The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Debate => Topic started by: fshy94 on February 25, 2008, 12:04:25 PM

http://www.perseus.gr/AstroLunarParallax.htm
So, what's the explanation for this?
Curvature/Noncurvature of the Earth cannot account for such a tremendous difference in distance, can it? So, what gives? Furthermore, this experiment can be carried out by anyone...no?

Did the author assume a Round Earth or a Flat Earth in his astronomical parallax calculations?
The baselines in Astronomical Parallax are different when you assume that the two distant points on earth are upon a sphere or on a plane; therefore the angle of a celestial body in the sky means something entirely different depending on which model we assume.
For example, we can use Astronomical Parallax with the assumption of a Flat Earth to calculate the distance to the sun: On March 2122 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.

http://www.perseus.gr/AstroLunarParallax.htm
So, what's the explanation for this?
Curvature/Noncurvature of the Earth cannot account for such a tremendous difference in distance, can it? So, what gives? Furthermore, this experiment can be carried out by anyone...no?
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Can you read? You're telling me a hundredfold difference is possible to explain via at most, via curvature of 500 miles over such a distance?
Look, and pay attention:
(http://img526.imageshack.us/img526/8672/fepx2fz2.th.jpg) (http://img526.imageshack.us/my.php?image=fepx2fz2.jpg)

He didn't even read the question, just answered with something that had to do with the title

Can you read? You're telling me a hundredfold difference is possible to explain via at most, via curvature of 500 miles over such a distance?
Yes. Astronomical Parallax means something entirely different on a Flat Earth than a round one.

Look, and pay attention:
You forgot that Astronomical Parallax requires two known points on the earth's surface to gauge an object's distance.
For example, if you saw a plane flying directly overhead could you judge how far away it is without knowing in advance how big the plane was? No. For all you know the plane you see could be a model plane at a low altitude. In order to determine the plane's precise altitude you would need to observe it from two different view points simultaneously. If you called up a friend a quarter of a mile away from your location and asked him to look at the plane directly over you he could tell you the angle it is from the horizon line. From there you could use simple triangles and a little geometry under the assumption of a Flat Earth and guesstimate precisely how high the plane is.
If your friend saw that the plane was 45 degrees in the sky, knowing that the plane is directly over you, you could create an imaginary isosceles right angled triangle. Since all isosceles right angled triangle must have arms of equal length, you could conclude that since your friend is a quarter of a mile away horizontally, the plane must therefore be a quarter of a mile in vertical altitude.
Now, since the sun is much higher than an airplane, the second observer would need to travel much much farther away in order to see a significant measurable decline in the sun's angle to the horizon. Over such distances the curvature of the earth, or lack thereof, must be assumed, since when you are traveling across a globe the base of a triangle is no longer flat. Accounting for the curvature of the earth with a bit of trigonometry and studying the sun's angle in the sky is exactly how astronomers have calculated the sun's distance from the earth in the RE model.
As another example, this link (http://www.serviastro.am.ub.es/engl/monographs/distance_earth_sun_transit_venus.html) shows us how modern Round Earth science calculated the distance between the Earth and Sun using the transit of Venus. You will immediately notice that the equations are highly dependent on the assumption of a Round Earth, needing two observers at far off distances. The math explicitly accounts for the earth's assumed curvature.
If we take those same triangulation equations in the link and use them under the assumption of a Flat Earth and do away with the compensation for the curvature of the Earth, assuming a flat surface, the final equation for the distance between the Earth and Sun becomes
rT = (PT / PV)2/3 (1  eT cos ET) / (1  eV cos Ev)
When we plug in the numbers from that link, the figure 'rT', the distance between the Earth and Sun, is equated out to an approximation close to 3,000 miles.
Assuming that the earth is flat creates a triangle with a flat base; giving us an easy straightforward proof for the distance to the sun.

You are truly incapable of reading. They did. And on the pic, the x=0 point was that on there...I want somebody to respond to this intelligently, and that excludes you, Tom. Look at the pic, Dogplatter, Username, or so on, and tell me how FE fixes this. And you're hypothetical claim is only a few miles off the ground, making such differences matter alot. However, the Moon is 385,000 km off the ground, so your point is moot.

So...anyone? Or does FE concede the point?

You are truly incapable of reading. They did. And on the pic, the x=0 point was that on there...I want somebody to respond to this intelligently, and that excludes you, Tom. Look at the pic, Dogplatter, Username, or so on, and tell me how FE fixes this. And you're hypothetical claim is only a few miles off the ground, making such differences matter alot. However, the Moon is 385,000 km off the ground, so your point is moot.
It makes a very big difference because the angle of the sun in the sky means something COMPLETELY different whether we are standing on a Flat Earth than a Round Earth. The method of Astronomical Parallax must be adjusted.
In your graph the blue square isn't "a few miles off of the ground." It's more like 500 miles off of the ground.

You are truly incapable of reading. They did. And on the pic, the x=0 point was that on there...I want somebody to respond to this intelligently, and that excludes you, Tom. Look at the pic, Dogplatter, Username, or so on, and tell me how FE fixes this. And you're hypothetical claim is only a few miles off the ground, making such differences matter alot. However, the Moon is 385,000 km off the ground, so your point is moot.
It makes a very big difference because the angle of the sun in the sky means something COMPLETELY different whether we are standing on a Flat Earth than a Round Earth. The method of Astronomical Parallax must be adjusted.
In your graph the blue square isn't "a few miles off of the ground." It's more like 500 miles off of the ground.
You are a bot. Why else would you speak of the sun? I'm speaking of the moon...
Furthermore, the angular difference is neglibile. I'll calculate the angular distance when I have time, but its definitely not within what you'd observe in FE.

Furthermore, the angular difference is neglibile. I'll calculate the angular distance when I have time, but its definitely not within what you'd observe in FE.
How do you refute the FE calculations I've made above?
http://theflatearthsociety.org/forum/index.php?topic=19986.msg372511#msg372511

Look above you. I did.
Now, is there anybody else who wants to answer this other than Tom? For the record, FE would observe a 5657 degree angle on observer two, if observe one was careful enough to get a 90 degree angle. However, even if the earth was flat, at 385,000, the angular distance would be around 89.5 degrees. So, we can assume, even if the Earth is flat, the Moon is over 300,000 km away.

Look above you. I did
Actually, you didn't. You failed to refute the calculations which demonstrate that the sun is 3,000 miles above the earth.

No, I never needed to. You did. Now get off my thread. I'm looking for somebody intelligent, and that by definition, excludes you.

No, I never needed to.
So you admit that there is no rebuttal to the calculations which demonstrate that the sun is 3,000 miles in altitude?
If you have no rebuttal, perhaps that means you are wrong.

Tom, you are purposely dragging this thread off topic. Answer the question at hand. This source specifically says:
"However, if a correction is applied so that Athens and Selsey are on the same physical plane when facing the moon, the computed distance is now 404,897 km or in error by only 2.37%."
This means the calculation was made with the two cities on a plane facing the moon, as in on Flat Earth.
So you admit that there is no rebuttal to the calculations which demonstrate that the sun is 3,000 miles in altitude?
Well, there is a significant flaw in your calculation. You do not take the atmospheric refraction that is so necessary in FE theory into account. Your calculation assumes that light travels in a straight line from the sun to the earth at 45 degrees latitude. Your own model says that this does not occur, but that light should bend quite significantly. Honestly, you could not determine the height of the sun in FE theory unless you knew the exact refracting properties of the earthsun medium.

Tom still doesn't realize we're talking about the moon ::)
At any rate, I don't think that's what that meant, but at any rate, the distinction is meaningless, as Tom still hasn't looked at my picture, apparently.
Secondly, if I'm not mistaken, you used hypothetical observers with hypothetical results. I'm showing real ones. Sure, if there was a FE and on opposite sides of the Earth, yes, the results would be the same. Not where I'm showing it.

Tom, you are purposely dragging this thread off topic. Answer the question at hand. This source specifically says:
"However, if a correction is applied so that Athens and Selsey are on the same physical plane when facing the moon, the computed distance is now 404,897 km or in error by only 2.37%."
This means the calculation was made with the two cities on a plane facing the moon, as in on Flat Earth.
Two points on the surface of a globe separated by a distance of 2370 km aren't on the same plane.
You lose.

Why can't they be on the same plane? The plane can pass through the surface of the globe.

Two points on the surface of a globe separated by a distance of 2370 km aren't on the same plane.
Read again Tom. It says that a better distance was calculated as if the two points were on the same plane facing the moon.
You lose.
A little more humility and openmindedness and a little less overconfidence wouldn't hurt either.

Tom, you are purposely dragging this thread off topic. Answer the question at hand. This source specifically says:
"However, if a correction is applied so that Athens and Selsey are on the same physical plane when facing the moon, the computed distance is now 404,897 km or in error by only 2.37%."
This means the calculation was made with the two cities on a plane facing the moon, as in on Flat Earth.
Two points on the surface of a globe separated by a distance of 2370 km aren't on the same plane.
You lose.
The calculations he spoke of were if the earth was flat.

Why can't they be on the same plane? The plane can pass through the surface of the globe.
Nope. If the second observer is on another part of the globe he is no longer standing on the same plane as the first observer.
If the moon is directly overhead of one observer, as the second observer recedes the moon will appear to become lower and lower in altitude due to the convexity of the earth.
Read again Tom. It says that a better distance was calculated as if the two points were on the same plane facing the moon.
While the first observer may be directly underneath the moon, the second observer is over 2,370 miles away behind the convexity of the globe, standing at a different angle to the first observer. His "up" is at a different vector. The angle of the moon in the sky is at an angle dependent on the earth's convexity.
The calculations he spoke of were if the earth was flat.
Actually, the calculations in the OP's original link assumes a Round Earth.

What's so hard about this that you can't understand it Tom? Read again. It says that a second calculation (the one with 2.37% error) was made as if the two cities were on the same plane i.e. as if they were on a flat earth. The roundness of the earth was not taken into account in this calculation. That's it.

Look at my picture again as well. At that scale, you can't even see the difference. Now are you telling me that the angular distance there is equal to approximately 30 degrees, like I showed before? 60 degrees puts it off at waaay out. So, I take it FE doesn't have an answer?

Note: The parallax angle in this particular example was computed to be 1113.6" (ie. 0.3093°) and, in conjunction with the physical distance of 2370 km between the two observers, suggests an EarthMoon distance of 438,988 km which is 10% higher than the actual distance (395,520 km) at the time these coordinated images were taken. However, if a correction is applied so that Athens and Selsey are on the same physical plane when facing the moon, the computed distance is now 404,897 km or in error by only 2.37%. Hipparchus of Rhodes (190120 BC) must be proud!
Am I misreading this?
It seems to say that the flat earth estimate is more accurate than a round earth estimate? It says when Athens and Selsey are on the same plane parallel to the moon (as in a flat earth) the error is 2.37% vs the RE estimate with 10% error?

You do know that planes can be set to go through the Earth you know...furthermore, are you admitting that the moon indeed is 385,000 miles away?

Well, in the model I'm working on it may well be. However, that is not a necessary implication from what I've said. It would be particularly damning to RE if a flat estimate yielded results that were more accurate than using a theoretically more correct round method.
So I did misread it then. How is the plane constructed then, without a third point or normal?

I assume they used the RE assumption, having no reason to doubt it in their minds. However, as I've pointed out, whether or not the Earth is flat or not makes little to no difference, being well within the margin of error of the calculations. The point, however, is that the Moon and Sun cannot be 3,000 miles away, which irked me more than the possibility of the Earth being flat actually.

My point being that if they used the normal of the "line" between Athens and Selsey it is assuming that the moon is far enough away that normal would be roughly accurate.
The difference though from the correction however is not the difference between a flat earth and a round earth.
I think I may be talking past what you are saying but I'm about to log since I'm leaving work. I'll have to read it later tonight.

Well, in the model I'm working on it may well be. However, that is not a necessary implication from what I've said. It would be particularly damning to RE if a flat estimate yielded results that were more accurate than using a theoretically more correct round method.
So I did misread it then. How is the plane constructed then, without a third point or normal?
Nothing in the flat earth model is more accurate than the round earth model.

Well, in the model I'm working on it may well be. However, that is not a necessary implication from what I've said. It would be particularly damning to RE if a flat estimate yielded results that were more accurate than using a theoretically more correct round method.
So I did misread it then. How is the plane constructed then, without a third point or normal?
Nothing in the flat earth model is more accurate than the round earth model.
Reread this post then <insert facepalm here>

What's so hard about this that you can't understand it Tom? Read again. It says that a second calculation (the one with 2.37% error) was made as if the two cities were on the same plane i.e. as if they were on a flat earth. The roundness of the earth was not taken into account in this calculation. That's it.
You can't assume a Flat Earth on a Round Earth. Two distant points on a RE in the experiment will have a different "down" and a different "up."
For example, if the second observer was standing a quarter the way around the world the moon would be at 90 degrees to the observer. The moon will be setting.
Nothing in the flat earth model is more accurate than the round earth model.
Actually, since the earth is flat, the FE model just happens to be the more accurate model.

Actually, since the earth is flat, the FE model just happens to be the more accurate model.
No, I was right when I said nothing in the FE model is more accurate than the RE model.

Actually, since the earth is flat, the FE model just happens to be the more accurate model.
Another classic circular argument...
FE > The moon 3000 miles away > ∴ Earth is flat
So why don't we see other parts of the moon (http://theflatearthsociety.org/forum/index.php?topic=20010.0)? I'm not trying to hijack this thread, its just that the whole moon/sun/3000 miles theory is falsifiable from your own armchair. (excluding a flat, shapeshifting?!? moon). Therefore, without an explanation, the whole theory can be thrown out.

Actually, since the earth is flat, the FE model just happens to be the more accurate model.
Another classic circular argument...
FE > The moon 3000 miles away > ∴ Earth is flat
So why don't we see other parts of the moon (http://theflatearthsociety.org/forum/index.php?topic=20010.0)? I'm not trying to hijack this thread, its just that the whole moon/sun/3000 miles theory is falsifiable from your own armchair. (excluding a flat, shapeshifting?!? moon). Therefore, without an explanation, the whole theory can be thrown out.
He doesn't know. Tom was owned so many times that he he learned to abandon arguments.

We cannot see other parts of the moon because the moon disappears below the horizon before it turns to any significant degree.

Your problem then occurs as to why it doesn't turn over a week's duration, and show a different side, if its rotation is so slow. Or, perhaps, does it do a little jiggle dance rotation, turning slowly one way, and then another way?

We cannot see other parts of the moon because the moon disappears below the horizon before it turns to any significant degree.
Sorry Tom, but as proven on the other thread people standing thousands of miles apart and looking up at the moon, even if one looks east to see it and the other looks west, will see the same face. If the moon were a mere 3000 miles away, that wouldn't be possible.

How can you make a plane off of two spots on a round earth without a normal or third point?

How can you make a plane off of two spots on a round earth without a normal or third point?
Good point. Presumably, the cities were placed on a hypothetical plane 500 miles apart. It's hard to judge these calculations unless we know exactly how the author made his calculation. We just don't have enough information.
We cannot see other parts of the moon because the moon disappears below the horizon before it turns to any significant degree.
Sorry Tom, but as proven on the other thread people standing thousands of miles apart and looking up at the moon, even if one looks east to see it and the other looks west, will see the same face. If the moon were a mere 3000 miles away, that wouldn't be possible.
He's right Tom. There's simply no way observers thousands of miles apart could see the same face of the moon if it was only 3000 miles away; just do the math. I'm afraid we have exposed a giant hole in your theory.

How can you make a plane off of two spots on a round earth without a normal or third point?
True, but what does it matter? Look at my graph, it doesn't really matter? Like I told you before, my best guess is probably that they just used the RE assumption, and tried to calculate it based on curvature of the Earth, and other math, etc., but unless they were ridiculously off, the Moon is over 300k kilometers off, just look at my original photos and try to manipulate them so that the parallax shows the observed results as well as have the moon only 3000 miles away. Not going to happen well...

I'd just like to the see the math they used to make their conclusion, especially when their wording is a bit confusing about said math. I don't have a lot of reason to trust two random guys on the internet.
I'm not trying to say they are part of the conspiracy or anything. Its just I'd like the actual data rather than a picture of the moon and the end result.

True, but we could do it, you know. Want to try it? Two locations, clear moon, no telescope required, only the ability to tell the angle, even roughly, would give us a ballpark figure.

Bumped for new answers from new people, and perhaps some old ones as well.