"Equator" problem

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cikljamas

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Re: "Equator" problem
« Reply #120 on: October 23, 2014, 04:27:22 AM »
@Alpha2Omega...



...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: " class="bbc_link" target="_blank" rel="noopener noreferrer">

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:



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:



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:



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!
« Last Edit: October 23, 2014, 06:46:47 AM by cikljamas »
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Rama Set

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Re: "Equator" problem
« Reply #121 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.
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

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QuQu

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Re: "Equator" problem
« Reply #122 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.

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cikljamas

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Re: "Equator" problem
« Reply #123 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/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
« Last Edit: October 23, 2014, 07:31:24 AM by cikljamas »
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Rama Set

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Re: "Equator" problem
« Reply #124 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/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


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.

« Last Edit: October 23, 2014, 07:51:58 AM by Rama Set »
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cikljamas

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Re: "Equator" problem
« Reply #125 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

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
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Re: "Equator" problem
« Reply #126 on: October 23, 2014, 09:05:03 AM »
@Alpha2Omega...



...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 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.



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:


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

Re: "Equator" problem
« Reply #127 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
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

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
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
« Last Edit: October 23, 2014, 11:22:06 AM by 29silhouette »

Re: "Equator" problem
« Reply #128 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.

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cikljamas

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Re: "Equator" problem
« Reply #129 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

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...
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Rama Set

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Re: "Equator" problem
« Reply #130 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.
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

Re: "Equator" problem
« Reply #131 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.
« Last Edit: October 23, 2014, 01:27:27 PM by 29silhouette »

Re: "Equator" problem
« Reply #132 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. 

Re: "Equator" problem
« Reply #133 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
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.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #134 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
FACT 2: http://www.energeticforum.com/265210-post527.html
FACT 3: http://www.energeticforum.com/263816-post278.html
FACT 4: http://www.energeticforum.com/263172-post251.html
FACT 5: http://www.energeticforum.com/263904-post281.html


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sokarul

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Re: "Equator" problem
« Reply #135 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
FACT 2: http://www.energeticforum.com/265210-post527.html
FACT 3: http://www.energeticforum.com/263816-post278.html
FACT 4: http://www.energeticforum.com/263172-post251.html
FACT 5: http://www.energeticforum.com/263904-post281.html
Hiding from all the hard arguments and posting the same incorrect links will not help you.
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Rama Set

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Re: "Equator" problem
« Reply #136 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?
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

Re: "Equator" problem
« Reply #137 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
FACT 2: http://www.energeticforum.com/265210-post527.html
FACT 3: http://www.energeticforum.com/263816-post278.html
FACT 4: http://www.energeticforum.com/263172-post251.html
FACT 5: 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.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

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Re: "Equator" problem
« Reply #138 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" class="bbc_link" target="_blank" rel="noopener noreferrer">#t=52m2s

DAWKINS - BEN STEIN interview : #t=3m26s" class="bbc_link" target="_blank" rel="noopener noreferrer">#t=3m26s

Richard Dawkins: Embarrassingly Bad Thinker, Incredibly Ignorant : " class="bbc_link" target="_blank" rel="noopener noreferrer">

The Science Delusion BANNED TED TALK: " class="bbc_link" target="_blank" rel="noopener noreferrer">

Moral consequences of a bad and wrong philosophy and politically determined "science" : #t=1h5m40s" class="bbc_link" target="_blank" rel="noopener noreferrer">#t=1h5m40s

Political consequences of a bad and wrong philosophy and politically determined "science" : Oliver Stone Tears Apart Obama's Empire : " class="bbc_link" target="_blank" rel="noopener noreferrer">

CNN Wants This Video Banned (SEE WHY) : " class="bbc_link" target="_blank" rel="noopener noreferrer">

These 5 Censored Books Tell a History the Establishment Wants Hidden " class="bbc_link" target="_blank" rel="noopener noreferrer">

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...
« Last Edit: October 24, 2014, 04:39:15 AM by cikljamas »
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #139 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.


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

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sokarul

  • 19303
  • Extra Racist
Re: "Equator" problem
« Reply #140 on: October 24, 2014, 09:29:20 AM »

The Science Delusion BANNED TED TALK: " class="bbc_link" target="_blank" rel="noopener noreferrer">

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.
« Last Edit: October 24, 2014, 09:31:50 AM by sokarul »
ANNIHILATOR OF  SHIFTER

It's no slur if it's fact.

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: "Equator" problem
« Reply #141 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!!!

« Last Edit: October 24, 2014, 11:55:35 AM by cikljamas »
"I can't breathe" George Floyd RIP

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: "Equator" problem
« Reply #142 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."
« Last Edit: October 24, 2014, 02:27:44 PM by cikljamas »
"I can't breathe" George Floyd RIP

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Misero

  • 1261
  • Of course it's flat. It looks that way up close.
Re: "Equator" problem
« Reply #143 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

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

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
I am the worst moderator ever.

Sometimes I wonder: "Why am  I on this site?"
Then I look at threads about clouds not existing and I go back to posting and lurking. Lurk moar.

Re: "Equator" problem
« Reply #144 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 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.
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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Rama Set

  • 6877
  • I am also an engineer
Re: "Equator" problem
« Reply #145 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.
Aether is the  characteristic of action or inaction of charged  & noncharged particals.

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: "Equator" problem
« Reply #146 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):




« Last Edit: October 25, 2014, 12:39:26 AM by cikljamas »
"I can't breathe" George Floyd RIP

Re: "Equator" problem
« Reply #147 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, and causes the 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".
"Everyone is entitled to his own opinion, but not to his own facts." - Daniel Patrick Moynihan

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cikljamas

  • 2432
  • Ex nihilo nihil fit
Re: "Equator" problem
« Reply #148 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...
« Last Edit: October 25, 2014, 02:15:10 PM by cikljamas »
"I can't breathe" George Floyd RIP

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Socratic Amusement

  • 636
  • An Exercise in Witty Exploration
Re: "Equator" problem
« Reply #149 on: October 25, 2014, 02:20:56 PM »



Man, Alpha2Omega is just killing cikljamas.

This is better than watching Spartacus.
"As for me, all I know is that I know nothing."