theflatearthsociety.org/forum/index.php?topic=59073.0#.VEkpfvLLKXU (http://theflatearthsociety.org/forum/index.php?topic=59073.0#.VEkpfvLLKXU)
It took a little looking to find the key piece of information:
12,431 ft at the peak.
I'm not sure what your point was posting that picture. 12,000' mountains are easily visible over the curvature of an 8,000-mi diameter sphere from 100 miles away.

Quote
Would it be so difficult to determine if the continents float and drift or not? What's more, i would say that it has been already determined, as i already have shown in my post #112
That post has several either unsubstantiated or obvious claims in it. What exactly has been determined? I can't say I've seen the spirit-level phenomenon asserted in support of "Fact 12." in years of working with instruments that are very sensitive to leveling, nor have I heard of the need to adjust for changing level with professional astronomical instruments (my dad was a professional astronomer, and I know many others). "Fact 16." has already been addressed in an earlier reply.

Quote
What would happen if you squeezed northern part of North America and southern part of South America simultaneously? If the Earth were round you would break up Mexico into two parts, if the Earth is flat, you would just cause tide...
What's going to do the squeezing? Even if possible, since the Earth isn't completely rigid on a large scale, depending on the time scale it might simply deform, but would, no doubt, have some "interesting" local effects.

At any rate, none of this is necessary to explain the tides. Gravitational attraction of the Moon is sufficient for most of the observed effect; a little additional effect from the Sun gets most of the rest.
Title: Re: "Equator" problem
Post by: cikljamas on October 23, 2014, 01:56:50 PM
That post has several either unsubstantiated or obvious claims in it. What exactly has been determined? I can't say I've seen the spirit-level phenomenon asserted in support of "Fact 12." in years of working with instruments that are very sensitive to leveling, nor have I heard of the need to adjust for changing level with professional astronomical instruments (my dad was a professional astronomer, and I know many others). "Fact 16." has already been addressed in an earlier reply.

So, your dad was a professional astronomer, nice to hear it, i send my greetings to your dad and if he were willing to tell you the real truth and how it really is above us, now could be the right time for you to ask him about next few remarkably interesting astronomical facts:

FACT 1: http://www.energeticforum.com/265053-post497.html (http://www.energeticforum.com/265053-post497.html)
FACT 2: http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)
FACT 3: http://www.energeticforum.com/263816-post278.html (http://www.energeticforum.com/263816-post278.html)
FACT 4: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)
FACT 5: http://www.energeticforum.com/263904-post281.html (http://www.energeticforum.com/263904-post281.html)


Title: Re: "Equator" problem
Post by: sokarul on October 23, 2014, 03:55:22 PM
That post has several either unsubstantiated or obvious claims in it. What exactly has been determined? I can't say I've seen the spirit-level phenomenon asserted in support of "Fact 12." in years of working with instruments that are very sensitive to leveling, nor have I heard of the need to adjust for changing level with professional astronomical instruments (my dad was a professional astronomer, and I know many others). "Fact 16." has already been addressed in an earlier reply.

So, your dad was a professional astronomer, nice to hear it, i send my greetings to your dad and if he were willing to tell you the real truth and how it really is above us, now could be the right time for you to ask him about next few remarkably interesting astronomical facts:

FACT 1: http://www.energeticforum.com/265053-post497.html (http://www.energeticforum.com/265053-post497.html)
FACT 2: http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)
FACT 3: http://www.energeticforum.com/263816-post278.html (http://www.energeticforum.com/263816-post278.html)
FACT 4: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)
FACT 5: http://www.energeticforum.com/263904-post281.html (http://www.energeticforum.com/263904-post281.html)
Hiding from all the hard arguments and posting the same incorrect links will not help you.
Title: Re: "Equator" problem
Post by: Rama Set on October 23, 2014, 04:27:37 PM
Do you know off-hand what the lower elevations are there?  If not, I might research it myself.
The last set of hills visible before the snow covered range is 2600ft iirc.  There are some islands a couple hundred feet high just off that point that aren't visible in the picture.

Sorry I was unclear. Do you know the possible elevation that photo was taken from?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 23, 2014, 08:34:18 PM
That post has several either unsubstantiated or obvious claims in it. What exactly has been determined? I can't say I've seen the spirit-level phenomenon asserted in support of "Fact 12." in years of working with instruments that are very sensitive to leveling, nor have I heard of the need to adjust for changing level with professional astronomical instruments (my dad was a professional astronomer, and I know many others). "Fact 16." has already been addressed in an earlier reply.

So, your dad was a professional astronomer, nice to hear it, i send my greetings to your dad and if he were willing to tell you the real truth and how it really is above us, now could be the right time for you to ask him about next few remarkably interesting astronomical facts:

FACT 1: http://www.energeticforum.com/265053-post497.html (http://www.energeticforum.com/265053-post497.html)
FACT 2: http://www.energeticforum.com/265210-post527.html (http://www.energeticforum.com/265210-post527.html)
FACT 3: http://www.energeticforum.com/263816-post278.html (http://www.energeticforum.com/263816-post278.html)
FACT 4: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)
FACT 5: http://www.energeticforum.com/263904-post281.html (http://www.energeticforum.com/263904-post281.html)
Um, well, if you notice, I said he was a professional astronomer. Not is, is a former, or used to be a professional astronomer. Do you want to take a wild guess why I used that particular tense? So, no, I'm not going to ask him about those completely nonsensical links. I wouldn't ask him about them even if I could, and there's no need to, anyway; they are complete, total, and unmitigated balderdash. I'm sure there must be something in at least one of them that might be correct, but, even if true, between the incoherent writing and the basic premise, it's not immediately apparent.

Beyond that, I'm actually offended by your suggestion that he wasn't telling "the real truth" all along. I don't offend easily, but the smug suggestion that scientists routinely lie to everyone (except, perhaps, their own cabal) does wear a bit thin, especially in this case.
Title: Re: "Equator" problem
Post by: cikljamas on October 24, 2014, 02:40:00 AM
Complete, total, and unmitigated balderdash you say! Your words damn well fit with heliocentric LIE which is the greatest hoax OF ALL TIME and the greatest insult for dignity and common sense of every sane human being who has lived in last 400 years, and FIVE FACTS which i have presented to you prove beyond any reasonable doubt that i am right! It seems that you haven't courage to face reality and IRREFUTABLE FACTS which everyone can easily understand! That is why you use offensive words, and this method of "disqualification without any basis in arguments" is very well known method. Let me show you just a few examples of that kind:

James Tour-evolution: #t=52m2s (http://#t=52m2s)

DAWKINS - BEN STEIN interview : #t=3m26s (http://#t=3m26s)

Richard Dawkins: Embarrassingly Bad Thinker, Incredibly Ignorant : (http://)

The Science Delusion BANNED TED TALK: (http://)

Moral consequences of a bad and wrong philosophy and politically determined "science" : #t=1h5m40s (http://#t=1h5m40s)

Political consequences of a bad and wrong philosophy and politically determined "science" : Oliver Stone Tears Apart Obama's Empire : (http://)

CNN Wants This Video Banned (SEE WHY) : (http://)

These 5 Censored Books Tell a History the Establishment Wants Hidden (http://)

http://www.energeticforum.com/263443-post264.html (http://www.energeticforum.com/263443-post264.html)

P.S. I am sorry about your dad, i had no idea that you use past tense for that reason...
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 24, 2014, 09:06:20 AM
Complete, total, and unmitigated balderdash you say! Your words damn well fit with heliocentric LIE which is the greatest hoax OF ALL TIME and the greatest insult for dignity and common sense of every sane human being who has lived in last 400 years, and FIVE FACTS which i have presented to you prove beyond any reasonable doubt that i am right! It seems that you haven't courage to face reality and IRREFUTABLE FACTS which everyone can easily understand! That is why you use offensive words, and this method of "disqualification without any basis in arguments" is very well known method. <Irrelevant links ignored>

P.S. I am sorry about your dad, i had no idea that you use past tense for that reason...
Starting at the top...
Quote
1. Heliocentrists claim that the stars and the sun are at rest, and that the Earth is in motion.
2. The fact is that the Earth is at rest, and the stars and the sun are in motion. Now we are going to prove this assertion.

If 1 then the rate (velocity) of annual motion of all the stars above the Earth has to be variable too, not just a velocity of Sun's ("apparent") motion across the sky, but the fact is that the rate (velocity) of annual motion of all the stars above the Earth is a constant.

We can not assign different velocities of Sun's ("apparent") motion across the sky to the different (variable) velocities of Earth's orbital motion and in the same time evade to apply different (variable) velocities of Earth's orbital motion to the steady (which then shouldn't be steady but variable) rate (velocity) of annual motion of all the stars above the Earth.

If 2 then the steady-even rate (velocity) of annual motion of all the stars above the Earth doesn't have to be variable, because in that case annual motion of the stars doesn't depend of any other motion, but presents and performs independent motion. In that case Sun's motion also presents independent motion and all that remains is to adjust (by some "Entity") these two motions in order to make them synchronous motions.

What do you mean by "annual motion of all the stars above the Earth"? There really isn't any, except for a very tiny amount of parallax in the position of the nearer ones when compared to more distant stars. This parallax is due, of course, to the size of earth's orbit about the barycenter (center of mass) of the Solar System. I suspect you mean diurnal (daily) motion - the apparent east to west motion of the stars across the sky - not annual motion. If that's the case, then yes, the apparent motion of each star is constant, due entirely to the rotation of the Earth about its axis and independent of its motion about the SS barycenter (neglecting the aforementioned parallax).

The apparent diurnal motion of the Sun depends mostly on the steady rotation of the Earth, but is also influenced by the Earth's orbit as described in this post from a different thread:

On a slightly different topic what is day length anyways? 23 hours 56 minutes or 24 hours?

Yes!  ;)

The earth rotates once in 23h 56m 4s.  This is called a sidereal day because it is measured with respect to the stars. During that period of rotation, earth also progressed along its orbit around the sun a little, so it has to rotate a little more to bring the sun back to the same position.  This is called a solar day.  The average length of the solar day is 24h, which is called the mean solar day, or civil day.

(http://burro.cwru.edu/Academics/Astr306/Coords/sidereal-v-solar-day.gif)
Image courtesy Case Western Reserve University

The length of the solar day varies slightly over the year because the axis of rotation is tilted with respect to the orbital plane and earth's orbit isn't a perfect circle. See Equation of Time (http://en.wikipedia.org/wiki/Equation_of_time) and Analemma (http://en.wikipedia.org/wiki/Analemma).

The apparent motion of the Sun does vary slightly through the year, moving eastward along the Ecliptic fastest in December and January and slowest in June and July. This is not inconsistent in any way with the Heliocentric Model; we certainly can have different rates for the sun and stars, and even a variable rate for the Sun, because the apparent motion of the sun is heavily influenced by earth's orbit and the stars are not. Your "proof" is invalid because its first premise is wrong.

This alone does not disprove geocentrism, but that small parallax observed in some stars, on the other hand, does provide strong evidence that the Earth's position is not fixed among the stars and does change on an annual basis in a way that is described by a (nearly) circular orbit about a point near the center of the sun.

Moving on...

Quote
Ernst, do you remember "Midnight sun" argument (you can't see the sun directly when the Earth turns 180 degree away from the Sun (which is midnight position let's say at 66 degree N...because 90-66 = 24; 90-23,5 = 66,5; 66,5 + 24 = 90,5)...It happens EVERY DAY...At midnight (position) at 66 degree N which imaginable (horizon) tangent can't reach the perpendicular line above the Tropic of Cancer (because it's angled more than 90 degree) we can't see the Sun DIRECTLY but it doesn't mean that we would be in darkness, we would see twilight.

Now, draw the diagonal line from midnight position 66 N towards the South and answer me: What would we see at 66 S at NOON position? The answer: We wouldn't be able to see the Sun (for the same reason as in above case at 66 N - MIDNIGHT position) but there wouldn't be darkness also. This is final geometrical proof against rotundity of the Earth! In another words: On the spherical Earth POLAR NIGHT would be pure IMPOSSIBILITY!!! Do you like my "POLAR NIGHT" argument?

So, now your main concern is: How long moderators are going to keep this thread up?

SINK INTO THE EARTH WITH SHAME!!!

And don't come back, you snake!!!
It's kind of difficult to see what the point of this mostly incoherent rant is.

The following ignores atmospheric refraction, and deals purely with geometric considerations.

In the first paragraph you argue that you can't see a "midnight sun" from 66 degrees north at the northern solstice. OK, so what? I don't know if anyone competent has claimed you could (ignoring refraction, that is). In the second, you seem to argue that even at noon on the northern solstice you couldn't see the Sun from 66 degrees south. This is not correct; the sun will appear briefly above the northern horizon at local solar noon. Why wouldn't it? 

How this means there can't be a polar night is a complete mystery. On that day, south of about 66.75 degrees south, the sun will not rise at all (since the sun has an apparent radius of about 1/4 degree, part of it will be visible up to 1/4 degree south of about 66.5 degrees south).

The rest of your "irrefutable facts" are similar, and easily refutable, unsubstantiated notions. As I said: balderdash.

Your apology is accepted. Thanks for that.
Title: Re: "Equator" problem
Post by: sokarul on October 24, 2014, 09:29:20 AM

The Science Delusion BANNED TED TALK: (http://)

I actually watched some of this. I will sum it up for everyone else. "Scientists are wrong because they don't like religion and magic." He also makes the speed of light changing claim.

So cikljamas, what is the difference between gold and silver? You never answered.
Title: Re: "Equator" problem
Post by: cikljamas on October 24, 2014, 11:51:40 AM
What do you mean by "annual motion of all the stars above the Earth"?

Sorry my friend, but you are just terribly wrong in both cases. It's a little bit harder to understand (at the first glance) a former astronomical fact that you have referred to, fortunately it is going to be much more easier for me to explain a latter astronomical fact.

As for the first case: This difference of 3 min 56 seconds which you get when you substract so called sideral time (the time required to complete one daily revolution of all stars above the Earth) from 24 hours (which is mean solar time), is a constant that i have emphasised as an exposing (HC lie) factor, so to say. This time "3 min 56 seconds" IS A CONSTANT!!! This constant (0,986 degree) is a daily rate of annual rotation of all the stars which means that the constellations make one rotation per year above the Earth. Beside one daily rotation there is also one annual rotation of all the stars above us, if you didn't know. But what is very important is to determine if this daily rate of annual rotation of the stars is perfectly steady or very erratic, so to be seen if Earth's supposed revolution around the Sun (which is very erratic according to HC theory) has anything to do (theoretically of course) with this daily rate of annual rotation of the stars or not? I claim that the Earth's supposed revolution around the Sun has nothing to do with a daily rate of annual rotation of all the stars.

What does this mean? This means that if a heliocentric theory was right then this time of daily rate of annual revolution of the stars wouldn't be a constant (not at all), this time would be equal to the so called "apparent solar time" which is another name for REAL solar time, which vary a lot through the year.

As for the second case: You have to think about this just a little bit harder, and everything will be very clear to you, believe me. First of all, you have to clarify to yourself the meaning of the NOON time and the MIDNIGHT time. A former one refers to the position of the certain spot on the Earth which is directed towards the Sun, a latter one refers to the position of the certain spot on the Earth which is directed away from the Sun for 180 degrees.

Now, when at the certain spot in the arctic circle is MIDNIGHT, in the same time at the another spot (which we determine so to draw diagonal line from our first northern-arctic point towards the point which is placed at the counterpart polar circle in the same southern latitude) is NOON.

For instance, in the most northern city of Sweden, Kiruna (located at the northern latitude of 67°51'), the polar night lasts for around 28 twenty-four hour periods, while the midnight sun lasts around 50 twenty-four hour periods.

A common misconception is that the polar shortest day is totally dark everywhere inside the polar circle. In places very close to the poles it is completely dark, but regions located at the inner border of the polar circles experience polar twilight instead of polar night.

Only, the amount of light which we can notice for example in this picture is of much lesser quantity than it would be the case if the Earth were round, since the Sun would be just beyond the horizon!!!

(http://i.imgur.com/Sf52IXQ.jpg)
Title: Re: "Equator" problem
Post by: cikljamas on October 24, 2014, 02:25:02 PM
One correction, in the last post i wrote by mistake:

What does this mean? This means that if a heliocentric theory was right then this time of daily rate of annual revolution of the stars wouldn't be a constant (not at all), this time would be equal to the so called "apparent solar time" which is another name for REAL solar time, which vary a lot through the year.

Underlined part of the last sentence should be corrected as follows:

This means that if a heliocentric theory was right then this time of daily rate of annual rotation of the stars wouldn't be a constant (not at all), this time would be equal to the difference of the "Mean Solar time" (24 hours) and "Apparent solar time" (Real Solar time)!

 
As for second case:

One of the main differences between the possible Polar Night in FET and impossible Polar Night in RET is the fact that in FET the Sun is much closer to the North Pole during the northern summer, and in the same time much further away from the Ice Wall, and vice versa. In addition, in FET the Sun is of incomparable lesser dimensions. These two important facts allow Polar Nights to happen on the Flat Earth!

Why would Polar Night be impossible on the Round Earth will be easier to understand after reading this short passage:


" The nearer the Sun gets to the Pole star, the earlier it rises, the higher it reaches at noon, and the later it sets. This apparenl independent motion of the Sun, therefore, seems to account for longer and shorter days, and the whole phenomena of the seasons, but why the Sun lags as described, or why it moves northerly and southerly at alternate periods, there ii no apparent evidence. On the supposition that the world is a globe rotating against the Sun, and revolving round that luminary, it is impossible to account for what Mr. Russell calls the lagging movement of the Sun. But, on a flat surface, like the world is known to be, there is no assumption needed to account for it. As I have shown the Earth is a stretched-out structure, which diverges from the Central north in all directions toward the south. The Equator, being mid-way between the north centre and the southern circumference, divides the course of the Sun into north and south declinations. The longest circle round the world which the Sun makes, is when it has reached its greatest southern declination. Gradually going northward the circle is contracted. In about three months after the southern extremity of its path has been reached, the Sun makes a circle round the Equator. Still pursuing a northerly course as it goes round and above the world, in another three months the greatest northern declination is reached, when the Sun again begins to go towards the south. In northern latitudes when the Sun is going north, it rises earlier each day, is higher at noon, and sets later; while in southern latitudes, at the same time, the Sun, as a matter of course, rises later; reaches a lesser altitude at noon and sets earlier. In northern latitudes during the southern summer, say from September to December, the Sun rises later each day, is lower at noon, and sets earlier; while in the south he rises earlier, reaches a higher altitude at noon, and sets later each day. This movement round the Earth daily is the cause of the alternation of day and night; while his northern and southern courses produce the Seasons. When the Sun is south of the Equator it is summer in the south and winter in the north, and vice-versa. The fact of the alternation of the Seasons flatly contradicts the Newtonian delusion that the Earth revolves in an orbit round the Sun. It 'is said that summer is caused by the Earth being nearer the Sun, and winter by its being farthest from the Sun. But, if the reader will follow the argument in any text-book, he will see that according to the theory, when the Earth is nearest the Sun there must be summer in both northern and southern latitudes; and in like manner when it is farthest from the Sun it must be winter all over the Earth at the same time, because the whole of the globe-earth would be farthest from the Sun ! ! ! In short it is impossible to account for the recurrence of the Seasons on the assumption that the Earth is globular, and that it revolves in an orbit round the Sun."
Title: Re: "Equator" problem
Post by: Misero on October 24, 2014, 02:27:15 PM
While you are searching for an example that Saros has asked you for, maybe next two arguments (one of which arguments has been named by the name of my friend Saros, because it consists from his own words) can be of some help in speeding up your search and make it (your futile and absurd search) more productive...

http://www.energeticforum.com/266032-post597.html (http://www.energeticforum.com/266032-post597.html)

Keep up, don't give up!  ;D
Keep up the good work. It's nice to see more people thinking for themselves. I had a topic similar to what you are talking about on the water running to the level of the sea.
Stick around as I'm enjoying your input as well as a few otehrs, like Legion, saros, etc. It makes a refreshing change to the tefal heads and their ready made scientific page turning MS answers.

Thanks for your words of encouragement, it's a pity that today's people are so brainwashed, so that thoughts of every trully free thinking man in today's world are precious as if they have been made of pure gold!!!

@BJ1234, according to you, rivers flow towards the centre of the Earth? Don't be ridiculous!

@sokarul, rivers on a globe would not have to run uphill, that is true, rivers on a globe would have to run downhill in both cases as it is described in my "Rivers" argument, only every day let's say at midnight downhill would be uphill, and at noon downhill would be again downhill as we know it from our everyday experience. And if you think that uphill would be impossibility because of inclinations of the river beds then i suggest you to read this for clarification:

Quote
" Vast areas exhibit a perfectly dead level, scarcely a rise existing through 1,500 miles from the Carpathians to the Urals, South of the Baltic the country is so flat that a prevailing north wind will drive the waters of the Stattiner Haf into the mouth of the Oder, and give the river a backward flow 30 or 40 miles."  Read more: http://www.energeticforum.com/265601-post587.html (http://www.energeticforum.com/265601-post587.html)

In fact, rivers on a globe wouldn't flow at all, they would be spilled out into space together with all other water on such imagined globe. Don't you see how insulting for sanity of every man on the Earth is that outrageously idiotic fairy tale about a imagined globular planet which our beautiful-true world supposed to be???

I think there is a difference between being brainwashed and having some faith in humanity.
Also, thanks for the stereotypical reaction as thought of people who believe in flat earth.  ;D
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 24, 2014, 03:32:30 PM
What do you mean by "annual motion of all the stars above the Earth"?

Sorry my friend, but you are just terribly wrong in both cases. It's a little bit harder to understand (at the first glance) a former astronomical fact that you have referred to, fortunately it is going to be much more easier for me to explain a latter astronomical fact.

As for the first case: This difference of 3 min 56 seconds which you get when you substract so called sideral time (the time required to complete one daily revolution of all stars above the Earth) from 24 hours (which is mean solar time), is a constant that i have emphasised as an exposing (HC lie) factor, so to say. This time "3 min 56 seconds" IS A CONSTANT!!! This constant (0,986 degree) is a daily rate of annual rotation of all the stars which means that the constellations make one rotation per year above the Earth. Beside one daily rotation there is also one annual rotation of all the stars above us, if you didn't know. But what is very important is to determine if this daily rate of annual rotation of the stars is perfectly steady or very erratic, so to be seen if Earth's supposed revolution around the Sun (which is very erratic according to HC theory) has anything to do (theoretically of course) with this daily rate of annual rotation of the stars or not? I claim that the Earth's supposed revolution around the Sun has nothing to do with a daily rate of annual rotation of all the stars.

What does this mean? This means that if a heliocentric theory was right then this time of daily rate of annual revolution of the stars wouldn't be a constant (not at all), this time would be equal to the so called "apparent solar time" which is another name for REAL solar time, which vary a lot through the year.
You treat the mean solar day - exactly 24 hours - as though it has some real astronomical meaning. It doesn't. "Mean" in this context is the average over a year, and the mean solar day is used for civil time as a convenience. The actual length of solar days (time from meridian crossing to meridian crossing of the Sun) varies from about 8 seconds less than 24 hours (early July) to about 8 seconds more than 24 hours (early January) due to the eccentricity of the Earth's orbit. There's an additional, somewhat larger, component to the variation in the length of the apparent solar day over the year due to the tilt of the axis of rotation with respect to the path the Sun appears to trace across the sky (the Ecliptic). See Equation of Time (http://en.wikipedia.org/wiki/Equation_of_time) for more information.

Calling this variation in the length of the solar day "erratic" is disingenuous. "Erratic" implies irregular and unpredictable; the equation of time is anything but. It does vary as the sum of two sinusoids with periods of one year and one-half year, but is smoothly changing and predictably follows the same pattern year after year after year...

At any rate, the length of the sidereal day is constant, irrespective of the length of the apparent solar day - that is, it has nothing to do with the difference between mean and apparent solar days - and this will be the same for both models. We can have the "Celestial Sphere" with the stars painted on it rotating about fixed earth at a constant rate of once in 23h 56m 04s, or have earth rotating once at the same constant rate among "fixed stars" at great distance. At this point, the two models behave exactly the same. Add to this the Sun migrating along the Ecliptic from west to east at a nearly, but not quite, constant rate or the Earth orbiting around the Sun at the same nearly, but not quite, constant rate. Again, we still can't distinguish between the two models.

Where the fixed-earth model fails is when explaining apparent retrograde motion of outer planets and parallax of some stars. That's why we can confidently pick one model over the other.
 
The second part will be a separate post.
Title: Re: "Equator" problem
Post by: Rama Set on October 24, 2014, 05:00:37 PM
I watched the Dawkins/Stein video and I just don't get how you can say he looked foolish unless you are part of the proverbial choir. He presented his views summarily, said it was dishonest to claim you know anything conclusively about the existence of God and then gave, what he considers a plausible scenario for ID. Did he stammer and muddle a bit? Sure. That says nothing of his intelligence.

You alternative thinkers can be so groupthink sometimes.
Title: Re: "Equator" problem
Post by: cikljamas on October 24, 2014, 11:35:38 PM
@Alpha2Omega,

It has been known since ancient times that the motion of the Sun along the Ecliptic is not uniform. Although it moves about a degree to the East (relative to the stars) each day, its motion gradually changes during the year, being faster in December and January, and slower in June and July. The actual change from day to day is very small, and not easily noticeable with the timekeeping methods available in ancient times, but during that part of the year when the Sun moves faster than normal on one day, it moves faster than normal every day, and over a month or so, the difference adds up in a very noticeable way.

An apparent solar day can be 20 seconds shorter or 30 seconds longer than a mean solar day. Long or short days occur in succession, so the difference builds up until mean time is ahead of apparent time by about 14 minutes near February 6 and behind apparent time by about 16 minutes near November 3. The equation of time is this difference, which is cyclical and does not accumulate from year to year.

One correction more (i am getting closer):

This means that if heliocentric theory was right then this time of daily rate of annual rotation of the stars wouldn't be a constant (not at all), this time would be equal to the difference of the "Mean Solar time" (24 hours) and "Apparent solar time" (Real Solar time)!

Underlined part of the last sentence should be corrected (once more) as follows:

Final version:  ...this time would be equal to the difference of the "Apparent solar time" (Real Solar time - which is irregular BUT IT'S TRUE (REAL) SOLAR TIME) and "23 h 56 m 4s" (time required for one daily rotation of all the stars above us - a.k.a. one HC alleged rotation of the Earth on it's axis)!!!

According to HC this time is equal to the difference of the "Mean Solar time" (24 hours- which is not irregular BUT IT'S NOT TRUE (REAL) SOLAR TIME (except during equinoxes)) and "23 h 56 m 4s" (sideral time = one rotation of the Earth on it's axis)

So, how can daily rate of annual rotation of the stars be a constant if two determining factors which produce this daily rate are not constants both???

Do you recognize this major flaw in HC theory, after this final version?

@Rama Set, did you notice these observations in one of the linked videos (third one):

(http://i.imgur.com/jLjNXyg.jpg)
(http://i.imgur.com/KytsJii.jpg)

Title: Re: "Equator" problem
Post by: Alpha2Omega on October 25, 2014, 12:04:07 PM
@Alpha2Omega,

It has been known since ancient times that the motion of the Sun along the Ecliptic is not uniform. Although it moves about a degree to the East (relative to the stars) each day, its motion gradually changes during the year, being faster in December and January, and slower in June and July. The actual change from day to day is very small, and not easily noticeable with the timekeeping methods available in ancient times, but during that part of the year when the Sun moves faster than normal on one day, it moves faster than normal every day, and over a month or so, the difference adds up in a very noticeable way.

An apparent solar day can be 20 seconds shorter or 30 seconds longer than a mean solar day. Long or short days occur in succession, so the difference builds up until mean time is ahead of apparent time by about 14 minutes near February 6 and behind apparent time by about 16 minutes near November 3. The equation of time is this difference, which is cyclical and does not accumulate from year to year.

One correction more (i am getting closer):

This means that if heliocentric theory was right then this time of daily rate of annual rotation of the stars wouldn't be a constant (not at all), this time would be equal to the difference of the "Mean Solar time" (24 hours) and "Apparent solar time" (Real Solar time)!

Underlined part of the last sentence should be corrected (once more) as follows:

Final version:  ...this time would be equal to the difference of the "Apparent solar time" (Real Solar time - which is irregular BUT IT'S TRUE (REAL) SOLAR TIME) and "23 h 56 m 4s" (time required for one daily rotation of all the stars above us - a.k.a. one HC alleged rotation of the Earth on it's axis)!!!

According to HC this time is equal to the difference of the "Mean Solar time" (24 hours- which is not irregular BUT IT'S NOT TRUE (REAL) SOLAR TIME (except during equinoxes)) and "23 h 56 m 4s" (sideral time = one rotation of the Earth on it's axis)

So, how can daily rate of annual rotation of the stars be a constant if two determining factors which produce this daily rate are not constants both???

Do you recognize this major flaw in HC theory, after this final version?

No. I recognize that there's still a major flaw in your understanding of HC theory even after that last correction. Your mistake has been bolded above. You're still not any closer.

The only factor determining the daily rate of rotation wrt the stars is the rotation of the earth. Period. This is a constant. You're getting tangled up in your  "daily rate of annual rotation", which is a convoluted way of describing the difference between the lengths of the sidereal day (constant) and apparent solar day (not quite constant) and thinking the length of the mean solar day (which is irrelevant here) is somehow mixed in here. It's not.

Sure, the accumulation of longer and shorter apparent solar days adds up to plus or minus about 15 minutes relative to mean solar time, but so what? The difference between mean solar time and apparent solar time is described by the Equation of Time (http://en.wikipedia.org/wiki/Equation_of_time), and causes the Analemma (http://en.wikipedia.org/wiki/Analemma). Read about these. They are not a problem for the Heliocentric Model of the Solar System - in fact, the HC model is used to explain why they happen. Further, that begs the question about how a geocentric model would explain the component of the EoT due to the eccentricity of earth's orbit. If it has the Sun orbiting the Earth in an elliptical orbit, then that model is the same as the HC model, just using a different frame of reference; if it just posits that the Sun speeds up and slows down on its path along the Ecliptic, then it needs to explain why this happens other than "it just does".
Title: Re: "Equator" problem
Post by: cikljamas on October 25, 2014, 02:12:40 PM
The only factor determining the daily rate of rotation wrt the stars is the rotation of the earth. Period. This is a constant. You're getting tangled up in your  "daily rate of annual rotation", which is a convoluted way of describing the difference between the lengths of the sidereal day (constant) and apparent solar day (not quite constant) and thinking the length of the mean solar day (which is irrelevant here) is somehow mixed in here. It's not.

Well, this is not an answer, and you know it. Now, i have to ask you this: If you are aware that what you are offering as an explanation doesn't explain anything, why then you do this?

The whole thing is very simple:

24 hours (mean solar day) MINUS 23h56m4s = 3m56s= 0,986 degree = daily rate of annual rotation of the stars above the Earth

Now, if you want to proceed to defend HC theory the only possible logical way out of this is this:

1. Either you claim that 0,986 degree is not equal to daily rate of annual roatation of the stars

2. Or you claim that it is constant indeed

If 1 then you disagree not just with FET, but with basic HC rules also, and you have to prove what you claim which will be very difficult...

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?

It is very strong argument against fraudulent HC theory, and there is no place to hide from it.

Sorry, but you are just busted!

And this is only the beginning...
Title: Re: "Equator" problem
Post by: Socratic Amusement on October 25, 2014, 02:20:56 PM
(http://persephonemagazine.com/wp-content/uploads/2013/01/jon-stewart-popcorn11.gif)


Man, Alpha2Omega is just killing cikljamas.

This is better than watching Spartacus.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 25, 2014, 07:47:33 PM
The only factor determining the daily rate of rotation wrt the stars is the rotation of the earth. Period. This is a constant. You're getting tangled up in your  "daily rate of annual rotation", which is a convoluted way of describing the difference between the lengths of the sidereal day (constant) and apparent solar day (not quite constant) and thinking the length of the mean solar day (which is irrelevant here) is somehow mixed in here. It's not.

Well, this is not an answer, and you know it. Now, i have to ask you this: If you are aware that what you are offering as an explanation doesn't explain anything, why then you do this?

The whole thing is very simple:

24 hours (mean solar day) MINUS 23h56m4s = 3m56s= 0,986 degree = daily rate of annual rotation of the stars above the Earth

Now, if you want to proceed to defend HC theory the only possible logical way out of this is this:

1. Either you claim that 0,986 degree is not equal to daily rate of annual roatation of the stars

2. Or you claim that it is constant indeed

If 1 then you disagree not just with FET, but with basic HC rules also, and you have to prove what you claim which will be very difficult...

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?

It is very strong argument against fraudulent HC theory, and there is no place to hide from it.

Sorry, but you are just busted!

And this is only the beginning...
It is an answer, and all evidence suggests strongly it's correct. it's just not the one you want to hear.

The length of the apparent solar day is not a constant through the year; it can vary by some seconds a day - I think we've agreed on that. The length of the mean solar day is, by definition, a constant since it's the length of all those apparent solar days averaged over the year - I think we've agreed on that, too. The length of the sidereal day is a constant - I think we also agree on that. If you subtract a constant from a constant, you get a constant. Any given distant star will culminate almost exactly 3m 56s earlier each civil day (civil days are based on the mean solar day, remember) due to the difference between the lengths of the mean solar day and sidereal day; the time between sun culminations will vary slightly from day to day due to eccentricity of earth's orbit and obliquity of the ecliptic. How does this disprove Heliocentric Solar System theory? Heliocentric theory isn't confounded by this; it explains why this happens.

It seems we agree on the basic premises (if you disagree with any of those first three statements in the paragraph above, please say clearly which one(s) you think is wrong, and, if possible, why), but your interpretation of them goes off the rails.

As I asked, and you ignored, how would you explain the varying length of the apparent solar day in a Geocentric model?
Title: Re: "Equator" problem
Post by: cikljamas on October 26, 2014, 01:43:17 AM
Quote
Sorry, but you are just busted!

And this is only the beginning...
It is an answer, and all evidence suggests strongly it's correct. it's just not the one you want to hear.

The length of the apparent solar day is not a constant through the year; it can vary by some seconds a day - I think we've agreed on that. The length of the mean solar day is, by definition, a constant since it's the length of all those apparent solar days averaged over the year - I think we've agreed on that, too. The length of the sidereal day is a constant - I think we also agree on that. If you subtract a constant from a constant, you get a constant. Any given distant star will culminate almost exactly 3m 56s earlier each civil day (civil days are based on the mean solar day, remember) due to the difference between the lengths of the mean solar day and sidereal day; the time between sun culminations will vary slightly from day to day due to eccentricity of earth's orbit and obliquity of the ecliptic. How does this disprove Heliocentric Solar System theory? Heliocentric theory isn't confounded by this; it explains why this happens.

I see, you vote for the second option, wise!

Well, now the only thing that remains to be solved is this question (i repeat):

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?

You partially answered it:

If you subtract a constant from a constant, you get a constant.

Bravo! That is the correct answer!

But the essence of my question wasn't about if we can get the constant by substracting one constant from another constant, since this is a self-evident logical inference, my question was about this:

Mean Solar Time doesn't happen every day, it happens only during equinoxes!

So, if every day of the year was an equinox then there would be no flaw in this matter.

But, the core of this flaw is in the fact that everyday is not an equinox!

So, if everyday is not an equinox, how can HC theory be defended by doing astronomical calculations as if everyday was an equinox indeed?

24 hours is a constant but an artificial one, and why this constant is an artificial one you know very well, don't pretend that you don't know.

It is an artificial one at any other day except at those equinox days.

You cannot just say: "here we have got three constants and everything stays in perfect order", and proceed as if nothing of great importance hasn't just happened (been changed), because something (replacement of identity) which makes big difference has happened and has been changed (replaced):

One out of these three constants doesn't present anything that exist in reality (except during equinox) and you still pretend that you are not aware of it!!!

And i repeat: Everyday is not an equinox!!!

So how can stars shift EVERYDAY EXACTLY 0,986 degrees if the time between sun culminations vary from day to day?

There you go: when you say "the time between sun culminations vary from day to day" what you really say is: One of these constants is not a REAL-ACTUAL constant at all!!!

Once again: How can you use an artificaial constant as if it were something that fits everyday reality?

Exit heliocentrism and join us, why would you support an obvious lie, you are too smart to be a heliocentrist (or a cop ), aren't you?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 26, 2014, 08:30:35 AM
So many words. So little understanding.

Quote
Sorry, but you are just busted!

And this is only the beginning...
It is an answer, and all evidence suggests strongly it's correct. it's just not the one you want to hear.

The length of the apparent solar day is not a constant through the year; it can vary by some seconds a day - I think we've agreed on that. The length of the mean solar day is, by definition, a constant since it's the length of all those apparent solar days averaged over the year - I think we've agreed on that, too. The length of the sidereal day is a constant - I think we also agree on that. If you subtract a constant from a constant, you get a constant. Any given distant star will culminate almost exactly 3m 56s earlier each civil day (civil days are based on the mean solar day, remember) due to the difference between the lengths of the mean solar day and sidereal day; the time between sun culminations will vary slightly from day to day due to eccentricity of earth's orbit and obliquity of the ecliptic. How does this disprove Heliocentric Solar System theory? Heliocentric theory isn't confounded by this; it explains why this happens.

I see, you vote for the second option, wise!

Well, now the only thing that remains to be solved is this question (i repeat):

If 2 then you agree with HC theory, but in this case you have to explain how can you (or any HC follower in the world) get 0,986 degree as a constant daily rate of annual rotation of the stars by substracting 23h56m4s from 24 hours (mean solar day) since Mean Solar Day is not a real time but an artificial time?
It depends on your definition of day, as in "constant daily rate of annual rotation". If, by "day" you mean exactly 24 hours, then the shift will be a constant. This appears to be the case here since that angle represents the "overshoot" of 360 degrees of rotation in exactly 24 hours. If your definition of "day" is local apparent noon to local apparent noon (solar culmination to solar culmination) and thus slightly variable, then the shift will be slightly variable. How hard is this to understand?

Quote
You partially answered it:

If you subtract a constant from a constant, you get a constant.

Bravo! That is the correct answer!

But the essence of my question wasn't about if we can get the constant by substracting one constant from another constant, since this is a self-evident logical inference, my question was about this:

Mean Solar Time doesn't happen every day, it happens only during equinoxes!
No, the difference between mean and apparent solar days is largest near the equinoxes and solstices. It would be largest exactly on those dates (longest apparent solar days on the solstices, shortest on the equinoxes)  if it weren't for the effect of the elliptical orbit, which shifts them a few days. The length of the apparent solar day is exactly 24 hours four times a year - roughly midway between the equinoxes and solstices. You either didn't look at the link for EoT (http://en.wikipedia.org/wiki/Equation_of_time), didn't understand it, or both.

Quote
So, if every day of the year was an equinox then there would be no flaw in this matter.

But, the core of this flaw is in the fact that everyday is not an equinox!

So, if everyday is not an equinox, how can HC theory be defended by doing astronomical calculations as if everyday was an equinox indeed?

24 hours is a constant but an artificial one, and why this constant is an artificial one you know very well, don't pretend that you don't know.

It is an artificial one at any other day except at those equinox days.
Ignoring your error about when those days occur, the core of the flaw is still your reasoning. You're the one insisting on using the mean solar day by insisting on a constant daily rate of change; the HC model doesn't require this. As I said earlier, the mean solar day has no astronomical meaning; it's used for everyday timekeeping as a convenience.

Quote
You cannot just say: "here we have got three constants and everything stays in perfect order", and proceed as if nothing of great importance hasn't just happened (been changed), because something (replacement of identity) which makes big difference has happened and has been changed (replaced):

One out of these three constants doesn't present anything that exist in reality (except during equinox) and you still pretend that you are not aware of it!!!
As I said earlier, the mean solar day has no astronomical meaning; it's used for everyday timekeeping as a convenience.

Quote
And i repeat: Everyday is not an equinox!!!

So how can stars shift EVERYDAY EXACTLY 0,986 degrees if the time between sun culminations vary from day to day?
It depends on your definition of day. If, by "day" as in "EVERYDAY" you mean exactly 24 hours, then the shift will be a constant. If your definition of "day" is local apparent noon to local apparent noon (solar culmination to solar culmination), then the shift will be slightly variable. How hard is this to understand?

Quote
There you go: when you say "the time between sun culminations vary from day to day" what you really say is: One of these constants is not a REAL-ACTUAL constant at all!!!

Once again: How can you use an artificaial constant as if it were something that fits everyday reality?

Exit heliocentrism and join us, why would you support an obvious lie, you are too smart to be a heliocentrist (or a cop ), aren't you?

Dude... will you lose the "lie" stuff, already? It's just a gratuitous insult that adds nothing to your argument (in fact, it weakens it, and your argument doesn't need that kind of "help"). It also makes it that much harder for you to back down once you recognize your mistake.

Why would I want to join you? You still haven't explained the Equation of Time in your model at all. I already have one that explains it well, and in detail, as well as other measured phenomena such as parallax, well, and in detail.
Title: Re: "Equator" problem
Post by: cikljamas on October 26, 2014, 09:59:40 AM
What determines a daily rate of annual rotation of the stars according to HC?

Answer: A daily rate of annual rotation of the stars around the Earth is determined by the the speed of the Earth's rotation on it's axis (which is a constant, in reality it is not a constant because it doesn't exist in the first place, but let's pretend that it exists and that it is a constant indeed), and by the orbital speed of the Earth which is not a constant (and which doesn't exist also)!

Mean time follows the "mean sun", best described by Meeus:

    "Consider a first fictitious Sun travelling along the ecliptic with a constant speed and coinciding with the true sun at the perigee and apogee (when the Earth is in perihelion and aphelion, respectively). Then consider a second fictitious Sun travelling along the celestial equator at a constant speed and coinciding with the first fictitious Sun at the equinoxes. This second fictitious sun is the mean Sun..."

Now, let's say that the Earth's moderate speed is around equinoxes (and it is so) and let's pretend that there is no axial tilt of the Earth (as it is in reality indeed), in order to simplify the whole thing a little bit.

During equinoxes the alleged orbital speed of the Earth is moderate, so let's call it The Mean Speed.
During northern solstice the alleged orbital speed of the Earth is lowest, so let's call it The Low Speed.
During southern solstice the alleged orbital speed of the Earth is highest, so let's call it The High Speed.

Apparent solar day depends on the current spatial orientation of the Earth due to Earth's alleged tilt and current orbital position with respect to the position of the Sun, and all this depends on the Earth's alleged orbital speed.

Now, for the sake of simplicity let's say that the longest apparent day coincides with The Low Speed (of the Earth), that the shortest apparent day coincides with The High Speed (of the Earth), and that the mean sun coincides with equinox a.k.a. The Mean Speed.

So, now you should finally understand what is the deal here.

As the speed of the Earth changes through the year, so the daily rate of an annual motion of the Stars has to change too, BUT IT DOESN'T CHANGE!!!

What i really meant by saying "Equinox is not everyday!" was: The orbital speed of the Earth is not the same through the year, and that is why a daily rate of an annual motion of the stars wouldn't be a constant if it were really depend on irregular orbital speed of the Earth.

But it doesn't depend on that, since the orbital speed of the Earth doesn't exist in the first place!!!
Title: Re: "Equator" problem
Post by: 29silhouette on October 26, 2014, 11:04:00 AM
Meanwhile for FET, the sun moves overhead twice as fast at different times of the year and different latitudes than others, but somehow this isn't noticed on the ground.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 26, 2014, 03:45:32 PM
What determines a daily rate of annual rotation of the stars according to HC?

Answer: A daily rate of annual rotation of the stars around the Earth is determined by the the speed of the Earth's rotation on it's axis (which is a constant

This is correct so far.

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, in reality it is not a constant because it doesn't exist in the first place

This is your opinion.

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, but let's pretend that it exists and that it is a constant indeed)

There's no need to pretend, but, as they say, for the sake of the argument...

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, and [also] by the orbital speed of the Earth which is not a constant

This depends on what you mean by "daily". If "daily" means apparent solar days, then yes; if "daily" means mean solar days, then no. Which do you prefer?

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(and which doesn't exist also)!

More opinion.

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Mean time follows the "mean sun", best described by Meeus:

    "Consider a first fictitious Sun travelling along the ecliptic with a constant speed and coinciding with the true sun at the perigee and apogee (when the Earth is in perihelion and aphelion, respectively). Then consider a second fictitious Sun travelling along the celestial equator at a constant speed and coinciding with the first fictitious Sun at the equinoxes. This second fictitious sun is the mean Sun..."

OK. Note that Meeus is not saying that the length of the apparent solar day equals the length of the mean solar day at the equinoxes. He is only setting his second fictitious Sun to coincide with the first fictitious Sun at those points.

You still haven't looked at a plot of the Equation of Time, have you? Yes or no, please? If yes, do you realize that the points on that plot where the curves level out (the four peaks and valleys on the graph are the times when apparent solar days are the same length as mean solar days and the steepest parts of the curve are when the lengths differ the most?

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Now, let's say that the Earth's moderate speed is around equinoxes (and it is so)

Are you referring to orbital speed? If so, then this is approximately correct because the line of apsides happens to very nearly coincide with the solstices.

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and let's pretend that there is no axial tilt of the Earth (as it is in reality indeed), in order to simplify the whole thing a little bit.

More opinion, but, for the sake of the argument, OK, no tilt.

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During equinoxes the alleged orbital speed of the Earth is moderate, so let's call it The Mean Speed.
During northern solstice the alleged orbital speed of the Earth is lowest, so let's call it The Low Speed.
During southern solstice the alleged orbital speed of the Earth is highest, so let's call it The High Speed.

A couple of points here. If there's no tilt, then there will be no discrete equinoxes and no solstices; every day will be an equinox. Can we restate these as:

At aphelion orbital speed of the Earth is lowest, so let's call it The Low Speed.
At perihelion orbital speed of the Earth is highest, so let's call it The High Speed.
Midway between these, orbital speed of the Earth is moderate, so let's call it The Mean Speed.

It happens that aphelion is near our (current) actual northern solstice, and perihelion is near our actual southern solstice, so your original statements can be thought of as fairly close, although the language used is imprecise since the solstices don't necessarily have to coincide with the apsides, and in our simplified model the solstices have been abolished, anyway.

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Apparent solar day depends on the current spatial orientation of the Earth due to Earth's alleged tilt and current orbital position with respect to the position of the Sun, and all this depends on the Earth's alleged orbital speed.

Wait... I thought we were simplifying by removing axial tilt. Now it's back, without ever using the simplification.

And that paragraph is kind of a hodge-podge of confused terminology and extraneous qualifiers. The length of the apparent solar day is affected by where the Earth is in its elliptical orbit, and thus its speed, and on how its axis of rotation is aligned wrt the Sun. The axial alignment does not depend on its orbital speed; in fact, the angle between the line of apsides and equinoxes is slowly changing due to precession.

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Now, for the sake of simplicity let's say that the longest apparent day coincides with The Low Speed (of the Earth), that the shortest apparent day coincides with The High Speed (of the Earth), and that the mean sun coincides with equinox a.k.a. The Mean Speed.

You have longest and shortest backwards, but, with that correction, OK. So are we back to no tilt? It looks like we are, but I'm losing track.

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So, now you should finally understand what is the deal here.

Absolutely! You are confusing yourself. That's been apparent all along.

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As the speed of the Earth changes through the year, so the daily rate of an annual motion of the Stars has to change too, BUT IT DOESN'T CHANGE!!!

I really think that the term "daily rate of an annual motion of the Stars" is a poor one and is causing you unneeded confusion. One part of the problem is that "daily" is ambiguous. "Daily" can refer to mean days or apparent days. PICK ONE! If you pick mean solar days, then, no the daily change of Right Ascension at the meridian at, say local noon each (mean solar) day does not change. If you pick apparent solar days, then the rate of change of the Sun's RA at (apparent solar) noon does change. You are mixing these definitions and claiming that proves something. It only proves you can't keep your terms straight.

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What i really meant by saying "Equinox is not everyday!" was: The orbital speed of the Earth is not the same through the year, and that is why a daily rate of an annual motion of the stars wouldn't be a constant if it were really depend on irregular orbital speed of the Earth.

The varying orbital speed is actually the lesser component in the varying length of the apparent solar day. The larger component is due to the obliquity of the ecliptic (since you're quoting Jean Meeus, I presume you know what that means.)

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But it doesn't depend on that, since the orbital speed of the Earth doesn't exist in the first place!!!

More opinion asserted as fact. So how's that non-rotating Geocentric explanation of the Equation of Time coming? And have you looked at stellar and planetary parallax yet?
Title: Re: "Equator" problem
Post by: cikljamas on October 26, 2014, 04:58:22 PM
Aha, thanks for the detailed post, it helped me a lot, now i can (at least) be pretty sure that you understand my argument.

Just a reminder:

If any of the determining factors is not the constant the result (function) can't be a constant!

Period!

Alleged orbital speed of the Earth (a main factor) is not a constant!

So, daily rate of annual rotation (a function) can't be a constant too!

You are busted! And you know that!

In addition:

Alleged orbital speed of the Earth would be the main factor (if it were exist) with respect to a daily rate of annual rotation!

Obliquity of the ecliptic is the main factor regarding the lenght of an apparent solar day!

Time for retraction!

Good night!

P.S. Tomorow i am going to teach you an important lesson about the stelar parallax hoax!
Title: Re: "Equator" problem
Post by: sokarul on October 26, 2014, 06:04:21 PM
Aha, thanks for the detailed post, it helped me a lot, now i can (at least) be pretty sure that you understand my argument.

Just a reminder:

If any of the determining factors is not the constant the result (function) can't be a constant!

Period!

Alleged orbital speed of the Earth (a main factor) is not a constant!

So, daily rate of annual rotation (a function) can't be a constant too!

You are busted! And you know that!

In addition:

Alleged orbital speed of the Earth would be the main factor (if it were exist) with respect to a daily rate of annual rotation!

Obliquity of the ecliptic is the main factor regarding the lenght of an apparent solar day!

Time for retraction!

Good night!

P.S. Tomorow i am going to teach you an important lesson about the stelar parallax hoax!
Can you teach us the "when things get further away they appear smaller hoax" as well?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 26, 2014, 06:40:54 PM
Aha, thanks for the detailed post, it helped me a lot, now i can (at least) be pretty sure that you understand my argument.

I think I understand what you're trying to argue but recognize errors in it.

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Just a reminder:

If any of the determining factors is not the constant the result (function) can't be a constant!

Period!

Not true.

If:
f(x) = x + 2
g(x) = 1 - x
h(x) = f(x) + g(x)

Then h(x) is a constant for any x even though f(x) and g(x) are not. Sure, this is a contrived example, but, remember, you said any. I only have to find one counter example to prove your assertion wong.

Here's another:
f(x) = 3x
g(x) = 6(x + 1) - 6x - 6
h(x) = f(x)g(x)

f(x) is not constant for any x, but h(x), which has f(x) as a factor is a constant for any x.

"If any of the determining factors is not the constant the result (function) can't be a constant!" is proven false by counterexample(s).

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Alleged orbital speed of the Earth (a main factor) is not a constant!

We've never disagreed on this. The bigger factor in the variation of the length of the apparent solar day, obliquity, is, I think, also accepted.

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So, daily rate of annual rotation (a function) can't be a constant too!

Why not? You still haven't shown that "daily rate of annual rotation" isn't a constant because you haven't defined what you mean by "daily". This means that "daily rate of annual rotation" may or may not depend on orbital speed of the Earth and obliquity of the ecliptic, so it may indeed be a constant. Even if you use a definition for "daily" that does depend on orbital speed of earth, you still haven't disproved the Heliocentric model, which was your original assertion, since, if you use that definition, you can no longer say "daily rate of annual rotation" is a constant. You gotta be consistent with your "days" or your argument is invalid.

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You are busted! And you know that!

Do you really still believe this, or are you just unwilling to admit you know you're wrong?

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In addition:

Alleged orbital speed of the Earth would be the main factor (if it were exist) with respect to a daily rate of annual rotation!

Obliquity of the ecliptic is the main factor regarding the lenght of an apparent solar day!

Wait...  what? Don't you understand what your quote from Meeus was saying?

First he posits a "first fictitious Sun" that travels along the ecliptic at a constant rate, coinciding with the real sun at the solstices. Sometimes this fictitious Sun will lead the real Sun and sometimes it will trail, but completes a circuit in exactly the same time as the real Sun, "smoothing out" the variations due to the elliptical orbit.

Next he posits a "second fictitious Sun"  that travels along the equator at a uniform rate, sometimes leading the first fictitious sun in RA and sometimes trailing it, but coinciding with it at the equinoxes. This smooths out the variations in change in RA per unit time due to obliquity of the ecliptic.

Are you suggesting there is a difference in how changes in the length of the apparent solar day due to eccentricity and obliquity affects the length of the apparent solar day or the apparent position of the stars relative to the Sun's RA? Why? Would these affect "daily rate of annual rotation" differently (if "daily" means apparent, not mean, solar days)? Please amplify this.

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Time for retraction!

Great! I've been waiting for days. It's tough to do, so I admire you for owning up that you were wrong.

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P.S. Tomorow i am going to teach you an important lesson about the stelar parallax hoax!

Ohboy! I can't wait! I trust that, unlike sceptimatic, you will actually deliver something as promised.

Still nothing on the Equation of Time? Not even to deny it exists?
Title: Re: "Equator" problem
Post by: cikljamas on October 27, 2014, 05:22:04 AM
Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice. When in the first point, the sun moves in 24 hours 1,01943 degree; in the second point, he moves only 0,95319 degree. The daily motion of the sun is constantly varying in every place of its orbit, between these two points. The medium of the two is 0,98632, or 59'11'', which is the daily motion of the sun about the beginning of October and April.

We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec

After one sidereal day passes away we have to wait additional 3min56sec for all the stars to move 0,986 degree!

This period of time never changes, it is always 3min56sec exactly!

Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

As for the equation of time: what that notion really presents?

It is about lining up ecliptic (irregular yearly motion of the Sun) with a steady-regular of the stars!

That is the manner in which we get our Mean Sun.

A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

The motion of the Stars, and the motion of the Sun are obviously two independent motions!

It has long been known that the Sun moves eastward relative to the fixed stars along the ecliptic. Since the middle of the first millennium BC the diurnal rotation of the fixed stars has been used to determine mean solar time, against which clocks were compared to determine their error rate.

Mechanical clocks did not achieve the accuracy of Earth's "star clock" until the beginning of the 20th century. Today's atomic clocks have a much more constant rate than the Earth, but its star clock is still used to determine mean solar time.

WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

http://www.energeticforum.com/264541-post421.html (http://www.energeticforum.com/264541-post421.html)

Now is maybe the right time for an introduction of the star parallax issue:

See the next post...

edit: I would rather wait your answer to this post, and then i am going to post "star parallax hoax" post...it will take me 1 sec to post it, since it is already finished...don't worry...i just want to keep this thread readable...
Title: Re: "Equator" problem
Post by: Rama Set on October 27, 2014, 05:39:16 AM
You know that the 3:56 seconds you are fixated on is not constant but rather becoming shorter as the rotation of the Earth slows down?  So it is in fact, not a constant at all.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 27, 2014, 07:09:21 AM
You know that the 3:56 seconds you are fixated on is not constant but rather becoming shorter as the rotation of the Earth slows down?  So it is in fact, not a constant at all.

He has much bigger issues than that, like a basic misunderstanding of what he's arguing about. I'm not sure he knows about this and I've been neglecting this very small variable as a needless complication. There are other small factors, such as precession and nutation, that also can be safely neglected in the discussion at hand (precession was touched on briefly in a late post in reference to the angle between apsides and the line of equinoxes slowly changing).

Also, the slowing spin will increase the difference between sidereal and solar days.

[Edit to add] And, actually, it's closer to 3:55.9084 according to Wikipedia, but 3:56 is close enough for the discussion.
Title: Re: "Equator" problem
Post by: cikljamas on October 27, 2014, 07:24:35 AM
1. If we accept the Copernican viewpoint and its unavoidable extrapolations with regard to the structure of the universe, we have to accept the consequences. Then we cannot hold on to the picture of a simple sun- centered cosmos, of which not even Newton was fully convinced, but which Bradley and Molyneux took for granted. Today the astronomers assure us that our Great Light is only an insignificant member of a spiral Milky Way galaxy, containing billions of stars. Our sun flies at a speed of about 250 km/sec around the center of this system. And that is not all, the ruling cosmology also tells us how the Milky Way itself whirls at 360,000 km/hr through the space occupied by the local group of galaxies. Now all these imposing particulars are theoretically gathered from observations assuming the speed of light to be 300,000 km/sec, at least, everywhere through our spatial neighborhood. But if this cosmological panorama is put through its paces, there is a hitch somewhere. The astronomical theorists cannot have their cake and eat it. If they accept— as all the textbooks still do!—Bradley's “proof” of the Copernican truth, then their cosmological extrapolations of that truth clash with a not-yet developed simple heliocentrism; that is to say, with the model of an earth orbiting a spatially unmoved sun.

The other way around, when holding on to their galactic conjectures, they are at a loss how to account for a steady 20”.5 stellar aberration. For in that scheme our earth, dragged along by the sun, joins in this minor star's 250 km/sec revolution around the center of the Milky Way. If, for instance, in March we indeed would be moving parallel to the sun's motion, our velocity would become 250+30 = 280 km/sec, and in September 250-30 = 220 km/sec. The “aberration of starlight,” according to post- Copernican doctrine, depends on the ratio of the velocity of the earth to the speed of light. As that velocity changes the ratio changes. Hence Bradley's 20”.496 should change, too. But it does not. Therefore, there is truly a fly in this astronomical ointment, paraded and promoted as a truth.

2.For those who maybe don't know:

1 sec. of an arc = An angle subtended by a U.S. dime coin at a distance of 4 km!!!
1 mas = 1/1000000 sec. of an arc
Tycho Brahe was able to measure angles about 0,3 minutes of an arc.
Casini was able to measure angles about 3,6 seconds of an arc.

The North Star, also known as Polaris, is known to stay fixed in our sky. It marks the location of the sky’s north pole, the point around which the whole sky turns. That’s why you can always use Polaris to find the direction north.

Just look at this insanity:

In a recent letter to the Astrophysical Journal, Turner et al.(2013) (TKUG from here on) suggested that the parallax as measured by Hipparcos (van Leeuwen 2007a,b) for Polaris (HIP 11767, HD 8890) is signi?cantly lower than it should be. The distance of 99 ± 2 pc suggested by TKUG on the basis of he assumed pulsation mode of Polaris is equivalent to a parallax of 10, 1 ± 0 2 mas, very di?erent from the parallax as measured by Hipparcos, 7, 54 ± 0 11 mas. Consequently, I have recently frequently been asked if it is at all possible for the Hipparcos parallax measurement to be so far o?. This letter shows the Hipparcos astrometric solution for Polaris in all detail as a means to assess the robustness of that solution, to assess whether its measurement of the parallax could be o?set by 23 times its standard error. It also brie?y discusses other arguments that have been used to suggest a signi?cantly shorter distance for Polaris than what has been measured by Hipparcos. READ MORE: http://arxiv.org/pdf/1301.0890v1.pdf (http://arxiv.org/pdf/1301.0890v1.pdf)

All that this "stars parallax" nonsense is about is this : throwing sand in the eyes of a sincere thinkers, nothing more than that!

3. http://www.realityreviewed.com/Negative%20parallax.htm (http://www.realityreviewed.com/Negative%20parallax.htm)

4 .Again and again have their theories been combated and exposed, but as often have the majority, who do not think for themselves, accepted the popular thing. No less an authority in his time than the celebrated Danish astronomer, Tycho Brahe, argued that if the earth revolves in an orbit round the sun, the change in the relative position of the stars thus necessarily occasioned, could not fail to be noticed. In the " History of the Conflict between Religion and Science," by Dr. Draper, pages 175 and 176, the matter is referred to m the following words :

" Among the arguments brought forward against the Copernican system at the time of its promulgation, was one by the great Danish astronomer, Tycho Brahe, originally urged by Aristarchus against the Pythagorean system, to the effect that if, as was alleged, the earth moves round the sun, there ought to he a change in the relative position of the stars ; they should seem to separate as we approach them, or to close together as we recede from them... At that time the sun's distance was greatly under-estimated. Had it been known, as it is now, that the distance exceeds 90 million miles, or that the diameter of the orbit is more than 180 million, that argument would doubtless have had very great weight. In reply to Tycho, it was said that, since the parallax of a body diminishes as its distance increases, a star may be so far off that its parallax may be imperceptible. THIS ANSWER PROVED TO BE CORRECT."

To the uninitiated, the words " this answer proved to be correct," might seem to settle the matter, and while it must be admitted that parallax is diminished or increased according as the star is distant or near, parallax and direction are very different terms and convey quite different meanings. Tycho stated that the direction of the stars would be altered ; his critics replied that the distance gave no sensible difference of parallax. This maybe set down as ingenious, but it is no answer to the proposition, which has remained unanswered to this hour, and is unanswerable.

If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."

5. This is not about the parallax but it gives us additional insight into a sanity of the heliocentric theory.: http://www.energeticforum.com/264118-post355.html (http://www.energeticforum.com/264118-post355.html)

Title: Re: "Equator" problem
Post by: markjo on October 27, 2014, 09:42:54 AM
Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice.
Which fits quite nicely with what one would expect if the earth has an elliptical orbit around the sun.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 27, 2014, 10:20:08 AM
...
In a recent letter to the Astrophysical Journal, Turner et al.(2013) (TKUG from here on) suggested that the parallax as measured by Hipparcos (van Leeuwen 2007a,b) for Polaris (HIP 11767, HD 8890) is signi?cantly lower than it should be. The distance of 99 ± 2 pc suggested by TKUG on the basis of he assumed pulsation mode of Polaris is equivalent to a parallax of 10, 1 ± 0 2 mas, very di?erent from the parallax as measured by Hipparcos, 7, 54 ± 0 11 mas. Consequently, I have recently frequently been asked if it is at all possible for the Hipparcos parallax measurement to be so far o?. This letter shows the Hipparcos astrometric solution for Polaris in all detail as a means to assess the robustness of that solution, to assess whether its measurement of the parallax could be o?set by 23 times its standard error. It also brie?y discusses other arguments that have been used to suggest a signi?cantly shorter distance for Polaris than what has been measured by Hipparcos. READ MORE: http://arxiv.org/pdf/1301.0890v1.pdf (http://arxiv.org/pdf/1301.0890v1.pdf)
...
Sorry about the time you spent putting that tome together. It doesn't address the question I asked.  Repeating it here:

How does the Geocentric model of the universe explain observed parallax of the distant stars and planets? 

Since you cite a case (quoted above) where newly-acquired high-precision parallax measurement of a star disagrees with an earlier estimate of distance, I presume that means you believe that stellar parallax actually exists. So how does it work in the GC model?

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.
Title: Re: "Equator" problem
Post by: cikljamas on October 27, 2014, 12:26:39 PM
Sorry about the time you spent putting that tome together.
What tome? (http://)

How come you skipped one whole post? Repeating it here:

Quote
Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice. When in the first point, the sun moves in 24 hours 1,01943 degree; in the second point, he moves only 0,95319 degree. The daily motion of the sun is constantly varying in every place of its orbit, between these two points. The medium of the two is 0,98632, or 59'11'', which is the daily motion of the sun about the beginning of October and April.

We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec

After one sidereal day passes away we have to wait additional 3min56sec for all the stars to move 0,986 degree!

This period of time never changes, it is always 3min56sec exactly!

Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

As for the equation of time: what that notion really presents?

It is about lining up the ecliptic (irregular yearly motion of the Sun) with a steady-regular motion of the stars!

That is the manner in which we get our Mean Sun.

A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

The motion of the Stars, and the motion of the Sun are obviously two independent motions!

It has long been known that the Sun moves eastward relative to the fixed stars along the ecliptic. Since the middle of the first millennium BC the diurnal rotation of the fixed stars has been used to determine mean solar time, against which clocks were compared to determine their error rate.

Mechanical clocks did not achieve the accuracy of Earth's "star clock" until the beginning of the 20th century. Today's atomic clocks have a much more constant rate than the Earth, but its star clock is still used to determine mean solar time.

WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

http://www.energeticforum.com/264541-post421.html (http://www.energeticforum.com/264541-post421.html)

Now is maybe the right time for an introduction of the star parallax issue:

See the next post...

edit: I would rather wait your answer to this post, and then i am going to post "star parallax hoax" post...it will take me 1 sec to post it, since it is already finished...don't worry...i just want to keep this thread readable...

How does the Geocentric model of the universe explain observed parallax of the distant stars and planets? 

Since you cite a case (quoted above) where newly-acquired high-precision parallax measurement of a star disagrees with an earlier estimate of distance, I presume that means you believe that stellar parallax actually exists. So how does it work in the GC model?

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.


No, i don't think that stellar parallax exists, not even geocentric one. There are some phenomena that HCs and GCs link with that parallax concept, but what these phenomena really are, is a big question. Micro arc seconds, tell me about it...thousands of light years, tell me about it...

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.

In which post? Last one (about stellar parallax hoax) or one that i just have repeated here?

Title: Re: "Equator" problem
Post by: Alpha2Omega on October 27, 2014, 12:35:53 PM
This didn't post. Fortunately, I saved a copy. Sorry for the confusion.

Numerous observations, made with precision, have ascertained, that the sun moves fastest in a point of his orbit situated near the winter solstice, and slowest in the opposite point of his orbit near the summer solstice. When in the first point, the sun moves in 24 hours 1,01943 degree; in the second point, he moves only 0,95319 degree.

OK. That's the apparent motion of the sun along the ecliptic, which is due to the eccentricity of the Earth's orbit.

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The daily motion of the sun is constantly varying in every place of its orbit, between these two points. The medium of the two is 0,98632, or 59'11'', which is the daily motion of the sun about the beginning of October and April.

You can't simply take the arithmetic average of fastest and slowest to get the mean here, but your answer is fairly close.

Note that this is not the only factor in the varying length of the apparent solar day, though. A larger factor is obliquity of the ecliptic, and this works to make the apparent solar days shortest at the equinoxes and longest at solstices, in case "apparent = mean at the equinoxes" is what you're getting at here; it's not.

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We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec

Hold it. Your ASTRONOMICAL DAY here is just the mean solar day we've been talking about all along (call it a civil day if you want a shorter name). The term 'ASTRONOMICAL DAY' would be better applied to the sidereal day, but we already have a perfectly good and well-accepted names for both of these, so why introduce a new term?

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After one sidereal day passes away we have to wait additional 3min56sec for all the stars to move 0,986 degree!

This period of time never changes, it is always 3min56sec exactly!

Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

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As for the equation of time: what that notion really presents?

It is about lining up ecliptic (irregular yearly motion of the Sun) with a steady-regular of the stars!

That is the manner in which we get our Mean Sun.

Actually, it shows the difference in RA between the steady-regular mean Sun and the sometimes-leading and sometimes-lagging true Sun. The stars aren't involved.

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A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical civil day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

The motion of the Stars, and the motion of the Sun are obviously two independent motions!

Bingo!

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It has long been known that the Sun moves eastward relative to the fixed stars along the ecliptic. Since the middle of the first millennium BC the diurnal rotation of the fixed stars has been used to determine mean solar time, against which clocks were compared to determine their error rate.

Mechanical clocks did not achieve the accuracy of Earth's "star clock" until the beginning of the 20th century. Today's atomic clocks have a much more constant rate than the Earth, but its star clock is still used to determine mean solar time.

Not sure how this adds anything to the discussion, but, OK, sure

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WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator." A slightly longer answer is that "Our celestial coordinate system is typically aligned with the Equator and Equinox of J2000, but the origin of this coordinate system is often the barycenter of the solar system, which is usually very close to the center of the Sun."

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<snip random link>

Now is maybe the right time for an introduction of the star parallax issue:

See the next post...

edit: I would rather wait your answer to this post, and then i am going to post "star parallax hoax" post...it will take me 1 sec to post it, since it is already finished...don't worry...i just want to keep this thread readable...

That's not a bad idea. [Oh, well...]
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 27, 2014, 02:52:50 PM
How come you skipped one whole post? Repeating it here:


Probably user error. Sorry about that.

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How does the Geocentric model of the universe explain observed parallax of the distant stars and planets? 

Since you cite a case (quoted above) where newly-acquired high-precision parallax measurement of a star disagrees with an earlier estimate of distance, I presume that means you believe that stellar parallax actually exists. So how does it work in the GC model?

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.

No, i don't think that stellar parallax exists, not even geocentric one. There are some phenomena that HCs and GCs link with that parallax concept, but what these phenomena really are, is a big question. Micro arc seconds, tell me about it...thousands of light years, tell me about it...

We can look at the mistakes in your post and examine the purported issues, if you still want to, after you answer the question asked.

In which post? Last one (about stellar parallax hoax) or one that i just have repeated here?

The one where you cited a study about parallax.

There wouldn't be parallax if the GC model were correct, so I can see why you don't want to believe it exists. I gather from your remarks that you don't think Aberration of Light exists, either. The nice thing about simply ignoring data is it doesn't get in the way of whatever model you want to believe in. This way you can "continue to seek the truth" without being bothered by actual observations and measurements. This won't get you very far discussing ideas with professional astronomers, however; you were asking me to get an opinion of this stuff from one earlier. It certainly won't get you very far in "finding the truth" either, but that's your problem.

Any thoughts about the apparent retrograde motion of the outer planets? This is easy to see and is attributed to parallax in the HC model, which is quite accurate in predicting where the planets will appear in the sky well into the future. I would like to see how you explain this without resorting to parallax, or is this a hoax, too?

Just for the record:

mas is milliarcsecond, not microarcsecond (micro is usually abbreviated with 'u' if the lower-case Greek letter 'mu' isn't available). But, hey, what's three orders of magnitude between friends?

Polaris is not exactly at the North Celestial Pole, and does not "stay fixed" in the sky. It's close, and for casual observation that's a good enough approximation, but the small circle it traces in the sky is evident in "star trail" photographs taken with even simple equipment. I'm not sure why you think this matters for the discussion of parallax; its position relative to the ecliptic pole is what counts in that case.

Title: Re: "Equator" problem
Post by: cikljamas on October 27, 2014, 03:40:18 PM
Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

You are very close to the truth, only it doesn't depend only on the rate of rotation of the Earth, it depends only on the rate of rotation of the Stars. But the problem is that it is not that simple in HC theory...Watch: (http://)

This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator."

So, if we lived on the Sun, which star would be our reference (fixed) star around which all other stars and clasters of stars would rotate in perfectly concentric circles?

Bingo!

What that supposed to mean? You are trying to be a smart guy?

Just for the record: mas is milliarcsecond, not microarcsecond

O.K., thanks for that correction! It is still small enough to be laughable, don't worry...

Apparent retrograde motion of the outer planets...that's another topic...


Title: Re: "Equator" problem
Post by: sokarul on October 27, 2014, 05:48:54 PM
You shotgun attacks on RE in this thread but won't answer one topic that goes against what you preach?  I am disappointed.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 27, 2014, 07:20:40 PM
Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

You are very close to the truth, only it doesn't depend only on the rate of rotation of the Earth, it depends only on the rate of rotation of the Stars. But the problem is that it is not that simple in HC theory...Watch: (http://)

No, you asked for a Heliocentric explanation so I gave a Heliocentric answer - one where the Earth spins. Here's the answer in context:
Quote
Quote
Since this period of time allegedly depends of the Earth's orbital speed, some heliocentrist should finally give us some sane explanation to the following question: how in the world these 3min56sec can keep up being a constant in this HC manner?

Answer: because it doesn't depend on the Earth's orbital speed. It depends only on the rate of rotation. The Earth rotates 0.986 degrees in 3m 56s, no matter where it is in its orbit or the length of the apparent solar day. It's really that easy.

Since the Earth being fixed and the distant stars rotating about it, or the distant stars being fixed and the Earth rotating under them is simply a change in the frame of reference, they are equivalent. So thanks for acknowledging that my answer from the rotating-earth frame of reference was correct. I'll take that.

Can you summarize what you think that youtube link explains? Sometimes these will provide a visual insight that isn't easily described in words or static diagrams, but much more often they're a complete waste of time.

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This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator."

So, if we lived on the Sun, which star would be our reference (fixed) star around which all other stars and clasters of stars would rotate in perfectly concentric circles?

Here's the whole answer in context:
Quote
Quote
WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

This is a different topic, but the short answer is "because we live on the Earth and not the Sun, and our reference is the Earth's axis of rotation, which is normal to the equator." A slightly longer answer is that "Our celestial coordinate system is typically aligned with the Equator and Equinox of J2000, but the origin of this coordinate system is often the barycenter of the solar system, which is usually very close to the center of the Sun."


If we lived on the sun, I seriously doubt the Earth's equator would play any part in solar-dweller cosmology. Note that modern cosmology doesn't expect perfect circles, nor, since it's based on an inertial universe (one that doesn't rotate), it doesn't have any point that everything revolves around. Assuming we could see any celestial objects at all (remember, if we lived on the Sun it would always be "daytime" because the "ground" would be literally as bright as the Sun), the solar-dweller celestial coordinate system would probably be based on the Sun's equator, but the origin for the RA coordinate would be up for grabs since there would be no significant point like the Vernal Equinox to tie it to. The origin would likely be the SS barycenter or the center of the Sun. What else would it be?

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Bingo!

What that supposed to mean? You are trying to be a smart guy?

It means "we have a winner". In this context it means you got that part right. It may be idiomatic enough that it didn't make sense to someone not from the US; if that's the case, sorry. It comes from a game of chance called "Bingo", usually played in large groups, and play continues until someone wins. The winner shouts "BINGO!" ending the game.

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Just for the record: mas is milliarcsecond, not microarcsecond

O.K., thanks for that correction! It is still small enough to be laughable, don't worry...

It's not that small. Instruments capable of measuring angles well below the arcsecond level were being built and used in the 19th century and have gotten better since then. Instead of laughing, do some research.

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Apparent retrograde motion of the outer planets...that's another topic...

It's not really, but since you've ruled out parallax, it may be a difficult problem. It's easily explained in a heliocentric solar system with parallax due to the Earth's orbit making the outer planets appear to move backward - and is strong evidence in favor of the Heliocentric model over the Geocentric one.  In a Geocentric system it's not so simple. Surely not impossible, but not so simple, either. I'm really looking forward to seeing your approach.
Title: Re: "Equator" problem
Post by: cikljamas on October 28, 2014, 10:07:36 AM
This is the problem: Astronomical day = Solar day

If we assumed that the Earth rotates but doesn't revolve then we could consider principally almost identical situations when comparing HC & GC. But since the HC supposes Earth's revolution around the Sun also, then you have to expect that you will be faced with unresolvable difficulties.

Whenever the Earth allegedly completes one sidereal rotation what follows right after is Our Famous Constant!

The difference between one Astronomical day and one Sidereal day is equal to Our Constant.

The difference between one Solar day and one Sidereal day is equal to Our Constant AGAIN.

Anything suspicious over here?

Why are these differences equal to the same period (amount) of time which is 3min56sec???

How do we get so perfect synchronization between one Astronomical and one Solar day, although the Earth's orbit is not a perfect circle, and the Earth's orbital speed is not a constant???

To make it even clearer for our audience let's pose this question:

How probable (to be accomplish in reality) do you think is this purely theoretical scenario:

The Earth makes exactly one SLOW (annual) rotation on it's axis after exactly one year, and each day of the year while traveling along the ECCENTRIC orbit around the Sun at different speeds, the Earth aligns with the Sun (Solar day) at the exact same time that takes for a completion of one Astronomical day???

Shall we talk about  the identically miraculous synchronization of the alleged one SLOW (monthly) rotation of the Moon while traveling along the eccentric orbit at different speeds around the Earth?

In addition:

1. What is the miraculous cause for the tilt of the Earth?
2: What maintains it?
3: What is the miraculous cause for the fixed spatial orientation of the Earth's axis?
4. What maintains it?
5. The Earth's axis is spatially fixed with respect to miraculous "what"?
6. Why there are no alternations regarding the relative positions of the Stars (Tycho Brahe stated that had the Earth traveled through space in it's orbit around the Sun he should have been able to observe such alternations.)?
7.Why are the parallaxes perfectly circular? (It shouldn't be so due to combination of the insanely fast journey of our "Solar" System around the "galactic" centre and the alleged Earth's orbit around the Sun, as W.V.D.Kampe had nicely explained)
8. What about the negative parallaxes?
9. The entire universe is centered to the Earth (not to the Sun)! What was your answer on this? Was it something like this: "Because we live on the Earth!" - Hm, the answer is correct, but it belongs to the realm of theology. So, how come that you are bringing in theological arguments into an astronomical discussion?

P.S. I know the meaning of the word "bingo", you missed the point of my question...it was about the irony that i wrongly noticed in that word...but it seems that there was no irony in your usage of that "bingo" either...so the only reasonable solution of this misunderstanding is that i was just paranoid without the reason, my apology!
Title: Re: "Equator" problem
Post by: Rama Set on October 28, 2014, 11:04:45 AM
You know a solar day is an arbitrary number? Just a sidereal day rounded up to the nearest hour?
Title: Re: "Equator" problem
Post by: cikljamas on October 28, 2014, 11:19:28 AM
You know a solar day is an arbitrary number? Just a sidereal day rounded up to the nearest hour?

It seems to me that a solar day in HC has to be much more than just an arbitrary number! If it turns out to be a correct inference then my argument becomes a winner of this game! I am going to refrain myself of yelling "Bingo" at this moment...
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 28, 2014, 01:53:55 PM
This is the problem: Astronomical day = Solar day
What is it that defines an "Astronomical day"? You said earlier that its length is 24 hours, which is the length of a mean solar day. So why is this a problem?

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If we assumed that the Earth rotates but doesn't revolve then we could consider principally almost identical situations when comparing HC & GC. But since the HC supposes Earth's revolution around the Sun also, then you have to expect that you will be faced with unresolvable difficulties.
Nope. No difficulties, unresolvable or otherwise.

Quote
Whenever the Earth allegedly completes one sidereal rotation what follows right after is Our Famous Constant!
Whenever the Earth completes one sidereal rotation it has to rotate an additional amount to bring the Sun back to the same position because it's progressed along its orbit. This additional rotation varies slightly from day to day but averages almost exactly 3m 54s over a year.

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The difference between one Astronomical day and one Sidereal day is equal to Our Constant.

The difference between one Solar day and one Sidereal day is equal to Our Constant AGAIN.

Anything suspicious over here?
Nope. Since they are both defined as the same thing, 24 hours, what else would you expect?  It does make one wonder why you think a new term is necessary.

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Why are these differences equal to the same period (amount) of time which is 3min56sec???
Because the length of the two types of "days" you're subtracting the length of the sidereal day from are the same?

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How do we get so perfect synchronization between one Astronomical and one Solar day, although the Earth's orbit is not a perfect circle, and the Earth's orbital speed is not a constant???
Because you defined "one Astronomical Day" as the same length of time as one mean solar day. Recall that the mean solar day has a constant length, unlike apparent solar days, which do depend on the season and orbital speed.

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To make it even clearer for our audience let's pose this question:

How probable (to be accomplish in reality) do you think is this purely theoretical scenario:

The Earth makes exactly one SLOW (annual) rotation on it's axis after exactly one year, and each day of the year while traveling along the ECCENTRIC orbit around the Sun at different speeds, the Earth aligns with the Sun (Solar day) at the exact same time that takes for a completion of one Astronomical day???
Since it's purely theoretical, it doesn't have to be probable at all. But your scenario doesn't make sense as written; you posit one rotation per year, and then talk about "each day of the year" and aligning with the sun each day. If earth rotates exactly one time per year, there will be zero "days"; the Sun will trace the Analemma in the sky and won't set at all in some locations, and will never rise at all in some others. Can you restate the question?

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Shall we talk about  the identically miraculous synchronization of the alleged one SLOW (monthly) rotation of the Moon while traveling along the eccentric orbit at different speeds around the Earth?
It's neither 'miraculous' nor a coincidence. The moon's period of rotation is equal to its period of orbit because it's tidally locked.   

Because the Moon's orbit is slightly eccentric, it appears to "wobble" a little as its rotation is alternately slightly faster and slower than its orbital speed. See  Libration (http://en.wikipedia.org/wiki/Libration).

Rather than lots of quote blocks, answers below will be in italics.

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In addition:

1. What is the miraculous cause for the tilt of the Earth?
I don't know. There's no reason to expect the Earth's axis of rotation to be exactly perpendicular to the plane of the orbit, though. It is what it is; why does what caused it matter?

2: What maintains it?
The Earth's spin.

3: What is the miraculous cause for the fixed spatial orientation of the Earth's axis?
See the answer to 2: above. Note that is isn't truly fixed, though; it precesses, completing a full circuit in about 26,000 years, and nutates (wobbles) by a few seconds of arc roughly every 19 years.

4. What maintains it?
See the answer to 2: above.

5. The Earth's axis is spatially fixed with respect to miraculous "what"?
The plane of the orbit is our most obvious reference for its orientation. But, actually, it's its orientation is fixed (neglecting precession and nutation) with respect to the inertial reference frame - that is, the universe as a whole

6. Why there are no alternations regarding the relative positions of the Stars (Tycho Brahe stated that had the Earth traveled through space in it's orbit around the Sun he should have been able to observe such alternations.)?
There are. It's called parallax. Tycho was mistaken about how far away the nearest stars were, so parallax was much smaller than he expected.

7.Why are the parallaxes perfectly circular? (It shouldn't be so due to combination of the insanely fast journey of our "Solar" System around the "galactic" centre and the alleged Earth's orbit around the Sun, as W.V.D.Kampe had nicely explained)
They're not. With no proper motion, parallax would be perfectly circular only at the ecliptic poles, a straight line on the ecliptic, and elliptical elsewhere. Most of the stars we can measure parallax for are moving in about the same direction at about the same speed as we are, so, combined with the great distances, their proper motion is low. Those speeds only seem "insanely fast" if you think the universe is small.

8. What about the negative parallaxes?
Binary systems can produce motion unrelated to, but mistaken for, parallax. Stellar parallax can be measured only by comparing positions wrt other stars assumed to be much further away; sometimes that assumption is not correct. Error margins when measuring small values of parallax can exceed the actual value itself and sometimes land on the wrong side of zero. This paper (http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1943AnDea...4....1L&db_key=AST&page_ind=0&data_type=GIF&type=SCREEN_VIEW&classic=YES) is a good analysis.

9. The entire universe is centered to the Earth (not to the Sun)! What was your answer on this? Was it something like this: "Because we live on the Earth!" - Hm, the answer is correct, but it belongs to the realm of theology. So, how come that you are bringing in theological arguments into an astronomical discussion?
If you reread my answer, I describe "references" and "coordinate systems" which are based on what is convenient to us. The universe is inertial (not spinning), and has no center, so any point (or direction) is as good as any other to use as a reference; we choose what makes things easiest for us.

[Edit] Clarification in answer to 5.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 28, 2014, 09:49:02 PM
To be sure I wasn't missing something obvious without waiting for your answer to:
This is the problem: Astronomical day = Solar day
What is it that defines an "Astronomical day"?

I found the post where the term was introduced. Here it is:
...
We can shape our argument in this way:

There are three constants:

1. ASTRONOMICAL DAY = 24 hours = sidereal day + our constant (0,986 degree)
2. SIDEREAL DAY = 23h56min4sec
3. OUR CONSTANT = 0,986 degree = 3min56sec
...

Let's use AD for "ASTRONOMICAL DAY", SD for "SIDEREAL DAY", C for "OUR CONSTANT", and MS for Mean Solar day, which is exactly 24 hours.

Now

MS - SD = C   (I think we can agree on this)

So

MS = SD + C   (simple algebra)

Your statement 1, above, says

AD = SD + C

Now, since

MS = SD + C

and

AD = SD + C

then

MS = AD.

According to your own definition of "ASTRONOMICAL DAY", it's identical to Mean Solar Day.

Further, you use "solar day", "mean solar day", "mean day", and "24 hours" interchangeably later in the same post
A sideral day is less than the solar day, for it is measured by 360 degrees, whereas the mean solar day is measured by 360 degrees 59'8'' nearly. If an astronomical day be = I, then a sidereal day is = 0,997269722; or the difference between the measures of a means solar day, and a sideral day, viz. 59'8'', reduced to time, at the rate of 24 hours to 360 degrees, gives 3'56''; from which we learn that a star which was on the meridian with the sun on one noon, will return to that meridian 3'56'' previous to the next noon (Mean noon i would add): therefore, a clock which measures mean days by 24 hours, will give 23h56m4sec. for the length of a sidereal day.

So, again, why, exactly, is

"Astronomical day = Solar day" a problem? You have defined them to be exactly the same thing: the length of a sidereal day plus the difference between mean solar day and sidereal day.
Title: Re: "Equator" problem
Post by: cikljamas on October 29, 2014, 06:42:17 AM
Astronomical day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Stars only!

Solar day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Sun only!

That is the difference!

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An apparent solar day can be 20 seconds shorter or 30 seconds longer than a mean solar day. Long or short days occur in succession, so the difference builds up until mean time is ahead of apparent time by about 14 minutes near February 6 and behind apparent time by about 16 minutes near November 3. The equation of time is this difference, which is cyclical and does not accumulate from year to year.

If the Earth's orbit were a perfect circle and if the Earth's orbital speed were a constant,  one additional (slow-annual)  rotation of the Earth (assuming that the Earth is placed at the perfect distance from the Sun) would perfectly match the Earth's annual sidereal rotation. That would be some synchronization...

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The problem is that in September the Sun takes less time (as measured by an accurate clock) to make an apparent revolution than it does in December; 24 "hours" of solar time can be 21 seconds less or 29 seconds more than 24 hours of clock time. As explained in the equation of time article, this is due to the ellipticity of the Earth's orbit and the fact that the Earth's axis is not perpendicular to the plane of its orbit.

What is even greater miracle is that although the Earth allegedly travels in the eccentric orbit at a different speeds, one additional annual rotation of the Earth still perfectly matches the Earth's annual sidereal rotation.

The apparent sun is the true sun as seen by an observer on Earth. Apparent solar time or true solar time is based on the apparent motion of the actual Sun. It is based on the apparent solar day, the interval between two successive returns of the Sun to the local meridian.

The question is this:

If the Earth's orbital speed is greater at a Perihelion (Northern Winter) how come that the interval between two successive returns of the Sun to the local meridian becomes shorter and shorter (20 sec per day) instead of being longer and longer when compared with an Aphelion (Northern Summer)???

Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?


In addition:

Quote
Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"

Does this make any difference?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 29, 2014, 11:44:44 AM
Astronomical day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Stars only!
Why do you think "Our Constant" is related to the stars? It's related only to the mean sun. Sidereal days alone are related to the distant stars.

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Solar day = 1 Sidereal day + Our Constant

In this case Our Constant is related to the Sun only!
This is true, but the "in this case" qualifier is unnecessary. This constant only applies to the mean solar day.

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That is the difference!
What difference? When you add the same two constants together you get the same answer, thus "Astronomical day" = Solar day. Didn't we already establish this? Why, again, do we need a new term, and a confusing one since it isn't really "astronomical"?

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An apparent solar day can be 20 seconds shorter or 30 seconds longer than a mean solar day. Long or short days occur in succession, so the difference builds up until mean time is ahead of apparent time by about 14 minutes near February 6 and behind apparent time by about 16 minutes near November 3. The equation of time is this difference, which is cyclical and does not accumulate from year to year.

If the Earth's orbit were a perfect circle and if the Earth's orbital speed were a constant,  one additional (slow-annual)  rotation of the Earth (assuming that the Earth is placed at the perfect distance from the Sun) would perfectly match the Earth's annual sidereal rotation. That would be some synchronization...
If you're contriving your example to be perfectly synchronized, why would you expect it to be other than synchronized?

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The problem is that in September the Sun takes less time (as measured by an accurate clock) to make an apparent revolution than it does in December; 24 "hours" of solar time can be 21 seconds less or 29 seconds more than 24 hours of clock time. As explained in the equation of time article, this is due to the ellipticity of the Earth's orbit and the fact that the Earth's axis is not perpendicular to the plane of its orbit.
Exactly like this:
(http://www.hartrao.ac.za/other/sundial/eqtime.gif)

The displacement of the solid black line above or below zero is the time the apparent sun leads (above zero) or lags (below zero) the mean sun. The slope of the solid black line is the difference between the lengths of the mean solar day and apparent solar day. When upward, apparent days are shorter (the apparent sun catches up with and passes the mean sun); when downward, apparent days are longer. At the four minima and maxima, where the slope momentarily flattens out, apparent and mean days are the same length. These descriptions also apply to the two component sine waves; note that the seasonal component (red dash-dot line) is somewhat larger than the orbital component (green dashed line) and has much higher maximum slopes. The latter means that the changes in the length of apparent solar days are more strongly affected by the seasonal component than the orbital component.

From the quote above, "in September the Sun takes less time (as measured by an accurate clock) to make an apparent revolution than it does in December". Note that the black line is sloping upward in September and even more steeply downward in December.

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What is even greater miracle is that although the Earth allegedly travels in the eccentric orbit at a different speeds, one additional annual rotation of the Earth still perfectly matches the Earth's annual sidereal rotation.
It's neither a miracle nor even mysterious. The Sun makes exactly one complete circuit of the ecliptic in exactly one year (that's what defines the year) even though it appeared to "speed up" and "slow down" a little along the way. That complete circuit of the Sun around the sky "uses up" exactly one complete [sidereal] rotation of the Earth and means that there will be exactly one less solar day in a year than sidereal days - regardless of the number of sidereal days in the year, or whether the motion was completely uniform or not. This is simple geometry and not magical or coincidental in any way. It really is that simple.

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The apparent sun is the true sun as seen by an observer on Earth. Apparent solar time or true solar time is based on the apparent motion of the actual Sun. It is based on the apparent solar day, the interval between two successive returns of the Sun to the local meridian.

The question is this:

If the Earth's orbital speed is greater at a Perihelion (Northern Winter) how come that the interval between two successive returns of the Sun to the local meridian becomes shorter and shorter (20 sec per day) instead of being longer and longer when compared with an Aphelion (Northern Summer)???
Are you sure that's right? I think you have it backward and apparent solar days will be longest near the southern solstice, where the perihelion of the orbit (earth moves faster, which lengthens the apparent solar days) nearly coincides with a solstice (which also lengthen the apparent solar days). 

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Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?

The latter. See the previous answer. Getting the previous assertion wrong has made you think there's a discrepancy where none actually exists.

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In addition:

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Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"
Citation needed.

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Does this make any difference?
Only if true.

[Edit] Typo.
Title: Re: "Equator" problem
Post by: cikljamas on October 29, 2014, 02:09:48 PM

 

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Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?

The latter. See the previous answer. Getting the previous assertion wrong has made you think there's a discrepancy where none actually exists.

Correct!

That is why this curve should look like this:

(http://i.imgur.com/KjI0tED.jpg)

When the Earth allegedly speeds up the above curve should go downwards, and vice versa...

Isn't that obvious to you?
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 29, 2014, 03:15:36 PM
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Imagine that the Earth travels in it's orbit around the Sun at a speed of just 10 km per hour, how long  would be the interval between two successive returns of the Sun to the local meridian in this case?

In this case this interval would depend almost solely on the Earth's rotational period which would completely overpower an effects of the Earth's orbital motion.

Now imagine that the Earth travels in it's orbit around the Sun 100 000 km per hour (alleged Earth's orbital speed is even greater than that). Have you imagined this picture and accompanying geometrical implications?

Now, in which of the two above cases we should have to wait longer for the arrival of the Sun to the local meridian?

The latter. See the previous answer. Getting the previous assertion wrong has made you think there's a discrepancy where none actually exists.

Correct!

That is why this curve should look like this:

(http://i.imgur.com/KjI0tED.jpg)

When the Earth allegedly speeds up the above curve should go downwards, and vice versa...

Isn't that obvious to you?
Um... yes. When is the Earth at perihelion and therefore going fastest? Early January, right? [Hint: the answer is yes.]

Yeah... you said it right here:
If the Earth's orbital speed is greater at a Perihelion (Northern Winter) how come that the interval between two successive returns of the Sun to the local meridian becomes shorter and shorter (20 sec per day) instead of being longer and longer when compared with an Aphelion (Northern Summer)???
You got the length-of-day effect backward in that, but the timing of perihelion and aphelion were correct. You corrected the length-of-day effect of orbital speed later in your post and confirmed it at the beginning of your reply quoted here.

Do you see the green dashed line in the drawing? That one represents the component of the Equation of Time due to the Earth's elliptical orbit. It's going down in early January as you say it should.

The dash-dot red line represents the component of the Equation of Time due to the obliquity of the ecliptic. It is independent of the green curve, but happens to also cross the zero line going downward in late December close to when the green curve crosses going downward. Since the obliquity causes the apparent days to be longest at the solstices and shortest at the equinoxes, we would expect this. The black curve is the sum of the other two, so it also crosses zero going downward in late December - early January.

So why did you think it necessary to butcher the original drawing? It was showing exactly what you said it should.

When the Earth allegedly speeds up the above curve should go downwards, and vice versa...
This is what the original plot showed.
Title: Re: "Equator" problem
Post by: rottingroom on October 29, 2014, 03:18:08 PM
WHY IS THE WHOLE UNIVERSE CENTERED TO THE EARTH'S EQUATOR (and not to the Sun's equator)???

Alpha2Omega is doing such a good job on you that there isn't much to say but I do want to respond to the quote above.

What do you mean when you say the universe is centered to the equator? Are you wondering why it appears as if the stars are all spinning around what geocentrists call the axial tilt instead of what we call the orbital plane / plane of the ecliptic?

(http://faculty.kutztown.edu/courtney/blackboard/Physical/21Seasons/tilt.jpg)

Care to take a guess why it looks like that?
Title: Re: "Equator" problem
Post by: cikljamas on October 30, 2014, 04:05:43 AM
This is perfectly in accordance with FET, and totally discordant with HC:

(http://i.imgur.com/7NrZgDN.jpg)

(http://i.imgur.com/HmWqy9k.jpg)

(http://i.imgur.com/qKrGL0N.jpg)

See also this: http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

Alpha2Omega, you are nice, inteligent and well educated person, and i must admit that i enjoyed to discuss with you, but we came to the point where you have to show that you are not just kind, smart and educated, now you have to show that you are a person with integrity also.

We have to follow the proofs whereever they lead us if we want to make out of this beutiful Earth good place for living for all human beings, for every one of them. We will never succeed to accomplish this task if we continue to hide the truth, and the first truth is that we don't live on a planet, but on the Earth!

Important lesson about a great importance of telling the truth all the time (no matter what):

http://www.energeticforum.com/265263-post549.html (http://www.energeticforum.com/265263-post549.html)
http://www.energeticforum.com/265264-post550.html (http://www.energeticforum.com/265264-post550.html)
http://www.energeticforum.com/265267-post553.html (http://www.energeticforum.com/265267-post553.html)

KNOWLEDGE IS THE BEGINNING: http://rutube.ru/video/9b8eec2d5b68ad6101657add1aef2287/ (http://rutube.ru/video/9b8eec2d5b68ad6101657add1aef2287/)
Title: Re: "Equator" problem
Post by: rottingroom on October 30, 2014, 06:09:41 AM
now you have to show that you are a person with integrity also.

oh the irony. I've never witnessed a beat down quite as severe on these forums as pungent as what Omega has done to you here.

(http://i.imgur.com/7NrZgDN.jpg)

This looks a lot like an analemma. Where did you get this from?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 30, 2014, 06:52:37 AM
oh the irony. I've never witnessed a beat down quite as severe on these forums
It's been like watching Mike Tyson vs Minnie Mouse.
Title: Re: "Equator" problem
Post by: cikljamas on October 30, 2014, 07:39:53 AM
Although he punches hard, he is not strong enough to knock me down, but you little pussies, what's the use of you?

(http://i.imgur.com/XCQXljP.jpg)

Can't believe that your ignorance is of such a legendary proportions so that you are completely unable to distinguish who have just knocked down whom?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 30, 2014, 07:41:31 AM
@cikljamas, if this were a fight the referee would have stepped in long ago for your own protection.
Title: Re: "Equator" problem
Post by: cikljamas on October 30, 2014, 07:46:19 AM
Good fighter needs to know to take the hits that eventually won. I just won, didn't you notice? Blind or stupid?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on October 30, 2014, 07:47:57 AM
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Good fighter needs to know to take the hits that eventually won.
eh, what?
Title: Re: "Equator" problem
Post by: rottingroom on October 30, 2014, 07:51:50 AM
Wow, much delusion. Very fantasy.
Title: Re: "Equator" problem
Post by: Socratic Amusement on October 30, 2014, 08:31:30 AM
cikljamas, the debate between yourself and Alpha2Omega has been, without hyperbole, akin to this:

cikljamas: "Two plus two equals fish!"
Alpha2Omega: "Fish are a biological organism, and the numbers that you are using are part of mathematics. While mathematics can be used to describe animals during an equation, the equation itself must be expressed as a number, not an animal."
cikljamas: "Then how come when four somethings come together, they form a fish?"
Alpha2Omega: "They don't. And I believe the word you were looking for was 'numbers,' but numbers never transform into aquatic life."
cikljamas: "You are so close! Aquatic life come about when...?"
Alpha2Omega: "When reproduction occurs. Which has nothing to do with mathematics, except when used to describe specific units during the mating process. But a nebulous 'two' has no part in that process."
cikljamas: "Exactly! Therefore I win! Two plus two equals fish!!!"

THE REST OF US: (https://c2.staticflickr.com/8/7206/6896176229_9f2ce22d7a_z.jpg)
Title: Re: "Equator" problem
Post by: theearthisrounddealwithit on October 30, 2014, 08:34:47 AM
cikljamas, the debate between yourself and Alpha2Omega has been, without hyperbole, akin to this:

cikljamas: "Two plus two equals fish!"
Alpha2Omega: "Fish are a biological organism, and the numbers that you are using are part of mathematics. While mathematics can be used to describe animals during an equation, the equation itself must be expressed as a number, not an animal."
cikljamas: "Then how come when four somethings come together, they form a fish?"
Alpha2Omega: "They don't. And I believe the word you were looking for was 'numbers,' but numbers never transform into aquatic life."
cikljamas: "You are so close! Aquatic life come about when...?"
Alpha2Omega: "When reproduction occurs. Which has nothing to do with mathematics, except when used to describe specific units during the mating process. But a nebulous 'two' has no part in that process."
cikljamas: "Exactly! therefore I win! Two plus two equals fish!!!"

THE REST OF US: (https://c2.staticflickr.com/8/7206/6896176229_9f2ce22d7a_z.jpg)

I know I'm not supposed to but.. LOL
Title: Re: "Equator" problem
Post by: cikljamas on October 30, 2014, 09:19:08 AM
Dear little children, it seems that you didn't understand what is speeding up, maintaining the speed, slowing down, maintaining the speed etc...No, it's not the Earth, it's the sun, it's really that simple, but knowing that you weren't able to infer this simple fact by yourself, i just let you know...Maybe now your little brains start to operate, at least just for a while, just a little bit, before you return to your playing court.

CAUTION, CHILDREN PLAYING!
Title: Re: "Equator" problem
Post by: rottingroom on October 30, 2014, 09:25:28 AM
Dear little children, it seems that you didn't understand what is speeding up, maintaining the speed, slowing down, maintaining the speed etc...No, it's not the Earth, it's the sun, it's really that simple, but knowing that you weren't able to infer this simple fact by yourself, i just let you know...Maybe now your little brains start to operate, at least just for a while, just a little bit, before you return to your playing court.

CAUTION, CHILDREN PLAYING!

Of course we (or at least I) knew that. Why else would I have mentioned an analemma?
Title: Re: "Equator" problem
Post by: Socratic Amusement on October 30, 2014, 09:39:40 AM
Dear little children, it seems that you didn't understand what is speeding up, maintaining the speed, slowing down, maintaining the speed etc...No, it's not the Earth, it's the sun, it's really that simple, but knowing that you weren't able to infer this simple fact by yourself, i just let you know...Maybe now your little brains start to operate, at least just for a while, just a little bit, before you return to your playing court.

CAUTION, CHILDREN PLAYING!

I love watching Flat Earthers flail in denial.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 30, 2014, 11:04:13 AM
This is perfectly in accordance with FET, and totally discordant with HC:

(http://i.imgur.com/7NrZgDN.jpg)

(http://i.imgur.com/HmWqy9k.jpg)

(http://i.imgur.com/qKrGL0N.jpg)
It's one thing to claim the Analemma is discordant with the Heliocentric model and another one entirely to demonstrate that it actually is. Your drawings here demonstrate nothing of the sort but do show that you really don't understand elliptical orbits at all, even in the most general sense.

In the embellished drawing of the orbit why is the speed "maintaining" at one of the apsides and changing from "maintaining" to "slowing" at the other? Shouldn't each be opposite the other? You really only need two of the colors shown - light blue "speeding up" (3. July to 3. January in the drawing) and green "slowing down" (3. January to 3. July), since "maintaining" only occurs for a moment at each of the apsides.

Also, saying it is in perfect accordance with FET isn't meaningful since there's no consistent FE model - at least none that I've seen - that makes useful predictions of the daily motion of the sun explaining the Analemma, among everything else we know about the apparent motion of the Sun. Most of them can't even explain sunsets and the rest only in the vaguest terms, and useless for making actual predictions. If you have, or know of one, please bring it forward. We've been begging for that.

This popped up after the rest of this post was composed. It answers a question I didn't think necessary to ask, but was wrong. Rather than re-write the above, consider the following as amendment.
<Gratuitous snarky comment> it seems that you didn't understand what is speeding up, maintaining the speed, slowing down, maintaining the speed etc...No, it's not the Earth, it's the sun, it's really that simple, but knowing that you weren't able to infer this simple fact by yourself, i just let you know... <more snark>.
No, I certainly didn't because you never said. Since you showed the entire northern loop of the Analemma as "maintaining speed" even though the apparent position of the Sun moved back and forth by about 4 degrees in azimuth (16 minutes of time) in 4 months, and at the bottom "maintaining speed" as it changed azimuth by about 5 degrees (20 minutes) at a variable rate in half that time, it sure as heck didn't look like you were referring to apparent speed of the Sun being constant! You did refer to varying orbital speed at varying points in the orbit recently in this discussion, and the diagram showed the position of earth at various points in its orbit, and the markings didn't make sense referring to either orbital speed or apparent speed of the sun, so I guessed (wrongly, it turns out), you were being consistent. Ill try to avoid that mistake in the future. You really need to use terms consistently, or at least explain how they're being used, if your aim is to educate rather than to confuse.

Anyway, in light of the clarification...

The apparent sun will be "moving slower than" the mean sun when apparent solar days are longer than mean solar days and "moving faster than" the apparent solar days are shorter than mean solar days. Considering only the effect of the elliptical orbit, apparent solar days are shortest at aphelion (3. July in your diagram) and longest at perihelion (3. January). Since the Sun "slows down", as the Earth "speeds up" in its orbit (July to January), and the Sun "speeds up" as the Earth "slows down" (January to July), then you still need only two colors, as above, but they're swapped. "Maintaining" is still just the moment they switch. Your embellished drawing is still wrong. Add in the effect of obliquity, and your spray-painted addition to the drawing is still wrong.

Everything I stated above your clarification except the dates for "slowing" and "speeding up" being swapped due to your ambiguous terminology still stands.

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See also this: http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)
What is in that link and why should I visit it? Every one of the links at that domain that I've read so far has been so filled with errors, misinformation, and misconceptions that reading them only points more issues to debunk. If this one's different, please describe briefly what it purports to explain and how it adds to the discussion at hand.

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Alpha2Omega, you are nice, inteligent and well educated person, and i must admit that i enjoyed to discuss with you, but we came to the point where you have to show that you are not just kind, smart and educated, now you have to show that you are a person with integrity also.
So now you're calling me a liar in nice terms. On what basis? Because I disagree with you on technical issues?

Since you brought up the integrity issue, can you provide the requested citation for this assertion...
In addition:

Quote
Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"
... or is this something you just made up to create FUD? If this is not the case, please provide a citation for the claim. If you did make it up, or even read it somewhere but just tossed it into the argument without any scrutiny at all or any intention to back it up, how honest is that?

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We have to follow the proofs whereever they lead us if we want to make out of this beutiful Earth good place for living for all human beings, for every one of them. We will never succeed to accomplish this task if we continue to hide the truth, and the first truth is that we don't live on a planet, but on the Earth!

Important lesson about a great importance of telling the truth all the time (no matter what):

<energeticforum links>

KNOWLEDGE IS THE BEGINNING:
<youtube link>
I'm fully aware of the importance of telling the truth. That's why I try to do so at all times. Sometimes I make mistakes - who doesn't - but don't intentionally lie.

You might consider heeding your own advice. The beginning of knowledge is actually looking at evidence. Simply dismissing evidence that doesn't match your preconceived notions is willful ignorance, not a way to gain knowledge.

Calling your opponent a liar and declaring victory without actually successfully defending your argument is not a viable debating technique. What does this say about the strength of your argument?
Title: Re: "Equator" problem
Post by: cikljamas on October 30, 2014, 01:11:34 PM
Shall we observe this illustration once more:

(http://i.imgur.com/qKrGL0N.jpg)

When the Earth allegedly speeds up, in reality the Sun speeds up instead, when the Earth allegedly slows down, it is the Sun which really slows down.

When the Earth allegedly speeds up (September - December) the apparent sun should be behind the mean sun, but it is not (it is ahead), and vice versa, when the Earth allegedly slows down (January - April) the apparent sun should be ahead the mean sun, but it is not (it is behind)!

A green dashed line must be replaced with a blue sprayed line which i subsequently added to show how it would really be if the Earth traveled around the Sun in the same direction in which she allegedly rotates on it's axis!

This is very powerful proof against the trueness of heliocentric theory, which proof strongly support validity of my claim "i won this game"!

Very similar fatal heliocentric error is shown in this link http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html) , don't be afraid to open it, why do you hesitate, you said you are afraid of finding an errors, funny reason for not to open this link, since if there is anything erroneous in it, you can use it against me, am i right?

Concerning your request for a citation the best i can do right now is this: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)

When you open above link just scroll down a little bit, and you will see a screenshot of the page that you are looking for...

P.S. I didn't call you a liar, not even in nice terms, i just appeal to you to find enough courage to admit the obvious truth. However, this is not an easy task, whatever someone could think of it, so there is no irony in my words...
Title: Re: "Equator" problem
Post by: rottingroom on October 30, 2014, 02:22:45 PM
You do realize that the shape of the analemma depends entirely on your location on earth. As a matter of fact it is UPSIDE DOWN when you are in the southern hemisphere. Note that this impossible on a flat earth.
Title: Re: "Equator" problem
Post by: Alpha2Omega on October 31, 2014, 03:44:09 PM
Shall we observe this illustration once more:

(http://i.imgur.com/qKrGL0N.jpg)

When the Earth allegedly speeds up, in reality the Sun speeds up instead, when the Earth allegedly slows down, it is the Sun which really slows down.
No, you still have it backwards. When the Earth speeds up in its orbit, the apparent solar day lengthens, i.e. the Sun appears to slow down relative to the mean sun. This happens because the Earth has moved a greater distance in a similar time (it's moving faster, remember?), so it needs more rotation (which takes longer since the rate of rotation is a constant, remember?) to bring the Sun back to the same meridian a day later. This is simple geometry - sketch it out if you need to.

If you prefer to think of the Earth as fixed while the whole freakin' universe spins once a day with the Sun orbiting inside it, then the effect is exactly the same. I've been saying all along the apparent motion of the Sun against the background stars won't distinguish between the two models. Other things do.

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When the Earth allegedly speeds up (September - December) the apparent sun should be behind the mean sun, but it is not (it is ahead), and vice versa, when the Earth allegedly slows down (January - April) the apparent sun should be ahead the mean sun, but it is not (it is behind)!

A green dashed line must be replaced with a blue sprayed line which i subsequently added to show how it would really be if the Earth traveled around the Sun in the same direction in which she allegedly rotates on it's axis!

This is very powerful proof against the trueness of heliocentric theory, which proof strongly support validity of my claim "i won this game"!

You've got the effect of orbital speed backwards. Since the premise your proof is built on is wrong, the proof is not valid.

See the dashed green line is concave upward January - June? This is because the Earth's orbital speed is slowing. See that it's concave downward July - December? That's because it's speeding up.

"When the Earth allegedly speeds up (September - December) the apparent sun should be behind the mean sun"

The Earth isn't speeding up only September-December. Are you basing that statement on the black line, which also includes obliquity, which is unrelated to the speed in orbit?

Quote
Very similar fatal heliocentric error is shown in this link http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html) , don't be afraid to open it, why do you hesitate, you said you are afraid of finding an errors, funny reason for not to open this link, since if there is anything erroneous in it, you can use it against me, am i right?
Can you describe the specific error here instead of just giving the link? Those links have loads of errors and misconceptions, and discussing each one would totally throw the discussion off track. Which may be what you want. We don't need another basket full of things to discuss; we've got enough already.

Quote
Concerning your request for a citation the best i can do right now is this: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)

When you open above link just scroll down a little bit, and you will see a screenshot of the page that you are looking for...

Since you give this as the citation I requested, I looked at it. Ignoring everything else in that link, here's this claim
Quote
Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"

in context:
(http://www.igreklik.com/slike/images/58895023636888751082.jpg)

The rather breathless (and still unattributed) text says that an early WAG at the length of Astronomical Unit (AU - the Sun's distance from the Earth) by Kepler in the early 17th century was about 12 million miles, and it's been subsequently revised significantly upward (to about 93 million miles). I don't know if the claimed 12 million mile estimate by Kepler is true or not (he had no way to determine this accurately), but, in fact, it is well established that until the mid-18th century the length of the AU was not known. In 1677 Edmond Halley proposed an elegant geometric method for determining the AU, but it was not until the 1761 transit of Venus that this could be accomplished. Here (http://www.exploratorium.edu/venus/question4.html) is a detailed description of the method.

What we had was cikljamas' quote taken out of context that "modern astronomers" have lengthened the AU by almost 100 million miles. What he didn't bother to mention was that the "modern astronomers" referred to did this 250 years ago and revised a much-too-small estimate of the AU (if there even was one) to close to the current value. By leaving this key piece of information out, the clear implication of "modern astronomers" is actually, like, "modern", as in, say, within the last 100 years or less, and this suggests that secret cabal "really" knows the AU is almost 200 million miles, not the 93 million mile "lie" fed to us "unwitting sheep". I doubt that cikljamas even knew any of this, he just accepted the claim, omitted key information (that the original value was 12 million miles for the AU) and dumped it here as FUD. Nice.
Quote
P.S. I didn't call you a liar, not even in nice terms, i just appeal to you to find enough courage to admit the obvious truth. However, this is not an easy task, whatever someone could think of it, so there is no irony in my words...
It sure reads like you did. The first definition of "integrity" is "the quality of being honest and having strong moral principles" and you glibly go on and on about "the Heliocentric lie" here. If you meant "courage" you should have used a term that meant that.

"Find enough courage to admit the obvious truth"? Back atcha. If you're getting tired of being shown wrong, and why, in detail, you can throw in the towel at any time.
Title: Re: "Equator" problem
Post by: cikljamas on November 01, 2014, 02:42:01 AM
What a waste of talent! What a pity!
You gave me a straight answer, so now i can respond to you in a straight manner also:

I really have hoped that you were going to pick up enough courage to admit the obviousness of the self-evident facts, but unfortunately my hopes were in vain.

Your entire last post (every word of it) is a bunch of shameful, deliberate lies. If you can live with them i can live with them too. You disappointed me a lot! Following  vigorous testimony of the great german writer fit so well with your integrity:

"It may be boldly asked where can the man be found,possessing the extraordinary gifts of Newton, who could suffer himself to be deluded by such a hocus-pocus, if he had not in the first instance wilfully deceived himself? Only those who know the strength of self-deception, and the extent to which it sometimes trenches on dishonesty, are in a condition to explain the conduct of Newton and of Newton's school. To support his unnatural theory Newton heaps fiction upon fiction, seeking to dazzle where he cannot convince."

In a Scientific Lecture, delivered in 1878, at Berlin by Dr. Schcepper, proving that the Earth neither rotates nor revolves, he quoted the following still stronger protest of Gothe against the delusions of Modern Astronomy. " In whatever way or manner may have occurred this business, I must still say that I curse this modern theory of Cosmogony, and hope that perchance there may appear, in due time, some young scientist of genius, who will pick up courage enough to upset this universally disseminated delirium of lunatics."


Even the great astronomer Humboldt had a big difficulties with finding enough courage to admit the first truth (HC is a brazen lie) let alone to go the whole hog (and admit that the Earth is flat)!

Modern science texts to this day, dominated by secular humanists, state that Galileo proved the Copernican sun-centered theory. The fact is, he proved nothing. Alexander von Humboldt (1769-1859), who sought to formulate the known facts about the universe into a uniform conception of nature in his Cosmos (5 Vols, 1845-1862), said quite candidly: "I have already known for a long time that we have no proof for the system of Copernicus . . .but I do not dare to be the first one to attack it."

I confess I do not understand how Humboldt could really have believed in the globularity of the world, when he penned the following passage, knowing, as a Cosmogonist, that water occupies, at the very lowest computation, at least three times the extent of the surface of the land "Among the causes which tend to lower the mean annual temperature, I include the following :—Elevation above the level of the sea, when not forming part of an extended plain."

" Cosmos," Vol. I., p. 326, Bohn's Edition.

Anyway, one thing that you should be aware of: You cannot win this fight because you fight against the Word of a living God who created Heaven and Earth! I have chosen to serve Him, you chose to serve one other guy, so you lose, i win. You are free to choose between lie and truth, but don't forget: You are responsible and you will be responsible for all your choices!!!

The Unpardonable Sin
"Therefore I say to you, any sin and blasphemy shall be forgiven people, but blasphemy against the Spirit shall not be forgiven. "Whoever speaks a word against the Son of Man, it shall be forgiven him; but whoever speaks against the Holy Spirit, it shall not be forgiven him, either in this age or in the age to come."
Title: Re: "Equator" problem
Post by: Socratic Amusement on November 01, 2014, 05:52:57 AM
Ahhhh cikljamas, so disappointing.

You bang on about integrity, and then once shown you are incorrect with numbers and charts, you screech about imaginary friends instead of dealing with reality. And then claim victory because you can quote some fairy tales.

Well, I can quote fictional books too, but my quote has more applicability to the real world.

'It is a capital mistake to theorize before one has data. Insensibly one begins to twist facts to suit theories, instead of theories to suit facts.'

Sherlock Holmes

-A Scandal in Bohemia
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 01, 2014, 11:14:00 AM
Your entire last post (every word of it) is a bunch of shameful, deliberate lies. If you can live with them i can live with them too.
Quote
Anyway, one thing that you should be aware of: You cannot win this fight because you fight against the Word of a living God who created Heaven and Earth! I have chosen to serve Him, you chose to serve one other guy, so you lose, i win. You are free to choose between lie and truth, but don't forget: You are responsible and you will be responsible for all your choices!!!
It's going to be a lot harder for you to deny that you're calling me a liar and declaring victory. As before, that's not an effective debating technique.

Since I'm not lying I have no trouble with anything I have said. I may be mistaken about the geometry of the solar system, perhaps, but I don't think so; if I am you have completely failed to show where.

Many people derive great comfort in religious faith and if it helps them get through difficulties and live honorable and productive lives, that's wonderful. I don't begrudge you your faith and hope that faith provides comfort to you. I do believe you are misapplying it here, however.

Since you've taken to philosophical quotations, I'll leave this topic with this, attributed to Caesar Baronius (http://www.oratoriosanfilippo.org/galileo-baronio-english.pdf), a 16th and 17th-Century Italian Cardinal and ecclesiastical historian:

"The Bible teaches us how to go to heaven, not how the heavens go."

We still have the discussion of the midnight Sun and a picture you provided of a particular Scandinavian town at twilight (but no other details) queued up. After your last few posts, I can presume that even if I provide patiently explained, detailed geometric reasons why your ideas are wrong, you will deny they're correct but fail to refute them, then just ignore them or call them lies, and insist that you know "the real truth" while I serve that "other guy". If that's the case, there may be little point in continuing that topic and we'll save both of us some time. I must say I have learned from this discussion (in particular, a better understanding of details about the Equation of Time ), so it wasn't a complete waste for me, and I can hope that explaining them here, at least someone else may benefit. For this reason, if you want to move on to your notion about the impossibility of the Midnight Sun if earth were spherical, just say so.

[Edit] Grammatical correction.
[Edit] Corrected correction.
[Edit] Moved last sentence in original second paragraph ahead, split into two sentences, and modified slightly to be the new second paragraph.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 01, 2014, 11:17:42 AM

You gave me a straight answer

....

Your entire last post (every word of it) is a bunch of shameful, deliberate lies.

Well, that's consistent....
Title: Re: "Equator" problem
Post by: ausGeoff on November 01, 2014, 07:00:30 PM
Anyway, one thing that you should be aware of: You cannot win this fight because you fight against the Word of a living God who created Heaven and Earth! I have chosen to serve Him, you chose to serve one other guy, so you lose, I win. You are free to choose between lie and truth, but don't forget: You are responsible and you will be responsible for all your choices!

You don't seem to be cognisant of what actually defines a debate cikljamas.  A debate is defined as "a formal contest in which the affirmative and negative sides of a proposition are advocated by opposing speakers".   It is NOT—as you seem to continually believe—a "fight".

There is no "winner" and no "loser" as you seem to think.

Incidentally, to bring your belief for the existence of a supernatural entity (your "god") to the table is one of the weakest and most self-damaging arguments you can offer.  True science is no place for fairy tales and ancient myths—they're in the purview of priests and medicine men, and churches and temples.

Personally, I for one am personally responsible for the consequences of my own choices, whereas you lay the ultimate responsibility of your choices at the feet of some imaginary friend of yours up in the sky.  If something does goes wrong in your little world, then you can always claim it was "God's will" and absolve yourself of any responsibility and assuage your conscience.  I'm sure you justify the horrendous Levite Massacre by claiming it was God's will that caused the deaths of thousands of innocent men, women and children?  Or it's your God's will that damns homosexuals to be put to death?

Then again, nobody relishes an unjust fight to the death more than a group of religious zealots.    >:(

Title: Re: "Equator" problem
Post by: cikljamas on November 02, 2014, 03:44:12 AM
@Alpha2Omega, don't be ridiculous!

Where do you think you can hide from this truth:

Quote
Shall we observe this illustration once more:

(http://i.imgur.com/qKrGL0N.jpg)

When the Earth allegedly speeds up, in reality the Sun speeds up instead, when the Earth allegedly slows down, it is the Sun which really slows down.

When the Earth allegedly speeds up (September - December) the apparent sun should be behind the mean sun, but it is not (it is ahead), and vice versa, when the Earth allegedly slows down (January - April) the apparent sun should be ahead the mean sun, but it is not (it is behind)!

A green dashed line must be replaced with a blue sprayed line which i subsequently added to show how it would really be if the Earth traveled around the Sun in the same direction in which she allegedly rotates on it's axis!

This is very powerful proof against the trueness of heliocentric theory, which proof strongly support validity of my claim "i won this game"!

Very similar fatal heliocentric error is shown in this link http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html) , don't be afraid to open it, why do you hesitate, you said you are afraid of finding an errors, funny reason for not to open this link, since if there is anything erroneous in it, you can use it against me, am i right?

Concerning your request for a citation the best i can do right now is this: http://www.energeticforum.com/263172-post251.html (http://www.energeticforum.com/263172-post251.html)

When you open above link just scroll down a little bit, and you will see a screenshot of the page that you are looking for...

P.S. I didn't call you a liar, not even in nice terms, i just appeal to you to find enough courage to admit the obvious truth. However, this is not an easy task, whatever someone could think of it, so there is no irony in my words...

Where is that place where you could hide from above truth? Tell me, where is that place? No, you don't have to tell me, I will tell you: the only place where you can hide from above truth is your deliberately self-deluded mind.

Once more:

What really represents a green dashed line is the changing speed of the Sun which is directly proportional to the apparent sun (or to be more precise: to the apparent sun as how it is related to the mean sun).

That is why the speed of the sun and the apparent sun can be described with one-same amplitude (a green dashed line).

On the other hand, if the Earth were in motion (as HC fraudulent theory claims), then an amplitude of a green dashed line wouldn't by any means be in accordance  with the amplitude of the apparent sun, because in that hypothetical case those two amplitudes would be in inversely proportional relation.

Hence, a blue sprayed line represents the hypothetical apparent sun as how it would look like if the HC fraudulent theory wouldn't be fraudulent but true!

How great idiot someone has to be to be unable to understand such a simple and such obvious argument against fraudulent HC theory?

How great liar someone has to be to be able to pretend that this simple, obvious argument doesn't show what it really shows?

You still haven't opened this link: http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

Well, i will open it for you:

Quote
Observational fact
The Sun in the sky during the summer in the Northern hemisphere (above the Tropic of Cancer) travels in a southern arc across the sky which is a West-West-South direction until noon and then a West-West-North direction until midnight as this illustration below shows:

http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

Heliocentric theory:
The Earth spins in an anti-clockwise direction (if viewed looking down from the North Pole). It spins on its axis just over 360° in 24 hours and travels around the sun in one year. It tilts 23.44° on its axis so that at the height of the summer (solstice), one hemisphere will be nearer to the sun than the other, and in 6 months on the other side of the sun, this same hemisphere will be further away (winter solstice). So, the heliocentric theory states that the Northern hemisphere (above the Tropic of Cancer) in the summer solstice tilts towards the sun at 23.44°.

So far so good. The sun is seen to travel in the sky East to West because the Earth is rotating in the opposite direction West to East. Now imagine any location in the Northern hemisphere (NH) above the Tropic of Cancer as it rotates anti-clockwise. At daybreak the NH is rotating in a downwards direction East-East-South until noon where it reverses and travels upwards East-East-North until midnight. The Sun is seen to travel in the sky in the opposite direction which is West-West-North until noon and then West-West-South until midnight. This is a northern arc, as the flipped illustration below demonstrates:

http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

As we can see, this is EXACTLY opposite to how the Sun is seen to traverse the sky. No matter what the season, the Sun in the Northern hemisphere above the Tropic of Cancer NEVER travels in a northern arc… EVER… not in winter, not in fall/spring, not in summer!

This is another valid, strong argument against the fraudulent HC lie, and i firmly stand behind it, because i checked the validity of this argument by doing personal observations of the motion of the Sun in the sky during different seasons!

After you admit the trueness of above two irrefutable arguments, there will be no need for answering to the next even more compelling argument against fraudulent HC theory:

Quote
According to RET when the Earth is closest to the Sun (in January (closer for 5 000 000 km than it is in July)) due to the alleged Earth's tilt, the Southern hemisphere is more exposed (than Northern hemisphere) to the Sun's sharp ("more vertical") rays, so we enjoy Summer in the South and Winter in the North and vice versa.

But, what scatters wet RET dreams is the fact that in January we have deadly synergy of the two important factors: the first factor: significant decrease of the distance between the Earth (which is closer for 5 000 000 km than it is in July) and the Sun; and the second factor: Sun's ("more vertical") rays hit the Southern Hemisphere under sharper angles comparing it with Northern hemisphere. But these sharper angles are the very same angles under which Sun's rays hit the Northern Hemisphere in July. So, why then in January in Southern hemisphere isn't 3 % (150 000 000 / 5 000 000) hotter than it is in July in Northern hemisphere? .........Don't forget: The angles are the same!!!

If the Earth were round and so far away from the Sun we would have to deal with the same problem (significant temperature difference between North and South) in Winter time also, that is to say, in July when the Earth is farthest from the Sun, Southern hemisphere this time should be tilted away from the Sun which would again have deadly impact for Southerners who would instantly freeze to death if southern-winter temperatures were this time 3 % lower comparing them with the northern-winter temperatures. Don't forget: The angles are still the same!!!

Just in case that you are not aware of the significance of that percentage (3%):

"If the Sun were 5% closer, then the water would boil up from the oceans and if the Sun were just 1% farther away, then the oceans would freeze, and that gives you just some idea of the knife edge we are on."

According to RET Southern Hemisphere should be completely uninhabitable!!!

Title: Re: "Equator" problem
Post by: Alpha2Omega on November 02, 2014, 10:32:47 AM
@Alpha2Omega, don't be ridiculous!

Where do you think you can hide from this truth:

<Quoted post replaced with link (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637115#msg1637115)>
I already answered that post here (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637299#msg1637299).

Your response was to call me a liar. Remember?

Your entire last post (every word of it) is a bunch of shameful, deliberate lies.
If you don't remember this, you really should get what's causing it checked. I'm not being snarky, I really mean it.

If you do remember this, then screw you! Go back and read the answers I gave the first time. They're the same now. And they're not lies, whether you want to believe they are or not.

Quote
Where is that place where you could hide from above truth? Tell me, where is that place? No, you don't have to tell me, I will tell you: the only place where you can hide from above truth is your deliberately self-deluded mind.

Once more:

What really represents a green dashed line is the changing speed of the Sun which is directly proportional to the apparent sun (or to be more precise: to the apparent sun as how it is related to the mean sun).

That is why the speed of the sun and the apparent sun can be described with one-same amplitude (a green dashed line).

On the other hand, if the Earth were in motion (as HC fraudulent theory claims), then an amplitude of a green dashed line wouldn't by any means be in accordance  with the amplitude of the apparent sun, because in that hypothetical case those two amplitudes would be in inversely proportional relation.

Hence, a blue sprayed line represents the hypothetical apparent sun as how it would look like if the HC fraudulent theory wouldn't be fraudulent but true!

How great idiot someone has to be to be unable to understand such a simple and such obvious argument against fraudulent HC theory?

How great liar someone has to be to be able to pretend that this simple, obvious argument doesn't show what it really shows?
Still going on with the "self-deluded", "fraudulent", "idiot", "liar" schtick because someone disagrees with you on a technical issue and backs up the reasoning? There is a lot of anger here. Chill, man.

Quote
You still haven't opened this link: http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

Well, i will open it for you:

Quote
Observational fact
The Sun in the sky during the summer in the Northern hemisphere (above the Tropic of Cancer) travels in a southern arc across the sky which is a West-West-South direction until noon and then a West-West-North direction until midnight as this illustration below shows:

http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

Heliocentric theory:
The Earth spins in an anti-clockwise direction (if viewed looking down from the North Pole). It spins on its axis just over 360° in 24 hours and travels around the sun in one year. It tilts 23.44° on its axis so that at the height of the summer (solstice), one hemisphere will be nearer to the sun than the other, and in 6 months on the other side of the sun, this same hemisphere will be further away (winter solstice). So, the heliocentric theory states that the Northern hemisphere (above the Tropic of Cancer) in the summer solstice tilts towards the sun at 23.44°.

So far so good. The sun is seen to travel in the sky East to West because the Earth is rotating in the opposite direction West to East. Now imagine any location in the Northern hemisphere (NH) above the Tropic of Cancer as it rotates anti-clockwise. At daybreak the NH is rotating in a downwards direction East-East-South until noon where it reverses and travels upwards East-East-North until midnight. The Sun is seen to travel in the sky in the opposite direction which is West-West-North until noon and then West-West-South until midnight. This is a northern arc, as the flipped illustration below demonstrates:

http://www.energeticforum.com/256670-post75.html (http://www.energeticforum.com/256670-post75.html)

As we can see, this is EXACTLY opposite to how the Sun is seen to traverse the sky. No matter what the season, the Sun in the Northern hemisphere above the Tropic of Cancer NEVER travels in a northern arc… EVER… not in winter, not in fall/spring, not in summer!

This is another valid, strong argument against the fraudulent HC lie, and i firmly stand behind it, because i checked the validity of this argument by doing personal observations of the motion of the Sun in the sky during different seasons!

After you admit the trueness of above two irrefutable arguments, there will be no need for answering to the next even more compelling argument against fraudulent HC theory:

Quote
According to RET when the Earth is closest to the Sun (in January (closer for 5 000 000 km than it is in July)) due to the alleged Earth's tilt, the Southern hemisphere is more exposed (than Northern hemisphere) to the Sun's sharp ("more vertical") rays, so we enjoy Summer in the South and Winter in the North and vice versa.

But, what scatters wet RET dreams is the fact that in January we have deadly synergy of the two important factors: the first factor: significant decrease of the distance between the Earth (which is closer for 5 000 000 km than it is in July) and the Sun; and the second factor: Sun's ("more vertical") rays hit the Southern Hemisphere under sharper angles comparing it with Northern hemisphere. But these sharper angles are the very same angles under which Sun's rays hit the Northern Hemisphere in July. So, why then in January in Southern hemisphere isn't 3 % (150 000 000 / 5 000 000) hotter than it is in July in Northern hemisphere? .........Don't forget: The angles are the same!!!

If the Earth were round and so far away from the Sun we would have to deal with the same problem (significant temperature difference between North and South) in Winter time also, that is to say, in July when the Earth is farthest from the Sun, Southern hemisphere this time should be tilted away from the Sun which would again have deadly impact for Southerners who would instantly freeze to death if southern-winter temperatures were this time 3 % lower comparing them with the northern-winter temperatures. Don't forget: The angles are still the same!!!

Just in case that you are not aware of the significance of that percentage (3%):

"If the Sun were 5% closer, then the water would boil up from the oceans and if the Sun were just 1% farther away, then the oceans would freeze, and that gives you just some idea of the knife edge we are on."

According to RET Southern Hemisphere should be completely uninhabitable!!!
And, as expected, that quoted link is full of errors and misconceptions. The first part is easily disproved, not irrefutable. All you have to remember is that, if you're north of the Tropic of Cancer, according to conventional HC thinking the noonday sun has to be south of you at any time of year. And it is. QED.

At the beginning of the second part:
"According to RET when the Earth is closest to the Sun (in January (closer for 5 000 000 km than it is in July)) due to the alleged Earth's tilt,"

Stop. Just, please stop.

The rest is no better.
Title: Re: "Equator" problem
Post by: cikljamas on November 02, 2014, 11:43:48 AM

I already answered that post here (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637299#msg1637299).

(http://www.zaslike.com/files/mj10yx37u9xjmkab269h.jpg)

And, as expected, that quoted link is full of errors and misconceptions. The first part is easily disproved, not irrefutable. All you have to remember is that, if you're north of the Tropic of Cancer, according to conventional HC thinking the noonday sun has to be south of you at any time of year. And it is. QED.

No, it is not!

At the beginning of the second part:
"According to RET when the Earth is closest to the Sun (in January (closer for 5 000 000 km than it is in July)) due to the alleged Earth's tilt,"

"due to the alleged Earth's tilt" is related to what comes after these words, not before, are you stupid or what?

Stop. Just, please stop.

The truth can be really painful, so i won't torture you any more, there is no need for that since every sane person can realize very easily how great and blatant lie HC really is, on the basis of already here presented arguments, but if you were a masochist we could provide even many more arguments against this universally disseminated delirium of lunatics...


You should know better tolerate your own defeats....
Title: Re: "Equator" problem
Post by: markjo on November 02, 2014, 12:09:35 PM
And, as expected, that quoted link is full of errors and misconceptions. The first part is easily disproved, not irrefutable. All you have to remember is that, if you're north of the Tropic of Cancer, according to conventional HC thinking the noonday sun has to be south of you at any time of year. And it is. QED.

No, it is not!
When is the noonday sun ever north of the Tropic of Cancer?  ???
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 02, 2014, 01:52:03 PM
Did you remember calling me a liar or not? If not, maybe someone hacked your account and is posting as you; if that's the case you should change your password at the very least. Given your preachy attitude, I presume you don't use drugs or alcohol (although you never know), so severe memory problems should be checked medically.

Is the cute "Facepalms" poster an effort to simply sidestep the answers to your earlier post without addressing them? Or, does the countdown mean that you're scoffing at the next point? If the latter, you never responded to the answers at all. Did you even read them? What, if any, do you have a problem with?

And, as expected, that quoted link is full of errors and misconceptions. The first part is easily disproved, not irrefutable. All you have to remember is that, if you're north of the Tropic of Cancer, according to conventional HC thinking the noonday sun has to be south of you at any time of year. And it is. QED.

No, it is not!
The Sun is not always south at noontime from north of the Tropic of Cancer? (I presume this is what you're disputing since you bolded that part of the quoted text.)

Didn't you say you observed this yourself, and you (again) think this somehow "disproved" the HC model?

Quote
Observational fact
...No matter what the season, the Sun in the Northern hemisphere above the Tropic of Cancer NEVER travels in a northern arc… EVER… not in winter, not in fall/spring, not in summer!

... and i firmly stand behind it, because i checked the validity of this argument by doing personal observations of the motion of the Sun in the sky during different seasons!

Since your descriptions are pretty hard to follow (see below), can you provide a diagram showing why you think the HC model demands a northerly sun at noontime from north of the Tropic of Cancer? If we can understand where you're confused, it might be possible to explain where your error is. Most likely, you'll just have another tantrum, stamp your feet and call me a liar again.

Quote
At the beginning of the second part:
"According to RET when the Earth is closest to the Sun (in January (closer for 5 000 000 km than it is in July)) due to the alleged Earth's tilt,"

"due to the alleged Earth's tilt" is related to what comes after these words, not before, are you stupid or what?
Then why is that comma where it is? You're saying we need to add "poorly written" to "full of errors and misconceptions".

Your statement says exactly what I am complaining about. If that's not what you meant, you should learn to punctuate.

"Let's eat grandma!"

"Let's eat, grandma!"

Do these mean the same thing? Punctuation: it matters.

Quote
Stop. Just, please stop.

The truth can be really painful, so i won't torture you any more, there is no need for that since every sane person can realize very easily how great and blatant lie HC really is, on the basis of already here presented arguments, but if you were a masochist we could provide even many more arguments against this universally disseminated delirium of lunatics...

What's really painful is trying to follow your tortured "facts" and decipher exactly what it is you're really trying to say.

Quote
You should know better tolerate your own defeats....

As predicted, here we go down another rabbit hole. Since you keep calling me a liar, I don't know why I continue with this. It's like picking at a scab, I guess, and it's become satisfying to point out where you're wrong (although daunting in quantity).


Title: Re: "Equator" problem
Post by: cikljamas on November 02, 2014, 02:09:06 PM
When is the noonday sun ever north of the Tropic of Cancer?  ???
Quote

Pardon me, i have overlooked that word (noonday), i thought he wants to say that the sun is always (at any time of the day) in the south, and the truth is that it is not. But if the HC theory were true, the sun should be generally always south for the observer at latitude 45 degree N (where i live). However, in the summer the sun rises NORTH-EAST, traverses the sky in southern arc, and at the end of the day the sun sets NORTH-WEST (although significantly less north in comparision with a sunrise)...The point of this argument is that the arc of the Sun (in the summer) should go in the direction SOUTH-NORTH-SOUTH, and from my own experience i can tell you with certainty that the Sun goes in a direction NORTH-SOUTH-NORTH... Totally opposite from what it should be if in the HC theory we could find a shred of truth !!!

@ Alpha2Omega, now you can respond to these words if you want... (and don't forget: you lose, i win)... No place to hide from the obvious truth, not even a rabbit hole, only deliberately self deluded HC mind...
Title: Re: "Equator" problem
Post by: sokarul on November 02, 2014, 02:17:14 PM
I had plants in my north facing window last summer. It's not a problem for RET, sorry.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 02, 2014, 04:49:44 PM
When is the noonday sun ever north of the Tropic of Cancer?  ???

Pardon me, i have overlooked that word (noonday), i thought he wants to say that the sun is always (at any time of the day) in the south, and the truth is that it is not. But if the HC theory were true, the sun should be generally always south for the observer at latitude 45 degree N (where i live). However, in the summer the sun rises NORTH-EAST, traverses the sky in southern arc, and at the end of the day the sun sets NORTH-WEST (although significantly less north in comparision with a sunrise)...The point of this argument is that the arc of the Sun (in the summer) should go in the direction SOUTH-NORTH-SOUTH, and from my own experience i can tell you with certainty that the Sun goes in a direction NORTH-SOUTH-NORTH... Totally opposite from what it should be if in the HC theory we could find a shred of truth !!!

@ Alpha2Omega, now you can respond to these words if you want... (and don't forget: you lose, i win)... No place to hide from the obvious truth, not even a rabbit hole, only deliberately self deluded HC mind...
I fixed the nested quote for you.

Answering markjo and "oh by the way" is a slick way to try to sidestep the rest of the points in my previous post. I know you read it because you're responding to part of it here. I presume you must be satisfied with the answers in the post about the EoT since you bring nothing up, and also no response to the explanation why your "closest to the Sun due to the tilt" statement doesn't actually say what you thought it did. Good!

Would you please lose the "deluded" crap and stick to your point? Ad-hominems make you look defensive and unsure of yourself, and your arguments are already weak enough. Also, you're better off holding the "I win" claim until you've actually at least scored some points somewhere other than your own mind. You are, of course, welcome to think that if it makes you feel better, but repeating it over and over here, without basis, makes you look desperate.

Your mistake is clear enough here. Where do you think the HC model says the Sun cannot rise north of due east or set north of due west if you're north of the Tropic of Cancer? It doesn't. The celestial equator intersects the ideal horizon due east and west of you; if the Sun is north of the Equator, it must rise north of due east and set north of due west, no matter where you are. In fact, near the Arctic Circle around the time of the June solstice, it will rise and set almost due north. This isn't a problem and will segue nicely into your "Midnight Sun" mistaken notion if you want to go there.

"The point of this argument is that the arc of the Sun (in the summer) should go in the direction SOUTH-NORTH-SOUTH [according to HC theory]". Why do you think HC says this? Your basis for this idea is simply mistaken due to the confused way you describe the motion of the Earth.

 Here's how you describe it:

Quote
Observational fact
...
At daybreak the NH is rotating in a downwards direction East-East-South until noon where it reverses and travels upwards East-East-North until midnight. The Sun is seen to travel in the sky in the opposite direction which is West-West-North until noon and then West-West-South until midnight. This is a northern arc...

As we can see, this is EXACTLY opposite to how the Sun is seen to traverse the sky. No matter what the season, the Sun in the Northern hemisphere above the Tropic of Cancer NEVER travels in a northern arc… EVER… not in winter, not in fall/spring, not in summer!

After you admit the trueness of above two irrefutable arguments, there will be no need for answering to the next even more compelling argument against fraudulent HC theory:

First of all, the northern hemisphere (NH) isn't rotating south at all. It only rotates eastward (so does the SH, but that doesn't matter here).

Let's presume we're talking about some point north of the Tropic of Cancer (call it "Point A") at around the northern solstice.

A point in the NH will be moving toward the equatorial plane at sunrise, but not southward. At sunrise in the NH, the parallel of latitude Point A is on is angled downward toward the plane of the orbit in the direction the earth is turning. Since the Sun's rays are arriving parallel to the plane of the orbit, they are arriving at Point A from "above" (north of) its latitude. At local solar noon, Point A is moving parallel to the plane of orbit for a moment, but is still north of it (remember, we're north of the Tropic of Cancer, and the plane of the orbit never intersects the surface of the Earth north of there). "Straight up" from Point A is a line from the center of the Earth through Point A, and since the center of the Earth is in the plane of earth's orbit and A is north of the plane, this line will pass north of the Sun; therefore, the Sun appears south of vertical, or "to the south" at noon. At sunset, the parallel is tilted upward in the direction of earth's rotation, away from the Sun, but the Sun is in the opposite direction, so, again, the rays are arriving from the north. So at Point A the Sun appears north of due east at sunrise, due south at noontime, and north of west at sunset. Exactly as we see.

So much for "irrefutable". You were just confused and disproved an incorrect model. Those later arguments no more compelling.

This is why you may want to avoid words like "irrefutable", "proof", etc. Show some humility. You may not be stupid, but you certainly don't seem to be as smart as you think you are, and isn't hubris a sin?
Title: Re: "Equator" problem
Post by: cikljamas on November 03, 2014, 05:32:59 AM
Alpha2Omega, you are such an awesome con artist, congratulations!
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 03, 2014, 07:30:50 AM
Alpha2Omega, you are such an awesome con artist, congratulations!

No answers to specific questions about your understanding of the Heliocentric model, just "you're lying".

That says it all.
Title: Re: "Equator" problem
Post by: ausGeoff on November 03, 2014, 08:56:35 AM
Alpha2Omega, you are such an awesome con artist, congratulations!

We all thank you for taking the time and effort to construct this astute, meaningful, insightful rebuttal of Alpha2Omega's comments.

You've well and truly shot him down haven't you?  And here's your gold star...

(http://i.kinja-img.com/gawker-media/image/upload/s--yT2pqFBF--/655227260230759600.jpg)

—See what you can do when you put your mind to it?  Good boy!

Title: Re: "Equator" problem
Post by: cikljamas on November 03, 2014, 09:05:28 AM
If english were my native tongue i wouldn't need to sacrifice too much time to answer to all your lies, but since it is not, i don't intend (whatsoever) to waste my time by answering to obvious, deliberate lies and misconceptions of an con artist such as you. If you want to hear my opinion on your lies just read again what i already wrote on this subject...

Here: some more stuff for a con artist's distortions:

We have seen that wherever the motions of the stars are carefully examined, it is found that all are connected, and move in relation to the northern centre of the earth. There is nowhere to be found a "break" in the general connection. Except, indeed, what is called the "proper motion" of certain stars and groups of stars all move in the same general direction, concentric with the north pole, and with velocities increasing with radial distance from it. To remove every possible doubt respecting the motions of the stars from the central north to the most extreme south, a number of special observers, each completely free from the bias of education respecting the supposed rotundity of the earth, might be placed in various southern localities, to observe and record the motions of the well known southern constellation, not in relation to a supposed south pole star, but to the meridian and latitude of each position. This would satisfy a certain number of those who cannot divest themselves of the idea of rotundity, but is not at all necessary for the satisfaction of those who are convinced that the earth is a plane, and that the extreme south is a vast circumference instead of a polar centre. To these the evidence already adduced will be sufficiently demonstrative.

The points of certainty are the following:--

1st.--Wherever the experiment is made the stars in the zenith do not rise, culminate, and set in the same straight line, or plane of latitude, as they would if the earth is a globe.

2nd.--The Southern Cross is not at all times visible from every point of the southern hemisphere, as the "Great Bear" is from every point in the northern, and as both must necessarily and equally be visible if the earth is globular. In reference to the several cases adduced of the Southern Cross not being visible until the observers had arrived in latitudes 8°, 14°, and 16° south, it cannot be said that they might not have cared to look for it, because we are assured that they "had long wished for it," and therefore must have been strictly on the look out as they advanced southwards. And when the traveller Humboldt saw it "the first time" it was "strongly inclined," and therefore low down on the eastern horizon, and therefore previously invisible, simply because it had not yet risen.

3rd.--The earth is a plane, with a northern centre, over which the stars (whether fixed in some peculiar substance or floating in some subtle medium is not yet known) move in concentric courses at different radial distances from the northern centre as far south as and wherever observations have been made. The evidence is the author's own experiments in Great Britain, Ireland, Isle of Man, Isle of Wight, and many other places; the statements of several unbiassed and truthful friends, who have resided in New Zealand, Australia, South Africa, Rio Janeiro, Valparaiso, and other southern localities, and the several incidental statements already quoted.

4th--The southern region of the earth is not central, but circumferential; and therefore there is no southern pole, no south pole star, and no southern circumpolar constellations; all statements to the contrary are doubtful, inconsistent with known facts, and therefore not admissible as evidence.
Title: Re: "Equator" problem
Post by: rottingroom on November 03, 2014, 10:19:13 AM
2nd.--The Southern Cross is not at all times visible from every point of the southern hemisphere, as the "Great Bear" is from every point in the northern, and as both must necessarily and equally be visible if the earth is globular. In reference to the several cases adduced of the Southern Cross not being visible until the observers had arrived in latitudes 8°, 14°, and 16° south, it cannot be said that they might not have cared to look for it, because we are assured that they "had long wished for it," and therefore must have been strictly on the look out as they advanced southwards. And when the traveller Humboldt saw it "the first time" it was "strongly inclined," and therefore low down on the eastern horizon, and therefore previously invisible, simply because it had not yet risen.


Well this is easy. The southern cross is not directly above(or below to northerners) the south pole.

3rd.--The earth is a plane, with a northern centre, over which the stars (whether fixed in some peculiar substance or floating in some subtle medium is not yet known) move in concentric courses at different radial distances from the northern centre as far south as and wherever observations have been made. The evidence is the author's own experiments in Great Britain, Ireland, Isle of Man, Isle of Wight, and many other places; the statements of several unbiassed and truthful friends, who have resided in New Zealand, Australia, South Africa, Rio Janeiro, Valparaiso, and other southern localities, and the several incidental statements already quoted.

Even easier as the stars in the southern hemisphere move in exactly the opposite direction.


4th--The southern region of the earth is not central, but circumferential; and therefore there is no southern pole, no south pole star, and no southern circumpolar constellations; all statements to the contrary are doubtful, inconsistent with known facts, and therefore not admissible as evidence.

Yet in the southern hemisphere, stars seems to "circle" around that southern cross just as they do around polaris in the north. There can't be 2 distinct locations in the sky that stars circle on a flat earth unless those stars are what is moving. The behavior they do exhibit are perfectly explained by HC though.
Title: Re: "Equator" problem
Post by: 29silhouette on November 03, 2014, 10:44:46 AM
We have seen that wherever the motions of the stars are carefully examined, it is found that all are connected, and move in relation to the northern centre of the earth. There is nowhere to be found a "break" in the general connection. Except, indeed, what is called the "proper motion" of certain stars and groups of stars all move in the same general direction, concentric with the north pole, and with velocities increasing with radial distance from it. To remove every possible doubt respecting the motions of the stars from the central north to the most extreme south, a number of special observers, each completely free from the bias of education respecting the supposed rotundity of the earth, might be placed in various southern localities, to observe and record the motions of the well known southern constellation, not in relation to a supposed south pole star, but to the meridian and latitude of each position. This would satisfy a certain number of those who cannot divest themselves of the idea of rotundity, but is not at all necessary for the satisfaction of those who are convinced that the earth is a plane, and that the extreme south is a vast circumference instead of a polar centre. To these the evidence already adduced will be sufficiently demonstrative.
Several observation are recorded in both video and photographic mediums.  Last I checked, cameras aren't biased.  Perhaps you could find some proof that the stars move in a pattern indicative of a flat disk instead of a globe.

Quote
The points of certainty are the following:--

1st.--Wherever the experiment is made the stars in the zenith do not rise, culminate, and set in the same straight line, or plane of latitude, as they would if the earth is a globe.
Evidence, either photographic or video of this?

Quote
2nd.--The Southern Cross is not at all times visible from every point of the southern hemisphere, as the "Great Bear" is from every point in the northern, and as both must necessarily and equally be visible if the earth is globular. In reference to the several cases adduced of the Southern Cross not being visible until the observers had arrived in latitudes 8°, 14°, and 16° south, it cannot be said that they might not have cared to look for it, because we are assured that they "had long wished for it," and therefore must have been strictly on the look out as they advanced southwards. And when the traveller Humboldt saw it "the first time" it was "strongly inclined," and therefore low down on the eastern horizon, and therefore previously invisible, simply because it had not yet risen.
So you are saying the Great Bear (Ursa Major) is visible from anywhwere north of the equator at any time of night?  The Southern Cross is at 60 degrees, not 89+ degrees like Polaris.  The great bear is located about 55 degrees and spans 20 degrees of the sky and crux is only about 7. 

Quote
3rd.--The earth is a plane, with a northern centre, over which the stars (whether fixed in some peculiar substance or floating in some subtle medium is not yet known) move in concentric courses at different radial distances from the northern centre as far south as and wherever observations have been made. The evidence is the author's own experiments in Great Britain, Ireland, Isle of Man, Isle of Wight, and many other places; the statements of several unbiassed and truthful friends, who have resided in New Zealand, Australia, South Africa, Rio Janeiro, Valparaiso, and other southern localities, and the several incidental statements already quoted.
Who is the author?  Any photographic or video evidence of these 'experiments'?

Quote
4th--The southern region of the earth is not central, but circumferential; and therefore there is no southern pole, no south pole star, and no southern circumpolar constellations; all statements to the contrary are doubtful, inconsistent with known facts, and therefore not admissible as evidence.
All known facts I find when searching indicate a south pole, south polar star, southern circumpolar constellations, etc.  Do you have any sources for 'known facts' that indicate otherwise?  Any long exposure photography or time-lapse video that shows the stars moving away into the distance, curving slightly, growing smaller, moving closer together, and a decrease in rate they 'set into', or 'rise out of' the horizon?
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 03, 2014, 11:11:54 AM
If english were my native tongue i wouldn't need to sacrifice too much time to answer to all your lies, but since it is not, i don't intend (whatsoever) to waste my time by answering to obvious, deliberate lies and misconceptions of an con artist such as you. If you want to hear my opinion on your lies just read again what i already wrote on this subject...

I've already carefully read it. Most of it is simply wrong. I'm not calling you a liar, but you definitely are mistaken about most of what you've written.

Quote
Here: some more stuff for a con artist's distortions:

We have seen that wherever the motions of the stars are carefully examined, it is found that all are connected, and move in relation to the northern centre of the earth. There is nowhere to be found a "break" in the general connection. Except, indeed, what is called the "proper motion" of certain stars and groups of stars all move in the same general direction, concentric with the north pole, and with velocities increasing with radial distance from it. To remove every possible doubt respecting the motions of the stars from the central north to the most extreme south, a number of special observers, each completely free from the bias of education respecting the supposed rotundity of the earth, might be placed in various southern localities, to observe and record the motions of the well known southern constellation, not in relation to a supposed south pole star, but to the meridian and latitude of each position. This would satisfy a certain number of those who cannot divest themselves of the idea of rotundity, but is not at all necessary for the satisfaction of those who are convinced that the earth is a plane, and that the extreme south is a vast circumference instead of a polar centre. To these the evidence already adduced will be sufficiently demonstrative.

The points of certainty are the following:--

1st.--Wherever the experiment is made the stars in the zenith do not rise, culminate, and set in the same straight line, or plane of latitude, as they would if the earth is a globe.

Any star that crosses the zenith at a fixed point will be at the same declination, equal to the latitude. They all will follow the same path in the sky (an arc, though, not a straight line). This would be the same for both the flat earth you propose and the spinning globe.

[Edit to add] Actually, it wouldn't be the same for your flat earth except fairly high northern latitudes. It wouldn't be at all similar in the southern hemisphere, so I stand corrected on that. What we actually see is consistent with a spherical earth, not the proposed flat one.

Quote
2nd.--The Southern Cross is not at all times visible from every point of the southern hemisphere, as the "Great Bear" is from every point in the northern, and as both must necessarily and equally be visible if the earth is globular. In reference to the several cases adduced of the Southern Cross not being visible until the observers had arrived in latitudes 8°, 14°, and 16° south, it cannot be said that they might not have cared to look for it, because we are assured that they "had long wished for it," and therefore must have been strictly on the look out as they advanced southwards. And when the traveller Humboldt saw it "the first time" it was "strongly inclined," and therefore low down on the eastern horizon, and therefore previously invisible, simply because it had not yet risen.

Gacrux, the northernmost star in the Southern Cross asterism, is at declination -57° 06', therefore it will sometimes drop below the horizon, meaning the whole asterism is not visible at all times, if your latitude is north of about 33° S (neglecting refraction). OK so far. The Big Dipper asterism (brightest and most-recognizable part of the constellation Ursa Major - The Great Bear) is not circumpolar (which means at least part of it sets) south of about 41˚ N latitude because Alkaid, at the end of the "handle of the dipper" (or the bear's tail), is at declination +49° 18′. If you really mean the "Great Bear", it extends even further south and isn't visible at all times south of about 59° N latitude since its southernmost star, Alula Australis is at declination about +31°. So, no, you're mistaken again. The "Great Bear" is not "at all times visible from every point of the northern hemisphere". At this time of year, early in the evening from my home in the northern hemisphere, parts of the "Great Bear" are well below the horizon.

Actually, the Southern Cross is circumpolar ("at all times visible") for more of the southern hemisphere than the Big Dipper is for the northern. The entire "Bear" is circumpolar for even less of the northern hemisphere.

Quote
3rd.--The earth is a plane, with a northern centre, over which the stars (whether fixed in some peculiar substance or floating in some subtle medium is not yet known) move in concentric courses at different radial distances from the northern centre as far south as and wherever observations have been made. The evidence is the author's own experiments in Great Britain, Ireland, Isle of Man, Isle of Wight, and many other places; the statements of several unbiassed and truthful friends, who have resided in New Zealand, Australia, South Africa, Rio Janeiro, Valparaiso, and other southern localities, and the several incidental statements already quoted.

Is "the author" you, or are you quoting someone else? This doesn't read like your writing style.

Quote
4th--The southern region of the earth is not central, but circumferential; and therefore there is no southern pole, no south pole star, and no southern circumpolar constellations; all statements to the contrary are doubtful, inconsistent with known facts, and therefore not admissible as evidence.

"No southern circumpolar constellations"? Really? Whoever reported this to you is, quite simply, wrong. If you really think you observed this yourself, then you either weren't paying attention or really do need a thorough medical checkup.

No one has to take my word for any of this. A lot can be witnessed for yourself (all of it, if you're willing to travel), and all of it can be corroborated from many reliable sources.

[Edit] Reply to 1st. "certainty" amended.
Title: Re: "Equator" problem
Post by: cikljamas on November 03, 2014, 12:02:24 PM
The author of above "successfully" refuted assertions is mr Rowbotham: This is the chapter from which i have extracted above excerpts: http://www.sacred-texts.com/earth/za/za48.htm (http://www.sacred-texts.com/earth/za/za48.htm)

As for my own examinations of "The Southern Cross (case)" you can also try to "successfully" analyze these words of mine:
http://www.energeticforum.com/264125-post357.html (http://www.energeticforum.com/264125-post357.html)

So, how come that the Southern Cross makes half of an alleged circumpolar nightly circle in the sky in just 6 hours or so (which is twice faster than the motion of the Cassiopeia and other northern (truely circumpolar) constellations)???
Title: Re: "Equator" problem
Post by: 29silhouette on November 03, 2014, 12:48:48 PM
The author of above "successfully" refuted assertions is mr Rowbotham: This is the chapter from which i have extracted above excerpts: http://www.sacred-texts.com/earth/za/za48.htm (http://www.sacred-texts.com/earth/za/za48.htm)
So why does all modern video and photography "successfully" refute Mr. Rowbotham's assertions?

Quote
As for my own examinations of "The Southern Cross (case)" you can also try to "successfully" analyze these words of mine:
http://www.energeticforum.com/264125-post357.html (http://www.energeticforum.com/264125-post357.html)

So, how come that the Southern Cross makes half of an alleged circumpolar nightly circle in the sky in just 6 hours or so (which is twice faster than the motion of the Cassiopeia and other northern (truely circumpolar) constellations)???
Who's quotes are those in that link?  I'll guess yours since you provided no other source or names.  In that case, what are the sources your quotes are based on?
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 03, 2014, 01:45:04 PM
The author of above "successfully" refuted assertions is mr Rowbotham: This is the chapter from which i have extracted above excerpts: http://www.sacred-texts.com/earth/za/za48.htm (http://www.sacred-texts.com/earth/za/za48.htm)

As for my own examinations of "The Southern Cross (case)" you can also try to "successfully" analyze these words of mine:
http://www.energeticforum.com/264125-post357.html (http://www.energeticforum.com/264125-post357.html)
Oh... Rowbotham. And you call me a con man.

Quote
So, how come that the Southern Cross makes half of an alleged circumpolar nightly circle in the sky in just 6 hours or so (which is twice faster than the motion of the Cassiopeia and other northern (truely circumpolar) constellations)???

Have you timed this yourself? If not, why do you think it's true?
Title: Re: "Equator" problem
Post by: ausGeoff on November 04, 2014, 07:19:43 AM
You can instantly dismiss the claims of anybody who cites the now debunked "experiments" of Samuel Rowbotham from 150 years ago.  For them to do so—to be so bereft of modern scientific evidence—merely confirms that they've been backed into a corner with not a shred of real evidence to support their claims, whatever they be.

Why is it that no flat earther can cite any experiments suggesting a flat earth from, say, ten years ago, or even fifty years ago?  Is 150-year-old "science" really the best they can come up with?  Does that not characterise desperation?

Just as his flat earth peers continually do, cikljamas is grasping at straws.


Title: Re: "Equator" problem
Post by: cikljamas on November 04, 2014, 09:49:45 AM
So, now your main objection is that Rowbotham's (or someone else's) experiments are just too old, ha? Hahahahhahaha...
Well, these experiments are just a little bit more recent: http://www.energeticforum.com/266709-post615.html (http://www.energeticforum.com/266709-post615.html)
Title: Re: "Equator" problem
Post by: inquisitive on November 04, 2014, 10:01:19 AM
So, now your main objection is that Rowbotham's (or someone else's) experiments are just too old, ha? Hahahahhahaha...
Well, these experiments are just a little bit more recent: http://www.energeticforum.com/266709-post615.html (http://www.energeticforum.com/266709-post615.html)
Please explain angles of sun observed from many places on earth at different times of day.
Title: Re: "Equator" problem
Post by: Son of Orospu on November 04, 2014, 10:06:40 AM
So, now your main objection is that Rowbotham's (or someone else's) experiments are just too old, ha? Hahahahhahaha...
Well, these experiments are just a little bit more recent: http://www.energeticforum.com/266709-post615.html (http://www.energeticforum.com/266709-post615.html)
Please explain angles of sun observed from many places on earth at different times of day.

Please post the data from your experiments.  Thanks. 
Title: Re: "Equator" problem
Post by: cikljamas on November 04, 2014, 11:16:32 AM
1. It IS commonly taught that the tides are caused by lunar attraction. Sir Robert Ball tells us that :

"The moon attracts the solid body of the earth with greater intensity than it attracts the water at the other side which lies more distant from it. The earth is thus drawn away from the water, which accordingly exhibits a high tide as well on the side of the earth away from the moon as on that toward the moon. The low tides occupy the intermediate positions."

No one who has the use of all his faculties and who dares to use them, need be told that this flimsy apology for what the learned cannot account for, contradicts itself. How could this attraction take place without disintegrating the globe? Besides, as the law of gravitation is said to operate according to the amount of matter of which each body consists, the statements of astronomers that the moon is 2,160 miles in diameter and the earth 8,000 miles in diameter flatly contradict their own other statements about the moon causing tides. How can the smaller body attract the larger We are informed in "Sun, Moon, and Stars," pages 160 to 163, that :

"The earth, it is true, attracts the moon. So also the moon attracts the earth ; THOUGH THE FAR GREATER WEIGHT OF THE EARTH MAKES HER ATTRACTION TO BE FAR GREATER."

How anyone can accept the current theory in face of the above is somewhat puzzling. Sir R. Ball says the moon attracts the solid body of the earth ; but the work from which I have just quoted states that :

"Her attraction (the moon's) draws up the yielding waters of the ocean in a vast wave."

Both these assertions cannot be true. Which is ? I say neither. And the astronomers' own theory of attraction also answers "neither," when it is taken into consideration that the moon cannot attract the earth, being a much smaller body. But if the moon lifted up the waters, it is evident that near the land, the water would be drawn away and low, instead of high tide, caused. Again, the velocity and path of the moon are uniform, and it follows that if she exerted any influence on the earth, that influence could only be a uniform influence. But the tides are not uniform. At Port Natal the rise and fall is about six feet, while at Beira, about 600 miles up the coast, the rise and fall is 26 feet. This effectually settles the matter that the moon has no influence on the tides.

How then are tides caused? The learned being as far from the truth in this as in every matter which we have brought to the test of the hard logic of facts, what is the
truth of the matter ?

The Leicester Daily Post, of 25th August, 1892, says :

" M. Bouquet de la Grye, an eminent hydrographical Engineer, has after long yearsof study calculated the atmospheric expansions and depressions which coincide with spring and neap tides. There have been cases in which air was moved in waves of 133 yards high, and in places where the barometrical pressure was seven-tentns ot an inch, ot six and a half miles. Near the upper surface of the earth's atmosphere condensations and dilations of this magnitude are trequent. The human nervous system may be said to register these air waves. We are only aware that they do so by the discomfort which we feel. The earth also registers them and to its very centre. The incandescent and fluid matter under the earth's crust acts in concert with the air and sea at the full of the moon. In 1889 a German Scientist, Dr. Rebeur Pachwitz, thought he noticed at Wilhelmshaven and Potsdam earth oscillations corresponding with the course of the moon. He wrote to the observatory at Tenerife asking for observations to be ma.de there in December, 1890 and April, 1891, which would be propitious times for them. From these observations and others simultanously made in the sandy plains round Berlin, IT WAS ESTABLISHED THAT THE Earth RISES AND FALLS LIKE THE OCEAN OR THE ATMOSPHERE. The movements, common to them all, may be likened to the chest in breathing. — Paris Correspondent Weekly Dispatch."

This is the answer to the question. Tides are caused by the gentle and gradual rise and fall of the earth on the bosom of the mighty deep. In inland lakes, there are no tides ; which also proves that the moon cannot attract either the earth or water to cause tides. But the fact that the basin of the lake is on the earth which rests on the waters of the deep, shows that no tides are possible, as the waters of the lakes together with the earth rise and fall, and thus the tides at the coast are caused ; while there are no tides on waters unconnected with the sea.


2. If, for example, the world be the globe of popular belief, it is impossible that there ever could have been a universal flood. For such a thing to have happened, it would be required to blot out the whole universe, to stop the revolution of the globe and to bring confusion and ruin to the whole of the "solar system."

http://www.energeticforum.com/255859-post9.html (http://www.energeticforum.com/255859-post9.html)

The most important geological discovery in the history of the world that has been covered up and still being covered up: ARK on Mt. Ararat: WHY the media BLACKOUT on the real history of Ararat? (http://#ws)
Title: Re: "Equator" problem
Post by: rottingroom on November 04, 2014, 11:34:52 AM
This

(http://i.stack.imgur.com/2zyMT.jpg)
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 04, 2014, 11:35:14 AM
1. It IS commonly taught that the tides are caused by lunar attraction. Sir Robert Ball tells us that :

...

Before we move on to the high tide of misunderstanding, can you answer my question about the six hours of Crux, please?

[Edit to add] Here is the question, in case you missed it:
So, how come that the Southern Cross makes half of an alleged circumpolar nightly circle in the sky in just 6 hours or so (which is twice faster than the motion of the Cassiopeia and other northern (truely circumpolar) constellations)???

Have you timed this yourself? If not, why do you think it's true?

One topic at a time, please. Focus!
Title: Re: "Equator" problem
Post by: cikljamas on November 05, 2014, 04:22:22 AM
(http://i.imgur.com/jaGd0Df.jpg)

Because this enormously important discovery proves that:

1. The Earth is flat

2. There was no f...ing moon-landings whatsoever....

3. The entire modern cosmology is a fairy tale at best

4. The Bible is a 100 % true - authentic Word of God

5. There is a huge conspiracy of world elites against the true Word of God and humanity

6. The theory of evolution of men is an utter LIE

7. The theory of evolution of cosmos is an utter LIE

8. The big bang theory is a fairy tale

9. The theory of relativity is a bull-s h i t (invented to cover up the fact that the Earth is at rest - Einstein even admitted it (between the lines))

10. The theory of gravitation is a bull-s h i t (invented with purpose to cover up the fact that the Earth is at rest (no orbital motion))

11. You currently live in a Truman reality show

12. Big Brother is your real enemy

13. Pope Francis is a mason and the worst antipope who has ever sited in St. Peters chair

IT IS TIME FOR WAKING UP, AND YES, I AM TALKING TO YOU PERSONALLY!!!

P.S. Alpha2Omega, focus on this!!! Use your talents in right direction! Time is limited!!! We should co-operate! It's about your ass too, believe it or not!
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 05, 2014, 06:25:39 AM

<Video about Noah's Ark and who knows what else>

P.S. Alpha2Omega, focus on this!!! Use your talents in right direction! Time is limited!!! We should co-operate! It's about your ass too, believe it or not!

You haven't answered the question about Crux yet!!! You're the one who brought it up.

In case you missed it originally and the last time I copied it:
So, how come that the Southern Cross makes half of an alleged circumpolar nightly circle in the sky in just 6 hours or so (which is twice faster than the motion of the Cassiopeia and other northern (truely circumpolar) constellations)???

Have you timed this yourself? If not, why do you think it's true?

After that was the impossibility of tides being caused by the Moon.

Now something about Noah's Ark proving or disproving a bunch of stuff.

Stay Calm and Focus!!!
Title: Re: "Equator" problem
Post by: cikljamas on November 05, 2014, 07:59:17 AM
Once and for all: i do not answer to idiotic questions...it doesn't mean that you are an idiot...but since you are not an idiot and you still pose an idiotic questions, we cannot not to be quite amazed with the way of conducting yourself...
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 05, 2014, 09:00:44 AM
Once and for all: i do not answer to idiotic questions...it doesn't mean that you are an idiot...but since you are not an idiot and you still pose an idiotic questions, we cannot not to be quite amazed with the way of conducting yourself...
Dude! You asked a question. I need more information to answer it. I can guess what you think is happening and what actually is, but getting important information about this observation from the one who asked how it can be explained would be better than guessing, don't you think?

Here is your question (bolded) in context:
As for my own examinations of "The Southern Cross (case)" you can also try to "successfully" analyze these words of mine:
http://www.energeticforum.com/264125-post357.html (http://www.energeticforum.com/264125-post357.html)

So, how come that the Southern Cross makes half of an alleged circumpolar nightly circle in the sky in just 6 hours or so (which is twice faster than the motion of the Cassiopeia and other northern (truely circumpolar) constellations)???
Before I can give an answer to your question I need more information. OK?

My question to you, phrased slightly differently, is:
Did you measure this six hours for Crux to travel halfway around its circumpolar path yourself? Yes or no?

Follow on, to save time since you seem to be in a panic:
Where was this observation made from (whether you made it yourself or someone else did)? It doesn't have to be exact; a country (and region if it's large) should be sufficient. If you don't know, you can say that.

So what are the answers?

If you're not interested in following up on this, I can only conclude that you or whoever claimed this is simply mistaken (again) about what he saw or thought he saw. Crux does not travel halfway around the pole in six hours. No evidence exists that suggests that it does. Right now there is only an unsubstantiated claim by an unknown person on the Internet.

If you're not interested in following up on this, then why did you ask the friggin' question in the first place?
Title: Re: "Equator" problem
Post by: cikljamas on November 05, 2014, 10:16:01 AM
This sentence..."Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight."...you can find here : http://www.csiro.au/helix/sciencemail/activities/crux.html (http://www.csiro.au/helix/sciencemail/activities/crux.html) ... and here: http://blackheathscouts.westernwarriorsjudo.com/index.php/navigating (http://blackheathscouts.westernwarriorsjudo.com/index.php/navigating)

These sentences..."Because the Southern Cross is so low in the sky and close to the South Celestial Pole, its path in the sky is short. From the time it rises to the time it begins to set, it is only in the sky for around six hours, whereas objects that rise closer to due east and set due west take approximately 12 hours to traverse the sky. In other words, don't expect to see the Southern Cross in the sky all evening." .... you can find here: http://the.honoluluadvertiser.com/article/2008/Feb/24/ln/hawaii802240344.html (http://the.honoluluadvertiser.com/article/2008/Feb/24/ln/hawaii802240344.html)

In this post you can find link to an amazing time-lapse video which proves that in just six hours the Crux changes it's position from being tilted to the left to the position in which the Crux is tilted over to the right : http://www.energeticforum.com/263712-post273.html (http://www.energeticforum.com/263712-post273.html)

1. The Crux is not visible in the late spring evenings (late september, october, early november) north of 34 degree South!!!

2. Casiopeia (northern alleged counterpart constellation) is visible on any clear night of the year from most location in the northern "hemisphere"!!!

Satisfied?
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 05, 2014, 12:21:42 PM
This sentence..."Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight."...you can find here : http://www.csiro.au/helix/sciencemail/activities/crux.html (http://www.csiro.au/helix/sciencemail/activities/crux.html) ... and here: http://blackheathscouts.westernwarriorsjudo.com/index.php/navigating (http://blackheathscouts.westernwarriorsjudo.com/index.php/navigating)

These sentences..."Because the Southern Cross is so low in the sky and close to the South Celestial Pole, its path in the sky is short. From the time it rises to the time it begins to set, it is only in the sky for around six hours, whereas objects that rise closer to due east and set due west take approximately 12 hours to traverse the sky. In other words, don't expect to see the Southern Cross in the sky all evening." .... you can find here: http://the.honoluluadvertiser.com/article/2008/Feb/24/ln/hawaii802240344.html (http://the.honoluluadvertiser.com/article/2008/Feb/24/ln/hawaii802240344.html)
I'll take that as a "no", you didn't time it yourself, so you're taking someone else's word for it. No matter, though, since we now know where this was happening and six hours sounds about right.

The answer to the second question appears to be "Hawaii". I had guessed "from about 20° North latitude", so would have been close.

The planetarium program Stellarium (http://www.stellarium.org/)* predicts on 14 May (guessing the "04" in "14/5/04" is the year 2004) the last star of the Southern Cross clears the horizon, with the cross leaning to the left, a bit before sunset from 20° N, culminates almost exactly 3 hours later (straight up and down), and sets 3 hours after that, leaning to the right, making the entire asterism visible for almost exactly six hours. Throw in some real-world atmospheric refraction and height above sea level, and you'll probably get that same six hours from a bit further north (like, say, Honolulu).

Now that this has been established, you need to know that the Southern Cross is not circumpolar at this latitude. It may be news to you, but in this context "circumpolar" does not mean "near the pole". It means "does not set" from a particular latitude. In fact, your own point below supports that Crux sets, and, thus, is not circumpolar, from Honolulu!!!

Quote
1. The Crux is not visible in the late spring evenings (late september, october, early november) north of 34 degree South!!!

It's above the horizon even less of the time as you go further north, too, reduced to about six hours per day by the time you reach 20° N or so. Exactly as you'd expect from a spherical earth!!! Whoever thought that its time above the horizon from Honolulu represented "half its circumpolar nightly circle" is wrong. It's only one-quarter of the circle; the rest of the circle is not visible because it's below the horizon.

Quote
In this post you can find link to an amazing time-lapse video which proves that in just six hours the Crux changes it's position from being tilted to the left to the position in which the Crux is tilted over to the right : http://www.energeticforum.com/263712-post273.html (http://www.energeticforum.com/263712-post273.html)
Tilting to the left to tilting to the right has never been an issue; you're just watching it rotate around the south pole. It doesn't turn anywhere close to 180°, though, so we're looking at far less than half its daily rotation about the pole.

Note that the linked time lapse is from 28° North latitude, but from high altitude, so Acrux is above the horizon further north than you'd expect from sea level with no atmospheric refraction. How can you tell the time lapse shows it for six hours, though? I think it's probably less.

Quote
2. Casiopeia (northern alleged counterpart constellation) is visible on any clear night of the year from most location in the northern "hemisphere"!!!
Citation needed.

The familiar 'W' or "throne" asterism of "Cass" has almost exactly the same range of declinations in the northern hemisphere as the Cross of Crux has in the southern. Thus, they will both become circumpolar at about the same latitudes in their respective hemispheres. All of "Cass" is visible all night on any clear night of the year only from north of about 33° N.

Quote
Satisfied?
Very, thanks.  You?


* I was having some trouble with the stellarium.org website today. If it fails, the program can be downloaded from http://stellarium.en.softonic.com/ (http://stellarium.en.softonic.com/), and probably elsewhere. It's an awesome program that lets you see what is in the sky and where it is, from any location and time. Set your location as Honolulu and the date to early May, speed up the time and watch Crux skim the southern horizon starting around sunset.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 05, 2014, 07:06:02 PM
These sentences..."Because the Southern Cross is so low in the sky and close to the South Celestial Pole, its path in the sky is short. From the time it rises to the time it begins to set, it is only in the sky for around six hours, whereas objects that rise closer to due east and set due west take approximately 12 hours to traverse the sky. In other words, don't expect to see the Southern Cross in the sky all evening." .... you can find here: http://the.honoluluadvertiser.com/article/2008/Feb/24/ln/hawaii802240344.html (http://the.honoluluadvertiser.com/article/2008/Feb/24/ln/hawaii802240344.html)

Oh, yeah, I almost forgot. The Honolulu Advertiser (now the Star-Advertiser) is a mainstream daily newspaper in a large American city. Is it OK to use "the media" for factual information now, or, since what they reported here, although accurate, didn't turn out to contradict conventional spherical-earth heliocentric models, they can't be trusted again?
Title: Re: "Equator" problem
Post by: markjo on November 05, 2014, 09:17:55 PM
Come now, you should know that the lamestream media is just a mouthpiece for the conspiracy.  ::)
Title: Re: "Equator" problem
Post by: cikljamas on November 06, 2014, 12:31:28 AM
I am tired of heliocentric lies!

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/ (http://www.constellation-guide.com/constellation-list/crux-constellation/))

2.  In the Northern hemisphere Cassiopeia never sets below the horizon, as a result it is visible all year in the night sky. (source : http://www.solarsystemquick.com/universe/cassiopeia-constellation.htm (http://www.solarsystemquick.com/universe/cassiopeia-constellation.htm))

3. Some stars within the far northern constellations, such as Cassiopeia, Cepheus, Ursa Major, and Ursa Minor, roughly north of the Tropic of Cancer (+23½°), will be circumpolar stars that never rise or set. (source : http://en.wikipedia.org/wiki/Circumpolar_star (http://en.wikipedia.org/wiki/Circumpolar_star))

4. The constellation is circumpolar south of 34 degrees S latitude and visible every night of the year, though it’s best viewed high overhead in the early evening from April through June.  During these months, south of the Tropic of Cancer (23.5 degrees N latitude), northern-hemisphere stargazers can glimpse the Southern Cross– just barely, above the southern horizon. - See more at: http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf (http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf)

5. Cassiopeia is visible on any clear night of the year from most locations in the northern hemisphere. For amateur stargazers, September through March are the optimal months for spotting Cassiopeia. In the southern hemisphere, the constellation is visible from May to August at locations north of the Tropic of Capricorn (23.5 degrees south latitude). (source : http://www.sciences360.com/index.php/how-to-identify-the-constellation-of-cassiopeia-in-the-night-sky-2648/ (http://www.sciences360.com/index.php/how-to-identify-the-constellation-of-cassiopeia-in-the-night-sky-2648/))

A) So, we can see Casiopeia all year long above the tropic of cancer (at least) which is 23,5 degree N, but north of 34 degree S Crux is not circumpolar constellation, how come?

B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?

C) "Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight.". This is not half a circle you say? Of course it's not, since Crux is not circumpolar constellation at all!

Now, stop lying and come back to the Earth!

Explain us how the great deluge could have happened on the round Earth?



Title: Re: "Equator" problem
Post by: ausGeoff on November 06, 2014, 01:53:22 AM
1. The Earth is flat

2. There was no f...ing moon-landings whatsoever....

3. The entire modern cosmology is a fairy tale at best

4. The Bible is a 100 % true - authentic Word of God

5. There is a huge conspiracy of world elites against the true Word of God and humanity

6. The theory of evolution of men is an utter LIE

7. The theory of evolution of cosmos is an utter LIE

8. The big bang theory is a fairy tale

9. The theory of relativity is a bull-s h i t (invented to cover up the fact that the Earth is at rest - Einstein even admitted it (between the lines))

10. The theory of gravitation is a bull-s h i t (invented with purpose to cover up the fact that the Earth is at rest (no orbital motion))

11. You currently live in a Truman reality show

12. Big Brother is your real enemy

13. Pope Francis is a mason and the worst antipope who has ever sited in St. Peters chair

IT IS TIME FOR WAKING UP, AND YES, I AM TALKING TO YOU PERSONALLY!!!

I'm thinking that by posting this ludicrous listing of fantasies, at the very least you've proven to everybody on these forums that you have absolutely zero knowledge of the earth sciences, astrophysics and geophysics, mathematics, biology, chemistry, mechanics, astronomy, meteorology, history etc.

You know; all the resources that educated, rational, enlightened, sane people draw upon to form accurate opinions of what's going on in our universe.  You appear to be living in some strange, distorted, and isolated world of your own making if you seriously believe even a couple of the delusional claims in your list.

Other than that?  I congratulate you for being possibly one of the most successful TROLLS on these forums.  Well done.  10/10.

Title: Re: "Equator" problem
Post by: cikljamas on November 06, 2014, 04:37:57 AM
Yes, i understand you, you participate in a Truman reality show from the very beginning of your life (as we all do), and it is not easy, by no means, to clear out your (or anyone's) mind from such a confusion which (heliocentrism + theory of evolution + international space programs + ufo hoax +  the rest of SF (hollywood) bull s h i t s) has been seeded into the minds of every innocent child who has had "a privilege" to be born in our modern "scientific" era of shameful, disgusting lies and deceptions.

LIE BRINGS FORTH DEATH! Just ask yourself which century was the bloodiest (BY FAR) in human history?

http://democraticpeace.wordpress.com/2009/01/27/why-the-20th-century-was-the-bloodiest-of-all/ (http://democraticpeace.wordpress.com/2009/01/27/why-the-20th-century-was-the-bloodiest-of-all/)

(http://www.hawaii.edu/powerkills/WAR.DEAD.16TO20C.JPG)

http://www.energeticforum.com/254609-post19.html (http://www.energeticforum.com/254609-post19.html) MUST READ

However, there is a way out of this universally disseminated delirium of lunatics: just follow the proofs wherever they lead you...

Oh, i know, i ask too much of you, sorry... In that case: stay where you are and enjoy perverted reality, one and only (no matter how illusive) reality which your deluded mind is able to accept!

You have nothing to say on "the great deluge" fact which is the last nail in the heliocentric coffin ?

Of course you don't!
Title: Re: "Equator" problem
Post by: Socratic Amusement on November 06, 2014, 07:07:45 AM
(http://forums.oce.leagueoflegends.com/board/attachment.php?attachmentid=1664)


I love when crazies start their posting in a seemingly intelligent, if misinformed, manner, and once instructed on their mistaken information, they devolve into full on pant-on-head crazy.
Title: Re: "Equator" problem
Post by: theearthisrounddealwithit on November 06, 2014, 07:36:41 AM
(http://forums.oce.leagueoflegends.com/board/attachment.php?attachmentid=1664)


I love when crazies start their posting in a seemingly intelligent, if misinformed, manner, and once instructed on their mistaken information, they devolve into full on pant-on-head crazy.

I've noticed this very well in my short time here so far and frankly I am disappointed. I thought that when confronted with a true challenge to their beliefs they would respond in a more rigorous fashion involving some form of data rather than the childish. "I'm right and you're wrong, that's all there is to it" (complete with tin foil hat and poutty face)
Title: Re: "Equator" problem
Post by: cikljamas on November 06, 2014, 10:10:24 AM
Mr. Laing  tells us:

" The conclusions of science are irresistible, and old forms of faith, however venerable and however endeared by a thousand associations, have no more chance in a collision with science than George Stephenson's cow had, if it stood on the rails and tried to stop the progress of a locomotive."

From purely practical data we have already seen that "the conclusions of science" are as unreasonable and fallacious as it is possible for the human mind to conceive. A mixture of infidel superstitions and gross absurdities constitute the most of present-day science respecting the world we live on. Its relation to truth is as darkness to light. Science has as much chance in a collision with TRUTH as a rotten ship would have in a collision with an ironclad.

Even professedly Christian people are hoodwinked and befogged by modern hypothetical science. A. Giberne in " Sun, Moon and Stars," says, when speaking of the Moon :

"All is dead, motionless, still. Is this verily a blasted world? Has it fallen under the breath of Almighty wrath, coming out scorched and seared ?"


The " lesser light " that God declares He made to " rule the night" is set down as a blasted world, and that by a professed Christian ! To this end the teaching of modern astronomy tends to " attract " all who receive its dicta, and cannot, therefore, be retained in the same mind with the Bible. A noteworthy feature of the present day is the fact that many so-called Christian ministers are joining hands with the enemies of the Bible to teach the people that the Old Book is so very unscientific that it can no longer be regarded in the light of a word from God at all.

In the Christian World Pulpit," of 14th June, 1893, the Rev. C .F. Aked is reported as saying, at Pembroke Chapel, Liverpool, that:

"No student of science is able to believe that any such flood as that recorded in the early chapters of Genesis ever took place in the history of the human race . , . . The Flood story IS A MYTH, 'not history'".


This gentleman has arrived at this conclusion by supposing that science is truth, and he is logically forced to believe that the Bible is a myth. Then what say the avowed
enemies of the Book of God ? Says the Freethinker, of 16th October, 1892 :

"There is something in Christianity calculated to make it hostile to science. Its sacred books are defaced by a puerile cosmogony, and a vast number of physical absurdities ; while its whole atmosphere,in the New as well as in the Old Testament, is in the highest degree unscientific. The Bible gives a false account of the origin of the world; a foolish account of the origin of man ; a ridiculous account of the origin of languages. It tells us of a universal flood which never happened. And all these falsities are bound up with essential doctrines, such as the fall of man and the atonement of Christ ; with important moral teachings and social regulations. It was therefore inevitable that the Church, deeming itself the divinely-appointed guardian of Revelation, should oppose such sciences as astronomy, geology, and biology, which could not add to the authority of the Scripture, but might very easily weaken it. Falsehood was in possession, and truth was in exile or a prisoner."

This is clinched by the Public Press which teaches people to think. Reynolds' Newspaper of 13th October, 1895, says:

" The most noteworthy feature of the British Association this year is that the assembled savants — representing religion, science, philosophy, politics — have surrendered hands down to views which, if accepted by anyone ten years ago, would be sneered at as a mark of disgrace. The Church has had to give in because geology and biology have been too strong for the Book of Genesis, which is no longer to be accepted as a real account of the Creation, but merely a symbolical one. The incontestable experiments and
experiences of the practical scientists have proved that Darwin was right, and that evolution is as certain a law as that of gravitation. What a number of the ' learned ' books of a few years ago opposing evolution must now be ignominiously withdrawn from circulation ? And how small must the controversial parson and the lay evangelist, who would prove to you in  two jiffies that science was all bosh,' feel at the thunders of competent scholars ! "


While the Press is filled with suchlike articles, the people who do not think for themselves take for granted that science is right, and as a consequence, reject the Bible.

In the "Earth Review" for January, 1893, the following is found :

"      HONEST AND NOBLE CONFESSIONS.
When we consider that the advocates of the earth's stationary and central position can account for, and explain the celestial phenomena as accurately, to their own thinking, as we can ours, in adition to which they have the evidence of their senses, and SCRIPTURE and FACTS in their favour. WHICH WE HAVE NOT : it is not without a show of reason that they maintain the superiority of their system .... However perfect our theory may appear in our estimation, and however simply and satisfactorily the Newtonian hypothesis may seem to us to account for all the celestial phenomena, yet we are here compelled to admit the astounding truth that, IF OUR PREMISES BE DISPUTED AND OUR FACTS CHALLENGED, THE WHOLE RANGE OF ASTRONOMY DOES NOT CONTAIN THE PROOFS OF ITS OWN ACCURACY.Dr. Woodhouse, a late Professor of Astronomy at Cambridge." 

Those who believe the plain and provable facts of the Bible are set down as lunatics, but the above shows where the lunacy really lies. John Wesley did not believe in the teachings of the men of the modern astronomical school, although most of his followers do. In his Journal he writes :

 "The more I consider them, the more I doubt of all systems of astronomy .... Even with regard to the distance of the sun from the earth, some affirm it lo be only three, and others ninety millions of miles."

Thus, this admittedly great and good man stands out in bold contrast with many of the present day " reverend " gentlemen. The Bishop of Peterborough is another notable  example. He says:

" I have no fear whatever, that the Bible will be found, In the long run, to contain more science than all the theories of philosophers put together."
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 06, 2014, 10:15:17 AM
Answers in italics.

I am tired of heliocentric lies!

Says the guy who can't refute the hard data he doesn't want to hear.

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/ (http://www.constellation-guide.com/constellation-list/crux-constellation/))

This is pretty close.

2.  In the Northern hemisphere Cassiopeia never sets below the horizon, as a result it is visible all year in the night sky. (source : http://www.solarsystemquick.com/universe/cassiopeia-constellation.htm (http://www.solarsystemquick.com/universe/cassiopeia-constellation.htm))

Yep, they say that, but not everything published is meaningful. The official constellation boundary extends as far north as about +78 declination, but the commonly-recognized asterism - the 'W' - is much further south, in the range of about +57° to +64° declination. There are no stars brighter than fourth magnitude in the constellation north of Segin (Epsilon Cas), northernmost star of the 'W' asterism. Technically, the northernmost part of the official constellation is circumpolar down to about 12° N latitude, but that is basically empty sky with very dim stars and dim Milky Way. Part of the 'W' sets if you're south of about 33° N and the whole thing will completely set at least briefly each day south of about 26° N.

3. Some stars within the far northern constellations, such as Cassiopeia, Cepheus, Ursa Major, and Ursa Minor, roughly north of the Tropic of Cancer (+23½°), will be circumpolar stars that never rise or set. (source : http://en.wikipedia.org/wiki/Circumpolar_star (http://en.wikipedia.org/wiki/Circumpolar_star))

Note that they say "some stars" within these constellations. See the above. It's interesting that they omit Draco and Camelopardalis, both of which officially extend further north than "Cass" and Ursa Major. Could it be that these dim constellations aren't well known because they are hard to see, so don't count? See the part above about no bright stars in "Cass" north of the 'W'.

4. The constellation is circumpolar south of 34 degrees S latitude and visible every night of the year, though it’s best viewed high overhead in the early evening from April through June.  During these months, south of the Tropic of Cancer (23.5 degrees N latitude), northern-hemisphere stargazers can glimpse the Southern Cross– just barely, above the southern horizon. - See more at: http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf (http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf)

The Southern Cross is visible evenings low in the south along the Tropic of Cancer April - June. It gets higher as you go further south, even while still in the northern hemisphere. It can be seen early in the morning much earlier in the year.

5. Cassiopeia is visible on any clear night of the year from most locations in the northern hemisphere. For amateur stargazers, September through March are the optimal months for spotting Cassiopeia. In the southern hemisphere, the constellation is visible from May to August at locations north of the Tropic of Capricorn (23.5 degrees south latitude). (source : http://www.sciences360.com/index.php/how-to-identify-the-constellation-of-cassiopeia-in-the-night-sky-2648/ (http://www.sciences360.com/index.php/how-to-identify-the-constellation-of-cassiopeia-in-the-night-sky-2648/))

The 'W' of "Cass" isn't visible in May or most of June from the Tropic of Capricorn; it doesn't fully rise before dawn drowns it out until late June, and isn't visible in the evening from there until November.

A) So, we can see Casiopeia all year long above the tropic of cancer (at least) which is 23,5 degree N, but north of 34 degree S Crux is not circumpolar constellation, how come?

Crux is the smallest constellation. Its official boundaries barely extend beyond the Cross asterism itself. Cassiopeia's official boundaries extend much closer to the pole, so you can technically say part of the constellation Cassiopeia is still circumpolar much further south than the asterism that's associated with the constellation is. If you're talking about what people recognize in the sky as Cassiopeia, it is not circumpolar south of about 33 N. See the answer to 2., above (again).

B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?

Where did you get those dates? Each of these constellations is visible from the opposite tropic at some time of night for much longer than that. Your dates for "Cass" from the Tropic of Capricorn are simply wrong. Crux rises just before dawn in December, and sets just after sunset in late June, from the Tropic of Cancer.

C) "Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight.". This is not half a circle you say? Of course it's not, since Crux is not circumpolar constellation at all!

Didn't we already both agree that Crux isn't circumpolar from the latitude of Honolulu? "Tilted to the left" doesn't mean laying on its side, it means it's tilted, as in, not vertical. Ditto for "tilted to the right". So, no, it's not half its circle. Crux is circumpolar south of about 34° S latitude. Didn't we also agree on this?

Now, stop lying and come back to the Earth!

Would you please cut the "you're lying" crap and discuss facts. Just ignoring mine and saying I'm lying makes it look like you have no facts to argue, which, I think actually is the case. Are there any other possibilities?

Explain us how the great deluge could have happened on the round Earth?

Let's not start on yet another topic until we've finished this one, then get finished with tides, OK?

[I see you just posted something else...]
Title: Re: "Equator" problem
Post by: Socratic Amusement on November 06, 2014, 10:15:28 AM
See what I mean? Like clockwork...
Title: Re: "Equator" problem
Post by: cikljamas on November 06, 2014, 10:27:25 AM
By the little help from my friend:

(http://i.imgur.com/7wanoRG.jpg)

Accompanying text:

This is a photo which I took last year from Batumi, Georgia. You can actually see in it Mt.Elbrus which is 202 km away !!! The photo was taken from the port, so it was taken from 1-2 meters above the sea level. Notice that you can see some buildings across the sea. They are more than 50 km away. The very next day I walked out and looked in the same direction, but neither Elbrus nor the buildings were visible. It appeared as if the horizon was much closer. There was no fog or anything, but nothing of what I saw the day before was visible. I wish I could spend more time there. I am sure one day I might have been able to actually take a picture of the coast on the other side...
Title: Re: "Equator" problem
Post by: cikljamas on November 06, 2014, 10:58:15 AM
@Alpha2Omega, i will study your post tomorrow, o.k.?... now i have to go...
Title: Re: "Equator" problem
Post by: ausGeoff on November 06, 2014, 11:08:29 AM
The above image(s) may well be genuine, but the person who wrote the accompanying text was either confused, drug or alcohol affected, or lying.

What one sees from a specific vantage point on one day doesn't—and cannot—change the following day (assuming swell and tide levels are constant).  Mt Erebus is 5,600m in height, and had obviously "disappeared" in the mist on the following day.  No mystery at all.  (Was that a pun?)
Title: Re: "Equator" problem
Post by: 29silhouette on November 06, 2014, 11:29:12 AM
A) So, we can see Casiopeia all year long above the tropic of cancer (at least) which is 23,5 degree N, but north of 34 degree S Crux is not circumpolar constellation, how come?
The observer is too far north.  Did you know that for an observer at either pole, pretty much all stars down to the celestial equator become circumpolar?

Quote
B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?
Maybe because one is much bigger than the other, not exactly opposite of each other, and those are probably 'optimal' viewing dates from two different sites.

Quote
C) "Tonight (14/5/04), the Southern Cross will be tilted to the left in the early evening, straight up and down at around 9:45 pm and tilted over to the right by midnight.". This is not half a circle you say? Of course it's not, since Crux is not circumpolar constellation at all!
Tilted how much?

Also, did you find any visual evidence yet, whether photographic or video, of the southern stars moving overhead in a manner indicative of being 3,000 miles high and traveling around the outer part of a giant disk?
Title: Re: "Equator" problem
Post by: 29silhouette on November 06, 2014, 11:38:04 AM
See what I mean? Like clockwork...
And just 12 minutes later... on to the next.  This must be a record for someone derailing their own derailings of their own thread.
Title: Re: "Equator" problem
Post by: Saros on November 06, 2014, 11:43:23 AM
The above image(s) may well be genuine, but the person who wrote the accompanying text was either confused, drug or alcohol affected, or lying.

What one sees from a specific vantage point on one day doesn't—and cannot—change the following day (assuming swell and tide levels are constant).  Mt Erebus is 5,600m in height, and had obviously "disappeared" in the mist on the following day.  No mystery at all.  (Was that a pun?)

You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions. You actually see parts of it which are much lower than 5642 m, you don't just see the very top 5 meters. Anyway, there are many more examples of long distance observations which are challenging the maximum possible visibility. By the way, forget about the peak, look at the building tops, so you think it is okay to see buildings 50 km into the sea? What about Earth's curvature, shouldn't it be much bigger for that distance? Yes, it should. In fact, you shouldn't see a 20-30 m tall building from 1-2 m height from farther than 20 km away, and we're talking seeing it at the horizon here, while in the photo they are above the horizon and the distance is about 50 km.

And in case you're wondering where Elbrus is follow the arrow in the photo below. The ship is actually the one on the left in the photo which Cikljamas submitted.
(http://i.imgur.com/L3y1Hgl.jpg?1)
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 06, 2014, 03:03:19 PM
B) In the southern "hemisphere" Casiopeia is visible from may to august at location north of 23,5 degree S, but in northern hemisphere Crux is visible from april to june (tropical regions), but from tropic of cancer only from may to june! Where this huge difference comes from?
Maybe because one is much bigger than the other, not exactly opposite of each other, and those are probably 'optimal' viewing dates from two different sites.

It's easier than that. His dates for Cassiopeia are simply wrong. The dates for Crux are approximately when it's visible in the evening from that far north, so you have a point about convenience. Crux is visible from the south coast of Kauai (about 22° N) in the morning before sunrise in January.
Title: Re: "Equator" problem
Post by: 29silhouette on November 06, 2014, 05:17:56 PM
You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions. You actually see parts of it which are much lower than 5642 m, you don't just see the very top 5 meters. Anyway, there are many more examples of long distance observations which are challenging the maximum possible visibility. By the way, forget about the peak, look at the building tops, so you think it is okay to see buildings 50 km into the sea? What about Earth's curvature, shouldn't it be much bigger for that distance? Yes, it should. In fact, you shouldn't see a 20-30 m tall building from 1-2 m height from farther than 20 km away, and we're talking seeing it at the horizon here, while in the photo they are above the horizon and the distance is about 50 km.

And in case you're wondering where Elbrus is follow the arrow in the photo below. The ship is actually the one on the left in the photo which Cikljamas submitted.
http://i.imgur.com/L3y1Hgl.jpg?1 (http://i.imgur.com/L3y1Hgl.jpg?1)
Measuring between the port of Batumi and Mt. Elbrus on google earth, I get 200km .  The coast is angled in relation to the line of sight, but, depending where in the port the picture was taken from, measures about about 31km (19 miles) away at the point directly in line with the peak and the port.

According to the chart in ENaG, the drop after 20 miles is 266 feet and 10 miles is 66 feet.  With a height of 2m (6 feet), the drop along the line of sight doesn't start for 3 miles (again, according to the chart and table in ENaG), so we are left with a curvature drop based on 16 miles for the opposite shore, and 117 out of 120 miles (200km) for the mountain (100 miles = drop of 6,666 feet, 120 miles = 9600 feet.)  If someone knows the math to figure out the drop for 16 and 117 miles, go for it.

The land along the coast there slopes upward with several small buildings and a few larger looking structures scattered up to a couple hundred feet above the shoreline. 

I'll throw and estimate of 3000 meters of drop along the line of sight to the mountain, which has an elevation of 5642m.  Refraction is a common occurance too.  Had a picture also been taken from higher up, any refraction and/or recovery of objects hidden beyond the curvature would have been seen.

As for being 15 degrees above the horizon, how did you come up with that?  How tall would something need to be to be 15 degrees above the horizon 200km away?  The original picture looks like magnification was used versus 1x.  The mountains wouldn't have been very high looking in a 1x shot.
Title: Re: "Equator" problem
Post by: Saros on November 06, 2014, 11:30:23 PM
You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions. You actually see parts of it which are much lower than 5642 m, you don't just see the very top 5 meters. Anyway, there are many more examples of long distance observations which are challenging the maximum possible visibility. By the way, forget about the peak, look at the building tops, so you think it is okay to see buildings 50 km into the sea? What about Earth's curvature, shouldn't it be much bigger for that distance? Yes, it should. In fact, you shouldn't see a 20-30 m tall building from 1-2 m height from farther than 20 km away, and we're talking seeing it at the horizon here, while in the photo they are above the horizon and the distance is about 50 km.

And in case you're wondering where Elbrus is follow the arrow in the photo below. The ship is actually the one on the left in the photo which Cikljamas submitted.
http://i.imgur.com/L3y1Hgl.jpg?1 (http://i.imgur.com/L3y1Hgl.jpg?1)
Measuring between the port of Batumi and Mt. Elbrus on google earth, I get 200km .  The coast is angled in relation to the line of sight, but, depending where in the port the picture was taken from, measures about about 31km (19 miles) away at the point directly in line with the peak and the port.

According to the chart in ENaG, the drop after 20 miles is 266 feet and 10 miles is 66 feet.  With a height of 2m (6 feet), the drop along the line of sight doesn't start for 3 miles (again, according to the chart and table in ENaG), so we are left with a curvature drop based on 16 miles for the opposite shore, and 117 out of 120 miles (200km) for the mountain (100 miles = drop of 6,666 feet, 120 miles = 9600 feet.)  If someone knows the math to figure out the drop for 16 and 117 miles, go for it.

The land along the coast there slopes upward with several small buildings and a few larger looking structures scattered up to a couple hundred feet above the shoreline. 

I'll throw and estimate of 3000 meters of drop along the line of sight to the mountain, which has an elevation of 5642m.  Refraction is a common occurance too.  Had a picture also been taken from higher up, any refraction and/or recovery of objects hidden beyond the curvature would have been seen.

As for being 15 degrees above the horizon, how did you come up with that?  How tall would something need to be to be 15 degrees above the horizon 200km away?  The original picture looks like magnification was used versus 1x.  The mountains wouldn't have been very high looking in a 1x shot.

Thanks for the analysis. Indeed there was 3x magnification, but anyway without magnification if you take a picture it looks much farther than looking at the object with the naked eye. You should remember that the eye has an average focal length of 40-50 mm while most cameras start at 20 mm. You actually need to go up to 2x even 3x approximately in order to see it as it really appears to you in real life.

Additionally, I am just pointing out that the maximum visible horizon is around 270 km for something 5642 m high if you're 2 m above the sea level. Then you would see it barely looming at the horizon. As you see in the photo it is not at the horizon at all. It is above it. It is not 270 km away either, but 202 km is far enough.

You're right about the distance of the buildings, the one in the cropped imaged I posted must be a little over 30 km away not 50 km away. But still it shouldn't be visible at all from 2 m height. The distance was wrongly computed due to the fact that those towns are indeed over 50 km away, but not when the distance is measured directly across the sea.

I would imagine though there are better days to take pictures of Elbrus when the visibility is much better. I was there for 3 days only, so I doubt I was so lucky to catch the best visibility ever. My guess is something is wrong with the horizon  distance formula or the Earth doesn't curve at the same rate everywhere if it does at all.

People often get it wrong when they use the horizon calculator and conclude it matches reality. They forget that when they see something they rarely see it only 1 degree above the horizon, but often much higher than that.

You might not believe me but the photo was really taken from about 1-2 meters above the sea level.

Take a look:
(http://i.imgur.com/Y3ro4i7.jpg?1)
Title: Re: "Equator" problem
Post by: cikljamas on November 07, 2014, 04:34:04 AM
@Alpha2Omega, remember one thing:

HELIOCENTRISM = SATANISM

I am very sorry, but as long as you defend this evil-perverted "science" of lying you cannot (by default) be anything but liar. I don't say that you deliberately lie, but since i am very aware how much intelligent and educated you are, you just don't leave me much choice...

Anyway, i will do my best to refrain myself of calling you "a liar"...

Answers in italics.

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/ (http://www.constellation-guide.com/constellation-list/crux-constellation/))

This is pretty close.

Doesn't this say it all?

Well, according to heliocentrists, Cassiopeia is circumpolar constellation, and Crux is ALMOST circumpolar constellation ! Why is this so, since they are exactly opposite positioned constellations? Now, who do you think you are fooling?

Once more, (quotes from more sources):

- Cassiopeia is circumpolar, and is visible all year round in the Northern Hemisphere. (http://www.eaas.co.uk/cms/index.php?option=com_content&view=article&id=24:andromeda-cassiopeia-and-perseus&catid=5:learning-zone&Itemid=8 (http://www.eaas.co.uk/cms/index.php?option=com_content&view=article&id=24:andromeda-cassiopeia-and-perseus&catid=5:learning-zone&Itemid=8))

- Cassiopeia, the Queen, is visible in the Northern Hemisphere all year long. (http://www.botproductions.com/stellar/cassiopeia.html (http://www.botproductions.com/stellar/cassiopeia.html))

- Cassiopeia is visible all year round, though the best time to see her is in the Autumn when she is high overhead. (http://www.skyguide.org.uk/constellations/cassiopeia/cassiopeia.htm (http://www.skyguide.org.uk/constellations/cassiopeia/cassiopeia.htm))

Now, this is going to be quite interesting for our readers:

The most southerly parts of Cassiopeia are visible from the North Island of New Zealand. They are highest in mid November at about 10.30 pm (NZDT). α Cas remains about 3° below the horizon as seen from Auckland and is immediately below (north of) M31, the Andromeda Galaxy. The most southerly bright star of the constellation, π Cas (mag 5.0), rises to 6° at Auckland, but is only just above the horizon at Wellington. Read more: http://www.rasnz.org.nz/Stars/Cassiopeia.htm (http://www.rasnz.org.nz/Stars/Cassiopeia.htm)

New Zealand latitude:

(http://i.imgur.com/cQblaJ7.jpg)

Now something very interesting about the picture provided by my friend Saros:

(http://i.imgur.com/2L0CXBz.jpg)

18,7 * 18,7 (miles) = 351,56 * 8 (inches) * 2,5 cm = 7031 cm / 100 = 70 meters (high hill of water)

If we assumed that this ship is right in the middle of this bay which is 18,7 miles long, and if we assumed that the deck of this tanker is 17,5 meters above the water (which is generous enough assumption) then we could with all right in the world ask this question:

Doesn't "horizon calculator formula" somehow mislead in favor of RET?

Maybe next illustration can help you to see what i mean:

(http://i.imgur.com/mE6q4dU.jpg)








Title: Re: "Equator" problem
Post by: Socratic Amusement on November 07, 2014, 05:58:19 AM
HELIOCENTRISM = SATANISM

You have failed to disprove the heliocentric model, so to claim it is merely a facet of the dogma of a sub-sect of Christians without providing any evidence for this claim is disingenuous at best.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 07, 2014, 07:45:56 AM
@Alpha2Omega, remember one thing:

HELIOCENTRISM = SATANISM

I am very sorry, but as long as you defend this evil-perverted "science" of lying you cannot (by default) be anything but liar. I don't say that you deliberately lie, but since i am very aware how much intelligent and educated you are, you just don't leave me much choice...

Anyway, i will do my best to refrain myself of calling you "a liar"...
I'm not going to argue about religion with you. I will stick with facts.

Quote

Answers in italics.

1. On the celestial sphere, Crux is exactly opposite the constellation Cassiopeia. (source : http://www.constellation-guide.com/constellation-list/crux-constellation/ (http://www.constellation-guide.com/constellation-list/crux-constellation/))

This is pretty close.

Doesn't this say it all?
It says when you can see all of one you can't see all (probably not any) of the other. What else do you think this says?

Quote
Well, according to heliocentrists, Cassiopeia is circumpolar constellation, and Crux is ALMOST circumpolar constellation ! Why is this so, since they are exactly opposite positioned constellations? Now, who do you think you are fooling?
In what context do astronomers say that Crux is "almost circumpolar"? Whether it is or not depends on the latitude, and all agree that it is circumpolar if you're far enough south. In fact, right here

4. The constellation is circumpolar south of 34 degrees S latitude and visible every night of the year, though it’s best viewed high overhead in the early evening from April through June.  During these months, south of the Tropic of Cancer (23.5 degrees N latitude), northern-hemisphere stargazers can glimpse the Southern Cross– just barely, above the southern horizon. - See more at: http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf (http://oneminuteastronomer.com/1976/legends-southern-cross/#sthash.ZmOW3e4S.dpuf)
you say it is circumpolar. I don't think anyone disagrees with that statement.

For that matter, Whether "Cass" is circumpolar or not also depends on where you are.

It would be helpful if, for Cassiopeia, you distinguished between the official constellation boundary, which is an area of sky defined in, I think, the 1870s, or the asterism (group of bright stars forming the familiar 'W' shape). This is less essential for Crux because its boundaries are barely larger than the familiar Southern Cross asterism.

Quote
Once more, (quotes from more sources):

- Cassiopeia is circumpolar, and is visible all year round in the Northern Hemisphere. (http://www.eaas.co.uk/cms/index.php?option=com_content&view=article&id=24:andromeda-cassiopeia-and-perseus&catid=5:learning-zone&Itemid=8 (http://www.eaas.co.uk/cms/index.php?option=com_content&view=article&id=24:andromeda-cassiopeia-and-perseus&catid=5:learning-zone&Itemid=8))

- Cassiopeia, the Queen, is visible in the Northern Hemisphere all year long. (http://www.botproductions.com/stellar/cassiopeia.html (http://www.botproductions.com/stellar/cassiopeia.html))

- Cassiopeia is visible all year round, though the best time to see her is in the Autumn when she is high overhead. (http://www.skyguide.org.uk/constellations/cassiopeia/cassiopeia.htm (http://www.skyguide.org.uk/constellations/cassiopeia/cassiopeia.htm))
Yes, yes. But these descriptions aren't very rigorous, so using them to try to prove a point in a technical discussion is weak. The second refers to "the Queen" suggesting they're talking about the group of stars including and immediately around the 'W' (or "throne"), so at least there's that, and it probably safe to assume the others are, as well. Note that none of these say that Cassiopeia is visible all night all year which would be necessary to meet the strict definition of circumpolar. For informal use, this might be good enough, but don't try to draw too many technical conclusions from the informal descriptive text.

Also note that the last link is clearly aimed at a British audience. All of Great Britain is far enough north that "Cass" is visible all night all year, but it makes no claim that this is true everywhere in the northern hemisphere.

Quote
Now, this is going to be quite interesting for our readers:

The most southerly parts of Cassiopeia are visible from the North Island of New Zealand. They are highest in mid November at about 10.30 pm (NZDT). α Cas remains about 3° below the horizon as seen from Auckland and is immediately below (north of) M31, the Andromeda Galaxy. The most southerly bright star of the constellation, π Cas (mag 5.0), rises to 6° at Auckland, but is only just above the horizon at Wellington. Read more: http://www.rasnz.org.nz/Stars/Cassiopeia.htm (http://www.rasnz.org.nz/Stars/Cassiopeia.htm)

New Zealand latitude:

<image showing New Zealand's North Island spanning about from about 34° S to 42° S latitude>

The first sentence in that link says:

Quote from: Royal Astronomical Society of New Zealand
Only the most southerly parts of Cassiopeia rise above the horizon in northern New Zealand. None of the brighter stars are visible.
We're clearly talking about the official constellation and not the familiar asterism ("None of the brighter stars..."). This is confirmed if you look at the included chart. Just as the official boundaries of Cassiopeia extend much farther north than the 'W' asterism, the southern boundary extends almost another 10° in the other direction, down to about +47°.

π Cas is just a smidgen north of declination +47°, Auckland and Wellington are at latitude 37.8° 36.8° S and 41.25° S, respectively. The reported maximum elevations of 6° and 1° for this star sound right for those latitudes.

All of this exactly matches what we would expect for a spherical earth. What is your point?

Quote
Now something very interesting about the picture provided by my friend Saros:

<off topic pictures and discussion containing some bad assumptions>
Later, if it's still active when we finish what's already on our plate.

[Edit] Correct Auckland latitude.
Title: Re: "Equator" problem
Post by: ausGeoff on November 07, 2014, 08:09:12 AM
You have no idea what you're talking about.

I thank you for this erudite answer LOL.  It proves absolutely my ignorance, and indisputably proves your vastly superior knowledge. [sic]

Where I lived in Victoria (Australia) I could more than easily see a mountain (LOL) with an elevation of 364 metres from a distance of over 90km.  This was across 52km of land and a 50km wide bay from my elevation of 3 metres at the shoreline of the bay.

Your guesstimations for the alleged visibility of Mt Erebus are laughably incorrect.  Please do better next time.

Title: Re: "Equator" problem
Post by: cikljamas on November 07, 2014, 08:44:33 AM
Cassiopeia Declination : 77,6923 - 48,6632
Auckland latitude : 36,84 S

36,84 + 6 degree = 42,84
48,6632 + 42,84 = 91,5 degree

What is my point?

This is my point:

(http://i.imgur.com/pcoZKYy.jpg)

Now, if you carefully observe above tangent lines and their geometrical implications for the observers on the alleged globular Earth you could easily figure that from the equator (if the Earth were really a globe) you wouldn't be able to see even Polaris, let alone being able to see any star of the Cassiopeia constellation from New Zealand...

This is another example of fooling people who haven't even begun to think for themselves!
Title: Re: "Equator" problem
Post by: markjo on November 07, 2014, 09:10:17 AM
What is my point?

This is my point:

(http://i.imgur.com/pcoZKYy.jpg)
Just as an FYI: the equator in your drawing is about 15 degrees too far north.
Title: Re: "Equator" problem
Post by: ausGeoff on November 07, 2014, 09:15:39 AM
Just as an FYI: the equator in your drawing is about 15 degrees too far north.

You should know by now that flat earthers never let the facts get in the way of a good story LOL.

Title: Re: "Equator" problem
Post by: cikljamas on November 07, 2014, 09:28:17 AM
What is my point?

This is my point:

(http://i.imgur.com/pcoZKYy.jpg)

Now, if you carefully observe above tangent lines and their geometrical implications for the observers on the alleged globular Earth you could easily figure that from the equator (if the Earth were really a globe) you wouldn't be able to see even Polaris, let alone being able to see any star of the Cassiopeia constellation from New Zealand...

This is another example of fooling people who haven't even begun to think for themselves!
Just as an FYI: the equator in your drawing is about 15 degrees too far north.

Thanks for your observation, this error (equator line drawn a little bit more toward north) makes my point even stronger and more plausible!
Title: Re: "Equator" problem
Post by: ausGeoff on November 07, 2014, 10:02:10 AM
Thanks for your observation, this error (equator line drawn a little bit more toward north) makes my point even stronger and more plausible!

LOL... typical flat earther.  Just shift the earth's equator to wherever you want it to be in order to improve your story.

Sad really.    ::)

Title: Re: "Equator" problem
Post by: Saros on November 07, 2014, 10:37:36 AM
You have no idea what you're talking about.

I thank you for this erudite answer LOL.  It proves absolutely my ignorance, and indisputably proves your vastly superior knowledge. [sic]

Where I lived in Victoria (Australia) I could more than easily see a mountain (LOL) with an elevation of 364 metres from a distance of over 90km.  This was across 52km of land and a 50km wide bay from my elevation of 3 metres at the shoreline of the bay.

Your guesstimations for the alleged visibility of Mt Erebus are laughably incorrect.  Please do better next time.

Mount Elbrus(not Erebus, are you Japanese?) can be seen maximum from 270 km if you're 2 m above the sea level.
Then it will be seen just barely at the very horizon. In the picture I showed it is not at the horizon, but above it, and it is not 270 km away, but only 202 km.

Anyway, that is not the point. I am not making up anything. I am just saying that the horizon distance calculators produce lower horizon distance results than what is observed in practice. The building tops also shouldn't be visible from over 30 km across the bay. You can check in the calculator that from a height of 2 meters you can see something which is 30 meters high maximum from 24 km away. You would only see the very top though. Do you understand? If you don't, study the horizon distance calculators first.

By the way, for your reference, your own example also contradicts the horizon distance formula.

Look it up http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)

From 3 meters the visual horizon, if you're looking at something 364 meters tall, is maximum 74 km!!! If you can see a 364 meters mountain from 90 km away just fine and you're only 3 meters above the sea level that means A. the horizon distance calculators are wrong B. the Earth is flat C. You are proving exactly what you wanted to refute D. You made it up

So, thank you very much for that great example, which if true, should point to the possibility that the Earth is flat. You're not even aware that what you think is normal is not. That is why I suggest use the calculator first. Then come back and tell me how exactly you can see a 364 m mountain from 90 km away, if you're at 3 m. You should have asked yourself this question a long time ago. Good you mentioned it here.
Title: Re: "Equator" problem
Post by: rottingroom on November 07, 2014, 10:52:50 AM
Those tangential lines in clicky's diagram are pointing to a spot just above the north pole / south pole. That makes no sense considering that polaris is 2.55009322 × 10^15 miles away.

As long as you can see below the horizon ever so slightly more than exactly 90° then you will be about to see farther and farther past the horizon for things that are farther away.
Title: Re: "Equator" problem
Post by: ausGeoff on November 07, 2014, 11:04:09 AM
So, thank you very much for that great example, which if true, should point to the possibility that the Earth is flat.

Of course there's a "possibility" that the earth is flat.  There's also a possibility that leprechauns exist too.  But science indicates that this possibility is so infinitesimally tiny that it's totally insignificant in any real-world scenario.

And it always amuses me how literally the flat earthers treat these distance to horizon calculations (when it suits them LOL).  So a 16km discrepancy (74km v. 90km) really proves the earth is flat.  Gee, then it must be true.

But... is that really the best you can do?    ::)
Title: Re: "Equator" problem
Post by: Saros on November 07, 2014, 11:08:38 AM
So, thank you very much for that great example, which if true, should point to the possibility that the Earth is flat.

Of course there's a "possibility" that the earth is flat.  There's also a possibility that leprechauns exist too.  But science indicates that this possibility is so infinitesimally tiny that it's totally insignificant in any real-world scenario.

And it always amuses me how literally the flat earthers treat these distance to horizon calculations (when it suits them LOL).  So a 16km discrepancy (74km v. 90km) really proves the earth is flat.  Gee, then it must be true.

But... is that really the best you can do?    ::)

I am not doing anything. Simply pointing out there is a discrepancy. It doesn't prove the Earth is flat, but makes you think why the formula doesn't really match reality. You choose how to interpret it. Explain why there is a discrepancy please. I would like to learn something new.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 07, 2014, 11:28:34 AM
Cassiopeia Declination : 77,6923 - 48,6632
Citation needed.

I get the southernmost boundary of Cassiopeia at declination about +46.75°. You're off by almost 2°.   
Sky & Telescope's Pocket Sky Atlas, ISBN 1-931559-31-7; Stellarium v.0.11.0.

Quote
Auckland latitude : 36,84 S

36,84 + 6 degree = 42,84
48,6632 + 42,84 = 91,5 degree

Oops... I had Auckland at 37.8°, not 36.8° S. Corrected.

The claim was that pi Cas was 6° above the horizon, not the dead southernmost boundary of Cassiopeia. pi Cas has declination +47° 06' (now; it was +47° 01' at epoch J2000) according to Stellarium and consistent with Pocket Sky Atlas.

36.84° + 6° = 42.84°
47.1° + 42.84° = 89.94°.

Pretty darn close.

The southern boundary of the constellation would be a bit higher than this.

Quote
What is my point?

This is my point:

<diagram based on incorrect data>

Now, if you carefully observe above tangent lines and their geometrical implications for the observers on the alleged globular Earth you could easily figure that from the equator (if the Earth were really a globe) you wouldn't be able to see even Polaris,
A north-south tangent line at the equator would be exactly parallel to the Earth's axis of rotation. Why wouldn't you see Polaris, or, more to the point, the North (and South) Celestial Pole(s) when viewing any distance (like, say, 5 feet) above the surface? This is neglecting refraction, which would help make it appear even higher above the horizon.

Quote
let alone being able to see any star of the Cassiopeia constellation from New Zealand...
Not true. See above.

Quote
This is another example of fooling people who haven't even begun to think for themselves!

I could come back with some smartass retort, but you are at least trying to think for yourself, which is good. What is not so good is that you're building your ideas on very shaky foundations (and I'm not referring to your religious beliefs, but instead to some of your "factual" ones), ignoring reliable data, and only accepting (blindly, apparently) whatever you can find that appears to support the idea you're already convinced of.
Title: Re: "Equator" problem
Post by: cikljamas on November 07, 2014, 02:46:45 PM
Those tangential lines in clicky's diagram are pointing to a spot just above the north pole / south pole. That makes no sense considering that polaris is 2.55009322 × 10^15 miles away.

As long as you can see below the horizon ever so slightly more than exactly 90° then you will be about to see farther and farther past the horizon for things that are farther away.

Prove that Polaris is so far away!

Before you make any effort in this direction maybe you would like to refresh your memory with just a few facts:

1. (http://www.zaslike.com/files/zqumw2ph80a6jojd60y.jpg)

2. Modern astronomers have lengthened the sun's distance by nearly a hundred millions of miles, which has necessarily increased the earth's supposed orbit more than 300 000 000 of MILES!!! But this extreme alteration is neither acknowledged nor permitted to detract from the great name of Kepler, lest it might also reflect upon the "science" of astronomy; for in this exact "science" the alteration of MILLIONS of MILES is "a mere detail!"

Copernicus wrote: " It is not necessary that hypotheses be true or even probable ; it is sufficient that they lead to results of calculation which agree with calculation. . . . Neither let any one, as far as hypotheses are concerned, expect anything certain from Astronomy, since that science can afford nothing of the kind, lest in case he should adopt for truth things feigned for another purpose, he should leave the science more foolish than when he came.. . . The hypothesis of the terrestrial motion was nothing but an hypothesis, valuable only so far as it explained phenomena not considered with reference to absolute truth or falsehood."

If such was the conviction of Copernicus, the reviver of the old Pagan system of Pythagoras, and of Newton, its chief expounder, what right have Modem Astronomers to assert that a theory, which was given only as a possibility, is a fact, especially when they differ so much among themselves even as regards the very first elements of the problem—the distance of the Sun from the Earth ? Copernicus computed it as being only three millions, while Meyer enlarged it to one hundred and four millions of miles, and there are many estimates between these two extremes.

John Wesley did not believe in the teachings of the men of the modern astronomical school, although most of his followers do. In his Journal he writes :

"The more I consider them, the more I doubt of all systems of astronomy .... Even with regard to the distance of the sun from the earth, some affirm it lo be only three, and others ninety millions of miles."

3.  The distance to the north star was recently downgraded from 424 light years to 324 years!

All this shows that scientists still hold no key to the universe, they have no clue about the real distances between celestial objects, it's all just guessing at best...

4. Again and again have their theories been combated and exposed, but as often have the majority, who do not think for themselves, accepted the popular thing. No less an authority in his time than the celebrated Danish astronomer, Tycho Brahe, argued that if the earth revolves in an orbit round the sun, the change in the relative position of the stars thus necessarily occasioned, could not fail to be noticed. In the " History of the Conflict between Religion and Science," by Dr. Draper, pages 175 and 176, the matter is referred to m the following words :

" Among the arguments brought forward against the Copernican system at the time of its promulgation, was one by the great Danish astronomer, Tycho Brahe, originally urged by Aristarchus against the Pythagorean system, to the effect that if, as was alleged, the earth moves round the sun, there ought to he a change in the relative position of the stars ; they should seem to separate as we approach them, or to close together as we recede from them... At that time the sun's distance was greatly under-estimated. Had it been known, as it is now, that the distance exceeds 90 million miles, or that the diameter of the orbit is more than 180 million, that argument would doubtless have had very great weight. In reply to Tycho, it was said that, since the parallax of a body diminishes as its distance increases, a star may be so far off that its parallax may be imperceptible. THIS ANSWER PROVED TO BE CORRECT."

To the uninitiated, the words " this answer proved to be correct," might seem to settle the matter, and while it must be admitted that parallax is diminished or increased according as the star is distant or near, parallax and direction are very different terms and convey quite different meanings. Tycho stated that the direction of the stars would be altered ; his critics replied that the distance gave no sensible difference of parallax. This maybe set down as ingenious, but it is no answer to the proposition, which has remained unanswered to this hour, and is unanswerable.

If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."

5. The other way around, when holding on to their galactic conjectures, they are at a loss how to account for a steady 20”.5 stellar aberration. For in that scheme our earth, dragged along by the sun, joins in this minor star's 250 km/sec revolution around the center of the Milky Way. If, for instance, in March we indeed would be moving parallel to the sun's motion, our velocity would become 250+30 = 280 km/sec, and in September 250-30 = 220 km/sec. The “aberration of starlight,” according to post- Copernican doctrine, depends on the ratio of the velocity of the earth to the speed of light. As that velocity changes the ratio changes. Hence Bradley's 20”.496 should change, too. But it does not. Therefore, there is truly a fly in this astronomical ointment, paraded and promoted as a truth.

So, in order to be able to explain how we could see celestial objects which are sometimes declined from the point of observation more than 90 degrees your last resort is a miraculous refraction, but before you could hide behind a miraculous refraction resort, you have to prove that these fabulous distances make any sense at all...


@Alpha2Omega, it goes for you too...

Fabulous distances doesn't add up when we look for something that supposed to be (but it is not since doesn't exist in the first place) curvature of water's surface in the pictures which show entire figures of 200 km distant mountains!!!
Title: Re: "Equator" problem
Post by: rottingroom on November 07, 2014, 03:00:00 PM
Is this an astrophysics class? I have no obligation to prove polaris' distance to you. You are making a claim with a faulty diagram showing tangential lines that incorrectly show a problem because those lines are pointing to the wrong spot above the celestial north pole. The claim you are trying to refute hinges on what scientists claim to be the distance, which is 2.55009322 × 10^15 miles or 433.8 light years away. So you need to show why the horizon is a problem GIVEN those massive distances. The distance to these objects is paramount to your claims.

By the way, again, I'm not going to type out a lengthy explanation about the distance to polaris or the sun but I can at least point you in the right direction.

The sun's distance is determined with the help of venus and polaris' distance can be determined using a technique called spectroscopic parallax.
Title: Re: "Equator" problem
Post by: cikljamas on November 07, 2014, 03:05:44 PM
You mean, spectroscopic my ass?
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 07, 2014, 03:27:43 PM
Oh, for Dog's sake!


<bunch of stuff already addressed, in detail, in this thread>


Since you've changed the subject again, I take it that this means you're satisfied with the analysis of when and where Crux and stars in Cassiopeia are visible. It's entirely consistent with mainstream cosmology, which includes a spinning spherical earth. Glad you finally agree!


Title: Re: "Equator" problem
Post by: 29silhouette on November 07, 2014, 04:20:48 PM
Thanks for the analysis. Indeed there was 3x magnification, but anyway without magnification if you take a picture it looks much farther than looking at the object with the naked eye. You should remember that the eye has an average focal length of 40-50 mm while most cameras start at 20 mm. You actually need to go up to 2x even 3x approximately in order to see it as it really appears to you in real life.

Additionally, I am just pointing out that the maximum visible horizon is around 270 km for something 5642 m high if you're 2 m above the sea level. Then you would see it barely looming at the horizon. As you see in the photo it is not at the horizon at all. It is above it. It is not 270 km away either, but 202 km is far enough.

You're right about the distance of the buildings, the one in the cropped imaged I posted must be a little over 30 km away not 50 km away. But still it shouldn't be visible at all from 2 m height. The distance was wrongly computed due to the fact that those towns are indeed over 50 km away, but not when the distance is measured directly across the sea.

I would imagine though there are better days to take pictures of Elbrus when the visibility is much better. I was there for 3 days only, so I doubt I was so lucky to catch the best visibility ever. My guess is something is wrong with the horizon  distance formula or the Earth doesn't curve at the same rate everywhere if it does at all.

People often get it wrong when they use the horizon calculator and conclude it matches reality. They forget that when they see something they rarely see it only 1 degree above the horizon, but often much higher than that.

You might not believe me but the photo was really taken from about 1-2 meters above the sea level.

Take a look:
http://i.imgur.com/Y3ro4i7.jpg?1 (http://i.imgur.com/Y3ro4i7.jpg?1)
Now that picture helps quite a bit.  Looking on GE and at a few random pictures people have posted, I found a shot of that same spot from a different angle.  The edge of the concrete looks to be about 1 meter.  If you were standing when you took the picture, then the camera is more than 2 meters high.  The raised part with the railing and the utility door looks to be an additional meter high, and since your camera lines up with the top of the railing instead of the bottom (the section running almost parallel to your line of sight) then you're probably more than 2 meters (6'6") but not quite 3 (9'10"), perhaps 8-9 feet.

Anyway, using this info  http://www.sacred-texts.com/earth/za/za05.htm#page_9  (http://www.sacred-texts.com/earth/za/za05.htm#page_9)

The first hilltop above the water just to the right of the line of sight to the peak is about 440 feet and is 23 miles away. Minus the 3 miles (at least, based of only 6 feet camera elevation) you're looking at a drop for 20 miles. According to ENaG, that is 266 feet.  Probably not even 200 feet, as your picture really looks to have been taken from higher than 6 feet.  About the structure, is it a building, or maybe a boat in the distance?

Again with the mountain, it's elevation is 18510 ft.  With a distance of 120 miles, the drop is based off of approx. 117 miles, but that's based off a height of only 6 feet, so probably even less than 117 as it looks like you're higher than 6.  Maybe 9000 feet of drop, if that.  An extra couple feet makes a big difference over long distances.  Add in refraction. 

Still curious how you got 15 degrees.  Did it look 15 degrees above the horizon when there in person with the naked eye?
Title: Re: "Equator" problem
Post by: 29silhouette on November 07, 2014, 04:29:07 PM
this error (equator line drawn a little bit more toward north) makes my point even stronger and more plausible!
No... not really.  With your red line perpendicular to where you should have drawn the equator, it would have been parallel to the axis, and thus lined up with both Polaris and Sigma Octantis.
Title: Re: "Equator" problem
Post by: Saros on November 07, 2014, 05:22:11 PM

Now that picture helps quite a bit.  Looking on GE and at a few random pictures people have posted, I found a shot of that same spot from a different angle.  The edge of the concrete looks to be about 1 meter.  If you were standing when you took the picture, then the camera is more than 2 meters high.  The raised part with the railing and the utility door looks to be an additional meter high, and since your camera lines up with the top of the railing instead of the bottom (the section running almost parallel to your line of sight) then you're probably more than 2 meters (6'6") but not quite 3 (9'10"), perhaps 8-9 feet.

Anyway, using this info  http://www.sacred-texts.com/earth/za/za05.htm#page_9  (http://www.sacred-texts.com/earth/za/za05.htm#page_9)

The first hilltop above the water just to the right of the line of sight to the peak is about 440 feet and is 23 miles away. Minus the 3 miles (at least, based of only 6 feet camera elevation) you're looking at a drop for 20 miles. According to ENaG, that is 266 feet.  Probably not even 200 feet, as your picture really looks to have been taken from higher than 6 feet.  About the structure, is it a building, or maybe a boat in the distance?

Again with the mountain, it's elevation is 18510 ft.  With a distance of 120 miles, the drop is based off of approx. 117 miles, but that's based off a height of only 6 feet, so probably even less than 117 as it looks like you're higher than 6.  Maybe 9000 feet of drop, if that.  An extra couple feet makes a big difference over long distances.  Add in refraction. 

Still curious how you got 15 degrees.  Did it look 15 degrees above the horizon when there in person with the naked eye?

Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only. AusGeoff, by the way, mentioned that when he lived in Victoria he could see a mountain 364 m high from 90 km away across the bay from the shore. That example also contradicts the calculated horizon distance...
By the way, distant mountain observations or anything observed from a great distance are governed by the law of perspective. I don't think the reason why something is not visible is because of the curvature at all. Most likely it is simply too far away for its size to be visible and the angle of view is too small. My point is that perhaps it is not the curvature that hides stuff but the perspective or the changing water surface level due to waves and so on. If something is not very big it is unlikely to be visible through so many layers of air. After all you couldn't possibly see anything 1000 km away even if the surface is totally flat.
 
Title: Re: "Equator" problem
Post by: 29silhouette on November 07, 2014, 06:00:12 PM
Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only. AusGeoff, by the way, mentioned that when he lived in Victoria he could see a mountain 364 m high from 90 km away across the bay from the shore. That example also contradicts the calculated horizon distance...
By the way, distant mountain observations or anything observed from a great distance are governed by the law of perspective. I don't think the reason why something is not visible is because of the curvature at all. Most likely it is simply too far away for its size to be visible and the angle of view is too small. My point is that perhaps it is not the curvature that hides stuff but the perspective or the changing water surface level due to waves and so on. If something is not very big it is unlikely to be visible through so many layers of air. After all you couldn't possibly see anything 1000 km away even if the surface is totally flat.
Here's a calculator I just found for figuring the drop over a specific distance.
http://www.cohp.org/local_curvature.html (http://www.cohp.org/local_curvature.html)
 I estimated 9000 feet over 117 miles, but it's 9,130feet.  115 miles gives us 8,820.  The mountain is 18,510 feet, so that's probably why you see a lot more than just the top 5 meters.

What was Ausgeoff's 'eye level' elevation?

I've got pictures of a bridge and hillside 'sinking' below the horizon as my viewing elevation dropped, yet houses along the shoreline were still visible (although compressed) due to refraction.

Title: Re: "Equator" problem
Post by: Saros on November 07, 2014, 06:36:53 PM
Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only. AusGeoff, by the way, mentioned that when he lived in Victoria he could see a mountain 364 m high from 90 km away across the bay from the shore. That example also contradicts the calculated horizon distance...
By the way, distant mountain observations or anything observed from a great distance are governed by the law of perspective. I don't think the reason why something is not visible is because of the curvature at all. Most likely it is simply too far away for its size to be visible and the angle of view is too small. My point is that perhaps it is not the curvature that hides stuff but the perspective or the changing water surface level due to waves and so on. If something is not very big it is unlikely to be visible through so many layers of air. After all you couldn't possibly see anything 1000 km away even if the surface is totally flat.
Here's a calculator I just found for figuring the drop over a specific distance.
http://www.cohp.org/local_curvature.html (http://www.cohp.org/local_curvature.html)
 I estimated 9000 feet over 117 miles, but it's 9,130feet.  115 miles gives us 8,820.  The mountain is 18,510 feet, so that's probably why you see a lot more than just the top 5 meters.

What was Ausgeoff's 'eye level' elevation?

I've got pictures of a bridge and hillside 'sinking' below the horizon as my viewing elevation dropped, yet houses along the shoreline were still visible (although compressed) due to refraction.

This is a better calculator -> http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)

Please be honest and admit you shouldn't see so much. Calculating the horizon distance doesn't mean you should see something from bottom to top, does it? It is 5642 meters, but even in my photos you can tell that the 2000 mark is above the horizon.

Ausgeoff's 'eye level' was 3 meters. You can read his comment. I hope he will provide more details about his observation though. Take a look at the new Elbrus picture I posted. I found it on the Internet. It was taken from Batumi beach. By the way, some of my photos can be found on the Internet on a website that deals with long distance observations.

(http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg)
 
Title: Re: "Equator" problem
Post by: cikljamas on November 08, 2014, 02:02:57 AM
@Alpha2Omega, i know that you know that you have no real arguments (just a mumbo jumbo type of arguments) to fight against obvious truth which says : The Earth is Flatly Flat! So, the final argument which is only capable to decide meritoriously about the real shape of the Earth is flatness of water's surface!

But before we get to the "TIDES ISSUE" & "NOAH'S ARK" A.K.A. "IMPOSSIBILITY OF A GREAT DELUGE TO HAPPEN ON THE GLOBULAR EARTH" ISSUE let's sum up what we have talked about so far:

1. In order to evade inevitable negative implications of lack of a real stellar parallax, heliocentric liars absolutely arbitrarily increased astronomical distances all the way up to idiotic numbers. Copernicus started with 3 000 000 miles, and now we should chew up 92 000 000 miles (alleged distance between the Earth and the Sun)!!! Chew it up if you like, but don't expect of any reasonable person to get over it as if it were a mere detail!

2. Has anybody ever tried to dispute this claim: "If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."

3. When Heliocentrists failed to disprove the geocentric nature that we live in, they resorted to inventing assumptions, many of which are so absurd that the inventors themselves admit that they are unfalsifiable (by implication unscientific) thought-experiments. Some of these assumptions include:

    -    the alleged tilt of the earth's axis,
     
    -    the so called Copernican principle,
     
    -    positive stellar parallax,
     
    -    uniformitiy of the speed of light,
     
    -    lengh contraction
     
    -    time dilation
     
    -    denial of inertia (only accepting an imaginary and isolated "chosen" inertial frame of reference)
     
    -    the earth supposedly moving at a various speeds (in order to account for the observed eclipses)

These and many other assumptions are presented as evidence to each other. In other words one assumption is used in order to prove another assumption. In fact these assumptions are so fundamentally dependent on each other that one becomes meaningless without the other, which shows that heliocentrists don't refrain from applying deceit (circular reasoning in this case) in order to make their assertions believable.

4. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's revolution around the Sun!
5. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's rotation on it's axis! http://www.energeticforum.com/256388-post62.html (http://www.energeticforum.com/256388-post62.html)
6. There is no axial tilt of the Earth, as well as no proof for such a nonsensical product (of deluded heliocentric mind) to be real or to be in accordance with anything in reality!
7. Whole universe is centred to the Earth! (Celestial equator is aligned with the Earth's equator). THERE IS NO SANE HELIOCENTRIC EXPLANATION FOR THIS ABSOLUTELY PROVEN SCIENTIFIC FACT!!!
8. There is no gravitation as such. Newton invented it with one and only purpose (to prove that the Earth revolves around the Sun)
9. http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636031#msg1636031 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636031#msg1636031)
10. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435)

Now, to refresh your memory:

A) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636172#msg1636172 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636172#msg1636172)

B) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695)

C) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759)

Are you ready for hitting the last nails ( TIDES & NOAH'S ARK issues) in heliocentric coffin?

edit:

Oh, i owe you conclusion on Crux - Casiopeia issue:

1. It is more than obvious that Casiopeia is a circumpolar constellation!
2. It is more than obvious that Crux is not a circumpolar constellation!
3. It is more than obvious that there is a lot of making up - inventing - adjusting data regarding "exact" numbers of different latitudes or "exact" positions of different stars, so it speaks for itself!!!
Title: Re: "Equator" problem
Post by: cikljamas on November 08, 2014, 06:02:13 AM
A north-south tangent line at the equator would be exactly parallel to the Earth's axis of rotation. Why wouldn't you see Polaris...?


(http://i.imgur.com/qfQjsfg.jpg)

Yes, why wouldn't we see Polaris if north-south tangent line at the equator would be exactly parallel to the Earth's axis of rotation?

Hm, strange question i must admit, pay attention to a green line in above picture and meditate some more on this issue...


Well, i just have finished my meditation on Polaris issue, and here is my conclusion:

Since Polaris declination is 89 degrees 19 ' even if we presumed that the distance between the Earth and Polaris is so idiotically great, we have to notice one problem associated with visibility of Polaris at the Equator:

Let's say that at midnight 1th January from the same point at the Equator we can see Polaris due to 0,8 degree (less) difference between 90 degree and 89 degree 19 ', this very same difference will be at midnight 1th June the reason with counter effect, am i right?

So, how come that there is no difference in visibility of Polaris from the same point at the Equator with respect to the constant half-annualy shifts of angles?

So, when someone says that we can see Polaris 1 or 2 degrees south of the Equator due to refraction, then that someone should take into account this 0,8 degree also!

Let alone seeing Polaris 12 degrees south of the Equator!
Title: Re: "Equator" problem
Post by: neimoka on November 08, 2014, 07:04:16 AM
2. Has anybody ever tried to dispute this claim: "If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."

Star positions do change measurably and this is well documented. Proper motion (that's something you might want to look up) of some stars changes as much as 10 arc seconds per year.

1. It is more than obvious that Casiopeia is a circumpolar constellation!
2. It is more than obvious that Crux is not a circumpolar constellation!
3. It is more than obvious that there is a lot of making up - inventing - adjusting data regarding "exact" numbers of different latitudes or "exact" positions of different stars, so it speaks for itself!!!
What is circumpolar and what is not depends on observer's location, nothing else. Both are circumpolar in some areas but not in some other areas.
Title: Re: "Equator" problem
Post by: 29silhouette on November 08, 2014, 09:52:31 AM
This is a better calculator -> http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)

Please be honest and admit you shouldn't see so much. Calculating the horizon distance doesn't mean you should see something from bottom to top, does it? It is 5642 meters, but even in my photos you can tell that the 2000 mark is above the horizon.

Ausgeoff's 'eye level' was 3 meters. You can read his comment. I hope he will provide more details about his observation though. Take a look at the new Elbrus picture I posted. I found it on the Internet. It was taken from Batumi beach. By the way, some of my photos can be found on the Internet on a website that deals with long distance observations.

http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg (http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg)
The calculator I linked shows the distance to the horizon from whatever height you enter, and just below that it shows how much Earth curves over whatever distance you enter.  I'm not sure how to get that info from the calculator you linked.

Here's a few pictures I took a while ago.  It's 9 miles to the bridge, and 12 to the buildings along the shore.  The left half was from an elevation of around 20 feet (I could be a couple feet off), and the right half was taken from about 7 inches.  Even from a height of 20 feet however, the horizon is 5.5 miles out, so there is still a drop over the remaining distance, and the buildings are still only visible due to refraction.  I would have had to be about 80 feet high to put the horizon at the shoreline and have a completely 'unsunken' view. 

(http://i1368.photobucket.com/albums/ag167/jeffro556/refractionandcurvature2_zps2a508228.jpg)

The bridge deck, tree-line, and hillside in general have visibly 'sunk' beyond the horizon/waterline.  Yet the buildings are still visible, although with a compressed appearance, including the landslide, due to refraction.  The top of the hillside is about 300 feet.

How many degrees above the horizon would you say the tops of the trees are? 
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 08, 2014, 12:54:14 PM
@Alpha2Omega, i know that you know that you have no real arguments (just a mumbo jumbo type of arguments) to fight against obvious truth which says : The Earth is Flatly Flat! So, the final argument which is only capable to decide meritoriously about the real shape of the Earth is flatness of water's surface!
"[ I] know that you know ..." No, you don't "know" that. It's what you want to believe, which is different. First thing in your post is a bogus claim. Not a good start.

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But before we get to the "TIDES ISSUE" & "NOAH'S ARK" A.K.A. "IMPOSSIBILITY OF A GREAT DELUGE TO HAPPEN ON THE GLOBULAR EARTH" ISSUE let's sum up what we have talked about so far:
Apparently there are still a lot of issues, though.

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1. In order to evade inevitable negative implications of lack of a real stellar parallax, heliocentric liars absolutely arbitrarily increased astronomical distances all the way up to idiotic numbers. Copernicus started with 3 000 000 miles, and now we should chew up 92 000 000 miles (alleged distance between the Earth and the Sun)!!! Chew it up if you like, but don't expect of any reasonable person to get over it as if it were a mere detail!
Note that increasing the radius of earth's orbit increases - not decreases - parallax. Clearly you have a misunderstanding what you're arguing about. Why would astronomers "arbitrarily" increase the length of the AU to "explain away" small values of stellar parallax? If that were their motive, they'd do the opposite. The first reasonably close measurement of the true value of the AU didn't occur until 1761 (as already discussed), well after Copernicus' (and Kepler's) time.

Oh, yes, you keep claiming that stellar parallax hasn't been measured. You're wrong.

"Liars". Again. In other words, you have no suitable explanation for this. Gotcha.  ::)

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2. Has anybody ever tried to dispute this claim: "If the earth is at a given point in space on say January 1st, and according to present-day science, at a distance of 90,000,000 miles from that point six months afterwards, it follows that the relative position and direction of the stars will have greatly changed, however small the angle of parallax maybe. THAT THIS GREAT CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED, incontestably proves that the earth is at rest — that it does not "move in an orbit round the sun."
"Has anybody ever tried to dispute this claim"? "Dispute"? All that's necessary is to point out that it's patently wrong.

Six months later earth will be 180,000,000 miles away, not 90,000 miles (using your approximate number). That 90,000,000 miles is the radius of the Earth's orbit, not its diameter, which is twice the radius; two times 90,000,000 miles is 180,000,000 miles. After half a year, the Earth has progressed halfway around it's orbit, so it's separated from that original point by the diameter of the orbit, not the radius. I've tried to avoid "mumbo-jumbo" here and explain it as best I can; if you don't understand this, I may be out of ideas - perhaps someone else can explain it better.

Anyway, if you had used the correct figure, it would have made your next point stronger (but still wrong).

What does "direction of the stars will have greatly changed, however small the angle of parallax maybe" mean? Isn't "direction ... will have greatly changed" the opposite of "however small the angle"?

"THAT THIS GREAT [or small] CHANGE IS NOWHERE APPARENT, AND HAS NEVER BEEN OBSERVED". Can you prove the assertion that stellar parallax "HAS NEVER BEEN OBSERVED"? You can't, of course. We'll talk about "proving" things in a bit. There are tons of published studies of stellar parallax, though, by many independent researchers, so plenty of evidence (not proof) that it can and has been measured. Whether you like it or not means exactly nothing.

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3. When Heliocentrists failed to disprove the geocentric nature that we live in, they resorted to inventing assumptions, many of which are so absurd that the inventors themselves admit that they are unfalsifiable (by implication unscientific) thought-experiments. [Citation needed] Some of these assumptions include:
 
    -    the alleged tilt of the earth's axis,
It explains apparent motion of the Sun and the seasons nicely. Also Precession of the equinoxes. This should be easily falsiable. Where's the evidence?   

    -    the so called Copernican principle,
Presuming otherwise has little value (other than, perhaps, spiritual, which is outside the realm of science) and makes many observations more difficult to explain.   

    -    positive stellar parallax,
Examined and answered in an earlier post.     

    -    uniformitiy of the speed of light,
     
    -    lengh contraction
     
    -    time dilation
These three are part of of the Theory of Relatively. It solved a 200-year mystery about the precession of the orbit of Mercury and many other observed and later-discovered phenomena. Although many have been tried, there have been no experiments that conclusively refute general relativity, but many that conclusively verify its predictions. It works so it is used.

    -    denial of inertia (only accepting an imaginary and isolated "chosen" inertial frame of reference)
The universe is thought to be inertial. This means that it is not, as a whole, accelerating, which, as a corollary, means that it isn't spinning. This does not mean that nothing in the universe accelerates (or spins), but the whole of the universe itself is not accelerating. We use reference frames that make calculations easiest. Sometimes we use earth-fixed reference frames, others, like when calculating orbits (which operate in the inertial frame) we use the inertial frame. There is no denial that inertia exists - you're again confusing terms.   

    -    the earth supposedly moving at a various speeds (in order to account for the observed eclipses)
And yet the eclipses happen exactly as predicted decades in advance. If the model those predictions are based is wrong, how likely would that be? It appears that it has merit and the proof of the pudding is in the eating (that adage is often misquoted).

These and many other assumptions are presented as evidence to each other. In other words one assumption is used in order to prove another assumption. In fact these assumptions are so fundamentally dependent on each other that one becomes meaningless without the other, which shows that heliocentrists don't refrain from applying deceit (circular reasoning in this case) in order to make their assertions believable.
These models (you call them "assumptions") are verified by actual observations and measurements as we go, though. Sometimes better observations result in better measurements that require earlier models to be revised or occasionally thrown out altogether. This is how science works. If it were a simple matter of "it says this, therefore this is so" then things like the revision of the distance to Polaris by a significant factor wouldn't happen even after the more-reliable parallax data was obtained. This isn't an "embarrassment" for science - it's a good example how it works: better measurements replace less-good or less-reliable measurements.

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4. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's revolution around the Sun!
5. There is not one single scientific proof that could corroborate alleged "scientific" fact of the alleged Earth's rotation on it's axis! http://www.energeticforum.com/256388-post62.html (http://www.energeticforum.com/256388-post62.html)
This is a good place to discuss the idea of "scientific proof". This is a term often tossed out by both sides, but, as a fact, nothing can be formally 'proven' scientifically, at least not in the same sense as a mathematical theorem is formally proven.

Quote from: Satoshi Kanazawa
Proofs have two features that do not exist in science:  They are final, and they are binary.  Once a theorem is proven, it will forever be true and there will be nothing in the future that will threaten its status as a proven theorem (unless a flaw is discovered in the proof).  Apart from a discovery of an error, a proven theorem will forever and always be a proven theorem.

In contrast, all scientific knowledge is tentative and provisional, and nothing is final.  There is no such thing as final proven knowledge in science.  The currently accepted theory of a phenomenon is simply the best explanation for it among all available alternatives.  Its status as the accepted theory is contingent on what other theories are available and might suddenly change tomorrow if there appears a better theory or new evidence that might challenge the accepted theory.  No knowledge or theory (which embodies scientific knowledge) is final.  That, by the way, is why science is so much fun.
From here (http://www.psychologytoday.com/blog/the-scientific-fundamentalist/200811/common-misconceptions-about-science-i-scientific-proof)

So you're technically right that neither of these things is 'proven', but, for the same reason, you'll never be able to 'prove' your model, either. While we can't 'prove' things, we can judge our models (or theories) based on how well they fit observations, and how well they can predict observations that haven't been made yet. A heliocentric solar system with elliptical Keplerian orbits (with a dash or general relativity when needed) is damn good at explaining the observed motion of solar system bodies and predicting future motion of these bodies. How's that explanation of retrograde motion of planets with a fixed earth and everything going around it coming along, by the way?

Many, many things are so well understood, characterized, and measured to within a micron of their lives that the likelihood that our understanding of them will change fundamentally is vanishingly small. Still a remote possibility, but exceedingly unlikely.
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6. There is no axial tilt of the Earth, as well as no proof for such a nonsensical product (of deluded heliocentric mind) to be real or to be in accordance with anything in reality!
Here we go with the "nonsensical" and "deluded" schtick again, and see the above about 'proof'. If you really think this is necessary to buck yourself up in your belief like this, then you need to examine what it is you believe in. How do you propose to explain the annual motion of the Sun if the equator isn't tilted with respect to the ecliptic?

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7. Whole universe is centred to the Earth! (Celestial equator is aligned with the Earth's equator). THERE IS NO SANE HELIOCENTRIC EXPLANATION FOR THIS ABSOLUTELY PROVEN SCIENTIFIC FACT!!!
From our point of view it's convenient to treat it as such because the universe appears to rotate about us; it's an illusion, though. From the point of view from elsewhere, it appears different. As explained earlier in this post, since the universe is inertial, we can use any point we want as a center of reference, so treating the Earth (or SS Barycenter) as the origin of a coordinate system makes our analysis easier.
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8. There is no gravitation as such. Newton invented it with one and only purpose (to prove that the Earth revolves around the Sun)
Newton deduced gravity while trying to understand why things in the heavens and on earth behaved as they do. He found a single force that explained both. Kepler applied Newton's discovery to explain, with great (but not perfect) accuracy, the motions of the planets.
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9. http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636031#msg1636031 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636031#msg1636031)
10. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435)

Now, to refresh your memory:

A) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636172#msg1636172 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636172#msg1636172)

B) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695)

C) http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759)

Are you ready for hitting the last nails ( TIDES & NOAH'S ARK issues) in heliocentric coffin?

A) already addressed in this thread.

B) and C) Let's finish this first before moving on. If you'll agree that you haven't disproved the Heliocentric model, and that people who believe that model is the best explanation for our observations aren't necessarily liars, cheats, idiots, Satan-worshippers, etc. because they of this, that would be a good start on wrapping this up. Note that doing so doesn't mean you agree with that model, only that you haven't disproved it and are willing to show some respect for those who disagree with you on this topic.

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edit:

Oh, i owe you conclusion on Crux - Casiopeia issue:

1. It is more than obvious that Casiopeia is a circumpolar constellation!
2. It is more than obvious that Crux is not a circumpolar constellation!
3. It is more than obvious that there is a lot of making up - inventing - adjusting data regarding "exact" numbers of different latitudes or "exact" positions of different stars, so it speaks for itself!!!
1. Is true if you're far enough north.

2. Is most assuredly not true. Even you yourself said it was circumpolar south of 34° S in point 4. here (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637881#msg1637881).

3. About the only "adjustments" to a simple spherical geometric solution for what stars are visible where and when are the observers height above the horizon, which is simply a little more geometry and often isn't necessary, and atmospheric refraction, which can be estimated accurately enough for most situations using a standard model of the atmosphere (1/2° at the horizon from near sea level is a good first-order approximation). For really precise measurements we need a more detailed model of the atmosphere based on current local conditions and should might also need to account for oblateness of earth, but details like these are really "down in the noise" in discussions like this; these weren't mentioned before now, because the math is much more difficult and it would look like "mumbo-jumbo" to most reasonably-well educated people, and simply aren't significant at this level of detail.

Insofar as your whining about needing to use "exact" latitudes and declinations, remember that, based on an error of about 1 1/2° in the declination of a star (pi Cas), you "found" a discrepancy of about 1 1/2° in the expected location of this star from Auckland. This discrepancy vanished when the correct declination was used. The need to use accurate positions if you want accurate answers should be self evident. It's also self evident that for arm-waving they can sometimes be a total nuisance, so I see why you want to dismiss their importance.

You keep going over the same ground again and again. The answers aren't going to change.

[Edit] Clarification in last long paragraph. Fix minor typos.
Title: Re: "Equator" problem
Post by: ausGeoff on November 08, 2014, 08:45:15 PM

           (http://www.zaslike.com/files/zqumw2ph80a6jojd60y.jpg)


Are you seriously tendering an evidential document that was written 150 years ago—about astrophysics?

Scientific knowledge has advanced logarithmically since then my friend.  You really need to catch up methinks LOL.
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 01:59:35 AM
Alpha2Omega, one day you will be one of the best FET apologist, this day is not so far away...

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"So you're technically right that neither of these things is 'proven', but, for the same reason, you'll never be able to 'prove' your model, either."

Sorry, but i am not just technically right, i am absolutely right!

Had there been any way to prove that the Earth is submitted to any kind of motion, scientists would have supplied us with these proofs up until now, and by doing this they would have provided immortal fame for themselves.

How about the proofs for the immobility of the Earth?

Allow me to present you some of them:

1. Observing the sun directly from the north pole the apparent motion of the sun would be straight line for days, and a camera should have to be slightly adjusted every few hours to cancel out scarcelly perceptible effect due to Earth's alleged rotation which speed is practically zero at North Pole.

Why haven't we seen until this day any such video which could have easily proved Earth's daily rotation? Lack of such video speaks for itself, and presents indirect but strong proof that there is no reason for making such video because such video would prove something else (that the Earth is at rest), and scientists are aware of that truth very well!


2. If the Earth was rotating about its axis, someone in Quito, Ecuador would be traveling twice as fast from west to east as someone in Oslo, Norway – at any moment, and at every moment. Meanwhile, someone looking at the proverbial North Pole, would hardly be moving at all! But is that reality?

Of course it is not reality, but this supposed fact of Earth's rotation now becomes deadliest error of all, concerning supposed differences of Earth rotational speeds at different latitudes.

If these differences were really the true fact then the speed of apparent motion of all celestial bodies would be twice greater for any observer on the equator than it would be for any observer on the latitude of Oslo.

How hard would be to make an experiment (measurement) of such kind???

3a) If the atmosphere were independent (non rotating but static) from Earth's daily rotation then we would have on the surface of the Earth permanent winds that blow 600 to 1600 km/h. Do you notice permanent winds which blow at such a speed?

3b) If the atmosphere were rotating along with the Earth the air flow at the surface of the Earth would have variable velocity (not the thermal), variable pressure (not the static), and variable density (not the normal). Such air flow and such air pressure regimes do not exist: http://www.energeticforum.com/256388-post62.html (http://www.energeticforum.com/256388-post62.html)

4. If the Earth were suffering a daily rotation it would generate an incredible deflection on all flying matters in air atmosphere. 5) The inertial motion is terminated in air atmosphere, and thus all dropped objects should land behind their starting positions on a rotating Earth. http://www.energeticforum.com/255938-post21.html (http://www.energeticforum.com/255938-post21.html)

5. To be pression or to be gravity? The choice of Earth’s rotation (the cause of pression), should repel the gravity from Earth. Consequently, the heliocentric model looses the most precious element. The choice of gravity should remove the concept of Earth’s rotation from the cosmos motion, consequently; the journey of the Earth around the sun becomes useless since half of the Earth should be always in darkness and the second half should be always in lightness.

6. No experiment has ever been performed with such excruciating persistence and meticulous precision, and in every conceivable manner, than that of trying to detect and measure the motion of the Earth. Yet they have all consistently and continually yielded a velocity for the Earth of exactly ZERO mph.

The toil of thousands of exasperated researchers, in the extremely varied experiments of Arago, De Coudre's induction, Fizeau, Fresnell drag, Hoek, Jaseja's lasers, Jenkins, Klinkerfuess, Michelson-Morley interferometry, Lord Rayleigh's polarimetry, Troughton-Noble torque, and the famous 'Airy's Failure' experiment, all conclusively failed to show any rotational or translational movement for the earth, whatsoever."

http://www.sacred-texts.com/earth/za/za21.htm (http://www.sacred-texts.com/earth/za/za21.htm)

7. NO CAUSE OF EARTH'S ROTATION WHATSOEVER: retains the state of illusion. The most important element in heliocentric model is the Earth’s rotation about its polar axis. What is the cause of Earth’s rotation? No one has attributed the cause of Earth’s rotation to any type of action or force even though they have attributed the cause of orbital motion (revolution) to Newton’s law of gravity.

8. NO CAUSE OF THE ROTATION OF THE AIR-LAYER: 2) The rotation of the air-layer next to the rigid Earth is without cause, and lacks a technique and tool. Perhaps, one may envision the whole rigid sphere undergoes a rotation about its polar axis. But, how one can envision the air atmosphere (the surface layer) rotates with the rigid sphere without an engineering method (e.g.air foil). In addition, what maintains the air’s rotation for tens of thousands of years (we are practical people) without stop. The rotation of the background air is the greatest hoax ever invented by mankind.

9. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435)


Arthur Eddington dared to contemplate that:

    "There was just one alternative; the earth's true velocity through space might happen to have been nil."

Lincoln Barnett agrees:

    "No physical experiment ever proved that the Earth actually is in motion."

So, when all attempts to prove any kind of motion of the Earth FAIL, what does it mean?

It means that the contrary is the fact: The Earth is at rest!

Every failure of all these attempts presents the proof to the contrary : The Earth is at rest!

If you don't want to make me laugh, then you cannot just say: O.K., we have failed to prove Earth's motion but there is still some chance that we could succeed to prove it in some distant future?

In how distant future? When we inhabit another galaxy? Come on, cut the crap, please!


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Anyway, if you had used the correct figure, it would have made your next point stronger (but still wrong).

EXACTLY! Thanks for helping me make it TWICE stronger!

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What does "direction of the stars will have greatly changed, however small the angle of parallax maybe" mean? Isn't "direction ... will have greatly changed" the opposite of "however small the angle"?

Here is the answer:

In the " History of the Conflict between Religion and Science," by Dr. Draper, pages 175 and 176, the matter is referred to m the following words :

" Among the arguments brought forward against the Copernican system at the time of its promulgation, was one by the great Danish astronomer, Tycho Brahe, originally urged by Aristarchus against the Pythagorean system, to the effect that if, as was alleged, the earth moves round the sun, there ought to he a change in the relative position of the stars ; they should seem to separate as we approach them, or to close together as we recede from them... At that time the sun's distance was greatly under-estimated. Had it been known, as it is now, that the distance exceeds 90 million miles, or that the diameter of the orbit is more than 180 million, that argument would doubtless have had very great weight. In reply to Tycho, it was said that, since the parallax of a body diminishes as its distance increases, a star may be so far off that its parallax may be imperceptible. THIS ANSWER PROVED TO BE CORRECT."

To the uninitiated, the words " this answer proved to be correct," might seem to settle the matter, and while it must be admitted that parallax is diminished or increased according as the star is distant or near, parallax and direction are very different terms and convey quite different meanings. Tycho stated that the direction of the stars would be altered ; his critics replied that the distance gave no sensible difference of parallax. This maybe set down as ingenious, but it is no answer to the proposition, which has remained unanswered to this hour, and is unanswerable.
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 02:05:07 AM
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These models (you call them "assumptions") are verified by actual observations and measurements as we go, though. Sometimes better observations result in better measurements that require earlier models to be revised or occasionally thrown out altogether. This is how science works.

Bull s h i t! Don't call this crap "science"? What actual observations are you talking about?

Gravitation is a clever illustration of the art of hocus-pocus—heads I win, tails you lose ; Newton won his fame, and the people lost their senses.

Lord Beaconsfield wisely said—" A subject or system that will not bear discussion is doomed." Both Copernicus himself, who revived the theory of the heathen philosopher Pythagoras, and his great exponent Sir Isaac Newton, confessed that their system of a revolving Earth was only a possibility, and could not be proved by facts. It is only their followers who have decorated it with the name of an " exact science," yea, according to them, " the most exact of all the sciences."

" We declare that this motion is all mere ' bosh,' and that the arguments which uphold it are, when examined by an eye that seeks Truth, mere nonsense and childish absurdity."


Now what confidence can any man place in a science which gives promissory notes of such extravagance as these? They are simply bankrupt bills, not worth the paper on which they are written. And yet, strange to say, many foolish people endorse them as if they were good, the reason being that they are too lazy to think for themselves, and, to their own sad cost, accept the bogus notes as if they had been issued by a Rothschild."

"      HONEST AND NOBLE CONFESSIONS.
When we consider that the advocates of the earth's stationary and central position can account for, and explain the celestial phenomena as accurately, to their own thinking, as we can ours, in adition to which they have the evidence of their senses, and SCRIPTURE and FACTS in their favour. WHICH WE HAVE NOT : it is not without a show of reason that they maintain the superiority of their system .... However perfect our theory may appear in our estimation, and however simply and satisfactorily the Newtonian hypothesis may seem to us to account for all the celestial phenomena, yet we are here compelled to admit the astounding truth that, IF OUR PREMISES BE DISPUTED AND OUR FACTS CHALLENGED, THE WHOLE RANGE OF ASTRONOMY DOES NOT CONTAIN THE PROOFS OF ITS OWN ACCURACY.— Dr. Woodhouse, a late Professor of Astronomy at Cambridge."

Those who believe the plain and provable facts of the Bible are set down as lunatics, but the above shows where the lunacy really lies. John Wesley did not believe in the teachings of the men of the modern astronomical school, although most of his followers do. In his Journal he writes :

 "The more I consider them, the more I doubt of all systems of astronomy .... Even with regard to the distance of the sun from the earth, some affirm it lo be only three, and others ninety millions of miles."



@ Ausgeoff, the testimonies of a noble men who had spotted Polaris at 12 degree South, and even all the way down to 23 degree South are still worthy of our attention, and always will be, no matter how far science of lying gets by developing various technics of deceiving and lying.

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Scientific knowledge has advanced logarithmically since then my friend.

Well, my friend, all you have to do is to read these two very educational and sobering texts:

http://www.energeticforum.com/253864-post240.html (http://www.energeticforum.com/253864-post240.html)

http://www.energeticforum.com/255678-post304.html (http://www.energeticforum.com/255678-post304.html)

It (reading it) will do good to you but only if you are not 200 % troll, we already know that you are 100 % troll!
Title: Re: "Equator" problem
Post by: Saros on November 09, 2014, 02:17:09 AM
This is a better calculator -> http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)

Please be honest and admit you shouldn't see so much. Calculating the horizon distance doesn't mean you should see something from bottom to top, does it? It is 5642 meters, but even in my photos you can tell that the 2000 mark is above the horizon.

Ausgeoff's 'eye level' was 3 meters. You can read his comment. I hope he will provide more details about his observation though. Take a look at the new Elbrus picture I posted. I found it on the Internet. It was taken from Batumi beach. By the way, some of my photos can be found on the Internet on a website that deals with long distance observations.

http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg (http://www.barents.pl/sites/default/files/550988_356863567725394_578677839_n.jpg)
The calculator I linked shows the distance to the horizon from whatever height you enter, and just below that it shows how much Earth curves over whatever distance you enter.  I'm not sure how to get that info from the calculator you linked.

Here's a few pictures I took a while ago.  It's 9 miles to the bridge, and 12 to the buildings along the shore.  The left half was from an elevation of around 20 feet (I could be a couple feet off), and the right half was taken from about 7 inches.  Even from a height of 20 feet however, the horizon is 5.5 miles out, so there is still a drop over the remaining distance, and the buildings are still only visible due to refraction.  I would have had to be about 80 feet high to put the horizon at the shoreline and have a completely 'unsunken' view. 

(http://i1368.photobucket.com/albums/ag167/jeffro556/refractionandcurvature2_zps2a508228.jpg)

The bridge deck, tree-line, and hillside in general have visibly 'sunk' beyond the horizon/waterline.  Yet the buildings are still visible, although with a compressed appearance, including the landslide, due to refraction.  The top of the hillside is about 300 feet.

How many degrees above the horizon would you say the tops of the trees are?
The calculator I linked is much better. You have the option to enter your height(H1) and then the height of the observed object (H2), and see the horizon distance as it changes. If you enter just H1 you would get the general horizon distance without taking into account the observed object's height, it basically assumes it's 0. Please stop attributing everything anomalous to refraction as it is not true. If you see something above the calculated horizon and it is not distorted, the refraction is not anomalous but within the normal range. Anomalous refraction causes distortion like in fata morgana or mirages. By the way, if I apply your calculations to my photo, the horizon distance from 7 ft is 3.24 miles, so bearing in mind that Elbrus is 5642 m (3.5 miles) does it mean that it is 0.26 miles (418 meters) above the horizon? I don't think it is calculated like  that, but if it were then Elbrus shouldn't be so much protruded above the horizon at all. It would literally mean that you can only see its top 418 meters. Are you sure this is what you're claiming? I don't think this makes sense, that is why you should enter both heights. The horizon distance from a height of 2 meters is 5 km, but the horizon distance from 5642 meters is close to 270 km. That is why there is a formula, and that is why it is necessary to enter both heights. Of course, the formula might be wrong as well. I am just showing you how the calculator works.
Title: Re: "Equator" problem
Post by: ausGeoff on November 09, 2014, 06:07:31 AM
@ Ausgeoff, the testimonies of a noble men who had spotted Polaris at 12 degree South, and even all the way down to 23 degree South are still worthy of our attention, and always will be, no matter how far science of lying gets by developing various technics of deceiving and lying.
Quote
Scientific knowledge has advanced logarithmically since then my friend.
Well, my friend, all you have to do is to read these two very educational and sobering texts:
http://www.energeticforum.com/253864-post240.html (http://www.energeticforum.com/253864-post240.html)

http://www.energeticforum.com/255678-post304.html (http://www.energeticforum.com/255678-post304.html)

It (reading it) will do good to you but only if you are not 200 % troll, we already know that you are 100 % troll!
Can you please refrain from repeatedly posting links to your delusional opinions on the Energetic Forum (http://bit.ly/1xbMcF5) site and referring to your own pseudo-scientific drivel as "educational".

And why is it that so many flat earthers—when confronted with actual contemporary science—start calling any opposition "trolls"—like petulant little schoolkids?
 
You can call me any names you like sonny, but it ain't gonna win you any friends here.    ;D


Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 09, 2014, 06:12:01 AM

Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only.
@Saros, where do you get 5 metres from?  You should be able to see the top 2700m of the mountain - which is an awful lot of mountain still.

You can try the calculation here:

http://www.cactus2000.de/uk/unit/masshor.shtml (http://www.cactus2000.de/uk/unit/masshor.shtml)


I entered the elevation as 3m, the distance as 200km.  This calculates the smallest object you can see as 2947m.  Which means if the mountain is 5642m we should still be able to see 2695m of mountain.  This is nearly 2.7km of mountain - ie loads.  I'm not sure what the problem is?
Title: Re: "Equator" problem
Post by: Saros on November 09, 2014, 07:53:43 AM

Yes, it looked much higher than the horizon. I am talking about Caucasus and especially Elbrus. I don't believe it should be visible that high from 200 km away. Seeing something at the horizon means exactly that. It means the very top of the thing is at the horizon. Clearly Mt.Elbrus is not at the horizon but much higher, and you definitely don't see its top 5 meters only.
@Saros, where do you get 5 metres from?  You should be able to see the top 2700m of the mountain - which is an awful lot of mountain still.

You can try the calculation here:

http://www.cactus2000.de/uk/unit/masshor.shtml (http://www.cactus2000.de/uk/unit/masshor.shtml)


I entered the elevation as 3m, the distance as 200km.  This calculates the smallest object you can see as 2947m.  Which means if the mountain is 5642m we should still be able to see 2695m of mountain.  This is nearly 2.7km of mountain - ie loads.  I'm not sure what the problem is?
Yeah, this is correct. I am not arguing against that. I am saying that if it were to be visible barely at the horizon you would only see its very top. In the photo it is only 200 km away, so it is not at the horizon but higher, which is normal indeed even though such long distance observations are not very common.

You can always argue how high the mountain should appear to be given the distance. I cannot provide any strong argument about this though.

What I am curious about is why the mountains observed from big distances are not inclined? I mean, assuming the Earth is round and they go straight up, if you look at them from say 200 km, they shouldn't be pointing upward 90 degrees from your perspective, but there should be an incline, no? Just making a point. At least the horizon distance diagrams appear to show just that. Do you know anything about this? Same should be true for tall buildings. If you look at one from a distance they shouldn't appear straight but like the leaning tower of Pisa, right?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 09, 2014, 08:33:18 AM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 09:17:54 AM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Well, you should live on a round planet in order to see how would it be if it were to be...But since you are clever boy you can try (and only try) to imagine such a nonsensical  world (in terms of functionality of basic physical laws as we (don't) know them)

In first link that i suggested for reading to AusGeoff you can find this description of hypothetical world of ours and how such hypothetical world would look like if a fairy tales made up by heliocentrists were true description of our reality:

According to the theory, there should be a distance beyond every edge of every galaxy and every star where the light behind is bent just the right amount to reach us here on Earth. All objects that we can see have other objects behind them. Every star we see has stars and/or galaxies behind it, and many objects we see are eclipsing objects of considerable brightness. If bending and lensing were true, we would expect every single object in the sky to be fully haloed. No, more than that: we should expect the entire sky to be filled with bent light.

Every object we see has an object behind it or near it, and every object has a distance of bending beyond every edge where the angle would be right to bend the light to us. Therefore the night sky should be filled from corner to corner with multiple images. According to the theory of light bending, there shouldn’t be a dark dot in the sky.


Read more: http://www.energeticforum.com/253864-post240.html (http://www.energeticforum.com/253864-post240.html)
Title: Re: "Equator" problem
Post by: 29silhouette on November 09, 2014, 09:40:36 AM
The calculator I linked is much better. You have the option to enter your height(H1) and then the height of the observed object (H2), and see the horizon distance as it changes. If you enter just H1 you would get the general horizon distance without taking into account the observed object's height, it basically assumes it's 0.
Using my picture of the bridge and hillside, I entered;
20ft for h1
300ft for h2 
It gives me;
radar horizon of 30.82 miles
visual horizon of 26.69 miles. 

How does that figure into my picture of a hillside 12 miles away?

Quote
Please stop attributing everything anomalous to refraction as it is not true. If you see something above the calculated horizon and it is not distorted, the refraction is not anomalous but within the normal range. Anomalous refraction causes distortion like in fata morgana or mirages.
The refraction is obvious in my picture since I took it from two different elevations.  The shoreline buildings that dropped below the horizon are still visible, but compressed, gradually diminishing the higher the object is.  It's quite common.  I've got other pictures taken from the same spot looking through a mirage.  The results are quite different.

Quote
By the way, if I apply your calculations to my photo, the horizon distance from 7 ft is 3.24 miles, so bearing in mind that Elbrus is 5642 m (3.5 miles) does it mean that it is 0.26 miles (418 meters) above the horizon? I don't think it is calculated like  that, but if it were then Elbrus shouldn't be so much protruded above the horizon at all. It would literally mean that you can only see its top 418 meters. Are you sure this is what you're claiming?
I thought you were the one claiming it should only be barely visible.  My calculations show a couple thousand meters should still be visible.

Quote
I don't think this makes sense, that is why you should enter both heights. The horizon distance from a height of 2 meters is 5 km, but the horizon distance from 5642 meters is close to 270 km. That is why there is a formula, and that is why it is necessary to enter both heights. Of course, the formula might be wrong as well. I am just showing you how the calculator works.
So an observer at the peak would see the horizon about 70km beyond Batumi. Observer's height, distance to horizon, curvature drop along the visual line of sight between the horizon and observed object, and common refraction all explain our pictures pretty well.
Title: Re: "Equator" problem
Post by: 29silhouette on November 09, 2014, 10:40:19 AM
How about the proofs for the immobility of the Earth?

Allow me to present you some of them:
1. At the equinox from the pole it would skim the horizon all day.  The camera would have to be spun around in a slow circle for 24 hours to track it.  There are videos such as this.  One in particular wasn't exactly at the pole though, so the sun moves up and down as it circles the horizon.

2.  Notice the 'star trail' pictures in which the stars closest to the celestial poles have the shortest trails versus those closer to the celestial equator? 

3a.  The air pretty much rotates with the surface.

3b.  Linking to your own flawed post with your own self quote doesn't mean much if you can't figure out that if a person or measuring device is moving along at the exact same speed as the wind to be measured, it's not going to measure much wind.

4.  Not sure exactly what you mean, but if you're trying to say things would fly off, no, one rotation every 24 hours isn't fast enough.

5.  The object would still have enertia as it falls. 

other 5.  Earth shouldn't have gravity because it rotates?

6.  No.  There are experiments to show it is rotating.

7.  Formation of the planet.

8.  Surface was rotating as atmosphere formed.  What is there outside atmosphere to slow it?

9.  I'll have to read more about this later.

Anyway, speaking of all these examples of evidence, did you ever find any visual proof yet southern stars moving overhead indicative of circling a big disk as opposed to circling a celestial pole of a globe as is commonly seen?
Title: Re: "Equator" problem
Post by: Saros on November 09, 2014, 10:47:16 AM
The calculator I linked is much better. You have the option to enter your height(H1) and then the height of the observed object (H2), and see the horizon distance as it changes. If you enter just H1 you would get the general horizon distance without taking into account the observed object's height, it basically assumes it's 0.
Using my picture of the bridge and hillside, I entered;
20ft for h1
300ft for h2 
It gives me;
radar horizon of 30.82 miles
visual horizon of 26.69 miles. 

How does that figure into my picture of a hillside 12 miles away?

Quote
Please stop attributing everything anomalous to refraction as it is not true. If you see something above the calculated horizon and it is not distorted, the refraction is not anomalous but within the normal range. Anomalous refraction causes distortion like in fata morgana or mirages.
The refraction is obvious in my picture since I took it from two different elevations.  The shoreline buildings that dropped below the horizon are still visible, but compressed, gradually diminishing the higher the object is.  It's quite common.  I've got other pictures taken from the same spot looking through a mirage.  The results are quite different.

Quote
By the way, if I apply your calculations to my photo, the horizon distance from 7 ft is 3.24 miles, so bearing in mind that Elbrus is 5642 m (3.5 miles) does it mean that it is 0.26 miles (418 meters) above the horizon? I don't think it is calculated like  that, but if it were then Elbrus shouldn't be so much protruded above the horizon at all. It would literally mean that you can only see its top 418 meters. Are you sure this is what you're claiming?
I thought you were the one claiming it should only be barely visible.  My calculations show a couple thousand meters should still be visible.

Quote
I don't think this makes sense, that is why you should enter both heights. The horizon distance from a height of 2 meters is 5 km, but the horizon distance from 5642 meters is close to 270 km. That is why there is a formula, and that is why it is necessary to enter both heights. Of course, the formula might be wrong as well. I am just showing you how the calculator works.
So an observer at the peak would see the horizon about 70km beyond Batumi. Observer's height, distance to horizon, curvature drop along the visual line of sight between the horizon and observed object, and common refraction all explain our pictures pretty well.

I am sorry, but I don't understand your point. You gave me the picture. I don't know anything about it. I just told you that the picture is fine. I didn't say you're looking at a mirage. There is no anomalous refraction in it at first glance. This is what I said. The refraction is normal. You're simply changing the angle of view. The refraction is completely normal as there is no distortion in any of the observed objects. You're the one who gave me another calculator claiming it does a better job when in fact it produces the same results. What is the argument about?  I am confused now, and I don't know what we're discussing here anymore. Please stop addressing every word I say, but focus on the big picture.
 
Title: Re: "Equator" problem
Post by: Saros on November 09, 2014, 10:51:29 AM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away? How about buildings which are smaller objects than peaks ;D? Shouldn't the leaning away be visible? From 15-20 km, there should be enough leaning away to be observed, right? I am just saying. Can't give you the exact math.
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 11:54:26 AM
Anyway, speaking of all these examples of evidence, did you ever find any visual proof yet southern stars moving overhead indicative of circling a big disk as opposed to circling a celestial pole of a globe as is commonly seen?

How come that you always turn discussion in wrong (celestial) direction whenever we point to the right direction which is Earth itself?

If we discuss shape of the Earth, why don't you stick with the Earth and her properties instead of hiding behind the celestial phenomena and thus diverging discussion in wrong direction? 

Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

(http://i.imgur.com/D4ZoqyH.jpg)
Title: Re: "Equator" problem
Post by: neimoka on November 09, 2014, 12:14:09 PM
topography not at all exaggerated  ::)
Title: Re: "Equator" problem
Post by: 29silhouette on November 09, 2014, 12:19:43 PM
I am sorry, but I don't understand your point. You gave me the picture. I don't know anything about it.
I gave details.  If you need to know more, just ask.

Quote
I just told you that the picture is fine. I didn't say you're looking at a mirage. There is no anomalous refraction in it at first glance. This is what I said. The refraction is normal. You're simply changing the angle of view.
Moving 20 feet will not change the angle of something 12 miles away enough to result in that much difference.

Quote
The refraction is completely normal as there is no distortion in any of the observed objects.
Except for the compression of the lower objects.

Quote
You're the one who gave me another calculator claiming it does a better job when in fact it produces the same results.
No, I provided a link to a site with calculators that shows the distance to the horizon based on elevation and the amount of curvature drop over a distance.  You're the one who provided another calculator claiming it does a better job when in fact it produces the same results for finding the distance to the horizon, but not the drop over the distance.

Quote
What is the argument about?  I am confused now, and I don't know what we're discussing here anymore. Please stop addressing every word I say,
Hmmm... ok. I'm not addressing every word, just the pertinent sentences.

Quote
but focus on the big picture.
Ok.  Your pictures are possible with a curved surface, and mine indicate a curved surface.
Title: Re: "Equator" problem
Post by: 29silhouette on November 09, 2014, 12:30:10 PM
Anyway, speaking of all these examples of evidence, did you ever find any visual proof yet southern stars moving overhead indicative of circling a big disk as opposed to circling a celestial pole of a globe as is commonly seen?

How come that you always turn discussion in wrong (celestial) direction whenever we point to the right direction which is Earth itself?

If we discuss shape of the Earth, why don't you stick with the Earth and her properties instead of hiding behind the celestial phenomena and thus diverging discussion in wrong direction?
Lol...Because you derailed your own thread onto the discussion of celestial objects with reply#10 back on the first page.   ::)

Quote
Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

http://i.imgur.com/D4ZoqyH.jpg (http://i.imgur.com/D4ZoqyH.jpg)
Go top off an ice-cube tray and get back to us.
Title: Re: "Equator" problem
Post by: ausGeoff on November 09, 2014, 12:46:44 PM
Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

(http://i.imgur.com/D4ZoqyH.jpg)

I'm pleased that you said something "like" this.  Your image is nothing more than a computer-generated rendition of the planet, with a scale perpendicular to its surface that's totally distorted.  And you still seem to think that the water in the oceans and lakes is "level" (or flat?).  It's not.  Its mean surface is curved.

—Please don't bother posting any more George Lucas-style cartoons.

Title: Re: "Equator" problem
Post by: Saros on November 09, 2014, 01:22:48 PM
Btw, have you ever asked yourself how did we get uniform water LEVEL on something like this:

(http://i.imgur.com/D4ZoqyH.jpg)

I'm pleased that you said something "like" this.  Your image is nothing more than a computer-generated rendition of the planet, with a scale perpendicular to its surface that's totally distorted.  And you still seem to think that the water in the oceans and lakes is "level" (or flat?).  It's not.  Its mean surface is curved.

—Please don't bother posting any more George Lucas-style cartoons.

OK, you are entitled to think so, but do you have any proof for that besides mere words? Has any experiment been done to confirm that the ocean surface is curved? For instance, a relatively easy experiment could be extending a straight metal rod across a bay. I have never heard of anything like that. All the evidence that it is curved is mathematical and observational(mainly astronomical). How about measuring the Earth itself?
Title: Re: "Equator" problem
Post by: ausGeoff on November 09, 2014, 01:43:10 PM

OK, you are entitled to think so, but do you have any proof for that besides mere words? Has any experiment been done to confirm that the ocean surface is curved? For instance, a relatively easy experiment could be extending a straight metal rod across a bay. I have never heard of anything like that. All the evidence that it is curved is mathematical and observational (mainly astronomical). How about measuring the Earth itself?

Yep.  The so-called "disappearing ship" illusion.  It proves unequivocally that the surface of the ocean is curved.

You also say yourself that all the evidence that it's curved is "mathematical and observational".  Or are you claiming that observation is insufficient evidence?  Please clarify.
Title: Re: "Equator" problem
Post by: Rama Set on November 09, 2014, 02:21:33 PM
It's called Geodetics and using many techniques they are constantly measuring and remeasing the dimensions of the Earth.

FYI a measurement is an observation.
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 02:24:32 PM
If, for example, the world be the globe of popular belief, it is impossible that there ever could have been a universal flood. For such a thing to have happened, it would be required to blot out the whole universe, to stop the revolution of the globe and to bring confusion and ruin to the whole of the "solar system."

"And the flood was forty days upon the earth; and the waters increased, and bare up the ark, and it was lift up above the earth. And the waters prevailed, and were increased greatly upon the earth; and the ark went upon the face of the waters. And the waters prevailed exceedingly upon the earth; and all the high hills, that were under the whole heaven, were covered. Fifteen cubits upward did the waters prevail; and the mountains were covered" (Gen. 7:17-20).

From these verses many say the Flood was worldwide, and that the tops of the highest mountains (about 29,000 feet) were covered with 15 cubits (about 22 feet) of water. This means the water level would have been five miles above the present sea level.

The water pressure would have been about 800 tons per square inch.


The Duration of the Flood


A second evidence of the global extent of the Flood is its duration. A careful study of the biblical data reveals the fact that the Flood lasted for 371 days. That the Flood continued for more than a year is entirely in keeping with its universality but cannot properly be reconciled with the local-flood theory.

Our imagination indeed staggers at the thought of a flood so gigantic as to overwhelm the highest mountains of the earth within a period of six weeks and then to continue prevailing over those mountains for an additional sixteen weeks, during which time the sole survivors of the human race drifted upon the face of a shoreless ocean! But if the biblical concept of a deluge covering the tops of mountains for sixteen consecutive weeks is hard to reconcile with the local-flood theory, what are we to say of the fact that Scripture records that an additional thirty-one weeks were required for the waters to subside sufficiently for Noah to disembark safely, somewhere in the mountains of Ararat?

Furthermore, it cannot be emphasized too strongly that it was not merely the top of the high mountain on which the Ark rested that was seen on the first day of the tenth month. The Scriptures inform us that on that day “were the tops of the mountains seen.” In other words, the flood waters must have subsided hundreds of feet in order for various mountain peaks of different altitudes to be seen by then (for the mountains were being pushed up from below).

The duration of the Flood in its assuaging, as well as in its prevailing, compels us to think of it as a global, not merely local catastrophe.


The Need for the Ark

Another indication that the Flood was universal is the necessity of the Ark. God told Noah to build the Ark “to keep seed alive upon the face of all the earth” (Genesis 7:3). The whole procedure of constructing this enormous Ark involving, no doubt, many years of planning and toil, simply to escape a local flood, can hardly be described as anything but utterly foolish and unnecessary. God could have told Noah to go on a vacation to Europe or Africa. And rather than sending the animals to the Ark, God could have sent the animals and birds out of the flood zone before the waters reached their highest point. The area could then have been repopulated by creatures spared outside the flood zone. But in a global Flood there would be no survivors among land animals and birds. The Ark was essential.

The Use of Universal Terms

Consider also the repetitive use of universal terms. Sixty times we find in the Flood account the use of such universal terms as “all,” “every,” “in whose nostrils is the breath of life,” and “everywhere under the heaven” (see for example, Genesis 7:19, (22)). While some of these terms are periodically used in the Old Testament in a limited and less than global or universal sense, it is the context that always indicates this. In Genesis 6–9 the context clearly does not limit the meaning of these universal terms. Their repetitious use is emphatic—this was a global Flood.

The Rainbow Covenant

For this short article, a final evidence of the universal extent of the Flood is the Rainbow Covenant in Genesis 9:8–17. Not only does it confirm the supernatural uniqueness of this global catastrophe, it proves its universality. This divine promise was made not just to Noah and his sons but to their families and all their descendants, to the animals and birds and all their offspring and to the earth itself. If the Flood was limited in geographical extent, the Rainbow Covenant has failed (i.e., God lied), for there have been hundreds of devastating local floods since then, which have killed millions of people and animals.

God could not have been more clear in Genesis—this was a unique global catastrophic Flood, an act of divine judgment against a sinful world and, as Jesus said in Matthew 24:37–39, a warning of the coming judgment when Christ returns.
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 02:27:12 PM
"Flood traditions" (the Gilgamesh Epic, the Atrahasis Epich, etc.), even though not as accurate as the Bible, all say the ark came to rest on a mountain. IMPOSSIBLE with a local flood.

Literary parallels to the biblical account

Which came first: the biblical narrative of the Flood, or the Mesopotamian epics? There are three choices:

    the epics were written first, and the writers of Scripture used them;
    the Bible was written first, and the epics copied them;
    both the Bible and the epics were dependent on a primitive original.

Most scholars insist that the writer of Genesis used elements from local epics, but this is impossible to prove. On the other hand, the theory of a primitive original is based on no evidence whatsoever and is simply an opinion of those who hold to it. Although difficult to prove, the preferred choice is that the biblical record came first and inspired the others.

    The Sumerian Deluge Story

    One of the oldest extrabiblical versions of the Flood story featured the survivor of the Flood, Ziusudra. Found in the Nippur excavations early in the twentieth century, it dates to 1600 BC.

    The Gilgamesh Epic-Tablet XI

    A well-known tale, found in Sumerian, Babylonian, Assyrian, Hittite, and Hurrian literature. Even in the Holy Land, a clay tablet (date ca. 1200 BC) was found with this man's name on it. He was the most popular hero in the Ancient Near East. Using the version from Ashurbanipal's library, in 1872, George Smith published the eleventh tablet of the Babylonian Gilgamesh Epic as The Chaldean Account of the Deluge. Gilgamesh's name appears among the kings in the Sumerian Kinglist (below). He was of the first dynasty of Uruk (Erech), the earliest period of Mesopotamian history. The Gilgamesh Epic indicates a close link with events immediately following the Flood. Someone who had survived the Flood still lived, possibly Ham. Gilgamesh visited him seeking immortality.

    Atrahasis Epic

    It has astonishing parallels with the biblical account.


Sumerian Kinglist Part II (Post-Flood).

"After the Flood had swept over the earth and when kingship was lowered again from heaven, kingship was first in Kish. . . . in Uruk (biblical Erech) the divine Gilgamesh . . . ruled 126 years . . . its kingship was removed to Ur" (at the peak of its glory). Note that Kish was the first city established after the Flood. Excavations there indicate it was founded about 3000 BC. "Divine" Gilgamesh listed above, actually visited a survivor of the Flood family (see Tablet XI of the Gilgamesh Epic). Therefore, Gilgamesh must have reigned shortly after the Flood regardless what the Kinglist says. There are many other worldwide records of the Flood story in: The Samaritan Pentateuch, Jewish Targums, Berossus, Josephus, the Sibylline Oracles, the Koran, etc.


Radioactive Dating Methods: How are they calibrated?

Although the equipment used to date radioactive materials has become more sophisticated through time, basic problems originally discovered by Willard Libby, inventor of the C14 dating method, still pertain. Calibrated using known dates of Egyptian tomb artifacts, it has proven somewhat accurate back to only about 2000 BC. This has created problems for radio carbon dating older than 5000 BP (Before Present). Dates earlier than that cannot be calibrated since there is no historical material older than 5000 BP. W. Libby himself said: "The first shock Dr. Arnold and I had was that our advisors informed us that history extended back only 5000 years. We had initially thought that we would be able to get samples along the curve back to 30,000 years, put the points in, and then our work would be finished . . . We learned rather abruptly that these numbers, these ancient ages are not known; in fact, it is about the time of the first dynasty in Egypt that the last [earliest] historical date of any real certainty has been established"(Libby 1958: 531). Furthermore, as Libby makes clear in his publication, all "dates" higher than 5000 years BP are not absolute dates, but only measure residual C14. Dendrochronology does not help, either, since under certain conditions trees can grow two and sometimes three rings a year.
Title: Re: "Equator" problem
Post by: cikljamas on November 09, 2014, 02:28:15 PM
River Deltas Begin Forming Worldwide about 3000 BC.

Only the worldwide Flood was such a stupendous catastrophe as to make it possible for rivers worldwide to begin flowing at about the same time water on the landmass subsided into deepened oceans, rain fell, and rivers began depositing sediments at their mouths to form deltas. Investigations of these deltas worldwide have revealed that they are only a few thousand years old. The Tigris and Euphrates delta is formed in the Persian Gulf. Many maps of the earliest periods of history show the shoreline as far north as Ur. That means the delta has filled in at least 150 miles during recorded times. Herodotus, the Greek historian, reported that Egyptian priests told him none of the land north of Lake Moeris was above water at the beginning of the First Dynasty (p. 104). The Mississippi River delta was investigated in 1850 and found to be only 40 feet in depth. It has not been flowing very long. One other time-measuring feature -- Niagara Falls -- began falling and receding from Lake Ontario toward Lake Erie, less than 10,000 years ago. The point is that none of these rivers could have been flowing for more than a few thousand years.


Problems with an Early Date

    If the Flood occurred as late as 10,000 BC, one cannot find a 7000 year (or larger) gap in Scripture, or in any of the literature of the Ancient Near East, for that matter, between the Flood and the beginning of historical records from 3000 BC.
    Nor can an explanation be found for the origin of families (nations) mentioned in Genesis 10-11.
    Cush was the grandson of Noah. The descendants of "Cush" built cities whose foundations date no earlier than 3000 BC in almost all cases (Genesis 10). Cities that are claimed to be older: Jericho (7000 BC), Jarmo (6000 BC), etc., were dated by C14 which cannot be calibrated by absolute dates earlier than 5000 years before the present. More caution should be used when considering these early dates.
    Ziggurats and pyramids are later than 3000 BC. If there were earlier civilizations, there is no trace of anything like ziggurats or pyramids at that time. A short time obviously elapsed between the Flood and their construction. But 7000 years? That is longer than the entire history of man since the Flood.
    Geneologies in Genesis 5 and 10 may be stretched slightly, but they cease to be geneologies if large gaps exist. Gaps of 7000 years make them meaningless for genealogical purposes.

Problem with a Late Date

The date of the Great Flood in relation to local floods in the Mesopotamian river basin is, at the present, impossible to determine since a universal Flood completely altered the surface of the earth. However, strong evidence given above suggests a date not long before 5000 BC.

Conclusion: The Flood Occurred 5000 years ago


    1.C14 is not useful in dating before 5000 B.P. according to the discoverer of the method.
    2.River deltas suggest a recent (ca. 3000 BC?) flood.
    3.All written history begins ca. 3000 BC.
    4.Foundations of cities began then.
    5.Families of mankind began then. Geneologies date back to it.
    6.A 10,000 BC (or earlier) flood wreaks havoc with geneologies.
    7.There is no record of a 10,000 BC flood in ANY of the literature.
    8.The Gilgamesh Epic (and other epics) fit well into a 3000 BC date.
    9.The biblical account did not derive from other literature. It is eyewitness testimony.
    10.It is clear from the biblical account that there was a universal flood about 3000 BC.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 09, 2014, 02:43:30 PM
Are you sure it should be leaning only slightly away? How about buildings which are smaller objects than peaks ;D? Shouldn't the leaning away be visible?
How?  From 200 km you just get an outline anyway - no depth perception.  As I say, if there is any lean, what would it look like?

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From 15-20 km, there should be enough leaning away to be observed, right?
No, the effect would be minute at that distance, that I'm sure about.

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I am just saying. Can't give you the exact math.
No, I can't be arsed to do the maths either, but I might have a crack later.  Or anyone else?
Title: Re: "Equator" problem
Post by: Socratic Amusement on November 09, 2014, 03:04:35 PM
cikljamas, could you focus on science, and leave your fairy tales for mythology fans?

Thanks.
Title: Re: "Equator" problem
Post by: ausGeoff on November 10, 2014, 02:15:22 AM
    1.C14 is not useful in dating before 5000 B.P. according to the discoverer of the method.
    2.River deltas suggest a recent (ca. 3000 BC?) flood.
    3.All written history begins ca. 3000 BC.
    4.Foundations of cities began then.
    5.Families of mankind began then. Geneologies date back to it.
    6.A 10,000 BC (or earlier) flood wreaks havoc with geneologies.
    7.There is no record of a 10,000 BC flood in ANY of the literature.
    8.The Gilgamesh Epic (and other epics) fit well into a 3000 BC date.
    9.The biblical account did not derive from other literature. It is eyewitness testimony.
    10.It is clear from the biblical account that there was a universal flood about 3000 BC.

Can you please refrain from posting repeated, irrelevant comments of a religious nature that have nothing to do with science—either the flat earth version and/or the round earth version.

There is NO place for discussions about supernatural entities or paranormal phenomena, or the Christian bible on any/all science-based forums.

If you insist on discussing your obsessive, unevidenced religious speculations about our planet, then can I suggest you join something like ReligiousForums.com (http://bit.ly/1yp1Slb) which is a group of like-minded, fundamentalist, creationist religious apologetics?

Your rigid, absolute and blind acceptance of the Abrahamic bible as being some sort of literal scientific [sic] record of the planet's geophysical  development is in reality absurd from the viewpoint of anybody with even a basic understanding of the earth sciences.

Title: Re: "Equator" problem
Post by: cikljamas on November 10, 2014, 02:55:06 AM
"The cosmos is all that is, or ever was, or ever will be. Come with me." With those words in 1980 the astronomer Carl Sagan launched "Cosmos," an epic 13-part TV series that brought science to the public like never before, and opened up all of space and time to exploration.

A generation later, Sagan's legacy lives again in "Cosmos: A Spacetime Odyssey," a 21st-century reboot premiering on Fox tonight (March 9). The new "Cosmos" (the original was billed as "A Personal Journey") updates its predecessor with a blend of spectacular visual effects and the latest astronomical discoveries."

Einstein, Carl Sagan, Neil deGrasse Tyson, Lawrence Krauss, Stephen Hawking, Brian Greene, etc. are dope dealers, and you are duped dope addicts! If you don't believe me watch this:

Fatal Errors of Big Bang Cosmology : (http://)
Cosmological Constant : Cosmological Constant (http://#)
Michio Kaku - Quantum Mechanics vs. General Relativity :  (http://)

If you have enough courage to face the real truth then just open above links, and smoke of marijuana will disperse! As simple as that!

Now, some real science:

http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637695#msg1637695)

The most important geological discovery in the history of the world that has been covered up and still being covered up: ARK on Mt. Ararat: WHY the media BLACKOUT on the real history of Ararat? (http://#ws)

Now, i believe that you are much better prepared to to grasp the trueness of these conclusions:

http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637759#msg1637759)

" M. Bouquet de la Grye, an eminent hydrographical Engineer, has after long yearsof study calculated the atmospheric expansions and depressions which coincide with spring and neap tides. There have been cases in which air was moved in waves of 133 yards high, and in places where the barometrical pressure was seven-tentns ot an inch, ot six and a half miles. Near the upper surface of the earth's atmosphere condensations and dilations of this magnitude are trequent. The human nervous system may be said to register these air waves. We are only aware that they do so by the discomfort which we feel. The earth also registers them and to its very centre. The incandescent and fluid matter under the earth's crust acts in concert with the air and sea at the full of the moon. In 1889 a German Scientist, Dr. Rebeur Pachwitz, thought he noticed at Wilhelmshaven and Potsdam earth oscillations corresponding with the course of the moon. He wrote to the observatory at Tenerife asking for observations to be ma.de there in December, 1890 and April, 1891, which would be propitious times for them. From these observations and others simultanously made in the sandy plains round Berlin, IT WAS ESTABLISHED THAT THE Earth RISES AND FALLS LIKE THE OCEAN OR THE ATMOSPHERE. The movements, common to them all, may be likened to the chest in breathing. — Paris Correspondent Weekly Dispatch."

This is the answer to the question. Tides are caused by the gentle and gradual rise and fall of the earth on the bosom of the mighty deep. In inland lakes, there are no tides ; which also proves that the moon cannot attract either the earth or water to cause tides. But the fact that the basin of the lake is on the earth which rests on the waters of the deep, shows that no tides are possible, as the waters of the lakes together with the earth rise and fall, and thus the tides at the coast are caused ; while there are no tides on waters unconnected with the sea.

If above description reveals the true mechanics of tides, could someone explain to me how could the alleged Globe have remained intact instead of being disintegrated under such (great deluge) cataclysmic circumstances (The water pressure would have been about 800 tons per square inch), and how about the alleged rotation of the Earth, how about anything that you still believe presents a true description of reality?

And you know, great deluge have happened, and Jesus Christ have lived and walked on the Earth, not on a damn Globe!

You don't have to belive me, just study and follow the hystorical and real scientific evidences!

STOP WEED SMOKING! And answer the questions if you can!!!

And the question is:

Either there wasn't great deluge (in which case you have to show us compelling historical and scientific evidences for that), or if such a cataclysmic event had happened then you have to be able to consistently conciliate next two assertions:

1. Great deluge had happened!
2. Great deluge could not have happened on a round Earth!

Do i ask too much of you?
Title: Re: "Equator" problem
Post by: ausGeoff on November 10, 2014, 03:14:20 AM
Einstein, Carl Sagan, Neil deGrasse Tyson, Lawrence Krauss, Stephen Hawking, Brian Greene, etc. are dope dealers, and you are duped dope addicts!
I can only assume that you've posted this puerile comment as some sort of joke?  Or are you really that far out of your depth that you have to resort to childish ad hominem attacks on some of the greatest scientists the world's ever known.

I also say this partly because you then quote Jean Jacques Anatole Bouquet de la Grye, an engineer who died more than 100 years ago as some sort of authority!

You're obviously unaware that science advances logarithmically over time LOL.


Title: Re: "Equator" problem
Post by: cikljamas on November 10, 2014, 03:41:00 AM
1. The obvious fudging of the data by Eddington and others is a blatant subversion of scientific process and may have misdirected scientific research for the better part of a century. It probably surpasses the Piltdown Man as the greatest hoax of 20th-century science. The BIPP asked, "Was this the hoax of the century?" and exclaimed, "Royal Society 1919 Eclipse Relativity Report Duped World for 80 Years!" McCausland stated that "In the author's opinion, the confident announcement of the decisive confirmation of Einstein's general theory in November 1919 was not a triumph of science, as it is often portrayed, but one of the most unfortunate incidents in the history of 20th-century science".

It cannot be emphasised enough that the Eclipse of 1919 made Einstein, Einstein.
It propelled him to international fame overnight, despite the fact that the data were fabricated and there was no support for general relativity whatsoever. This perversion of history has been known about for over 80 years and is still supported by people like Stephen Hawking and David Levy.

2.
So, after conning us with the faked Moon missions we are being conned again with coming faked Mars missions, after all black military projects are great devourers of money, and in order to be better monitored by our slave owners we have to provide them (NASA and co.) with stupendous amount of money so that they can establish absolute control over us! Read more: http://www.energeticforum.com/renewable-energy/17050-north-south-3.html (http://www.energeticforum.com/renewable-energy/17050-north-south-3.html)

Universe The Cosmology Quest Part 1 of 2.wmv : (http://)

STOP WEED SMOKING!

Title: Re: "Equator" problem
Post by: ausGeoff on November 10, 2014, 03:47:07 AM
STOP WEED SMOKING!

Man..... this guy never gives up does he?

    ;D    ;D    ;D

I guess the laughs are worth it.  Or maybe not?

Title: Re: "Equator" problem
Post by: cikljamas on November 10, 2014, 05:03:07 AM
Smoking, joking, that's all fine with me, but you still have to face the truth, and after that we can all keep smoking and joking as long as you wish...  ;)

Since the great deluge is to hard nut for you, how about one another hard nut:

The visual obstacle from Tunguska measures 7463 km; we are told that the rays of light from the Sun (and it was morning over Siberia on June 30, at 7:20 am) cannot reach, for example, London, at the same time, due to the curvature; then NOTHING could have been observed/seen from Tunguska as well on a globe; an explosion on one side of a globe could not possibly influence in any way visual observations on the other side of the same globe; the visual range limit for the Tunguska explosion, on that cloudless day, is just 400 km.

Newspapers could be read at midnight in London, photographs could be taken outdoors in Stockholm without flash apparatus; no other meteorological/astronomical phenomenon occurred at that time in the world, no such records exist.

That is why this is the very best proof that the surface of the Earth is actually flat.

Talking about the aurora borealis, which stems from the solar particles hitting the upper atmosphere, presents such a terrible, awful example...  The sun is a much larger object than the particular body which caused the event.  The event occurred in only one area, and the light spread outwards.  light only, not particles like solar flares.

You are trying to change the subject from the event to the aurora borealis where you just say, "see, it has lights."

Instead of focusing on a sun with a greater distance from earth (seen by everyone in FE and RE theory alike) be realistic and focus on the explosion that occurred at 7 km.  That should not have been seen.  Neither should the light from the area have been seen.

Auroras are unrelated.

The Aurora and this event are so unrelated and different that your comparison makes no sense.  Please note the different colors of an aurora borealis, please note the wave like shapes.  Also notice, how an aurora never gets as bright as that.

Once again, solar particles, which hit a large area from solar flares, and impact areas with temperatures totaling millions of degrees stand completely different from each other.  The sun, being seen by everyone, can emit particles that affect large areas.  That scenario just poses a big DUH...  Tunguska was just one localized event within the earth's atmosphere and close to the ground.

Provide just one shred of evidence that an impact from an object and solar particles hitting the upper atmosphere are the same scenario.

The explosion came from just one area...  Solar particles bombard many different portions of the atmosphere at the same time...

..."The sun lies in the open to everyone.  Its rays and flares touch all directions." No, they do not. The particles that create aurora follow specific paths along the Earth's magnetic field lines. Just because sunLIGHT touches all directions, does not mean aurora is a diffuse phenomena. That reasoning is erroneous.

You mean it could not have been visible on a RE, not FE, but I know what you meant.

"In addition, the event lacked UV light and radiation.  No solar radiation = no similarity." UV light and radiation are the same thing...radiation IS light. The way you are discussing this is imprecise. You believe solar radiation consists of UV light only? You should learn what solar radiation IS, and I would be very interested to hear what you know about the spectra of aurora and how THAT SPECIFIC spectra compares with the corresponding waveband of THAT SPECIFIC spectra in the explosion. You must compare the total energy, its distribution, and propagation in this waveband. You must be specific, avoiding qualitative comments such as "it doesn't add up" and instead using quantitative comments like: "if you compare the spatial distribution of the total energy in the range corresponding to ____ to ____ nm, etc..."

This is how we should compare it. Only by addressing the specifics can we lend any credibility to the FE view.

Read more: http://www.theflatearthsociety.org/forum/index.php?topic=60042.125;wap2 (http://www.theflatearthsociety.org/forum/index.php?topic=60042.125;wap2)
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 10, 2014, 09:28:21 AM
I recognize that several posts have been made between your post and this response. This response applies to the post as originally made.

Alpha2Omega, one day you will be one of the best FET apologist, this day is not so far away...

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"So you're technically right that neither of these things is 'proven', but, for the same reason, you'll never be able to 'prove' your model, either."

Sorry, but i am not just technically right, i am absolutely right!

Had there been any way to prove that the Earth is submitted to any kind of motion, scientists would have supplied us with these proofs up until now, and by doing this they would have provided immortal fame for themselves.
I kind of figured you'd latch on to that sentence and ignore the rest.

Fine. You're right that these can't be 'proven'; that's what I said. By the same token you can't 'prove' earth doesn't move, either. You can offer evidence, but you can't 'prove' anything. To see why, read the rest of that part of the post.

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How about the proofs for the immobility of the Earth?
There are none. But you're going to offer what you consider evidence.

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Allow me to present you some of them:

1. Observing the sun directly from the north pole the apparent motion of the sun would be straight line for days, and a camera should have to be slightly adjusted every few hours to cancel out scarcelly perceptible effect due to Earth's alleged rotation which speed is practically zero at North Pole.

Why haven't we seen until this day any such video which could have easily proved Earth's daily rotation? Lack of such video speaks for itself, and presents indirect but strong proof that there is no reason for making such video because such video would prove something else (that the Earth is at rest), and scientists are aware of that truth very well!
Really? Have you tried Google? It's pretty cool and very useful for finding stuff like

midnight sun video

First hit is a link to a youtube video of 24 hours of direct sunlight from the Arctic.

The earth spins at exactly the same angular rate everywhere (360° / sidereal day); the tangential velocity depends on latitude, but the angular rotation is what matters here, and is everywhere the same.

I suppose I am grateful that you didn't simply say "all scientists are liars", so thank you for that. You obviously still mean it, and are still wrong, but this is an improvement, I think.

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2. If the Earth was rotating about its axis, someone in Quito, Ecuador would be traveling twice as fast from west to east as someone in Oslo, Norway – at any moment, and at every moment. Meanwhile, someone looking at from the proverbial North Pole, would hardly be moving at all! But is that reality?
Is the correction above (removed replaced) what you meant to say? If not, please advise.

Yes, it is real if you're talking about tangential velocity. If you're talking about angular velocity, then no, they are all the same. When viewing distant celestial objects, the tangential velocity is insignificant, but angular velocity is very significant.

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Of course it is not reality, but this supposed fact of Earth's rotation now becomes deadliest error of all, concerning supposed differences of Earth rotational speeds at different latitudes.

If these differences were really the true fact then the speed of apparent motion of all celestial bodies would be twice greater for any observer on the equator than it would be for any observer on the latitude of Oslo.
No. Apparent motion of celestial bodies doesn't depend on the tangential velocity of the observer. It depends only on the angular rate the observer's frame of reference is changing with respect to what he's looking at. This, as already noted, is the same everywhere on earth - Oslo, Quito, north pole.

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How hard would be to make an experiment (measurement) of such kind???
It's not hard at all and been done many time with the expected result: the earth rotates exactly as proposed.

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3a) If the atmosphere were independent (non rotating but static) from Earth's daily rotation then we would have on the surface of the Earth permanent winds that blow 600 to 1600 km/h. Do you notice permanent winds which blow at such a speed?
No such winds are observed, as you well know, therefore the conjecture is false.

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3b) If the atmosphere were rotating along with the Earth the air flow at the surface of the Earth would have variable velocity (not the thermal), variable pressure (not the static), and variable density (not the normal). Such air flow and such air pressure regimes do not exist: http://www.energeticforum.com/256388-post62.html (http://www.energeticforum.com/256388-post62.html)
In that linked post, you seem to posit supersonic winds near the equator, Mach 1 winds near 45° latitude, and subsonic winds at higher latitudes. Isn't that the same model as 3a), which we both agree is not correct?

If your point in that post is that relative to inertial space, that air coupled to the rotation of the Earth is physically moving at various Mach numbers, well, so what? The ground (and your measuring device) is also moving at the same velocity, so the relative velocity between them (earth and air), which is what matters, is typically a gentle zephyr due to thermal effects. Your pitotstatic tube is going to register the gentle zephyr, not a supersonic gale. Since the instrument illustrated in Fig. 2.1, attached to the Earth, is moving in tandem with the air, which is coupled to and not moving with respect to the earth (neglecting slight winds), there is no Flow, and V = 0.

BTW the term "perpendicular to motion" at the top of Fig. 2.1 of the linked post is wrong. It's "parallel to motion". Note that the white arrows assigned to the former are parallel to the latter, so one can't be perpendicular to motion and the other parallel to motion; they are both going the same way. If you want anyone competent to buy into what you're saying, you're at least going to have to get the basics right, which takes some effort and understanding. If you're just looking to "sell" something to unknowing people, please stop questioning other people's motives.

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4. If the Earth were suffering a daily rotation it would generate an incredible deflection on all flying matters in air atmosphere. 5) The inertial motion is terminated in air atmosphere, and thus all dropped objects should land behind their starting positions on a rotating Earth. http://www.energeticforum.com/255938-post21.html (http://www.energeticforum.com/255938-post21.html)
"inertial motion is terminated in air atmosphere" Why do you think this is true? Did you read it somewhere? Do you have any evidence that this is actually the case?

Air applies drag against objects moving within it, which resists the motion. Inertia is preserved, but slowly transferred from the moving object to the air.

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5. To be pression or to be gravity? The choice of Earth’s rotation (the cause of pression), should repel the gravity from Earth. Consequently, the heliocentric model looses the most precious element. The choice of gravity should remove the concept of Earth’s rotation from the cosmos motion, consequently; the journey of the Earth around the sun becomes useless since half of the Earth should be always in darkness and the second half should be always in lightness.
Presson? What does this mean? It's French for "pressure" (I think), but even with that substitution this still reads like nonsense. Can you restate this?

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6. No experiment has ever been performed with such excruciating persistence and meticulous precision, and in every conceivable manner, than that of trying to detect and measure the motion of the Earth. Yet they have all consistently and continually yielded a velocity for the Earth of exactly ZERO mph.
Stating this, even using florid language, doesn't mean it's true. Are you quoting Rowbotham again?

There are many experiments that support a rotating earth, not least, the variation of gravity as a function of latitude.

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The toil of thousands of exasperated researchers, in the extremely varied experiments of Arago, De Coudre's induction, Fizeau, Fresnell drag, Hoek, Jaseja's lasers, Jenkins, Klinkerfuess, Michelson-Morley interferometry, Lord Rayleigh's polarimetry, Troughton-Noble torque, and the famous 'Airy's Failure' experiment, all conclusively failed to show any rotational or translational movement for the earth, whatsoever."

http://www.sacred-texts.com/earth/za/za21.htm (http://www.sacred-texts.com/earth/za/za21.htm)
I don't see any of those names in the cited link.

The linked article is so wrong it would be funny, except way too many people might accept it because of the Victorian-Era language and a desire to disbelieve what competent high-school physics shows, perhaps because they weren't any good at high-school physics.

The fundamental flaw is here (from your linked text):
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Let the ball be thrown upwards from the mast-head of a stationary ship, and it will fall back to the mast-head, and pass downwards to the foot of the mast. The same result would follow if the ball were thrown upwards from the mouth of a mine, or the top of a tower, on a stationary earth. Now put the ship in motion, and let the ball be thrown upwards. It will, as in the first instance, partake of the two motions--the upward or vertical, A, C, and the horizontal, A, B, as shown in fig. 47; but

FIG. 47.

because the two motions act conjointly, the ball will take the diagonal direction, A, D [actually, it will follow a parabolic path a from A to D despite the incorrect assertion in the final paragraph in this section]. By the time the ball has arrived at D, the ship will have reached the position, [ B]; and now, as the two forces will have been expended, [hold it right there! The upward motion will be 'expended' because gravity is causing a downward acceleration, but the horizontal component continues as before - what is resisting it?] the ball will begin to fall, by the force of gravity alone, in the vertical direction, D, B, H; [no, it won't; it will accelerate downward, but still continues to the right because nothing (except drag, which is gradual, not abrupt) is slowing the motion in that direction] but during its fall towards H, the ship will have passed on to the position S, leaving the ball at H, a given distance behind it [no, it will meet the ship at S, because there's no reason that motion to the right will suddenly stop at D - why would it?].

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7. NO CAUSE OF EARTH'S ROTATION WHATSOEVER: retains the state of illusion. The most important element in heliocentric model is the Earth’s rotation about its polar axis. What is the cause of Earth’s rotation? No one has attributed the cause of Earth’s rotation to any type of action or force even though they have attributed the cause of orbital motion (revolution) to Newton’s law of gravity.
Conservation of momentum from the formation of the solar system.

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8. NO CAUSE OF THE ROTATION OF THE AIR-LAYER: 2) The rotation of the air-layer next to the rigid Earth is without cause, and lacks a technique and tool. Perhaps, one may envision the whole rigid sphere undergoes a rotation about its polar axis. But, how one can envision the air atmosphere (the surface layer) rotates with the rigid sphere without an engineering method (e.g.air foil). In addition, what maintains the air’s rotation for tens of thousands of years (we are practical people) without stop. The rotation of the background air is the greatest hoax ever invented by mankind.
Viscous drag from the surface. If you take a glass of water and rotate the glass a few times, the water will overcome its inertia and rotate with it. This isn't mysterious and easily seen.

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9. CONCLUSIVE INFERENCE ABOUT THE EQUATION OF TIME ISSUE: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637435#msg1637435)
You're still confused about what that graph means. Already refuted here (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1637299#msg1637299). Stamping your foot and insisting "I'm right, all the rest of you are wrong" is called a tantrum, not proof.

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Arthur Eddington dared to contemplate that:

    "There was just one alternative; the earth's true velocity through space might happen to have been nil."

Lincoln Barnett agrees:

    "No physical experiment ever proved that the Earth actually is in motion."
Where are these quotes from? They sound like they're taken out of context.

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So, when all attempts to prove any kind of motion of the Earth FAIL, what does it mean?

It means that the contrary is the fact: The Earth is at rest!
It means no such thing. Failure to 'prove' the earth moves is not itself 'proof' that earth is at rest. See why scientists refrain from saying they can prove something a few posts ago (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1638279#msg1638279) after your points 4. and 5. The evidence (not 'proof') is strongly in favor of a moving earth, and weak to nonexistent for a fixed earth.

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Every failure of all these attempts presents the proof to the contrary : The Earth is at rest!
Negative. Failure to prove something doesn't automatically prove an alternative. You should know that.

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If you don't want to make me laugh, then you cannot just say: O.K., we have failed to prove Earth's motion but there is still some chance that we could succeed to prove it in some distant future?

In how distant future? When we inhabit another galaxy? Come on, cut the crap, please![/b][/u]
See why scientists refrain from saying they can prove something a few posts ago (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1638279#msg1638279) after your points 4. and 5..
 
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Anyway, if you had used the correct figure, it would have made your next point stronger (but still wrong).

EXACTLY! Thanks for helping me make it TWICE stronger!

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What does "direction of the stars will have greatly changed, however small the angle of parallax maybe" mean? Isn't "direction ... will have greatly changed" the opposite of "however small the angle"?

Here is the answer:

In the " History of the Conflict between Religion and Science," by Dr. Draper, pages 175 and 176, the matter is referred to m the following words :

" Among the arguments brought forward against the Copernican system at the time of its promulgation, was one by the great Danish astronomer, Tycho Brahe, originally urged by Aristarchus against the Pythagorean system, to the effect that if, as was alleged, the earth moves round the sun, there ought to he a change in the relative position of the stars ; they should seem to separate as we approach them, or to close together as we recede from them... At that time the sun's distance was greatly under-estimated. Had it been known, as it is now, that the distance exceeds 90 million miles, or that the diameter of the orbit is more than 180 million, that argument would doubtless have had very great weight. In reply to Tycho, it was said that, since the parallax of a body diminishes as its distance increases, a star may be so far off that its parallax may be imperceptible. THIS ANSWER PROVED TO BE CORRECT."

To the uninitiated, the words " this answer proved to be correct," might seem to settle the matter, and while it must be admitted that parallax is diminished or increased according as the star is distant or near, parallax and direction are very different terms and convey quite different meanings. Tycho stated that the direction of the stars would be altered ; his critics replied that the distance gave no sensible difference of parallax. This maybe set down as ingenious, but it is no answer to the proposition, which has remained unanswered to this hour, and is unanswerable.
What's the problem? You're still wrong. Even with a 180-million-mile baseline (instead of your erroneous 90 million), the stars are so distant that parallax was imperceptible to Tycho. With better instruments and better techniques, stellar parallax is no longer imperceptible. That last paragraph makes little sense. Exactly what is meant by 'direction' that is different from 'parallax' in this context? Their positions appear to shift as we move in orbit; that's parallax.

Why is this difficult to understand?

Stamping your foot and insisting "I'm right, all the rest of you are wrong" is called a tantrum, not proof.

I guess we're done with your objection to Crux being circumpolar. Remember, you said yourself that it was.

Have you solved the apparent retrograde motion of the outer planets yet?

[Edit] Minor corrections and formatting fixes needed due to hitting Post instead of Preview.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 10, 2014, 10:46:22 AM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away?
Yes, and I've done the maths now.  Using this formula:

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)

so using a distance of 200 km and earth radius of 6371 km:

angle = 2 x arcsin (0.5 x 200 / 6371)

angle = 1.9 degrees


This would obviously be imperceptible whilst looking at a mountain range 200km away.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 10, 2014, 11:08:26 AM
Quote
These models (you call them "assumptions") are verified by actual observations and measurements as we go, though. Sometimes better observations result in better measurements that require earlier models to be revised or occasionally thrown out altogether. This is how science works.

Bull s h i t! Don't call this crap "science"? What actual observations are you talking about?

Gravitation is a clever illustration of the art of hocus-pocus—heads I win, tails you lose ; Newton won his fame, and the people lost their senses.

You have presented no evidence what I said was wrong, instead you just call it "BS", "crap", "hocus-pocus" and (again) declare victory. This is followed by more quotes out of context and the repetition of an unsubstantiated report that Polaris was visible to "a noble men" [sic] from well south of the equator.

Get a grip.
Title: Re: "Equator" problem
Post by: Saros on November 10, 2014, 01:57:49 PM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away?
Yes, and I've done the maths now.  Using this formula:

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)

so using a distance of 200 km and earth radius of 6371 km:

angle = 2 x arcsin (0.5 x 200 / 6371)

angle = 1.9 degrees


This would obviously be imperceptible whilst looking at a mountain range 200km away.

I think you're mistaken. In your opinion the height of the observed objects doesn't matter? It is not imperceptible. This sounds counterintuitive.
Title: Re: "Equator" problem
Post by: Socratic Amusement on November 10, 2014, 03:47:10 PM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away?
Yes, and I've done the maths now.  Using this formula:

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)

so using a distance of 200 km and earth radius of 6371 km:

angle = 2 x arcsin (0.5 x 200 / 6371)

angle = 1.9 degrees


This would obviously be imperceptible whilst looking at a mountain range 200km away.

I think you're mistaken. In your opinion the height of the observed objects doesn't matter? It is not imperceptible. This sounds counterintuitive.

Lots of things are counter-intuitive.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 10, 2014, 04:31:23 PM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away?
Yes, and I've done the maths now.  Using this formula:

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)

so using a distance of 200 km and earth radius of 6371 km:

angle = 2 x arcsin (0.5 x 200 / 6371)

angle = 1.9 degrees


This would obviously be imperceptible whilst looking at a mountain range 200km away.
It's easier than that.

angle = d / r
 = 200 / 6371
 = 0.0314 radians

Since a full 360 degrees is 2 pi radians, 180 degrees is pi radians. To convert from radians to degrees, multiply by 180 and divide by pi (3.14159...).

angle = 0.0314 radians * 180 degrees/pi radians
 = 1.80 degrees

I got 1.8 degrees using your formula, too, not 1.9. This uses 200 km as the distance along the surface; your formula uses 200 km as the chord. The difference is about 8 meters in this case, so the difference is waaayyyyy down in the noise. How good is that "200 km", after all, and 6371 is an average value for earth radius.
Title: Re: "Equator" problem
Post by: 29silhouette on November 10, 2014, 06:01:38 PM
How good is that "200 km", after all,
The picture showing the pier, building, etc, shows where the spot is on GE to within a meter or two, and I get 200.22km with the ruler tool, measuring to the mountain 'info icon' at the peak.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 10, 2014, 06:13:48 PM
How good is that "200 km", after all,
The picture showing the pier, building, etc, shows where the spot is on GE to within a meter or two, and I get 200.22km with the ruler tool, measuring to the mountain 'info icon' at the peak.

Thanks. 8 meters is definitely "in the noise".
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 11, 2014, 01:57:52 AM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away?
Yes, and I've done the maths now.  Using this formula:

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)

so using a distance of 200 km and earth radius of 6371 km:

angle = 2 x arcsin (0.5 x 200 / 6371)

angle = 1.9 degrees


This would obviously be imperceptible whilst looking at a mountain range 200km away.
It's easier than that.

angle = d / r
 = 200 / 6371
 = 0.0314 radians

Since a full 360 degrees is 2 pi radians, 180 degrees is pi radians. To convert from radians to degrees, multiply by 180 and divide by pi (3.14159...).

angle = 0.0314 radians * 180 degrees/pi radians
 = 1.80 degrees

I got 1.8 degrees using your formula, too, not 1.9.
Ooop, my mistake - typo.

Title: Re: "Equator" problem
Post by: Saros on November 11, 2014, 02:11:55 AM
@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?
.
Are you sure it should be leaning only slightly away?
Yes, and I've done the maths now.  Using this formula:

angle = 2 x arcsin (0.5 x |P1 - P2| / radius)

so using a distance of 200 km and earth radius of 6371 km:

angle = 2 x arcsin (0.5 x 200 / 6371)

angle = 1.9 degrees


This would obviously be imperceptible whilst looking at a mountain range 200km away.
It's easier than that.

angle = d / r
 = 200 / 6371
 = 0.0314 radians

Since a full 360 degrees is 2 pi radians, 180 degrees is pi radians. To convert from radians to degrees, multiply by 180 and divide by pi (3.14159...).

angle = 0.0314 radians * 180 degrees/pi radians
 = 1.80 degrees

I got 1.8 degrees using your formula, too, not 1.9.
Ooop, my mistake - typo.

You guys are very confused. Using a formula you get some results and of course you believe them immediately. It is mathematics after all. This formula is wrong, and it is common sense that if there is a curvature the objects will lean more especially if they are tall enough when observed from a big distance. You go tell this 1.9% to someone else, I am not buying that math, as it is wrong to anyone with half a brain, but you apparently. That might be useful in high school to get an A but not in real life. Not to mention, that I was misleading you by claiming that you can't see they are leaning away. I just wanted to see how confused you are. You can see they are leaning away, but you guys are so big fans of math that you just proved you know nothing about reality, and supposedly you just proved that any leaning away shouldn't be visible, when in fact it is. Good job! Great math. Great logic. Congratulations!!! Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet, because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 11, 2014, 02:35:12 AM


I think you're mistaken.
Then demonstrate as much.

Quote
In your opinion the height of the observed objects doesn't matter? It is not imperceptible. This sounds counterintuitive.
Intuition is a terrible tool to use in these situations - which is why I used maths.  The human brain simply isn't evolved to deal with problems like this intuitively.
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 11, 2014, 02:38:25 AM
You guys are very confused. Using a formula you get some results and of course you believe them immediately. It is mathematics after all. This formula is wrong, and it is common sense that if there is a curvature the objects will lean more especially if they are tall enough when observed from a big distance. You go tell this 1.9% to someone else, I am not buying that math, as it is wrong to anyone with half a brain, but you apparently. That might be useful in high school to get an A but not in real life. Not to mention, that I was misleading you by claiming that you can't see they are leaning away. I just wanted to see how confused you are. You can see they are leaning away, but you guys are so big fans of math that you just proved you know nothing about reality, and supposedly you just proved that any leaning away shouldn't be visible, when in fact it is. Good job! Great math. Great logic. Congratulations!!! Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet, because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.
You lose the debate, so you throw your toys out the pram?

If you think 1.8% is wrong for the predicated lean, then offer another figure, obviously based on "on real, first-hand observations, experiments and analysis".
Title: Re: "Equator" problem
Post by: Saros on November 11, 2014, 02:41:19 AM


I think you're mistaken.
Then demonstrate as much.

Quote
In your opinion the height of the observed objects doesn't matter? It is not imperceptible. This sounds counterintuitive.
Intuition is a terrible tool to use in these situations - which is why I used maths.  The human brain simply isn't evolved to deal with problems like this intuitively.

So the human brain is not evolved enough. Amazing revelation which, of course, your brain made about itself.

Does mathematics come from an alien brain somehow then?

Are you saying that when you use mathematics you're not using your brain(because it is not evolved enough), because that is what it sounds like.

You do realize that without your brain you can use neither math nor intuition?
Title: Re: "Equator" problem
Post by: neimoka on November 11, 2014, 03:27:48 AM
Quote from: JimmyTheCrab
If you think 1.8% is wrong for the predicated lean, then offer another figure, obviously based on "on real, first-hand observations, experiments and analysis".
That will happen  ::)
Title: Re: "Equator" problem
Post by: Goth on November 11, 2014, 04:59:21 AM
What's done can't be undone.    ;D
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 11, 2014, 05:24:39 AM


I think you're mistaken.
Then demonstrate as much.

Quote
In your opinion the height of the observed objects doesn't matter? It is not imperceptible. This sounds counterintuitive.
Intuition is a terrible tool to use in these situations - which is why I used maths.  The human brain simply isn't evolved to deal with problems like this intuitively.

So the human brain is not evolved enough. Amazing revelation which, of course, your brain made about itself.
You seem to have terrible problems with reading comprehension.   I said:

"The human brain simply isn't evolved to deal with problems like this intuitively"

Which is why we invented maths, logic and science.


Now instead of ranting, why don't you address the point we were debating?  If you think 1.8% is wrong, then show us what you think it should be and why we are wrong.
Title: Re: "Equator" problem
Post by: cikljamas on November 11, 2014, 05:58:11 AM
On GOD:

"The cosmos is all that is, or ever was, or ever will be." Carl Sagan

According to above logic we should assume that one of next two options must be true:

1. The cosmos is eternal
2. The cosmos is not eternal, but still is "all that is, or ever was, or ever will be"

If 1 then the cosmos is just another word for God! - What is wrong with such an idiotic identification? This is what is wrong: - The true meaning of a word "God" is: A being which cannot not to be!!! Such Being can't be created since God has no need to have been created, He exists outside time, but the cosmos HAS NEED TO HAVE BEEN CREATED and IS submitted to the second law of thermodynamics!

If 2 then the cosmos popped into existence out of nothing, and one of the most basic philosophical principles is EX NIHILO NIHIL FIT which means that there is no such thing as "creation caused by nothing", or as "created entity that comes out of nothing."
Of course, there is one necessary "exception": God himself! God is Absolute, Uncontingent Being, Essence of Existence, Actus Purus, First - Necessary Cause, Unmoved Mover, God exists forever and ever, He is a Giver of all contingent existences, but He Himself is Unlimited! God had introduced Himself to Moses using perfectly consistent philosophical definition: "I AM THAT I AM"! (IHWH).

In Ex. 3:13-14, Moses asks God, “Whom should I say has sent me?” and God responds by saying, “I AM that I AM… You must say this to the Israelites, ‘I AM has sent me to you.’” However, it could be awkward for Moses to go to the Israelites and Pharaoh and say, “I am has sent me.” So, in Ex. 3:15 God revises this phrase and changes it to the third person by saying, “Tell them that ‘He is’ has sent you.”

The word “He is” comes from the Hebrew root word haya, which means, “to be.” It is the third person form of this word, “He is,” that becomes the name Yahweh.

It was not God's intention to be Hidden from us, to be Hidden God, He revealed Himself to us:

- In the philosophy
- In "the book of nature" (creation)
- In the Bible (in history)

God even became one of us, and took our sins away! It is hard to understand such a great love, but it is possible for us to believe that such a perfect love can exist! Creation itself is a token of God's perfect love towards us!

What we think about the world - our Weltanschauung - cannot legitimately be excluded from the domain of religion. As St. Thomas Aquinas writes in the Summa Contra Gentiles (Bk. II, ch. 3): "It is absolutely false to maintain, with reference to the truths of our faith, that what we believe regarding the creation is of no consequence, so long as one has an exact conception concerning God; because an error regarding the nature of creation always gives rise to a false idea about God." I would add that I perceive the contemporary penchant for accommodating the teachings of Christianity to the so-called truths of science as a striking confirmation of this Thomistic principle: a case, almost invariably, of scientistic errors begetting flawed theological ideas. I will go so far as to contend that religion goes astray the moment it relinquishes its just rights in the so-called natural domain nowadays occupied by science.

I believe that the contemporary crisis of faith and the ongoing de-Christianization of Western society have much to do with the fact that for centuries the material world has been left to the mercy of the scientists. This has of course been said many times before (YET NOT NEARLY OFTEN ENOUGH!)


Theodore Roszak, for one, has put it exceptionally well: "SCIENCE IS OUR RELIGION, BECAUSE WE CANNOT, MOST OF US, WITH ANY LIVING CONVICTION SEE AROUND IT!

Oskar Milosz (1877-1939), a European writer said: "UNLESS A MAN'S CONCEPT OF THE PHYSICAL UNIVERSE ACCORDS WITH REALITY, HIS SPIRITUAL LIFE WILL BE CRIPPLED AT ITS ROOTS, WITH DEVASTATING CONSEQUENCES FOR EVERY OTHER ASPECT OF HIS LIFE." It could not have been better said!

Nowadays there are some who say that the method of St. Thomas is too scholastic and artificial, that it is not sufficiently historical and real. It is, so they say, too much an a priori method, almost always a process of deduction and analysis, or else in the analysis itself there is too much abstraction. It even seems at times to confound logical abstractions with the objectivity of things.

For them, the abstract object not only is not concrete, but it is not real. Thus the essence of man, of virtue, of society, and such things, would not be anything real, and the whole of metaphysics, not excepting the principle of contradiction, would be reduced to logic, logical abstractions, logical being, or, as they say, to extreme intellectualism that is without reality and lifeless. They would not dare to say explicitly that the abstract principle of contradiction (that some thing cannot at the same time be and not be) is not a law of real being but only a logical law governing the operations of the mind, as the laws of syllogism are. To such extreme admission, however, is one brought by this silly and at the present day common enough objection.


ON MOTION:

Every failure of all these attempts (to prove any kind of motion of the Earth) presents the proof to the contrary: The Earth is at rest!

Why is that so?

- If something is not white, doesn't mean that it is black, it can also be variety of colors.
- If something is not circular, doesn't mean that it is triangular, it can also be square, pentagonal etc...
- But, if something is not in motion, IT DOES MEAN THAT IT IS AT REST! 

Time lapse photography proves it: http://theflatearthsociety.org/forum/index.php?topic=62199.msg1635830#msg1635830 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1635830#msg1635830)


ON THE SHAPE OF THE EARTH:

1. Rowbotham has EXPERIMENTALLY proved tht the surface of all waters on the Earth is flat!

2. Photographs of a distant mountains proves it - horizon calculator is wrong: http://www.energeticforum.com/258148-post188.html (http://www.energeticforum.com/258148-post188.html)

3. Plane Sailing proves it: http://www.energeticforum.com/265962-post590.html (http://www.energeticforum.com/265962-post590.html)

4. River beds prove it: http://www.energeticforum.com/265601-post587.html (http://www.energeticforum.com/265601-post587.html)

5. Eye level horizon (no matter how high we ascend) proves it: http://www.energeticforum.com/266709-post615.html (http://www.energeticforum.com/266709-post615.html)

6. Engineering proves it: http://www.energeticforum.com/265582-post583.html (http://www.energeticforum.com/265582-post583.html)

7. The real mechanics of tides proves it: http://www.energeticforum.com/266629-post613.html (http://www.energeticforum.com/266629-post613.html)

8. "SAROS" ARGUMENT PROVES IT:

The whole thing with the water encircling the Earth is kind of nonsensical. We know that rivers flow down to the oceans. On a globe there can't be up and down (so they claim), so why exactly water flows down to the oceans if the oceans also have an incline and are not flat. Actually of course a round sphere has a top and a bottom. Doesn't make sense at all. The oceans would destroy the land completely if the Earth were round and somehow the water managed to stick to the surface. They would constantly push on to land till they cut through it. That is how water behaves if it is on a slope and meets a barrier on its way. If there is an incline to water then you wouldn't need wind to sail, you would just go down with the flow. A round surface gives you the incline, so it doesn't make sense.

9. Great deluge proves it!

10. Tunguska proves it!


On INTUITION:

Saros, you are absolutely right!

Title: Re: "Equator" problem
Post by: cikljamas on November 11, 2014, 06:04:22 AM
ON GREAT FLOOD:

Did Noah's flood cover the whole earth or just areas where man lived?

Often when this question comes up it is because many, perhaps most, scientists these days are trying to say that if there was a flood of cataclysmic proportions, it was only a localized, or regional flood. That is, it occurred only where there was life, in particular, human life.

The scientists and others can't deny there was a great flood, because: 1) There is sea fossil life on even the summits of some of the world's highest mountains; and 2) Cave drawings, and other recorded art and written works by primitive civilizations around the globe record in their legends a great flood of some sort.

God's Word says:

"And the flood was forty days upon the earth; and the waters increased, and bare up the ark, and it was lift up above the earth. And the waters prevailed, and were increased greatly upon the earth; and the ark went upon the face of the waters. And the waters prevailed exceedingly upon the earth; and all the high hills, that were under the whole heaven, were covered. Fifteen cubits upward did the waters prevail; and the mountains were covered" (Gen. 7:17-20).

The Bible says further:

"And all flesh died that moved upon the earth, both of fowl, and of cattle, and of beast, and of every creeping thing that creepeth upon the earth, and every man: All in whose nostrils was the breath of life, of all that was in the dry land, died. And every living substance was destroyed which was upon the face of the ground, both man, and cattle, and the creeping things, and the fowl of the heaven; and they were destroyed from the earth: and Noah only remained alive, and they that were with him in the ark"
(Gen. 7:21-23).

Peter the apostle forewarns that the skeptics would be talking about such things in the last days.

"Knowing this first, that there shall come in the last days scoffers, walking after their own lusts, And saying, Where is the promise of his coming? for since the fathers fell asleep, all things continue as they were from the beginning of the creation. For this they willingly are ignorant of, that by the word of God the heavens were of old, and the earth standing out of the water and in the water: Whereby the world that then was, being overflowed with water, perished"[/b][/u] (2 Pet. 3: 3-6).

Believe the scientists and the other scoffers, or believe God, who cannot lie. We weren't there, so we must believe one or the other. It's a matter of faith, but the geological record is on God's side. The whole earth was covered at least 15 cubits above the highest mountain peak.
Title: Re: "Equator" problem
Post by: Rama Set on November 11, 2014, 06:25:40 AM
On GOD:

"The cosmos is all that is, or ever was, or ever will be." Carl Sagan

According to above logic we should assume that one of next two options must be true:

1. The cosmos is eternal
2. The cosmos is not eternal, but still is "all that is, or ever was, or ever will be"

I choose to insert another option:

3. The cosmos is finite and before that nothing existed.

Quote
If 1 then the cosmos is just another word for God! - What is wrong with such an idiotic identification?

No it isn't.  God is purported to be a sentient being whereas the cosmos has no such requirement.

Quote
This is what is wrong: - The true meaning of a word "God" is: A being which cannot not to be!!!

Who declared this the true definition?  It sounds like you are asserting it is the truth for the expedience of your argument.

Quote
Such Being can't be created since God has no need to have been created, He exists outside time, but the cosmos HAS NEED TO HAVE BEEN CREATED and IS submitted to the second law of thermodynamics!

No one knows if this is the truth or not.  This is merely conjecture.  Please be honest and treat it as such.

Quote
If 2 then the cosmos popped into existence out of nothing, and one of the most basic philosophical principles is EX NIHILO NIHIL FIT which means that there is no such thing as "creation caused by nothing", or as "created entity that comes out of nothing."

That may not be true.  Please refer to the work of Lawrence Krauss.  At this time, Ex Nihilo creation cannot be definitively ruled out for creation of the universe.

Quote
Of course, there is one necessary "exception": God himself!

Only necessary because you have asserted as such.  You literally have no evidence to show this.

Quote
God is Absolute, Uncontingent Being, Essence of Existence, Actus Purus, First - Necessary Cause, Unmoved Mover, God exists forever and ever, He is a Giver of all contingent existences, but He Himself is Unlimited! God had introduced Himself to Moses using perfectly consistent philosophical definition: "I AM THAT I AM"! (IHWH).

God told you this himself, or you read it in a thousands of years old book of questionable authorship?

Quote
...

tl;dr
Title: Re: "Equator" problem
Post by: ausGeoff on November 11, 2014, 08:20:17 AM
It's actually quite astounding that in a scientifically enlightened 21st century, we still find the occasional diehard, fundamentalist god-botherer such as cikljamas attempting—in vain—to discredit every one of the natural sciences whose confirmatory data set has been built up over centuries by guys certainly smarter than any of us here.

The typically smug, condescending attitude of the bible-bashers is certainly repugnant to any rational person in 2014 that's for sure.  Bear in mind that these religious apologists use as a reference book a document cobbled together by a disparate group of ill-educated desert nomads more than 2,000 years ago over a 400-year period.  And those same guys believed in supernatural entities (such as their "god") and paranormal phenomena (such as "miracles").

And what's even more disconcerting is that someone like cikljamas—who should know better, and be better informed—still believes in the existence of gods and angels, and devils and miracles.

That being the case, it's virtually impossible to engage in any sort of meaningful, contemporary scientific debate with people like cikljamas who either choose to willfully ignore the scientific evidence, or who lack the necessary technical and/or logic skills to evaluate it.

Title: Re: "Equator" problem
Post by: cikljamas on November 11, 2014, 09:53:46 AM
@ Rama Set, watch this :
Can God's Existence be Demonstrated? (William Lane Craig) : (http://)

It's actually quite astounding that in a scientifically enlightened 21st century, we still find the occasional diehard, fundamentalist god-botherer such as cikljamas attempting—in vain—to discredit every one of the natural sciences whose confirmatory data set has been built up over centuries by guys certainly smarter than any of us here.

The typically smug, condescending attitude of the bible-bashers is certainly repugnant to any rational person in 2014 that's for sure.  Bear in mind that these religious apologists use as a reference book a document cobbled together by a disparate group of ill-educated desert nomads more than 2,000 years ago over a 400-year period.  And those same guys believed in supernatural entities (such as their "god") and paranormal phenomena (such as "miracles").

And what's even more disconcerting is that someone like cikljamas—who should know better, and be better informed—still believes in the existence of gods and angels, and devils and miracles.

That being the case, it's virtually impossible to engage in any sort of meaningful, contemporary scientific debate with people like cikljamas who either choose to willfully ignore the scientific evidence, or who lack the necessary technical and/or logic skills to evaluate it.
(http://3.bp.blogspot.com/-rj8NyHL1SpE/UqR1FcFI2jI/AAAAAAAABcQ/VsT9QNznItg/s1600/whtbj+laugh.gif)

Evaluate this : Proof The Earth Is Young and Noah's Flood : (http://)
Title: Re: "Equator" problem
Post by: 29silhouette on November 11, 2014, 10:02:30 AM
You guys are very confused.
I am confused now, and I don't know what we're discussing here anymore.

Quote
You can see they are leaning away,
Can you point the indicators of this in your pictures?

Quote
because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.
I conducted an experiment by taking pictures from different elevations, and then analyzed the results.  It shows me Earth isn't flat.

Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet,
Would it help if the formula and/or calculation was in the form of a calculator?  You seem to be ok with those.

What do you believe the shape of the planet to be?  Flat or globe?
Title: Re: "Equator" problem
Post by: ausGeoff on November 11, 2014, 10:05:34 AM
Evaluate this : Proof The Earth Is Young and Noah's Flood : (http://)

I'm afraid I don't have the time or the inclination to sit through a fairy story proposed by William Lane Craig.  I've already seen more than enough of his religious drivel during an encounter with the late Christopher Hitchens, and during which Hitchens shot the guy down in flames.  Same goes for nearly an hour of the alleged "young age" of the earth video.

All these young-earth creationists must've come off the one assembly line LOL.  They're all certifiable whack-jobs.   
Title: Re: "Equator" problem
Post by: Socratic Amusement on November 11, 2014, 10:07:39 AM
HA! William Lane Craig...

Wow, those are some pretty old, thoroughly destroyed arguments you are linking to, cikljamas.

Jeez. I haven't seen someone use WLC since 2008. I thought everyone knew how intellectually bankrupt he is, including Christians?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 11, 2014, 10:35:50 AM
Would it help if the formula and/or calculation was in the form of a calculator?  You seem to be ok with those.

There is a calculator here:

http://www.1728.org/radians.htm (http://www.1728.org/radians.htm)
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 11, 2014, 11:25:11 AM
You guys are very confused. Using a formula you get some results and of course you believe them immediately. It is mathematics after all. This formula is wrong, and it is common sense that if there is a curvature the objects will lean more especially if they are tall enough when observed from a big distance. You go tell this 1.9% to someone else, I am not buying that math, as it is wrong to anyone with half a brain, but you apparently. That might be useful in high school to get an A but not in real life.
You seem to be the one who's confused.

Since cikljamas has gone off to a completely religious diatribe which I wouldn't touch with a 10-foot pole, there's time for this sub-discussion. The question at hand seems to have originated with this:

@Saros, I understand what you mean about them "leaning away", but how much do you think this would be by? 

More importantly, how would you ever perceive it?  How would a large object 200km away leaning slightly away from you look any different from one that wasn't?

Are you sure it should be leaning only slightly away? How about buildings which are smaller objects than peaks ;D? Shouldn't the leaning away be visible? From 15-20 km, there should be enough leaning away to be observed, right? I am just saying. Can't give you the exact math.
Here Saros surmises that buildings 15-20 km distant should lean away from the observer enough to be perceived and suggests that math can give the answer, but says he doesn't know how to do the math. That's fair enough; most people don't know how to approach this problem. But now, after being shown how to do the math - two different ways - he has a temper tantrum and complains that it isn't a math problem. Who's confused here?

Revisiting Saros' specific question about whether buildings 15-20 km away should be visibly tilting or not, and his assertion that they should, let's see what we get:

a = 20 km / 6371 km = 0.00314 radians
0.00314 radians * 180° / pi radians = 0.18°

A plumb building 20 km from you will be leaning directly away from you by 0.18°. If this building is 100m tall, its top will be 31 cm further from you than if it were square with your sightline. (hint: you can multiply that radian measure of the angle by the height and arrive directly at the answer here). Do you think you can detect that change in distance - directly away from you - in a 30-story building from 20 km away? Considering the building itself would subtend only a bit more than 1/4° (if you could see the whole thing), this seems exceedingly unlikely.

How is that formula wrong? You said you didn't know the exact math, didn't you. If you've learned how to do this since your original post, what do you think the correct formula is?

Not to mention, that I was misleading you by claiming that you can't see they are leaning away. I just wanted to see how confused you are. You can see they are leaning away, but you guys are so big fans of math that you just proved you know nothing about reality, and supposedly you just proved that any leaning away shouldn't be visible, when in fact it is. Good job! Great math. Great logic. Congratulations!!! Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet, because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.
If they are leaning away, doesn't that indicate the Earth is curved? Or did I miss something while trying to catch up with this branch of the thread? So, how much do your first-hand observations, experiments and analysis show they are leaning away from you? Can you describe the analysis? Does it involve math?

As an aside, I remember reading about building the Verrazano Narrows Bridge (http://en.wikipedia.org/wiki/Verrazano%E2%80%93Narrows_Bridge) in NYC when I was a kid. It had the longest suspension-bridge span in the world when it was built and the fact that they had to account for the curvature of the Earth was mentioned at the time. It's not much, but, apparently, was enough to be considered in the engineering.

Quote from: Wikipedia
Because of the height of the towers (693 ft or 211 m) and their distance apart (4,260 ft or 1,298 m), the curvature of the Earth's surface had to be taken into account when designing the bridge—the towers are 1 5⁄8 inches (41.275 mm) farther apart at their tops than at their bases.
I get a bit less than 43 mm using the technique above, but they may be using a slightly different value for the radius (6371 km is the volumetric mean), and the difference is only about 1/16", so may mostly be round-off error.
Title: Re: "Equator" problem
Post by: cikljamas on November 11, 2014, 12:06:55 PM
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.

(http://i26.photobucket.com/albums/c118/FromVegaButNotVegan/tfesMtnsInDistance3_zpsc828fd5f.png)

Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 11, 2014, 01:36:32 PM
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.

(http://i26.photobucket.com/albums/c118/FromVegaButNotVegan/tfesMtnsInDistance3_zpsc828fd5f.png)

Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
What's wrong with the explanation? Where does it not match the diagram?
Title: Re: "Equator" problem
Post by: cikljamas on November 11, 2014, 02:39:12 PM
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.

(http://i26.photobucket.com/albums/c118/FromVegaButNotVegan/tfesMtnsInDistance3_zpsc828fd5f.png)

Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
What's wrong with the explanation? Where does it not match the diagram?

Here:

(http://i.imgur.com/7wanoRG.jpg)

Zoomed shot of the same mountains taken from the same spot as in the first picture above:

(http://i.imgur.com/L3y1Hgl.jpg?1)

Accompanying words of Saros:

You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it. Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions.

Now, try to conciliate your diagram with what is clearly shown in above picture and with what implies my question which i repeat once more:

1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Show me the "brow" of my "hill of water" in above pictures!

ONE PICTURE IS WORTH MORE THAN A THOUSAND WORDS!!!
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 11, 2014, 08:35:31 PM
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?

Alpha2Omega in post #126 writes:

See that sharp line "cutting off" bottom of the mountains in the distance, commonly called the horizon? That's the "brow" of your "hill of water". It's fairly close (and so not as "high") to the photographer since he's at a relatively low height above the water. It's further from the mountains in the distance and higher than their bases by your 2000' since they're farther away. You still see their tops because they're higher than the "hill of water". This diagram may explain it better.

(http://i26.photobucket.com/albums/c118/FromVegaButNotVegan/tfesMtnsInDistance3_zpsc828fd5f.png)

Alpha2Omega offers us totally wrong explanation (as always), but his diagram is much closer to the supposed reality of allegedly globular Earth!
What's wrong with the explanation? Where does it not match the diagram?

Here:

(http://i.imgur.com/7wanoRG.jpg)

Zoomed shot of the same mountains taken from the same spot as in the first picture above:

(http://i.imgur.com/L3y1Hgl.jpg?1)
These aren't the picture the diagram was drawn to represent.

Please place a link to posts you want to reference rather than a reply number. Navigating around long threads like this one is a pain (there is a shortcut instead of using the page numbers, but direct links are easier). Here's how to do that:

The Very top line of each post shows the title of the post, probably Re: <thread name>. That title is a link to the post itself. This one is
Re: "Equator" problem

In some browsers on Windows systems, if you right click that title, it will give you the option to "Copy link address" or something like that (Chrome does). If it's available, just select it. If you don't have the option to do that, just click the link and copy the URL from your browser's address line. Either way, paste it into your reply and make it a URL by highlighting it and clicking the little SPHERICAL EARTH and document icon  (http://i26.photobucket.com/albums/c118/FromVegaButNotVegan/TFESurlIcon_zpse98c3571.gif)  above the reply area. Ha! They really should do something about that! :)
like this:
From http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636303#msg1636303 (http://theflatearthsociety.org/forum/index.php?topic=62199.msg1636303#msg1636303)

instead of
Alpha2Omega in post #126 writes:

Anyway...
That picture showed a mountain peak dropping directly below the sea-level horizon. This shows a distant peak behind a nearer ridge which is behind an even nearer low ridge for part of it, both of which fall below the sea-level horizon. The same principle applies in general, though.

Quote
Accompanying words of Saros:

You have no idea what you're talking about. Normally even without any fog and clear skies you cannot see anything into the sea which is 50 km away. The mountains indeed can be seen sometimes due to their height, but in this particular picture you're actually seeing Mt.Elbrus from 202 km while at the same time the photo was taken from 1-2 m, do you see the difference? Even with its height of 5642 m, Mt.Elbrus is physically impossible to be seen from more than 270 km away under most perfect conditions and observation height of 2 meters, and you should see it then barely at the horizon. Does it look it's getting close to the horizon? It is at least 15 degrees above it.
How did he determine that 15°? I recall seeing this question before, but don't think he answered it. If the Earth were flat and something 200 km appeared 15° up, it would have to be more than 50 km high. Does he think this mountain is that high? Angles like this are notoriously hard to estimate; I think Saros is simply wrong when he says "at least 15 degrees".
Quote
Even if it were 2000 meters high though it would still be seen at the horizon under those atmospheric conditions.
What?

Quote
Now, try to conciliate your diagram with what is clearly shown in above picture and with what implies my question which i repeat once more:

1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?
It looks like there is a ridgeline between the photographer and the distant peak. That keeps you from seeing anything but the highest elevations of the more distant peak. It's hard to tell from these pictures what's going on with the lower reaches of the closer ridge, or how far away it is.  At any rate, that ridge appears to drop below the horizon.

Quote
Show me the "brow" of my "hill of water" in above pictures!

ONE PICTURE IS WORTH MORE THAN A THOUSAND WORDS!!!
Are you kidding? The "brow" of the "hill of water" is the sharp line between dark blue (water) and lighter blue (distant hillsides) where the bottom of the ships disappear. What else would it be?

The words conveyed by the picture don't mean anything if you don't understand what the picture is saying.
Title: Re: "Equator" problem
Post by: 29silhouette on November 11, 2014, 10:17:54 PM
Accompanying words of Saros:
You two sure were arguing though on the energeticforum.  And now you're quoting him to support your argument? 

Quote
1. How about being able to see a whole figure of 200 km distant mountains from bottom to top? How come that more than a half of these distant mountains is not out of range of our sight due to alleged curvature of the Earth?
You're not seeing the whole mountain.

Quote
Show me the "brow" of my "hill of water" in above pictures!
Do you see the waterline where the hills appear?

Quote
ONE PICTURE IS WORTH MORE THAN A THOUSAND WORDS!!!
My pictures shows the hills sinking below the curvature.

Title: Re: "Equator" problem
Post by: 29silhouette on November 11, 2014, 10:45:52 PM
How did he determine that 15°? I recall seeing this question before, but don't think he answered it. If the Earth were flat and something 200 km appeared 15° up, it would have to be more than 50 km high. Does he think this mountain is that high? Angles like this are notoriously hard to estimate; I think Saros is simply wrong when he says "at least 15 degrees".
I never did see an answer after I asked.  I even asked how high he figured the tree line to be in my picture.

Quote
It looks like there is a ridgeline between the photographer and the distant peak. That keeps you from seeing anything but the highest elevations of the more distant peak. It's hard to tell from these pictures what's going on with the lower reaches of the closer ridge, or how far away it is.  At any rate, that ridge appears to drop below the horizon.
  There's a small dark spot just below the peak (the arrow is pointing directly toward it).  It's at the 10,000 foot level (10,300ft total just to the right) of some peaks 37 miles closer.  Mt Elbrus just pokes above it.  Just below this ridge is another ridge visible.  There's a lighter area (devoid of forest), sort of oblong shaped with the right end in line with the arrow.  The bottom is 5400 feet and the top is 6600 feet.  It's about another 5 miles closer.

Measuring the few distinct features on these ridges that match up in Ge, their full height extends below the waterline/horizon/peak of the hill of water quite a bit.  It's hard to get an exact measurement, as the lower elevations are closer than the higher elevations, which means the higher elevations are dropped that much more due to the long distance.  Beside which, just how close were these actually to the horizon and so how affected by refraction is everything?  And, is the waterline really exactly where it appears, or is it a tad higher in reality?  Judging by the way the hill slopes down to meet the water, and the small white dot, I think there's a tiny bit of mirage. 
Title: Re: "Equator" problem
Post by: rottingroom on November 12, 2014, 06:19:26 AM
You guys are very confused. Using a formula you get some results and of course you believe them immediately. It is mathematics after all. This formula is wrong, and it is common sense that if there is a curvature the objects will lean more especially if they are tall enough when observed from a big distance. You go tell this 1.9% to someone else, I am not buying that math, as it is wrong to anyone with half a brain, but you apparently. That might be useful in high school to get an A but not in real life. Not to mention, that I was misleading you by claiming that you can't see they are leaning away. I just wanted to see how confused you are. You can see they are leaning away, but you guys are so big fans of math that you just proved you know nothing about reality, and supposedly you just proved that any leaning away shouldn't be visible, when in fact it is. Good job! Great math. Great logic. Congratulations!!! Enjoy your math world, which has nothing to do with reality. Now I am finally convinced you know nothing about the shape of the planet, because your knowledge is not based on real, first-hand observations, experiments and analysis, but on mathematical equations, which assume unproven facts.

We can do this so many ways and we keep coming up with the same number. Here is another way.

A circle is 360°.

The circumference of the round earth is 40,000km.

200km/40,000km = x/360°

Solve for x.

0.005 = x/360°

0.005*360° = x

1.8° = x

---------------------------------------

If you have a problem with the math then let's check it for something intuitive. Suppose there is a skyscraper at both the north pole and at the equator. If the earth is round then the skyscraper at the equator should be tilted 90° away from the skyscraper at the north pole.

According to this google search result (https://www.google.com/search?q=distance+from+north+pole+to+the+equator (https://www.google.com/search?q=distance+from+north+pole+to+the+equator)), the distance from the north pole to the equator is roughly 10,000 km, which incidentally, is what originally defined the metric system.

Again, the circumference of earth is 40,000km and there are 360° in a circle. If these numbers are plugged in it should all check out, right?

10,000km/40,000km = 90°/360°

1/4 = 1/4

Looks good!





Title: Re: "Equator" problem
Post by: cikljamas on November 12, 2014, 08:57:53 AM

(http://i.imgur.com/Vy9KjEi.jpg)

(http://i.imgur.com/1efjZSb.jpg)
Title: Re: "Equator" problem
Post by: rottingroom on November 12, 2014, 09:04:05 AM
what part of your intuition is irrelevant is hard for you to grasp?
Title: Re: "Equator" problem
Post by: rottingroom on November 12, 2014, 09:38:13 AM
clickijamas,

A while back a forum member by the name of Scintific Method made a thread called measure a mountain. Here is the thread.

http://theflatearthsociety.org/forum/index.php?topic=59240.0#.VGOZ8sljWmw (http://theflatearthsociety.org/forum/index.php?topic=59240.0#.VGOZ8sljWmw)

I invite you to do his experiment and see what you come up with. Here is the relevant parts of the original post.

Equipment: 1 spirit level, 1 set square, and 1 ruler. You might also want something to set the spirit level up on, and some packing to level it.

Method: pick a mountain that is easily identifiable from a distance (I have a few to pick from where I am). Find a point on the map 30km away, 40km away, and 50km away (or any other largish distances, as long as you can measure them accurately. This is part of what negates any doctoring of the maps: picking arbitrary points to measure from). Set up your spirit level so that it points toward the mountain you picked, ensuring it is level. Set your ruler at a specific distance along the top of the level, standing up at right angles to it (that's what the set square is for). Sight from the edge of the level furthest from the mountain to the mountain top, and note where your line of sight crosses the ruler. The ratio of the measurement on the ruler to it's distance along the top of the level would be equal to the ratio of the mountain's height above your observation point to your distance to it (simple triangular geometry).

The height of the location the measurement is taken from is important, as you will get different results from different elevations at a given distance. Another reason why this cannot be faked or doctored.

In the below diagram, the ratio of h/d should be equal to r/s if the earth is flat, and should always be that way, no matter how far away from the mountain you are.

(http://img94.imageshack.us/img94/4084/mountainmeasuring.png)

When I did this experiment though, there was a difference between h/d and r/s, and it got bigger and bigger the further away I got from the mountain. I did some extra maths to work out where the mountain top would be for a round earth:

apparent height on a round earth = square root of ((earth's radius + h)2 - d2) - (earth's radius + altitude of your position)

When I tested the results against those calculations, they matched perfectly. Remember: maths and geometry don't care what shape the earth is; they will always tell you the truth.

Results:

Mountain (Castle Top, in the Nandewar Ranges): 1075m AMSL
Point 1 (my back yard): 30.44km from peak, 215m AMSL
Point 2 (Kamilaroi Hwy): 40.46km from peak, 205m ASMSL
Point 3 (Wee Waa levee): 55.57km from peak, 192m AMSL

Using a 1100mm baseline on the spirit level, the following apparent heights were recorded:
Point 1: 28mm
Point 2: 18mm
Point 3: 11mm
Let's be pessimistic and say they're only within 1mm either way.

Here's what they should have been for a flat earth:
Point 1: 31mm
Point 2: 21mm
Point 3: 16mm
Note the increasing difference. This is due to the mountain 'going over the edge'.

Conclusion: The data collected disagrees with FE predictions*, and almost exactly matches RE predictions. The experiment is easy to conduct, and not prone to significant error, so the results can be considered reliable.

* Rowbotham's 'perspective' has been cited as a possible cause of the results. However, a passing familiarity with geometry should be sufficient to show that this 'effect' is not responsible for the rate at which the actual measurements diverge from what they should be for a flat earth.

Interesting Note: if you use the following equation, you can work out the difference between the apparent height for a round earth and a flat earth (with respect to the diagram):

rf - rr = s * tan((d / ce) * 360 / 2)

where rf is the apparent height on a flat earth, rr is the apparent height on a round earth, and ce is the circumference of the round earth. d and ce need to be in the same units as each other, and rf - rr will be in the same units as s.


Sorry for the long post! I thought it would be best to give plenty of detail, so that others can reproduce this experiment and obtain accurate results.

Title: Re: "Equator" problem
Post by: Alpha2Omega on November 12, 2014, 11:42:53 AM

(http://i.imgur.com/Vy9KjEi.jpg)

(http://i.imgur.com/1efjZSb.jpg)
It's Mt. Elbrus, not Mt. Erebus, isn't it?

I think those lines are at about the right places on your annotated picture. They are saying that a point a bit more than halfway between sea level and the top of Elbrus is about horizon level from your vantage point 200 km away (2989 is 53%  of 5642, approximately half). Extrapolating that downward gives an idea where sea level directly below the peak of Elbrus is below your local horizon.

What's funny about it, and where else would you expect it to be on a spherical planet?
Title: Re: "Equator" problem
Post by: Saros on November 12, 2014, 01:02:46 PM

It's Mt. Elbrus, not Mt. Erebus, isn't it?

I think those lines are at about the right places on your annotated picture. They are saying that a point a bit more than halfway between sea level and the top of Elbrus is about horizon level from your vantage point 200 km away (2989 is 53%  of 5642, approximately half). Extrapolating that downward gives an idea where sea level directly below the peak of Elbrus is below your local horizon.

What's funny about it, and where else would you expect it to be on a spherical planet?

Here is what it should look like from the same coordinates and 20 meters elevation.
(http://i.imgur.com/Tq1YhaE.jpg?1)

Source:http://www.peakfinder.org/?lat=41.6487&lng=41.645&off=20&ele=4 (http://www.peakfinder.org/?lat=41.6487&lng=41.645&off=20&ele=4)

And this is what it looked like with absolutely no optical zoom(note that with the naked eye it appeared about twice bigger than what you see in the photo), and  the camera had only 3x optical zoom.
(http://i.imgur.com/IbhhL85.jpg?1)
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 12, 2014, 01:51:38 PM
@Saros, I've no idea what your point is??

"And this is what it looked like with absolutely no zoom"

So what was the focal length?
Title: Re: "Equator" problem
Post by: neimoka on November 12, 2014, 01:57:50 PM
And this is what it looked like with absolutely no zoom(note that with the naked eye it appeared about twice bigger), and  the camera had only 3x zoom.
"3x zoom" means absolutely nothing. It's like saying that something is three times bigger. Also, cameras do not have zoom. Ever. It's a feature of the lens. With the naked eye it appeared about twice bigger than what? Are you looking at the picture on a cell phone or a 50" screen or or what?
Title: Re: "Equator" problem
Post by: cikljamas on November 12, 2014, 01:59:16 PM

(http://i.imgur.com/Vy9KjEi.jpg)

(http://i.imgur.com/1efjZSb.jpg)
It's Mt. Elbrus, not Mt. Erebus, isn't it?

Whatever, it doesn't make any difference however! It is still VERY INTUITIVE!!!

I am absolutely amazed with your ability to perceive and discern between large and small scale features.

Do you notice any difference between these two dogs:

(http://i.imgur.com/Ioogzfd.jpg)

(http://i.imgur.com/2QBV973.jpg)

Your attitude, not your aptitude, will determine your altitude.


This is a bad attitude:

(http://i.imgur.com/30KYJlh.jpg)

(http://i.imgur.com/6ZKRB3j.jpg)

Title: Re: "Equator" problem
Post by: Rama Set on November 12, 2014, 02:01:55 PM
Well that was unexpected.
Title: Re: "Equator" problem
Post by: cikljamas on November 12, 2014, 02:06:24 PM
Well that was unexpected.

You mean Einstein?

Well...

(http://3.bp.blogspot.com/-rj8NyHL1SpE/UqR1FcFI2jI/AAAAAAAABcQ/VsT9QNznItg/s1600/whtbj+laugh.gif)
Title: Re: "Equator" problem
Post by: Saros on November 12, 2014, 02:16:48 PM
@Saros, I've no idea what your point is??

"And this is what it looked like with absolutely no zoom"

So what was the focal length?

This is the camera used http://www.manualslib.com/manual/26037/Casio-Exilim-Ex-Z19.html?page=135 (http://www.manualslib.com/manual/26037/Casio-Exilim-Ex-Z19.html?page=135)

Lens/Focal Distance
F2.8 (W) to 5.2 (T) f= 6.2 to 18.6 mm(equivalent to 37.5 to 112.5 mm in 35 mm format)Six lenses in five groups, including aspherical lens.

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?


Title: Re: "Equator" problem
Post by: neimoka on November 12, 2014, 02:33:52 PM
@Saros, I've no idea what your point is??

"And this is what it looked like with absolutely no zoom"

So what was the focal length?

This is the camera used http://www.manualslib.com/manual/26037/Casio-Exilim-Ex-Z19.html?page=135 (http://www.manualslib.com/manual/26037/Casio-Exilim-Ex-Z19.html?page=135)

Lens/Focal Distance
F2.8 (W) to 5.2 (T) f= 6.2 to 18.6 mm(equivalent to 37.5 to 112.5 mm in 35 mm format)Six lenses in five groups, including aspherical lens.

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
Focal length in itself doesn't say much either; size of the imaging device is important. It's easiest to state either the field of view in degrees or use the 35mm format equivalents - those are stated in that spec sheet and most people understand them fairly well.
Title: Re: "Equator" problem
Post by: sokarul on November 12, 2014, 02:42:23 PM

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
The peak in the picture that should be way under 2989m?
Title: Re: "Equator" problem
Post by: Saros on November 12, 2014, 02:58:44 PM

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
The peak in the picture that should be way under 2989m?

(http://i.imgur.com/wBcAwCy.jpg?1)

Enhanced image:

(http://i.imgur.com/U1DCzZ1.jpg?1)

(http://i.imgur.com/tmuelf8.jpg?1)

 (http://i.imgur.com/k6fm4bl.jpg?1)
Title: Re: "Equator" problem
Post by: sokarul on November 12, 2014, 04:46:56 PM

By the way for your information the peak just in front of Elbrus is 3103 m - Gora Digdali-Dudi. Actually, I have no idea why you're still looking at these pictures?! I thought you believe the Earth is 100% round? Why bother? Why are you wasting your time here?
The peak in the picture that should be way under 2989m?

(http://i.imgur.com/wBcAwCy.jpg?1)

Enhanced image:

(http://i.imgur.com/U1DCzZ1.jpg?1)

(http://i.imgur.com/tmuelf8.jpg?1)

 (http://i.imgur.com/k6fm4bl.jpg?1)
So how is the distance between the two peaks 2,400m when 1,000m is halfway down from the peak?
Title: Re: "Equator" problem
Post by: 29silhouette on November 12, 2014, 06:06:12 PM
Here's the view using 20 meters on that neat PeakFinder site. 

Gora Tsulashi over to the left is only half a mile closer than Kvira.  Kvira is the ridge with the clearing I mentioned a few posts ago.

Tsulashi is 3596 ft.  Kvira is 6686 ft.  The difference is 3090 ft.  Taking that same distance below the top of Tsulashi (for a total of 6180 ft) we see that it reaches well below the waterline/horizon.  An additional 506 ft is required to give us the total height of Kvira.  This is a distance of about 80.2-80.7 miles (129.8 km )away.

Looks like there's a hill of water there.
(http://i1368.photobucket.com/albums/ag167/jeffro556/kavira_zpsae2c10ad.png)
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 02:14:16 AM
Here's the view using 20 meters on that neat PeakFinder site. 

Gora Tsulashi over to the left is only half a mile closer than Kvira.  Kvira is the ridge with the clearing I mentioned a few posts ago.

Tsulashi is 3596 ft.  Kvira is 6686 ft.  The difference is 3090 ft.  Taking that same distance below the top of Tsulashi (for a total of 6180 ft) we see that it reaches well below the waterline/horizon.  An additional 506 ft is required to give us the total height of Kvira.  This is a distance of about 80.2-80.7 miles (129.8 km )away.

Looks like there's a hill of water there.
(http://i1368.photobucket.com/albums/ag167/jeffro556/kavira_zpsae2c10ad.png)

 

(http://i.imgur.com/HY6VWEm.jpg?1)

Do you see the blue line? There is nothing higher than 400 meters all the way to Mt Kvira.

(http://i.imgur.com/U1DCzZ1.jpg?1)
Title: Re: "Equator" problem
Post by: cikljamas on November 13, 2014, 02:38:20 AM
Even more enhanced picture:

(http://i.imgur.com/Iehi9iP.jpg)
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 04:20:40 AM
Here is something more to think about:

(http://i.imgur.com/TOqN8D9.jpg?1)

The hills along the coastline are only between 70 and 120 m high.

(http://i.imgur.com/unIdH7Z.jpg?1)

Source:http://www.udeuschle.selfhost.pro/panoramas/panqueryfull.aspx?mode=newstandard&data=lon%3A41.64471%24%24%24lat%3A41.64906%24%24%24alt%3Aauto%24%24%24altcam%3A10%24%24%24hialt%3Afalse%24%24%24resolution%3A200%24%24%24azimut%3A360%24%24%24sweep%3A60%24%24%24leftbound%3A16.55000000000001%24%24%24rightbound%3A22.55000000000001%24%24%24split%3A60%24%24%24splitnr%3A1%24%24%24tilt%3A0.47916666666666696%24%24%24tiltsplit%3Afalse%24%24%24elexagg%3A1.2%24%24%24range%3A300%24%24%24colorcoding%3Afalse%24%24%24colorcodinglimit%3A231%24%24%24title%3AZugspitze%20%5BTelescope%5D%24%24%24description%3A%24%24%24email%3A%24%24%24language%3Aen%24%24%24screenwidth%3A1366%24%24%24screenheight%3A728 (http://www.udeuschle.selfhost.pro/panoramas/panqueryfull.aspx?mode=newstandard&data=lon%3A41.64471%24%24%24lat%3A41.64906%24%24%24alt%3Aauto%24%24%24altcam%3A10%24%24%24hialt%3Afalse%24%24%24resolution%3A200%24%24%24azimut%3A360%24%24%24sweep%3A60%24%24%24leftbound%3A16.55000000000001%24%24%24rightbound%3A22.55000000000001%24%24%24split%3A60%24%24%24splitnr%3A1%24%24%24tilt%3A0.47916666666666696%24%24%24tiltsplit%3Afalse%24%24%24elexagg%3A1.2%24%24%24range%3A300%24%24%24colorcoding%3Afalse%24%24%24colorcodinglimit%3A231%24%24%24title%3AZugspitze%20%5BTelescope%5D%24%24%24description%3A%24%24%24email%3A%24%24%24language%3Aen%24%24%24screenwidth%3A1366%24%24%24screenheight%3A728)
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 04:47:26 AM
It almost seems like the further away something is, the less dramatic its height. Wonder why that is.
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 04:50:52 AM
It almost seems like the further away something is, the less dramatic its height. Wonder why that is.

For starters, how about perspective?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 13, 2014, 05:35:58 AM
It almost seems like the further away something is, the less dramatic its height. Wonder why that is.
Strange, isn't it?
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 06:06:40 AM
It almost seems like the further away something is, the less dramatic its height. Wonder why that is.

For starters, how about perspective?

So you spend all this time trying to disprove the earth being round by showing us measurements and numbers and when you realize that what we see is exactly what's expected on a round earth, actual observed heights are no longer applicable and you essentially move the goal posts to perspective?

Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 06:31:25 AM
It almost seems like the further away something is, the less dramatic its height. Wonder why that is.

For starters, how about perspective?

So you spend all this time trying to disprove the earth being round by showing us measurements and numbers and when you realize that what we see is exactly what's expected on a round earth, actual observed heights are no longer applicable and you essentially move the goal posts to perspective?

Should I tell you or you're going to figure it out eventually by yourself that you guys are classic trolls and refuse to accept or even consider anything against your predetermined beliefs(or should I say agenda) while simultaneously trying to disrupt the discussion and disregard any valid arguments? I am obviously not posting the photos for ganders like you, so if it bothers you please don't look at them.

The photos actually prove just the opposite of what you have assumed, but since you're apparently blinkered you might continue shouting forever that they match the calculations. And are you going to deny that perspective exists when objects get far away? What else do you expect to happen to something which gets far away? You want it to get bigger?

And by the way, why are you here in this thread to begin with if you're not contributing? You don't need to answer. That was a rhetorical question.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 06:36:58 AM
It almost seems like the further away something is, the less dramatic its height. Wonder why that is.

For starters, how about perspective?

So you spend all this time trying to disprove the earth being round by showing us measurements and numbers and when you realize that what we see is exactly what's expected on a round earth, actual observed heights are no longer applicable and you essentially move the goal posts to perspective?

Should I tell you or you're going to figure it out eventually by yourself that you guys are classic trolls and refuse to accept or even consider anything against your predetermined beliefs(or should I say agenda) while simultaneously trying to disrupt the discussion and disregard any valid arguments? I am obviously not posting the photos for ganders like you, so if it bothers you please don't look at them.

The photos actually prove just the opposite of what you have assumed, but since you're apparently blinkered you might continue shouting forever that they match the calculations. And are you going to deny that perspective exists when objects get far away? What else do you expect to happen to something which gets far away? You want it to get bigger?

And by the way, why are you here in this thread to begin with if you're not contributing? You don't need to answer. That was a rhetorical question.

Was your intent in showing these pictures and diagrams to show a discrepancy in RET? I'm simply pointing out that you've utterly failed. And so it goes with FE'rs.
Title: Re: "Equator" problem
Post by: cikljamas on November 13, 2014, 06:52:23 AM
Should I tell you or you're going to figure it out eventually by yourself that you guys are classic trolls and refuse to accept or even consider anything against your predetermined beliefs(or should I say agenda) while simultaneously trying to disrupt the discussion and disregard any valid arguments? I am obviously not posting the photos for ganders like you, so if it bothers you please don't look at them.

The photos actually prove just the opposite of what you have assumed, but since you're apparently blinkered you might continue shouting forever that they match the calculations. And are you going to deny that perspective exists when objects get far away? What else do you expect to happen to something which gets far away? You want it to get bigger?

If it is needed to become bigger in order to make the Earth round then yes, we expect it to become bigger, why not? hahahaha...

(http://i.imgur.com/8V7mFWx.jpg)

If someone can figure out the exact distances in this photo we would be very pleased if you let us know correct numbers:

(http://i.imgur.com/DpMp5fz.jpg)

Use this link: http://www.amazingnz.com/HoneymoonHeaven-English.html (http://www.amazingnz.com/HoneymoonHeaven-English.html)
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 07:30:20 AM
the numbers you've plugged into that calculator make no sense. A height of a staggering 130 km?

I've found the calculator you are using here: http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)

According to it, for a height of 5642 m (Mt Elburus) the distance to it's horizon is 268 km, and for a height of 3103 m (Gora-Didgali Dudy), the distance to the horizon is 199 km. I'm really not sure how those measurements help anyone here. We need the apparent height for the observer at the distances you provided and the horizon calculator exceeds those numbers.

The failure here for you, is any attempt at showing anyone else's error. Perspective can and will cause things to be smaller the further away they get and if the earth is round, then that effect will be even more dramatic. You've provided data but none that shows one way or the other.
Title: Re: "Equator" problem
Post by: cikljamas on November 13, 2014, 08:16:39 AM
I have already corrected it, but the horizon line can still remain where i put it the first time.

However, the horizon line on that picture shows correct altitude if we take Kvira as a reference point, but of course we won't get in this manner correct result for Mt Elbrus, or for other (more distant than Kvira is) mountains...

If we were to calculate visual horizon for more distant mountains (than Kvira is) then we should have to raise the horizon line for certain amount of mm or cm, in this way the difference between FET and RET version would become even more striking, but i am content even with this soft version since the striking difference between the reality and RET dreams is huge even if we don't raise the horizon line to higher altitudes...
Title: Re: "Equator" problem
Post by: 29silhouette on November 13, 2014, 08:57:24 AM
Do you see the blue line? There is nothing higher than 400 meters all the way to Mt Kvira.
Your point being?  The other mountain I referenced isn't exactly along the line of sight, but it's distance is nearly identical to Kvira. 

Anyway, if you draw a line to Kvira itself, there are some hills 119km away that reach up to 405 meters high.

Here is something more to think about:

(http://i.imgur.com/TOqN8D9.jpg?1)
The hills along the coastline are only between 70 and 120 m high.
And they're only 30-40km away.  So where are the markers for 10-60 meter elevations?  Looks like they'd be below the waterline/horizon because of the curvature. 
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 09:03:58 AM
I have already corrected it, but the horizon line can still remain where i put it the first time.

However, the horizon line on that picture shows correct altitude if we take Kvira as a reference point, but of course we won't get in this manner correct result for Mt Elbrus, or for other (more distant than Kvira is) mountains...

If we were to calculate visual horizon for more distant mountains (than Kvira is) then we should have to raise the horizon line for certain amount of mm or cm, in this way the difference between FET and RET version would become even more striking, but i am content even with this soft version since the striking difference between the reality and RET dreams is huge even if we don't raise the horizon line to higher altitudes...

You used the wrong numbers again. You said Kvira is 2038 m high. This puts the horizon line at 166 km.

What, if anything does that tell us in reference to this picture? Kvira is closer than the horizon.
Title: Re: "Equator" problem
Post by: 29silhouette on November 13, 2014, 09:06:26 AM
I have already corrected it, but the horizon line can still remain where i put it the first time.
You keep wanting to move the horizon.  Why?  If the observer is 6 feet, the horizon/waterline would be 3 miles away.  The calculator I linked to shows it, the calculator Saros linked shows it, and ENaG shows it.  How does the distance and elevation of something beyond that actually change what happens between the observer and the horizon? 

*Yes, the 'horizon' distance (how far away it can be seen) for the highest point of the object in the distance will change depending on it's height and that of the observer, but the physical 'horizon' (the waterline in this case) will only be affected by the observer's elevation.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 09:17:53 AM
I have already corrected it, but the horizon line can still remain where i put it the first time.
You keep wanting to move the horizon.  Why?  If the observer is 6 feet, the horizon/waterline would be 3 miles away.  The calculator I linked to shows it, the calculator Saros linked shows it, and ENaG shows it.  How does the distance and elevation of something beyond that actually change what happens between the observer and the horizon?

The horizon he keeps moving is not the observer's horizon. He keeps moving the horizon for the object we are looking at, Kvira. I have no idea why. Kvira's horizon is larger than the distance between the horizon observer and Kvira. On top of that, he's using the wrong numbers to get Kvira's horizon.

Edit: clarification
Title: Re: "Equator" problem
Post by: 29silhouette on November 13, 2014, 09:38:10 AM
The horizon he keeps moving is not the observer's horizon. He keeps moving the horizon for the object we are looking at, Kvira. I have no idea why. Kvira's horizon is larger than the distance between the horizon observer and Kvira. On top of that, he's using the wrong numbers to get Kvira's horizon.

Edit: clarification
Edited my post for clarification too.  Anyway, using that calculator, I plug in 7 feet for the observer, 6686 ft for Kvira, and get a visual horizon of 103 miles.  Since Kvira is only 80 miles away, and it appears well above the waterline/horizon, I don't see a problem. 
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 09:52:22 AM
The horizon he keeps moving is not the observer's horizon. He keeps moving the horizon for the object we are looking at, Kvira. I have no idea why. Kvira's horizon is larger than the distance between the horizon observer and Kvira. On top of that, he's using the wrong numbers to get Kvira's horizon.

Edit: clarification
Edited my post for clarification too.  Anyway, using that calculator, I plug in 7 feet for the observer, 6686 ft for Kvira, and get a visual horizon of 103 miles.  Since Kvira is only 80 miles away, and it appears well above the waterline/horizon, I don't see a problem.

So what you just said is just absurd. You see no problem if a mountain which is around 6686 ft high and like you said 'well above the horizon' 80 miles away to disappear completely if it were 23-30 miles further away? Are you pretending or yes? What you just said suggests that it takes ~30 miles for a whole mountain of which currently 6686 ft is seen to get from 'well above the horizon' to below it. You see no problem with that?
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 09:58:44 AM
The horizon he keeps moving is not the observer's horizon. He keeps moving the horizon for the object we are looking at, Kvira. I have no idea why. Kvira's horizon is larger than the distance between the horizon observer and Kvira. On top of that, he's using the wrong numbers to get Kvira's horizon.

Edit: clarification
Edited my post for clarification too.  Anyway, using that calculator, I plug in 7 feet for the observer, 6686 ft for Kvira, and get a visual horizon of 103 miles.  Since Kvira is only 80 miles away, and it appears well above the waterline/horizon, I don't see a problem.

So what you just said is just absurd. You see no problem if a mountain which is around 6686 ft high and like you said 'well above the horizon' 80 miles away to disappear completely if it were 23-30 miles further away? Are you pretending or yes? What you just said suggests that it takes ~30 miles for a whole mountain of which currently 6686 ft is seen to get from 'well above the horizon' to below it. You see no problem with that?

Not the whole moutain. Part of the mountain, at 80 miles away, is already gone. 30 more miles would take care of the rest. To add further insult to injury.... you can test this yourself. Take a guess about the math that is used in these calculators. You can bet that pi is involved (which implies that a circle is involved) and you can bet that if the calculator says that an object in the distance would disappear from view at a particular distance away from it, that it will actually do just that.
Title: Re: "Equator" problem
Post by: 29silhouette on November 13, 2014, 10:10:01 AM
The horizon he keeps moving is not the observer's horizon. He keeps moving the horizon for the object we are looking at, Kvira. I have no idea why. Kvira's horizon is larger than the distance between the horizon observer and Kvira. On top of that, he's using the wrong numbers to get Kvira's horizon.

Edit: clarification
Edited my post for clarification too.  Anyway, using that calculator, I plug in 7 feet for the observer, 6686 ft for Kvira, and get a visual horizon of 103 miles.  Since Kvira is only 80 miles away, and it appears well above the waterline/horizon, I don't see a problem.

So what you just said is just absurd. You see no problem if a mountain which is around 6686 ft high and like you said 'well above the horizon' 80 miles away to disappear completely if it were 23-30 miles further away? Are you pretending or yes? What you just said suggests that it takes ~30 miles for a whole mountain of which currently 6686 ft is seen to get from 'well above the horizon' to below it. You see no problem with that?
Much of that 6686 feet is already below the waterline.  Are you saying the calculator you linked us to, and claimed is better, is wrong?
Title: Re: "Equator" problem
Post by: cikljamas on November 13, 2014, 10:12:42 AM
29silhouette, you don't see the problem? Maybe you should close your eyes in order to see better...Eyes wide closed, is that it?

Well, from the beginning blue horizon line has been drawn perfectly correct (it must be due to my perfect intuition  :)), but since you were right about the distance of Mt Kvira, here is improved version which cannot be in better accordance with a position of the horizon line, i would say:

(http://i.imgur.com/jNHFh3F.jpg)

So, if the Earth were round then you would be able to see just last 610 meters of Mt Kvira, that is, first 1428 meters would be below the horizon!!!

If i tried to explain this simple equation to you in to chinese would you understand it easier?

Why do you play the fool?

So to look smarter?
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 10:33:29 AM
Clickajamas, your hypothetic 1428 m high mountain would also be mostly covered by the horizon. As the distance becomes larger, more of an object is covered by the horizon. You're not getting this.
Title: Re: "Equator" problem
Post by: 29silhouette on November 13, 2014, 11:33:07 AM
29silhouette, you don't see the problem? Maybe you should close your eyes in order to see better...Eyes wide closed, is that it?

Well, from the beginning blue horizon line has been drawn perfectly correct (it must be due to my perfect intuition  :)), but since you were right about the distance of Mt Kvira, here is improved version which cannot be in better accordance with a position of the horizon line, i would say:

(http://i.imgur.com/jNHFh3F.jpg)

So, if the Earth were round then you would be able to see just last 610 meters of Mt Kvira, that is, first 1428 meters would be below the horizon!!!

If i tried to explain this simple equation to you in to chinese would you understand it easier?

Why do you play the fool?

So to look smarter?
'Eyes wide shut' if you're refering to the movie. 

Your picture shows the blue line at an area around 125km away at an elevation in the area of 1400 meters.  the peak of Kvira is about 130km.  Why did you decide on 1428m, which is visible out to 140km according to your calculator results, to subtract from 2038m for a peak 130km away, which has a visible horizon of 166km, and then claim those 610 meter should be the only portion visible above the waterline/horizon, yet show that portion (area above blue line) well above the water/horizon?
Title: Re: "Equator" problem
Post by: JimmyTheCrab on November 13, 2014, 12:06:00 PM
I don't understand why they can't understand.
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 02:30:49 PM
I don't understand why they can't understand.

Mainly because you're wrong or pretend you don't understand what you see.
 
Check this out:

H1 = 2 m
H2=  1100 m

Maximum horizon distance = 123 km (H2 seen well above the horizon from 125 km away, but according to the formula it should already be below it, if the distance is bigger than 123 km)

In fact, you can still see the 900 m mark of Mt Kvira from 125 km away above the horizon, and it should be well below it when observed from 2 m height.

H1=2 m
H2=900 m

Maximum horizon distance = 112 km

Now if you still don't understand that what you see in the photo simply doesn't match the calculated horizon distance, I can't help you out ...If you could explain why there are constant discrepancies in what the formula produces and the reality, I would appreciate it. These are discrepancies over small distances, but they add up and over a distance of 1000 km the discrepancy may be significant. Do we not know the correct Earth's radius? Why exactly is there a discrepancy all the time?

(http://i.imgur.com/CRtUqHU.jpg?1)

Horizon distance calculator used: http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)

Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 02:36:04 PM
Saros, where are you getting the idea that we can see the 900m mark?
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 02:42:32 PM
Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 13, 2014, 02:57:33 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 03:10:36 PM
Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.

We use the Earth's radius in the formula. A circle is a circle and has a radius. By the way, if the Earth is flat there still will be a horizon.

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/ (http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/)

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon. Why are you still arguing when you don't know basic facts?
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 03:13:01 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 13, 2014, 03:19:07 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 03:30:08 PM
Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.

We use the Earth's radius in the formula. A circle is a circle and has a radius. By the way, if the Earth is flat there still will be a horizon.

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/ (http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/)

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon. Why are you still arguing when you don't know basic facts?

What you said earlier implied that you think the formulas use of pi is because of a flat earths planar circle. Nothing I said implied that I don't know the earth's radius is used. It is a side profile.

What you are arguing is hearsay. You are throwing numbers around without any source. The only numbers we know are the heights of the mountains. Saying that you know the height of some point halfway down the mountain is a lie.
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 03:34:03 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?


Did you read the link? There will be a horizon even if the Earth is flat.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 03:36:22 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?


Did you read the link? There will be a horizon even if the Earth is flat.

Clearly we disagree. We can at least agree that it wouldn't be 5 km away for 2 meter height. Yet it apparently is.
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 03:39:16 PM
Also, the circle that is used is a side profile/slice of the sphere. Not the planar circle that you are implying.

We use the Earth's radius in the formula. A circle is a circle and has a radius. By the way, if the Earth is flat there still will be a horizon.

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/ (http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/)

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon. Why are you still arguing when you don't know basic facts?

What you said earlier implied that you think the formulas use of pi is because of a flat earths planar circle. Nothing I said implied that I don't know the earth's radius is used. It is a side profile.

What you are arguing is hearsay. You are throwing numbers around without any source. The only numbers we know are the heights of the mountains. Saying that you know the height of some point halfway down the mountain is a lie.

Well, it is not a lie. You're not paying attention.
http://www.udeuschle.selfhost.pro/panoramas/panqueryfull.aspx?mode=newstandard&data=lon%3A41.64471%24%24%24lat%3A41.64906%24%24%24alt%3Aauto%24%24%24altcam%3A10%24%24%24hialt%3Afalse%24%24%24resolution%3A200%24%24%24azimut%3A360%24%24%24sweep%3A60%24%24%24leftbound%3A16.55000000000001%24%24%24rightbound%3A22.55000000000001%24%24%24split%3A60%24%24%24splitnr%3A1%24%24%24tilt%3A0.47916666666666696%24%24%24tiltsplit%3Afalse%24%24%24elexagg%3A1.2%24%24%24range%3A300%24%24%24colorcoding%3Afalse%24%24%24colorcodinglimit%3A231%24%24%24title%3AZugspitze%20%5BTelescope%5D%24%24%24description%3A%24%24%24email%3A%24%24%24language%3Aen%24%24%24screenwidth%3A1366%24%24%24screenheight%3A728 (http://www.udeuschle.selfhost.pro/panoramas/panqueryfull.aspx?mode=newstandard&data=lon%3A41.64471%24%24%24lat%3A41.64906%24%24%24alt%3Aauto%24%24%24altcam%3A10%24%24%24hialt%3Afalse%24%24%24resolution%3A200%24%24%24azimut%3A360%24%24%24sweep%3A60%24%24%24leftbound%3A16.55000000000001%24%24%24rightbound%3A22.55000000000001%24%24%24split%3A60%24%24%24splitnr%3A1%24%24%24tilt%3A0.47916666666666696%24%24%24tiltsplit%3Afalse%24%24%24elexagg%3A1.2%24%24%24range%3A300%24%24%24colorcoding%3Afalse%24%24%24colorcodinglimit%3A231%24%24%24title%3AZugspitze%20%5BTelescope%5D%24%24%24description%3A%24%24%24email%3A%24%24%24language%3Aen%24%24%24screenwidth%3A1366%24%24%24screenheight%3A728)
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 03:40:54 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?


Did you read the link? There will be a horizon even if the Earth is flat.

Clearly we disagree. We can at least agree that it wouldn't be 5 km away for 2 meter height. Yet it apparently is.

Haha, so you disagree with the link I gave you? Good job. Write him an e-mail. The guy is a mathematician.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 03:47:32 PM
You must be drunk.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 13, 2014, 04:06:01 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?


Did you read the link? There will be a horizon even if the Earth is flat.

It says the horizon would be at approximately the same angle (except if it isn't, then it would appear to rise until directly overhead). What he doesn't say is anything about a sharp line about 5 km away - instead he says it would be slightly higher (by almost 0.04 degree and way off in the distance).
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 04:07:49 PM
You must be drunk.

Did you read the link?

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/ (http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/)

This is what it says, because apparently you can't read.

"Q: If Earth was flat, would there be a horizon? If so, what would it look like? If the Earth was flat and had infinite area, would that change the answer?

Physicist: There’d definitely still be a horizon if the Earth were flat.  It would be in almost exactly the same place, and look essentially identical.  "

So, you completely ignored my other arguments that even the calculator doesn't show correct results which match reality, now you don't want to accept this information. Great.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 04:25:21 PM
You must be drunk.

Did you read the link?

http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/ (http://www.askamathematician.com/2012/08/q-if-earth-was-flat-would-there-be-the-horizon-if-so-what-would-it-look-like-if-the-earth-was-flat-and-had-infinite-area-would-that-change-the-answer/)

This is what it says, because apparently you can't read.

"Q: If Earth was flat, would there be a horizon? If so, what would it look like? If the Earth was flat and had infinite area, would that change the answer?

Physicist: There’d definitely still be a horizon if the Earth were flat.  It would be in almost exactly the same place, and look essentially identical.  "

So, you completely ignored my other arguments that even the calculator doesn't show correct results which match reality, now you don't want to accept this information. Great.

I've never commented on your link. Get a grip.
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 04:26:51 PM
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.

Do you really think this is an issue, or are you just generating more chaff?

Your sightline and the center of the Earth define a plane, since a line and point not on the line define a plane. The intersection of any plane with a sphere is a circle. Since the plane contains the center of the sphere in this case, it's a great circle - that is, a circle with the same radius as the sphere. That's why a circle is all you need to calculate distance to the horizon.

Don't they teach geometry in school any more? Or do they teach it but the ones who never paid attention are the ones arguing the Earth is flat?

Well, your whole horizon argument proving RE due to observed curvature falls apart when we take into account the fact that on a flat Earth there still will be a horizon pretty much in the same place. How about that?
Why would there be a horizon at all on a flat Earth? Wouldn't things just fade into the distance when looking over, say, the open ocean instead of disappearing from the bottom up?

You didn't answer the question about whether your "circle" question was real or not. Were you sleeping through geometry?


Did you read the link? There will be a horizon even if the Earth is flat.

It says the horizon would be at approximately the same angle (except if it isn't, then it would appear to rise until directly overhead). What he doesn't say is anything about a sharp line about 5 km away - instead he says it would be slightly higher (by almost 0.04 degree and way off in the distance).

Haha, you continue trolling. So now your problem is that there is a horizon about 5 km away(according to your math only) as if the photos here were not enough proof for the contrary? Didn't you see that the calculators are off? Of course there is an observed drop, but it has nothing to do with the Earth's curvature. And by the way, go to any lake and try to find a horizon 5 km away. Stop showing off  and go out and check for yourself. As for the guy in the link, he clearly stated that the horizon will be essentially identical in both scenarios.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 04:32:39 PM
I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.
Title: Re: "Equator" problem
Post by: Saros on November 13, 2014, 04:58:19 PM
I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

So you agree that on a flat Earth there is a horizon too, but at the same time you keep arguing that since the ships appear to 'sink' below the horizon it proves that the Earth is round? What conclusions have I drawn? What are you talking about? I showed you that objects appear higher in real life than what is predicted by the formula. You didn't agree with the numbers. I gave you a link to check it for yourself. There you can see the altitude for any given point of any mountain not only the peaks. You disregarded that. The whole time everyone here was arguing that since there is a visible dip below the horizon this proves the Earth is round. I showed you that even with flat Earth we still have a horizon and it is pretty much the same, I also showed you that the calculations for the position of the RE horizon are wrong when checked in practice.
Title: Re: "Equator" problem
Post by: rottingroom on November 13, 2014, 05:09:43 PM
I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

So you agree that on a flat Earth there is a horizon too, but at the same time you keep arguing that since the ships appear to 'sink' below the horizon it proves that the Earth is round? What conclusions have I drawn? What are you talking about? I showed you that objects appear higher in real life than what is predicted by the formula. You didn't agree with the numbers. I gave you a link to check it for yourself. There you can see the altitude for any given point of any mountain not only the peaks. You disregarded that. The whole time everyone here was arguing that since there is a visible dip below the horizon this proves the Earth is round. I showed you that even with flat Earth we still have a horizon and it is pretty much the same, I also showed you that the calculations for the position of the RE horizon are wrong when checked in practice.

This is why I think you are drunk. I see links that prove the opposite of what you are saying. WTF are you talking about?
Title: Re: "Equator" problem
Post by: Rama Set on November 13, 2014, 05:31:19 PM
I mean I agree with everything in the article. I don't see where you've drawn any of the conclusions you've made.

So you agree that on a flat Earth there is a horizon too, but at the same time you keep arguing that since the ships appear to 'sink' below the horizon it proves that the Earth is round? What conclusions have I drawn? What are you talking about? I showed you that objects appear higher in real life than what is predicted by the formula. You didn't agree with the numbers. I gave you a link to check it for yourself. There you can see the altitude for any given point of any mountain not only the peaks. You disregarded that. The whole time everyone here was arguing that since there is a visible dip below the horizon this proves the Earth is round. I showed you that even with flat Earth we still have a horizon and it is pretty much the same, I also showed you that the calculations for the position of the RE horizon are wrong when checked in practice.

On a FE, an object would not disappear from the bottom up on the horizon, it would get smaller and smaller until it vanished all together.
Title: Re: "Equator" problem
Post by: 29silhouette on November 13, 2014, 07:10:36 PM
In fact, you can still see the 900 m mark of Mt Kvira from 125 km away above the horizon, and it should be well below it when observed from 2 m height.

H1=2 m
H2=900 m

Maximum horizon distance = 112 km

Now if you still don't understand that what you see in the photo simply doesn't match the calculated horizon distance, I can't help you out ...If you could explain why there are constant discrepancies in what the formula produces and the reality, I would appreciate it.
That's easy.  Refraction.... the same reason the shoreline buildings are visible in both shots of the bridge and hillside I posted.

Quote
These are discrepancies over small distances, but they add up and over a distance of 1000 km the discrepancy may be significant. Do we not know the correct Earth's radius? Why exactly is there a discrepancy all the time?

(http://i.imgur.com/CRtUqHU.jpg?1)
And where do you intend to place the marks for 100-800 meters?

Quote
Horizon distance calculator used: http://members.home.nl/7seas/radcalc.htm (http://members.home.nl/7seas/radcalc.htm)
Are you saying the calculator that is "better" is wrong?

Quote
Also, you might have noticed that the formula for calculating horizon distance doesn't use a sphere, but a circle. I guess it doesn't matter that a circle is flat.
  Comedy gold.  This is Sceptimatic level material right here.
Title: Re: "Equator" problem
Post by: cikljamas on November 14, 2014, 02:58:41 AM
Quote
Physicist says:

"There’d definitely still be a horizon if the Earth were flat.  It would be in almost exactly the same place, and look essentially identical. "

I cannot open the link, but above sentence will do solely....All we need here is to use a little bit logic...

This sentence looks to me as if a perpetrator has confessed to the crime.

Why?

Because, what this scientist claims in this sentence is an obvious lie!

Discrepancies between the reality and horizon calculators proves it.
Quote
Saros says:

What I am arguing about here is that the calculators estimate the horizon distance wrongly, and even though the discrepancy is small, it happens to be in favor of the flat Earth horizon.

You can rely on these horizon calculators only in the fog, because in the fog you can't see anything anyway, but as soon as fog dissapears you can discard calculator estimations because they are simply wrong.

Now, you have to ask yourself why would someone claim such an obvious lie (as all physicists do when they have to make references to "horizon issue")?

If the Earth were round nobody would ever came up to the idea to claim such a stupidity, but all physicists do exactly that all the time when they are faced with "horizon issue"!

Try to imagine quite opposite situation:

Imagine that a mainstream science finally embraces FET and admits it's trueness!

Now imagine that someone ask question like this:

If the world was actually round, what would the horizon look like?

You know, and i know that no one would ever come up to the idea to answer in this manner:

"There’d definitely still be a horizon if the Earth were round.  It would be in almost exactly the same place, and look essentially identical."

You just have to start to think, but i know that you can't do that because you cannot afford to yourself that kind of luxury, by doing that you would allow to yourself to think openly, and then you would have to discard all your prejudices, and then you would become a flat earthers, but you would rather die than confess that the Earth is flat, or that God exists! Wouldn't you?

So you are a prisoners of your closed minds who serve a life sentence!

Quote
Another perpetrator confessing to the crime:

On Earth, the land you see will drop off at a rate of about 0.000126 miles per mile (well, for the first mile anyway). That is small enough to make the Earth appear flat by human observation, at least when you consider that mountains and valleys prevent us from seeing infinitely far. The angle of the horizon and clouds would appear remarkably similar, as in both round and flat Earth they would appear to converge toward a single line in very similar ways. On a flat Earth, you would see more clouds as they shrink toward the horizon (assuming ideal cloud conditions), but that is already the way clouds appear to behave anyway, more or less.

That small distance adds up fairly quickly if you are looking over an ocean however. A normal human at sea level can see about 3 miles or so. On a flat plane, you could see as far as the height of the waves allowed, basically. Very far, in ideal circumstances. So you can tell the Earth is round at flat zones like oceans or big lakes.

http://www.reddit.com/r/askscience/comments/kjrbt/if_the_world_was_actually_flat_what_would_the/ (http://www.reddit.com/r/askscience/comments/kjrbt/if_the_world_was_actually_flat_what_would_the/)

Gotcha!!!
Title: Re: "Equator" problem
Post by: 29silhouette on November 14, 2014, 08:54:35 AM
Quote
Another perpetrator confessing to the crime:

On Earth, the land you see will drop off at a rate of about 0.000126 miles per mile (well, for the first mile anyway). That is small enough to make the Earth appear flat by human observation, at least when you consider that mountains and valleys prevent us from seeing infinitely far. The angle of the horizon and clouds would appear remarkably similar, as in both round and flat Earth they would appear to converge toward a single line in very similar ways. On a flat Earth, you would see more clouds as they shrink toward the horizon (assuming ideal cloud conditions), but that is already the way clouds appear to behave anyway, more or less.

That small distance adds up fairly quickly if you are looking over an ocean however. A normal human at sea level can see about 3 miles or so. On a flat plane, you could see as far as the height of the waves allowed, basically. Very far, in ideal circumstances. So you can tell the Earth is round at flat zones like oceans or big lakes.

http://www.reddit.com/r/askscience/comments/kjrbt/if_the_world_was_actually_flat_what_would_the/ (http://www.reddit.com/r/askscience/comments/kjrbt/if_the_world_was_actually_flat_what_would_the/)

Gotcha!!!
Of course clouds would grow smaller and get closer to the horizon on the FE.  There's still going to be perspective, and 'flat zones' means areas with a constant elevation, no hills, mountains, obstructions, etc.  If you want to take it literally, "flat zones" means there would be opposing 'zones' that aren't flat, therefore "curved zones", which means Earth is both flat and spherical. 
Title: Re: "Equator" problem
Post by: cikljamas on November 15, 2014, 11:06:52 AM
(http://i.imgur.com/9U2g34Z.jpg)

Let  us  turn  to  Fig.  14  to  illustrate  this  fact.  Let  the point  E   represent  the  position  of  the  observer  on  the  sea-level  ;  his  line  of  sight  would  be  a  tangent  to  the  sphere  at the  place  of  observation,  as  shewn  by  the  line  E  H,  and  the  dip  of  an  object  at  J  would  be  represented  by  the  line  H  J. Now  raise  the  observer  to  the  height  of  the  telescope  at  F  ; his  line  of  sight  is  still a  horizontal  line  in  the  direction  of G, and  parallel  to  E  H,  therefore  the  dip  from  G  to  J  is manifestly  greater  than  that  from  H  to  J.  And  this  is  true whether  we  reckon  the  dip  towards  the  centre  of  the  globe in  the  direction  of  G  L,  or  at  right-angles  from  the  line  of sight  G  M.
Title: Re: "Equator" problem
Post by: ausGeoff on November 15, 2014, 12:39:53 PM
(http://i.imgur.com/9U2g34Z.jpg)

Why is it that flat earthers invariably rely on "science" texts written a century or more ago?  Why is it that they have no texts written even in the 20th century?

The above extract is from a book written by an anonymous flat-earth author called "Zetetes" who actually believed in the existence of the Christian god, and that this imaginary god created the heavens and the earth.  Obviously then Zetetes was no scientist LOL.

It's sad that people like cikljamas just can't—or won't—accept the obvious fact that scientific knowledge and awareness of our cosmos has advance logarithmically since the period of this book's writing.  Most flat earthers seem convinced that the advancement of science ground to a halt in 1899.

So... the extract above—and all the contents of the book "The Sea-Earth Globe and its Monstrous Hypothetical Motions or Modern Theoretical Astronomy" are nothing more than imaginative fiction, or an amusing, historical literary artifact of times long gone.

—Science it is not.
Title: Re: "Equator" problem
Post by: 29silhouette on November 15, 2014, 01:25:46 PM
Now  raise  the  observer  to  the  height  of  the  telescope  at  F  ; his  line  of  sight  is  still a  horizontal  line  in  the  direction  of G, and  parallel  to  E  H,  therefore  the  dip  from  G  to  J  is manifestly  greater  than  that  from  H  to  J. 
For parallel level lines of sight from two different heights, the difference between G-J and H-J is the same as the height of the telescope regardless of distance according to the diagram.

*edit for clarification- Actually, if the difference between H and G is based of a distant point that is 90 degrees to the ground/center of the globe, then yes, it will increase slightly, but not much (or even be noticeable) for an observer on the ground.

Quote
...the height of the observer should in strictness be added to the amount of the dip
No, the dip is going to be the same.  The height of the observer allows viewing of objects between H and J and reduces the amount hidden from view below the 'dip'.
Title: Re: "Equator" problem
Post by: Alpha2Omega on November 15, 2014, 07:42:35 PM
(http://i.imgur.com/9U2g34Z.jpg)

Let  us  turn  to  Fig.  14  to  illustrate  this  fact.  Let  the point  E   represent  the  position  of  the  observer  on  the  sea-level  ;  his  line  of  sight  would  be  a  tangent  to  the  sphere  at the  place  of  observation,  as  shewn  by  the  line  E  H,  and  the  dip  of  an  object  at  J  would  be  represented  by  the  line  H  J. Now  raise  the  observer  to  the  height  of  the  telescope  at  F  ; his  line  of  sight  is  still a  horizontal  line  in  the  direction  of G, and  parallel  to  E  H,  therefore  the  dip  from  G  to  J  is manifestly  greater  than  that  from  H  to  J.  And  this  is  true whether  we  reckon  the  dip  towards  the  centre  of  the  globe in  the  direction  of  G  L,  or  at  right-angles  from  the  line  of sight  G  M.
"on the sea-level"? "shewn"? How quaint. Let's get our 19th-century on here, shall we!

What are you and the original author trying to show with this? Where did this come from, anyway? When you paste text like this, or quote it, it's at least a courtesy to the original author to give attribution. It also helps your readers see any omitted context or supporting information, too. For instance, where is Fig. 11? It's referenced in the text, so you need to show it.

To the accuracy of his approximation "ignoring some small decimal points", the straight-line distance EH in Fig. 14 is the same as the length of the curved line EJ. If we accept 3 digits of precision, you can go about as far as about 625 miles (9 degrees of angle) before the error in the length of HJ starts to blow up.

The height EF for "a telescope" is enormously exaggerated here - probably 10% the radius of the circle. The height EF is going to be a trifle compared with EC (C at the center of the circle) for any real terrestrial instrument, and for HJ at significant distance.

If you want to do this for an object in space, where EF becomes a significant factor (or even a better answer if earthbound), use the correct formulas, not the approximation described. We've had pocket calculators for 40 years that can easily give much more exact answers using the correct formulas instead of the approximations that were needed as a practical matter 150 years ago.

Title: Re: "Equator" problem
Post by: ausGeoff on November 15, 2014, 07:59:34 PM
Where did this come from, anyway? When you paste text like this, or quote it, it's at least a courtesy to the original author to give attribution. It also helps your readers see any omitted context or supporting information, too. For instance, where is Fig. 11? It's referenced in the text, so you need to show it.

Check out the whole document HERE (http://bit.ly/1EL7Zmb).  It's a true comedy of errors.    ;D
Title: Re: "Equator" problem
Post by: cikljamas on November 16, 2014, 08:32:37 AM
What are you and the original author trying to show with this?

1. 1660 km/h(supposed speed of Earth's rotation) : 3600 = 0,46 km/s

0,46 * 60 = 27,6 km/min .... 27,6 : 1,5 = 18,4 miles

Now, using Rowbotham's (verified) formula we calculate like this:
18,4 * 18,4 = 338,56 * 8 (inches) = 2708 * 2,5 = 6771,2 cm = 67 m roughly

So, if the Earth is round the Sun should drop off behind horizon 67m/min which is equivalent of climbing uphill 67m/min...

8880 m (Everest altitude) : 67 = 132,5 min = 2 hours, 12,5 minutes

So, if the Earth is rotund one guy standing on the top of the Everest should be able to see sunrise 2,12 hours earlier/sunset 2,12 hours longer than the other guy who stands at the bottom of the Everest...

As we all very well know it is not the case, not even closely to that...

Disprove this if you can!

2.

(http://i.imgur.com/SIbKZEz.jpg)

Accompanying video : (http://)
Title: Re: "Equator" problem
Post by: Rama Set on November 16, 2014, 09:27:06 AM
What are you and the original author trying to show with this?

1. 1660 km/h(supposed speed of Earth's rotation) : 3600 = 0,46 km/s

0,46 * 60 = 27,6 km/min .... 27,6 : 1,5 = 18,4 miles

Now, using Rowbotham's (verified) formula we calculate like this:
18,4 * 18,4 = 338,56 * 8 (inches) = 2708 * 2,5 = 6771,2 cm = 67 m roughly

So, if the Earth is round the Sun should drop off behind horizon 67m/min which is equivalent of climbing uphill 67m/min...

8880 m (Everest altitude) : 67 = 132,5 min = 2 hours, 12,5 minutes

So, if the Earth is rotund one guy standing on the top of the Everest should be able to see sunrise 2,12 hours earlier/sunset 2,12 hours longer than the other guy who stands at the bottom of the Everest...

As we all very well know it is not the case, not even closely to that...

Disprove this if you can!

2.

(http://i.imgur.com/SIbKZEz.jpg)

Accompanying video : (http://