The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Q&A => Topic started by: areyouguysserious on January 24, 2012, 06:20:19 PM
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"The Sun is a sphere. It has a diameter of 32 miles and is located approximately 3000 miles above the surface of the earth." (this is according to the flat earth wiki)
3000 miles is really really far. A disc that is 32 miles across is very very small, comparatively.
3000 miles is roughly the distance from New York to LA, so if you were to stand in New York with a clear and unobstructed line of site between you and LA, and there was a building that was 32 miles long, than you would be able to see it very clearly.
Right?
Right??
WRONG!!! Theres no way. The sheer distance involved would render any such object so miniscule that it would be impossible to detect with the human eye.
And yet we are expected to believe that the sun and moon are only 32 miles across yet 3000 miles away, and yet they can be clearly seen and they provide much light for us earthlings.
Baloney!!!
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This is the most terrible argument I have ever had the unfortunate circumstance of skimming over.
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Is it really? So I guess you have no trouble seeing objects 3000 miles away with your superman vision?
Pray, enlighten me on the specifics as to why my argument is so "terrible".
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a building that was 32 miles long
wat
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This is the most terrible argument I have ever had the unfortunate circumstance of skimming over.
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a building that was 32 miles long
wat
Forgive my example....but a theoretical building that is 32 miles long is alot more plausible than a 32 mile in diameter spotlight that glows on us from 3000 miles away.
a building that was 32 miles long
wat
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I still havent heard why my argument is so terrible.
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I still havent heard why my argument is so terrible.
Have you actually done the math to see how big a 32 mile diameter sun would appear at a distance of 3000 miles and then compare that to the actual observed size of the sun? The results may surprise you.
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Im afraid i dont know the math to figure that out. What is the equation?
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Im afraid i dont know the math to figure that out. What is the equation?
I am no FE'er, but there are two parts to determining if you can or cannot see something at a distance
1.)size
2.)brightness
We can see galaxies that are billions of light-years away because they are extremely bright. The same can be said for an FE sun. it is extremely bright, and can be visible from extreme distances, even though it is relatively small. Not to mention that a thirty two mile diameter object would be visible from 3000 miles away, even if it was not extremely bright, like the moon.
The real issues is that there is no known mechanism that could power such a small object for such extreme lengths of time. But any FE'er will tell you that just because they do not know what the powers the sun doesn't mean that the earth is round, they just do not have enough money to conduct sufficient experiments.
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Im afraid i dont know the math to figure that out. What is the equation?
It's simple trigonometry. I'm sure that you can figure it out, if you put your mind to it.
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I'm a little rusty with my math, but by my calculations a round object 32 miles in diameter at 3000 miles would subtend an angle of about .611 degrees. The sun and moon are usually measured to subtend an angle of about .52 degrees.
Being off by a tenth of a degree doesn't sound like much, but the human eye (with clear 20/20 vision) can can discern a resolution angle of one arc minute (1/60 a degree). If there were two moons in the sky, one the normal size, and one the size predicted by flat earth numbers, you could literally see the difference.
But, hell... Cut the flat earth society some slack. It's not like they care about details.
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I'm a little rusty with my math, but by my calculations a round object 32 miles in diameter at 3000 miles would subtend an angle of about .611 degrees. The sun and moon are usually measured to subtend an angle of about .52 degrees.
I'm afraid you've made a mistake somewhere in your calculations. The result should be very close to 0.52 degrees, maybe 0.53. Give me a moment, I'll write it out and scan the calculations.
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I'm a little rusty with my math, but by my calculations a round object 32 miles in diameter at 3000 miles would subtend an angle of about .611 degrees. The sun and moon are usually measured to subtend an angle of about .52 degrees.
I'm afraid you've made a mistake somewhere in your calculations. The result should be very close to 0.52 degrees, maybe 0.53. Give me a moment, I'll write it out and scan the calculations.
I would love to see it.
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Hey, what do you know, it does come out as about 0.61 degrees. Clearly the distance of the Sun of these dimensions from the Earth would have to be about 3500 miles for this to work. Sorry for causing unnecessary confusion.
Interestingly enough, when I did my first quick calculation (that made me say you're wrong in the first place), I made an elementary mistake that made then numbers just right. For a moment, I assumed we're talking about an equilateral triangle, so to get the unknown side, I divided 3000 (the height) by sqrt(3) and multiplied by 2. That gives us a side of 3464.1. Continuing with this mistake, I got the right numbers. I'm a bit stumped by my own stupidity here, but it seems like I wasn't the only one who made this mistake!
Thank you for showing this to me. I'll bring it up with the FAQ team soon.
EDIT: I'll re-write my calculations in a cleaner form and post them for review, just to make sure I'm not being a moron again.
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Okay, I've re-read this a thousand times and I'm pretty sure it's right. Please, could others have a look and confirm? If this is correct, welp, we need to let the 3000-miles-away-Sun proponents know, and I'll probably have to adjust the Wiki FAQ.
I would especially appreciate feedback from RE'ers, and proponents of the 32 mile Sun, but others are welcome to have a say too.
(http://omgomg.eu/images/sun_angle_s.png) (http://omgomg.eu/images/sun_angle.png)
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How do we know a = 32, again?
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I agree, PP's calculation is completely backwards, starting with 32 miles as an initial given. The process is:
1. triangulate the distance to the sun
2. determine its size using the distance and apparent size
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The sun is 27 miles across if 3000 miles high. (Voliva)
(http://www.lhup.edu/~dsimanek/flat/flatmap.jpg)
Its 32 miles across if 700 miles high. (Robowtham).
http://www.sacred-texts.com/earth/za/za23.htm
I think most people here subscribe to Voliva. Certainly Tom Bishop and I both use Voliva's numbers.
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How do we know a = 32, again?
http://www.theflatearthsociety.org/forum/index.php?topic=1324.msg566287#msg566287
I agree, PP's calculation is completely backwards, starting with 32 miles as an initial given. The process is:
1. triangulate the distance to the sun
2. determine its size using the distance and apparent size
We're not determining the numbers anew. We are testing if the numbers from the FAQ make sense. We take two numbers from the FAQ as granted, and expect to obtain the third given. Since we did not get the third given, we've disproved the model. This will work for whichever two numbers we'll take of the three: distance, size, angular size.
Simultaneously, we're checking if AnonConda is right (which he almost certainly is):
I'm a little rusty with my math, but by my calculations a round object 32 miles in diameter at 3000 miles would subtend an angle of about .611 degrees. The sun and moon are usually measured to subtend an angle of about .52 degrees.
Can I repeat my request to check the correctness of my calculations? We can tackle the mutual understanding of the logic behind it once that's done. It's just that I've made a very bad mistake right before writing this, so I want to make sure there aren't any others.
The sun is 27 miles across if 3000 miles high. (Voliva)
This matches the size observable from the Earth. Perhaps that's what whoever wrote the old FAQ meant.
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We're not determining the numbers anew. We are testing if the numbers from the FAQ make sense.
Ah I see. Aren't you reinventing the wheel though? We already know the FAQ is full of shit. The triangulation gives different results at different degrees, which means either a) the bottom of the triangle changes position (i.e. the Earth is not flat); or b) [insert this week's flavor of bendy light bullshit here]. Either way, it's already been established that the algorithm for determining the sun's distance (and by extension, size) is unreliable.
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3000 miles is roughly the distance from New York to LA, so if you were to stand in New York with a clear and unobstructed line of site between you and LA, and there was a building that was 32 miles long, than you would be able to see it very clearly.
Last night I went to walmart and bought these sick x100 eagle vision goggles. I only heard in fairy tales that the sun was a tiny yellow circle. BUT NOW I KNOW ITS TRUE!!!!!!!!!!!!!
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Hey, what do you know, it does come out as about 0.61 degrees. Clearly the distance of the Sun of these dimensions from the Earth would have to be about 3500 miles for this to work. Sorry for causing unnecessary confusion.
Your mistake war really quite simple. First of all, the sun being 3000 miles above the flat earth is not the same as the sun being 3000 miles away from the observer (unless you are directly below the sun). Also, don't forget that the sun's image is being projected on to the upper layers of the atmoplane and atmospheric magnification will cause the sun to appear a consistent size regardless of the sun's elevation above the horizon right up to the point where perspective causes the sun to appear to sink below the horizon.
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Hey, what do you know, it does come out as about 0.61 degrees. Clearly the distance of the Sun of these dimensions from the Earth would have to be about 3500 miles for this to work. Sorry for causing unnecessary confusion.
Your mistake war really quite simple. First of all, the sun being 3000 miles above the flat earth is not the same as the sun being 3000 miles away from the observer (unless you are directly below the sun). Also, don't forget that the sun's image is being projected on to the upper layers of the atmoplane and atmospheric magnification will cause the sun to appear a consistent size regardless of the sun's elevation above the horizon right up to the point where perspective causes the sun to appear to sink below the horizon.
Index of refraction of air at sea level = 1.000293
Thats not enough to visibly bend light, let a lone magnify an image that much. On top of that, the atmosphere would have to be like a lens (bent), implying roundness. Care to tell me the radius of curvature of this magical atmosphere lens?
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Ah I see. Aren't you reinventing the wheel though? We already know the FAQ is full of shit.
We've recently started a project to try and make it a bit less shit. It's a child of the Wiki project. Hopefully, sooner or later (likely later) we'll get some of the issues resolved and then replace the old FAQ.
The triangulation gives different results at different degrees, which means either a) the bottom of the triangle changes position (i.e. the Earth is not flat); or b) [insert this week's flavor of bendy light bullshit here]. Either way, it's already been established that the algorithm for determining the sun's distance (and by extension, size) is unreliable.
Or c) markjo's reply, which is a pretty good point. I may have been too hasty in conceding with AnonConda's point.
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Actually, Tom Bishop's nonsense about the sun changing apparent size doesn't help explain the results of the triangulation; quite the opposite, it makes them make even less sense. I suspect that's why markjo mentioned it in the first place. It's just another piece to add to the growing mound of self-contradictory claims that is FET. Good thing you don't believe in it.
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Actually, Tom Bishop's nonsense about the sun changing apparent size doesn't help explain the results of the triangulation; quite the opposite, it makes them make even less sense.
This is why you will never see Tom (or pretty much any other FE'er) try to create a scale model that includes the FE, sun, moon and any other celestial objects needed to demonstrate such basic observations as sunrise, sunset, moonrise, moonset, the phases of the moon, the sun's path as the seasons change and eclipses, despite numerous requests. He knows that there is no way to create one that works. In fact, it seems that the only way that FE can appear to work is to through as much obfuscation as possible.
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Does the FET mob presume the heat essential to life as we know it, radiates from the spot light? If so, what is the heat generated?
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Does the FET mob presume the heat essential to life as we know it, radiates from the spot light? If so, what is the heat generated?
Yes they believe the sun heats the earth, no they have no clue what powers the sun.
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What heat does a spotlight give out 3 thousand mile away? How hot is the bulb, I guess i'm asking
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What heat does a spotlight give out 3 thousand mile away? How hot is the bulb, I guess i'm asking
The sun is not a spotlight. The sun is a sphere which shines light from all points of its surface. Its light is limited to a spot of light upon the earth, which is what we mean by spotlight.
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How can a glowing sphere that circles above a flat surface, focus it's light? Plug a light bulb into the socket of the room you're in now without a shade and tell me where the darkness is.
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How can a glowing sphere that circles above a flat surface, focus it's light? Plug a light bulb into the socket of the room you're in now without a shade and tell me where the darkness is.
Actually he is right that a spherical sun would cast a circular area of light on a plane. However, the area would have a different shape than the one mapped from the surface of a sphere with the same light shining on it. Tom Bishop acknowledges the fact that the sunlight pattern would be different on a sphere:
The answer is no, the area of sunlight would not be the same on a Flat Earth as it would be on a Round Earth. The earth is not round, it's flat.
Okay, so what you're telling me then is that the daylight maps, like this one (http://www.die.net/earth/), are wrong. That means that high-altitude photos that include the day-night terminator must show it in a different place or having a different curvature than expected. Please provide evidence that this has ever happened.
Instead of responding to the request in bold, Tom uses this copout:
The values in both of those links are from calculators based on RET, not direct observations of reality.
(He thinks calculators aren't based on direct observations of reality.)
I repeated my request for evidence twice:
The fact that the images are computer-generated is irrelevant. Computer programs use data. The veracity of the data is what we're talking about here.
You think the data doesn't represent reality. Okay. I already acknowledged that that is your belief. Now answer the rest of my post:
That means that high-altitude photos that include the day-night terminator must show it in a different place or having a different curvature than expected. Please provide evidence that this has ever happened.
Or for that matter, provide any evidence that reality ever disagrees with this data.
You see, if your claim (that the calculators are wrong) is the correct one, then your proof would be the easiest to produce. We can, and have, given you several repeatable demonstrations that the predictions are correct, but these could conceivably be coincidences. For you, however, all you need to do is provide one repeatable demonstration that the predictions are ever wrong, and you will have proved your claim. Why can't you do this?
Of course, it was at this point that Tom abandoned the topic.
So just to save you some time, this is where you inquiry ultimately leads to: Tom beating around the bush and then running away.
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How can a glowing sphere that circles above a flat surface, focus it's light? Plug a light bulb into the socket of the room you're in now without a shade and tell me where the darkness is.
Actually he is right that a spherical sun would cast a circular area of light on a plane. However, the area would have a different shape than the one mapped from the surface of a sphere with the same light shining on it. Tom Bishop acknowledges the fact that the sunlight pattern would be different on a sphere:
Is it not another issue that if the sun does not act as a spotlight it should cast light on the moon?
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"The Sun is a sphere. It has a diameter of 32 miles and is located approximately 3000 miles above the surface of the earth." (this is according to the flat earth wiki)
3000 miles is really really far. A disc that is 32 miles across is very very small, comparatively.
3000 miles is roughly the distance from New York to LA, so if you were to stand in New York with a clear and unobstructed line of site between you and LA, and there was a building that was 32 miles long, than you would be able to see it very clearly.
Right?
Right??
WRONG!!! Theres no way. The sheer distance involved would render any such object so miniscule that it would be impossible to detect with the human eye.
And yet we are expected to believe that the sun and moon are only 32 miles across yet 3000 miles away, and yet they can be clearly seen and they provide much light for us earthlings.
Baloney!!!
you said that you would not be able to see the sun from 3000 miles away ??? ... does it makes more sense seeing something from 93 million miles away ...ha ha ha ... right ... so do you think that the sun could be closer then 3,000 miles then???
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I know this is 3 years too late but how about using:
Beta = 2*ATAN(16/3000)?
Xi
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Let's break it down to a scale most of us are familiar with...the view from and airliner window.
The 3000:32 ratio is roughly the same as the size of a football field from the cruising altitude of an airliner. So, if the size of the field from this height is the same as the size of the moon the field should be very easy to find from your airliner window, no? It is tiny. If not for the running track that surrounds most school football fields in the US the field would be nearly impossible to find, From that height the field sure looks a lot smaller than the moon or sun to me.. What about you?
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Just to let you know, the 3000 mile and 32 mile measurements do actually work out. Even Vsauce said that mathematically, it is correct, as long as you assume the Earth is flat and not round. I don't know how you figure a 100 to one ratio is like looking at a football from an airplane or whatever jumbled thought you were trying to convey to us.
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These measurements do not work if several observations are made from different locations at the same time.
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These measurements do not work if several observations are made from different locations at the same time.
Obviously they would be closest to the observer directly underneath and further from observers who were further away.
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These measurements do not work if several observations are made from different locations at the same time.
Obviously they would be closest to the observer directly underneath and further from observers who were further away.
The next equinox is on the 22rd of September. During an equinox, the sun rises directly in the east on every location on the planet. Also on this day, every location on earth receives 12 hours of light and 12 hours of darkness. Obviously, on the flat earth this could never happen, so that's just blown the FE model to bits right then and there.
(http://www.eclipsegeeks.com/communities/3/004/009/983/113//images/4598933829.jpg)
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Just to let you know, the 3000 mile and 32 mile measurements do actually work out.
In some respects, in many fundamental ones they don't. For a start, the sun will never meet the horizon, let alone set below it.
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Obviously, on the flat earth this could never happen, so that's just blown the FE model to bits right then and there.
Your response seems like something for the debate section rather than FE Q&A.