"Equator" problem

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"Equator" problem
« 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

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!
« Last Edit: October 15, 2014, 06:12:13 AM by cikljamas »

Re: "Equator" problem
« Reply #1 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

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.

Re: "Equator" problem
« Reply #2 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

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markjo

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Re: "Equator" problem
« Reply #3 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?
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

?

BJ1234

  • 1931
Re: "Equator" problem
« Reply #4 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
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.

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Rama Set

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Re: "Equator" problem
« Reply #5 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

The circumference measured around the North-South circumference is less that the circumference around the Equator. 
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

Re: "Equator" problem
« Reply #6 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

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!!!
« Last Edit: October 15, 2014, 07:06:06 AM by cikljamas »

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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #7 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.

Re: "Equator" problem
« Reply #8 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

Re: "Equator" problem
« Reply #9 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

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.

Re: "Equator" problem
« Reply #10 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
"Daily rate of annual motion of the stars - a constant" argument : http://www.energeticforum.com/265053-post497.html
"Lighthouses" argument : http://www.energeticforum.com/264766-post457.html
"No rotation - No revolution" argument : http://www.energeticforum.com/264738-post450.html
"Midnight Sun - Polaris - Southern Cross" argument : http://www.energeticforum.com/264204-post367.html
"Antarctica nothing alike Arctic" argument : http://www.energeticforum.com/264212-post368.html
"Water level experiments" argument : http://www.energeticforum.com/264258-post373.html
"Southern Hemisphere" argument : http://www.energeticforum.com/263904-post281.html
"Moon" argument: http://www.energeticforum.com/263172-post251.html
"Sextant" argument : http://www.energeticforum.com/264972-post486.html
"Triangulation" argument : 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!

Re: "Equator" problem
« Reply #11 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

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!!!

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sokarul

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Re: "Equator" problem
« Reply #12 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
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
Solar time vs sidereal time, learn it.

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

Quote
"No rotation - No revolution" argument : 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
Shows the Earth to be round, thanks?

Quote
"Antarctica nothing alike Arctic" argument : 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?
Sokarul

ANNIHILATOR OF  SHIFTER

Re: "Equator" problem
« Reply #13 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!

Re: "Equator" problem
« Reply #14 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.

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sokarul

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Re: "Equator" problem
« Reply #15 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.
Sokarul

ANNIHILATOR OF  SHIFTER

Re: "Equator" problem
« Reply #16 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???
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: "Equator" problem
« Reply #17 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" 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:



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.
« Last Edit: October 15, 2014, 07:20:20 PM by Alpha2Omega »
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

Re: "Equator" problem
« Reply #18 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)



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

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

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!!!

Re: "Equator" problem
« Reply #19 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)



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

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

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.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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markjo

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  • The Elder Ones
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Re: "Equator" problem
« Reply #20 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)



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.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

Re: "Equator" problem
« Reply #21 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.  ??? ??? ???
« Last Edit: October 16, 2014, 10:02:48 AM by cikljamas »

Re: "Equator" problem
« Reply #22 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...

Re: "Equator" problem
« Reply #23 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?

« Last Edit: October 17, 2014, 02:47:55 AM by cikljamas »

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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #24 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!

Re: "Equator" problem
« Reply #25 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.

Re: "Equator" problem
« Reply #26 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?

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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #27 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.



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.


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markjo

  • Content Nazi
  • The Elder Ones
  • 37581
Re: "Equator" problem
« Reply #28 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.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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ausGeoff

  • 6091
Re: "Equator" problem
« Reply #29 on: October 16, 2014, 11:31:36 PM »
Also a convex meniscus in a pipette...





EDIT:  [img] coding corrected.



« Last Edit: October 18, 2014, 05:21:52 AM by ausGeoff »