I believe this topic should be allowed in the lime-light once again, if only for question number 2:
This is a pretty big problem, especially with the Lake Michigan-Chicago thing, that thing being that only the upper portions of the skyline is visible from a park across Lake Michigan. Are there really 100-meter swells in the lake? Why would the lower, thicker parts of buildings vanish before the spire of the Willis Tower?
I believe this topic should be allowed in the lime-light once again, if only for question number 2:
This is a pretty big problem, especially with the Lake Michigan-Chicago thing, that thing being that only the upper portions of the skyline is visible from a park across Lake Michigan. Are there really 100-meter swells in the lake? Why would the lower, thicker parts of buildings vanish before the spire of the Willis Tower?
It would not take 100m swells to obscure part of a building that, due to perspective, would appear to be very small.
If the earth was round none of those towers would be visible :'(
It is already in another thread, I forget which.If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
Feel free to present it at your leisure.It is already in another thread, I forget which.If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
If my maths are correct (see below for an explanation of the equation--feel free to correct me if I'm wrong and/or provide me with the correct equations), then round-earth theory predicts the lower 200 meters of the Willis Tower would be covered. Compare the sunset photo with this diagram: (http://upload.wikimedia.org/wikipedia/commons/thumb/4/4e/Dubai-CN-Sears-towers.svg/293px-Dubai-CN-Sears-towers.svg.png)
In the sunset photo, you can clearly see where the tower's middle section ends and the top begins. There is no likewise break between the middle and bottom, implying the bottom is mostly, if not entirely, covered. The bottom section is almost exactly 200 meters tall. This coincides perfectly with that round-earth predicts.
EQUATION:
Indiana Dunes is 50 kilometers from Chicago. That is about 0.00125 the circumference of the Earth, or about .4499 degrees around the circle. The (average) radius of the Earth being 6,371.0 kilometers, we can solve a triangle with the hypotenuse going form the center of the Earth to the base of the Willis Tower and the long leg going toward, but not quite touching, the base of the camera.
With an angle between the hypotenuse and long leg of .4499 degrees, the long leg is 6,371.0 * cos(.4499) = 6370.8 kilometers. This means the base of the Willis Tower is .2 kilometers below what the camera would view as 'horizontal'.
I factored in the fact that the tower isn't going straight 'up' according to the camera, but is leaning away at .4499 degrees. However, this changed the amount covered by a few millimeters at best, so I ignored them.
http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU (http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU)Feel free to present it at your leisure.It is already in another thread, I forget which.If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU (http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU)Feel free to present it at your leisure.It is already in another thread, I forget which.If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
You really should lurk more, but here is the short version from memory. The expected RE drop over 45.5 miles was something like 1400', and most of those buildings are shorter than that.http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU (http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU)Feel free to present it at your leisure.It is already in another thread, I forget which.If the earth was round none of those towers would be visible :'(
Care to take after Alex's example, and present the math?
I see Lolflatdiscs's calculations, but none of your own.
So because you couldn't be bothered to direct me toward the math used, you pull numbers out of nowhere and tell me to find how they were gotten myself? That's not my issue, I'm not the one claiming the buildings would be covered entirely by the curvature of the Earth. I didn't even claim they wouldn't, only asked for the equations which found that result.You really should lurk more, but here is the short version from memory. The expected RE drop over 45.5 miles was something like 1400', and most of those buildings are shorter than that.http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU (http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU)Feel free to present it at your leisure.Care to take after Alex's example, and present the math?It is already in another thread, I forget which.
I see Lolflatdiscs's calculations, but none of your own.
This quote is the result of a previous study.So because you couldn't be bothered to direct me toward the math used, you pull numbers out of nowhere and tell me to find how they were gotten myself? That's not my issue, I'm not the one claiming the buildings would be covered entirely by the curvature of the Earth. I didn't even claim they wouldn't, only asked for the equations which found that result.You really should lurk more, but here is the short version from memory. The expected RE drop over 45.5 miles was something like 1400', and most of those buildings are shorter than that.http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU (http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU)Feel free to present it at your leisure.Care to take after Alex's example, and present the math?It is already in another thread, I forget which.
I see Lolflatdiscs's calculations, but none of your own.
If you want a proof that they wouldn't, address Alex's post. Until then, feel free to present the math at you leisure. The actual equations, please, and not just the resultant numbers.
This quote is the result of a previous study.So because you couldn't be bothered to direct me toward the math used, you pull numbers out of nowhere and tell me to find how they were gotten myself? That's not my issue, I'm not the one claiming the buildings would be covered entirely by the curvature of the Earth. I didn't even claim they wouldn't, only asked for the equations which found that result.You really should lurk more, but here is the short version from memory. The expected RE drop over 45.5 miles was something like 1400', and most of those buildings are shorter than that.http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU (http://www.theflatearthsociety.org/forum/index.php/topic,58738.80.html#.UfUF9OBCRYU)Feel free to present it at your leisure.Care to take after Alex's example, and present the math?It is already in another thread, I forget which.
I see Lolflatdiscs's calculations, but none of your own.
If you want a proof that they wouldn't, address Alex's post. Until then, feel free to present the math at you leisure. The actual equations, please, and not just the resultant numbers.
The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.
Feel free to review the thread I linked earlier. Those 600' buildings would be under a lot of water, covering way past their tops. You will learn more doing your own work, rather than FE'ers spoon feeding you.
This quote is the result of a previous study.
The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.
Feel free to review the thread I linked earlier. Those 600' buildings would be under a lot of water, covering way past their tops. You will learn more doing your own work, rather than FE'ers spoon feeding you.
The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
The linked chart below is an accepted view on RE drop, it uses the pythagorean theorem. I have personally verified it's accuracy, you are welcome to also.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
The linked chart below is an accepted view on RE drop, it uses the pythagorean theorem. I have personally verified it's accuracy, you are welcome to also.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
The expected drop over 48.5 miles is 1398'.
(http://img211.imageshack.us/img211/7277/skyscrapers.jpg) (http://imageshack.us/photo/my-images/211/skyscrapers.jpg/)
Here is all the data on the office buildings in downtown Chicago. Willis tower is 1482' high. Aon tower is 1136' high, the second highest building. You should not be able to see any of this Aon Center tower.1136 - 1398 = -262. This means Aon tower should be 262' under water by RE reckoning. However you see it, and many other smaller buildings. This could not be possible on a round earth.
Click on the other building to obtain their height and other data.
http://en.wikipedia.org/wiki/Willis_Tower (http://en.wikipedia.org/wiki/Willis_Tower)
I hope this helps.
The linked chart below is an accepted view on RE drop, it uses the pythagorean theorem. I have personally verified it's accuracy, you are welcome to also.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
The expected drop over 48.5 miles is 1398'.
(http://img211.imageshack.us/img211/7277/skyscrapers.jpg) (http://imageshack.us/photo/my-images/211/skyscrapers.jpg/)
Here is all the data on the office buildings in downtown Chicago. Willis tower is 1482' high. Aon tower is 1136' high, the second highest building. You should not be able to see any of this Aon Center tower.1136 - 1398 = -262. This means Aon tower should be 262' under water by RE reckoning. However you see it, and many other smaller buildings. This could not be possible on a round earth.
Click on the other building to obtain their height and other data.
http://en.wikipedia.org/wiki/Willis_Tower (http://en.wikipedia.org/wiki/Willis_Tower)
I hope this helps.
Look at the data presented and study it. If you want to keep denying the obvious, you are certainly welcome to.
Here is all the data on the office buildings in downtown Chicago. Willis tower is 1482' high. Aon tower is 1136' high, the second highest building. You should not be able to see any of this Aon Center tower.1136 - 1398 = -262. This means Aon tower should be 262' under water by RE reckoning. However you see it, and many other smaller buildings. This could not be possible on a round earth.
It seems you are making this more complicated than need be. The drop over 48.5 miles is 1398' according to RET. Look at the heights of the buildings that are seen, some are 600' high. They should be way under water, but as you can see they are not. Subtract the building height from 1398, that tells you how far under water the building should be. Surely you can tell that those buildings are not under water.
Already factored 20' on the dunes. Makes it 1378, still taller than almost all of those buildings. I'm done here.It seems you are making this more complicated than need be. The drop over 48.5 miles is 1398' according to RET. Look at the heights of the buildings that are seen, some are 600' high. They should be way under water, but as you can see they are not. Subtract the building height from 1398, that tells you how far under water the building should be. Surely you can tell that those buildings are not under water.
Once again, you disregard refraction and, as Alex pointed out, height of the viewer. It's more complicated than you think it needs to be because those are factors that should be included in calculating viewing distance.
Already factored 20' on the dunes. Makes it 1378, still taller than almost all of those buildings. I'm done here.It seems you are making this more complicated than need be. The drop over 48.5 miles is 1398' according to RET. Look at the heights of the buildings that are seen, some are 600' high. They should be way under water, but as you can see they are not. Subtract the building height from 1398, that tells you how far under water the building should be. Surely you can tell that those buildings are not under water.
Once again, you disregard refraction and, as Alex pointed out, height of the viewer. It's more complicated than you think it needs to be because those are factors that should be included in calculating viewing distance.
Already factored 20' on the dunes. Makes it 1378, still taller than almost all of those buildings. I'm done here.It seems you are making this more complicated than need be. The drop over 48.5 miles is 1398' according to RET. Look at the heights of the buildings that are seen, some are 600' high. They should be way under water, but as you can see they are not. Subtract the building height from 1398, that tells you how far under water the building should be. Surely you can tell that those buildings are not under water.
Once again, you disregard refraction and, as Alex pointed out, height of the viewer. It's more complicated than you think it needs to be because those are factors that should be included in calculating viewing distance.
If the camera and buildings were all at water level, but they aren't.The expected drop over 48.5 miles is 1398'.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
It seems you are making this more complicated than need be. The drop over 48.5 miles is 1398' according to RET. Look at the heights of the buildings that are seen, some are 600' high. They should be way under water, but as you can see they are not. Subtract the building height from 1398, that tells you how far under water the building should be. Surely you can tell that those buildings are not under water.
Once again, you disregard refraction
Bendy light?! The irony and hubris is amazing...
Well, I don't believe in electromagnetic acceleration, either, so forgive me if I remain unimpressed when you run to refraction everytime observation does not match your expectation after deriding it so thoroughly elsewhere.
It seems you are making this more complicated than need be. The drop over 48.5 miles is 1398' according to RET. Look at the heights of the buildings that are seen, some are 600' high. They should be way under water, but as you can see they are not. Subtract the building height from 1398, that tells you how far under water the building should be. Surely you can tell that those buildings are not under water.
Once again, you disregard refraction
Bendy light?! The irony and hubris is amazing...
Well, I don't believe in electromagnetic acceleration, either, so forgive me if I remain unimpressed when you run to refraction everytime observation does not match your expectation after deriding it so thoroughly elsewhere.Atmospheric refraction is a fact that can not be ignored just because you're sick of hearing about it.
Well, I don't believe in electromagnetic acceleration, either, so forgive me if I remain unimpressed when you run to refraction everytime observation does not match your expectation after deriding it so thoroughly elsewhere.
I find it hard to believe that the lot of you are so obtuse: I believe in refraction. What I find hard to believe (but shouldn't) is the lengths a globularist will go to invoke refraction at everyturn where you meet the otherwise unexplainable, but then hold the truth to an entirely different standard.
I find it hard to believe that the lot of you are so obtuse: I believe in refraction. What I find hard to believe (but shouldn't) is the lengths a globularist will go to invoke refraction at everyturn where you meet the otherwise unexplainable, but then hold the truth to an entirely different standard.Perhaps it's because atmospheric refraction is often a perfectly reasonable explanation for otherwise unexplainable phenomena. The problem is that atmospheric conditions are generally not documented well enough to determine whether or not refraction would be a significant factor in the observation. However, the fact that certain refractive events will occur under certain conditions means that atmospheric refraction can not be automatically invoked or dismissed without knowing what conditions were present at the time of the observation.
The math doesn't work that way. Take the total distance, find the drop over the total distance. Then subtract 20' for being on the dunes, subtract a couple of feet, because downtown Chicago is a couple of feet above sea level. 1398' - 20' - 2' = 1376', which is the drop on a round earthIf the camera and buildings were all at water level, but they aren't.The expected drop over 48.5 miles is 1398'.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
Having the camera higher effectively increases the distance before it's line of site intersects the water.
According that chart in ENaG, There is a 20 ft drop after 5.5 miles. With the camera at 20 feet, the curvature that actually blocks the view doesn't start for 5.5 miles. The building is also 20 feet above the lake, giving it a 5.5 mile advantage at the other end.
Measuring from where that picture was taken from to downtown I get about 46 miles - 11= 35 miles worth of drop along the line of sight, which is about 850 feet according to ENaG. Someone who knows the math can figure it out better.
Compared to other photos, it looks like the bottom of the buildings are compressed a little, which would be refraction making more of them visible.
The math doesn't work that way. Take the total distance, find the drop over the total distance. Then subtract 20' for being on the dunes, subtract a couple of feet, because downtown Chicago is a couple of feet above sea level. 1398' - 20' - 2' = 1376', which is the drop on a round earthIf the camera and buildings were all at water level, but they aren't.The expected drop over 48.5 miles is 1398'.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
Having the camera higher effectively increases the distance before it's line of site intersects the water.
According that chart in ENaG, There is a 20 ft drop after 5.5 miles. With the camera at 20 feet, the curvature that actually blocks the view doesn't start for 5.5 miles. The building is also 20 feet above the lake, giving it a 5.5 mile advantage at the other end.
Measuring from where that picture was taken from to downtown I get about 46 miles - 11= 35 miles worth of drop along the line of sight, which is about 850 feet according to ENaG. Someone who knows the math can figure it out better.
Compared to other photos, it looks like the bottom of the buildings are compressed a little, which would be refraction making more of them visible.
1376' is higher than all but 1 of those downtown Chicago buildings. 1136' is the height of the the second tallest building. However you can see many more of them.
Yes it is a losing battle. 29silohette is a good guy, so i responded to him. If you want to add anything go ahead.
I drew a triangle and found that your math doesn't appear to add up. I had two legs of equal length, connected them with an arc, and drew two different length lines from the legs. I then did a tangent line connecting the bottom of the shorter line to the long one, then repeated for the top line. I found that the difference in height of the short line was smaller than the line of sight difference of the tall line.The math doesn't work that way. Take the total distance, find the drop over the total distance. Then subtract 20' for being on the dunes, subtract a couple of feet, because downtown Chicago is a couple of feet above sea level. 1398' - 20' - 2' = 1376', which is the drop on a round earthIf the camera and buildings were all at water level, but they aren't.The expected drop over 48.5 miles is 1398'.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
Having the camera higher effectively increases the distance before it's line of site intersects the water.
According that chart in ENaG, There is a 20 ft drop after 5.5 miles. With the camera at 20 feet, the curvature that actually blocks the view doesn't start for 5.5 miles. The building is also 20 feet above the lake, giving it a 5.5 mile advantage at the other end.
Measuring from where that picture was taken from to downtown I get about 46 miles - 11= 35 miles worth of drop along the line of sight, which is about 850 feet according to ENaG. Someone who knows the math can figure it out better.
Compared to other photos, it looks like the bottom of the buildings are compressed a little, which would be refraction making more of them visible.
1376' is higher than all but 1 of those downtown Chicago buildings. 1136' is the height of the the second tallest building. However you can see many more of them.
There exist certain experiments that I possess a deep knowledge as to how to perform them. The type of experiments that even if refraction did occur, there appears the opportunity to still acquire data. It involves measuring across two points in the distance. When you take an island that measures approximately 22 plus miles long or two ships twelve miles separate, the horizon you receive is always level when using a straight edge to compare. Even if refraction does occur with the two points, their relativism will remain unchanged. The earth is flat and remember gravity does bend light as well...
Yes it is a losing battle. 29silohette is a good guy, so i responded to him. If you want to add anything go ahead.
I will make a .dwg file tonight and we can settle this while letting anyone who cares also look at the raw data. I will also post a screen shot, so you people do not have to worry about format.
The linked chart below is an accepted view on RE drop, it uses the pythagorean theorem. I have personally verified it's accuracy, you are welcome to also.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
The expected drop over 48.5 miles is 1398'.
(http://img211.imageshack.us/img211/7277/skyscrapers.jpg) (http://imageshack.us/photo/my-images/211/skyscrapers.jpg/)
Here is all the data on the office buildings in downtown Chicago. Willis tower is 1482' high. Aon tower is 1136' high, the second highest building. You should not be able to see any of this Aon Center tower.1136 - 1398 = -262. This means Aon tower should be 262' under water by RE reckoning. However you see it, and many other smaller buildings. This could not be possible on a round earth.
Click on the other building to obtain their height and other data.
http://en.wikipedia.org/wiki/Willis_Tower (http://en.wikipedia.org/wiki/Willis_Tower)
I hope this helps.
DQ only the top of the Willis Tower could poke above a RE horizon. Clearly you can see many more buildings which should be hidden if the earth were round.
The math doesn't work that way. Take the total distance, find the drop over the total distance. Then subtract 20' for being on the dunes, subtract a couple of feet, because downtown Chicago is a couple of feet above sea level. 1398' - 20' - 2' = 1376', which is the drop on a round earthIf the camera and buildings were all at water level, but they aren't.The expected drop over 48.5 miles is 1398'.The RE expected drop is 1398' at 45.8 miles. Plus 20' of the dunes. This means you could see nothing at all under 1378' if the earth is round. You can see several buildings that are 600' high. There would be no way to see the other buildings at all. The earth is flat.The Willis Tower is about 20 ft above the lake. The photographer is about 20 ft above the lake. This reduces the amount of line of sight blocking curvature to perhaps 37 miles. The drop for 37 miles is maybe 850 ft. (If someone wants to do the math for this, go for it)
Refraction would allow for a compressed view of some more that would be hidden below the water.
Having the camera higher effectively increases the distance before it's line of site intersects the water.
According that chart in ENaG, There is a 20 ft drop after 5.5 miles. With the camera at 20 feet, the curvature that actually blocks the view doesn't start for 5.5 miles. The building is also 20 feet above the lake, giving it a 5.5 mile advantage at the other end.
Measuring from where that picture was taken from to downtown I get about 46 miles - 11= 35 miles worth of drop along the line of sight, which is about 850 feet according to ENaG. Someone who knows the math can figure it out better.
Compared to other photos, it looks like the bottom of the buildings are compressed a little, which would be refraction making more of them visible.
1376' is higher than all but 1 of those downtown Chicago buildings. 1136' is the height of the the second tallest building. However you can see many more of them.
As explained earlier, it is easiest and most correct to find the total horizontal distance. Calculate the RE drop, then subtract the camera height and the building heights. Our vision and camera resolution make it nearly impossible to make out details from nearly 50 miles. However we can clearly see the buildings are not behind the horizon, rather sitting on it.
Incorrect, the math and pictures are included in this thread. The buildings are not covered by water, this conclusively proves that the earth is not a sphere.
Lurk more, read and study this thread.Incorrect, the math and pictures are included in this thread. The buildings are not covered by water, this conclusively proves that the earth is not a sphere.
Why do all you FErs use such blatant non sequiturs? How does it follow that since the buildings are not covered by water, the earth is not a sphere.
Lurk more, read and study this thread.Incorrect, the math and pictures are included in this thread. The buildings are not covered by water, this conclusively proves that the earth is not a sphere.
Why do all you FErs use such blatant non sequiturs? How does it follow that since the buildings are not covered by water, the earth is not a sphere.
It does follow that if the earth drops nearly 1400', that you would not see a 600' building.
Incorrect, the math and pictures are included in this thread. The buildings are not covered by water, this conclusively proves that the earth is not a sphere.
Incorrect, the math and pictures are included in this thread. The buildings are not covered by water, this conclusively proves that the earth is not a sphere.
I suggest you buy some glasses then. There are buildings in the picture, several hundred feet tall, and yet you see only the very top. Only possible on a spherical earth. End of discussion.
Your way may be easier, but it's certainly not correct.
However, if you want to draw conclusions based on incorrect observations, then be my guest.Observations are neither correct or incorrect. It's the conclusions that you draw from from those observations that are subject to debate.
Incorrect, the math and pictures are included in this thread. The buildings are not covered by water, this conclusively proves that the earth is not a sphere.
I suggest you buy some glasses then. There are buildings in the picture, several hundred feet tall, and yet you see only the very top. Only possible on a spherical earth. End of discussion.
I wish I could see the world through childlike eyes sometimes. It would make life a lot simpler. However, if you want to draw conclusions based on incorrect observations, then be my guest.
I will make a .dwg file tonight and we can settle this while letting anyone who cares also look at the raw data. I will also post a screen shot, so you people do not have to worry about format.
Yes it is a losing battle. 29silohette is a good guy, so i responded to him. If you want to add anything go ahead.
I will make a .dwg file tonight and we can settle this while letting anyone who cares also look at the raw data. I will also post a screen shot, so you people do not have to worry about format.
Yes it is a losing battle. 29silohette is a good guy, so i responded to him. If you want to add anything go ahead.
I will make a .dwg file tonight and we can settle this while letting anyone who cares also look at the raw data. I will also post a screen shot, so you people do not have to worry about format.
Any update on that file, Jroa? It'd be helpful to deciding the thread.
Yes it is a losing battle. 29silohette is a good guy, so i responded to him. If you want to add anything go ahead.
I will make a .dwg file tonight and we can settle this while letting anyone who cares also look at the raw data. I will also post a screen shot, so you people do not have to worry about format.
Any update on that file, Jroa? It'd be helpful to deciding the thread.
Been a three week long night, is there anything going with that file or do we need to wait a month?
Sorry but the Earth is flat. Want proof? Look outside your window.
Sorry but the Earth is flat. Want proof? Look outside your window.
Sorry but the Earth is flat. Want proof? Look outside your window.
Funny, when I look out my window I see so much evidence supporting a round earth, and next to nothing to support a flat earth, apart from the appearance of flatness, which could just be (is) due to the earth being a really big ball.
Sorry but the Earth is flat. Want proof? Look outside your window.
Funny, when I look out my window I see so much evidence supporting a round earth, and next to nothing to support a flat earth, apart from the appearance of flatness, which could just be (is) due to the earth being a really big ball.
so what evidence do you see that supports the RET when you look our your window?
Check your image link. Diagram's not showing up for me.Uploaded as a bmp instead and added a link. Maybe that will work.
Check your image link. Diagram's not showing up for me.Uploaded as a bmp instead and added a link. Maybe that will work.
image is squished...It's 200x30,000 pixels. Start scrolling to the right and you'll see the black start to drop away and the lines of sight separating.
image is squished...It's 200x30,000 pixels. Start scrolling to the right and you'll see the black start to drop away and the lines of sight separating.
If it appears as a thin black line, zoom in, then scroll.
Sorry but the Earth is flat. Want proof? Look outside your window.
Funny, when I look out my window I see so much evidence supporting a round earth, and next to nothing to support a flat earth, apart from the appearance of flatness, which could just be (is) due to the earth being a really big ball.
so what evidence do you see that supports the RET when you look our your window?
While we wait for jroa's diagram, here's the rough 200x30,000 pixel diagram I did. I plotted out points along the curvature from a straight and level line working from the table in ENaG, and then went from point to point connecting them.
Red= zero elevation line of sight
Green = slight elevation line of sight
Blue = higher elevation line of sight
These lines are straight and don't take refraction into account.
(scroll to the right)
http://img10.imageshack.us/img10/1945/fbdl.png (http://img10.imageshack.us/img10/1945/fbdl.png)
How is yours coming along jroa?
What table in ENaG is this?This one.
This is because the buildings are so close together. If they were very,very,very far away from each other than you would be able to see them lean away from each other.
This is because the buildings are so close together. If they were very,very,very far away from each other than you would be able to see them lean away from each other.
This is because the buildings are so close together. If they were very,very,very far away from each other than you would be able to see them lean away from each other.
I disagree with you phoenix. Its just too far away too notice and the difference in curvature isn't enough to notice such an effect.
The reason why the horizon doesn't appear to curve is because well... that really doesn't make any sense. No matter where you look at that horizon you are equidistant away from it. If what scepti was proposing were to occur then the horizon behind you would have to be much lower. You can imagine an ant on a basketball and notice that from the ants view, the horizon is always equidistant from the ant. Alex created a nice model in a geometry program that shows the altitudes which you should be able to see the curvature. You can find it the debates section.
You're right, they should. But how much? Assuming each building is built perfectly perpendicular to the Earth's surface, than each should appear to another to be listing away just a little bit.(http://krfoto.zenfolio.com/img/s9/v17/p345511614-3.jpg)
I have a query about this picture.
Ok, as we can see, half the buildings disappear and this is attributed to the curvature of the earth, right?
Well here's my problem with this.
If that is the case, we also are told that the earth is round, so it should curve from right to left but that line looks perfectly straight, yet my real query is.
If the buildings disappear by half, over that distance, then they should also angle away from each other, for instance, they should be slightly leaning left and right from that distance, also.
They may be spirit level straight up when your a mile away or a bit more, but not from that distance. Any reason why they are bolt upright, if the earth is a globe?
You're right, they should. But how much? Assuming each building is built perfectly perpendicular to the Earth's surface, than each should appear to another to be listing away just a little bit.(http://krfoto.zenfolio.com/img/s9/v17/p345511614-3.jpg)
I have a query about this picture.
Ok, as we can see, half the buildings disappear and this is attributed to the curvature of the earth, right?
Well here's my problem with this.
If that is the case, we also are told that the earth is round, so it should curve from right to left but that line looks perfectly straight, yet my real query is.
If the buildings disappear by half, over that distance, then they should also angle away from each other, for instance, they should be slightly leaning left and right from that distance, also.
They may be spirit level straight up when your a mile away or a bit more, but not from that distance. Any reason why they are bolt upright, if the earth is a globe?
Let's calculate this list between the Willis Tower (second from left) and the camera. The distance is 74.3 km, or 0.0018557 times the circumference of the Earth. Multiply by 360° and we get a list of .668°. Since that list is directly away from the camera (it's leaning backwards) really the only indication this would be happening is it would appear shorter--3.6 cm shorter to be exact. Hardly noticeable at 74.3 km.
Now, the Willis Tower and the John Hancock Center (that's the tallest building on the right, number 10 from the left). also list away from each-other, which, as you say, would be seen from the camera's perspective as them, well, leaning away from each-other. They are 2.4 kilometers apart, or .00005996 times the circumference of the Earth. Multiply by 360° for a total angular difference of 0.0216° or 1.3'. That means that the difference in the tops of a JHC parallel to the Willis Tower and the current JHC would be about 17 centimeters.
Once again, not noticeable at the 73.5 distance between the camera and the JHC.
So you're right that they should be tilted away from each-other and from the camera, but the tilt isn't visible on such a small scale. On a flat earth the Willis Tower would appear less than 4 cm taller, and the JHC would appear 17 cm closer to the Willis Tower. At this distance, such tiny variations are invisible.
Edit: For reference as to what a JHC 17 cm closer to the Willis Tower would look like, the width of the JHC visible from this photo is about 32 meters, making each pixel about 10 meters wide. The tilt away would be about 1/10th the width of a pixel. In other words, undetectable.
The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.(http://krfoto.zenfolio.com/img/s9/v17/p345511614-3.jpg)
I have a query about this picture.
Ok, as we can see, half the buildings disappear and this is attributed to the curvature of the earth, right?
Well here's my problem with this.
If that is the case, we also are told that the earth is round, so it should curve from right to left but that line looks perfectly straight, yet my real query is.
If the buildings disappear by half, over that distance, then they should also angle away from each other, for instance, they should be slightly leaning left and right from that distance, also.
They may be spirit level straight up when your a mile away or a bit more, but not from that distance. Any reason why they are bolt upright, if the earth is a globe?
The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.(http://krfoto.zenfolio.com/img/s9/v17/p345511614-3.jpg)
I have a query about this picture.
Ok, as we can see, half the buildings disappear and this is attributed to the curvature of the earth, right?
Well here's my problem with this.
If that is the case, we also are told that the earth is round, so it should curve from right to left but that line looks perfectly straight, yet my real query is.
If the buildings disappear by half, over that distance, then they should also angle away from each other, for instance, they should be slightly leaning left and right from that distance, also.
They may be spirit level straight up when your a mile away or a bit more, but not from that distance. Any reason why they are bolt upright, if the earth is a globe?
The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
Because RET states that they are under water.Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
I have no idea why you think they should be under water.
Besides, if you insist on stuff disappearing then it makes no sense that the top of the building is still there.
Because RET states that they are under water.Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
I have no idea why you think they should be under water.
Besides, if you insist on stuff disappearing then it makes no sense that the top of the building is still there.
It makes no sense to see any part of those buildings, if you think the earth is a sphere.It is similar to ships disappearing in the distance, the bottom disappears first and you can still see the top.
You do know that that is because of distance right? The curve makes the bottoms disappear before the tops. As you get further away the ground drops more and more and you see less of the buildings (from the bottom up) until you can't see the buildings at all.Because RET states that they are under water.Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
I have no idea why you think they should be under water.
Besides, if you insist on stuff disappearing then it makes no sense that the top of the building is still there.
It makes no sense to see any part of those buildings, if you think the earth is a sphere.It is similar to ships disappearing in the distance, the bottom disappears first and you can still see the top.
Because RET states that they are under water.
It makes no sense to see any part of those buildings, if you think the earth is a sphere.It is similar to ships disappearing in the distance, the bottom disappears first and you can still see the top.
The only problem with the curve covering up the bottom, is that same curve should have covered the whole building, if it was there.You do know that that is because of distance right? The curve makes the bottoms disappear before the tops. As you get further away the ground drops more and more and you see less of the buildings (from the bottom up) until you can't see the buildings at all.Because RET states that they are under water.Things disappear into the distance. You watch a plane fly by, a couple of minutes later you can't see it. You saw stay up in the air, it just disappears because our vision is not good enough to see something the size of a plane at that distance. Those buildings that you can plainly see are supposed to be under hundreds of feet of water.The biggest problem for me is the fact that, all but one of those buildings should be completely under water. If the earth were a sphere.
The biggest problem for me is (and always has been with these photos) that on a flat earth, the entire skyline should be clearly visible, not just this handful of buildings, which are clearly partially obscured, despite the viewer's elevated position (which would allow him/her to see more of the building than you might think, as has been explained).
I have no idea why you think they should be under water.
Besides, if you insist on stuff disappearing then it makes no sense that the top of the building is still there.
It makes no sense to see any part of those buildings, if you think the earth is a sphere.It is similar to ships disappearing in the distance, the bottom disappears first and you can still see the top.
I already have done the math here, I don't know where you made your mistake. According to RET those buildings would not be visible at all.Because RET states that they are under water.
It makes no sense to see any part of those buildings, if you think the earth is a sphere.It is similar to ships disappearing in the distance, the bottom disappears first and you can still see the top.
Actually, Hoppy, I recently did a post that conclusively shows that at least five buildings should be clearly visible above the water (http://www.theflatearthsociety.org/forum/index.php/topic,674.msg1539175.html#msg1539175). If you have problems with the model please bring them up. The math has been explained about 4 pages back, so you can look there, too, for ideas on where I've obviously messed up.
If buildings disappeared because they were too small to see, why are the thinner tops visible when the thick bottoms have vanished?
(http://i.imgur.com/oNKwSL9.jpg)(http://i.imgur.com/ddjkBEI.png)
Building Identities
It should be noted that the third building from the left (Franklin Center) appears shorter than the first from the left (311 South Wacker Drive) even though the diagram positions it the other way around. This means that either I have the identities very wrong or there exists some optical effect that distorts the true building sizes.
I already have done the math here, I don't know where you made your mistake. According to RET those buildings would not be visible at all.Thanks for posting that so we can check it! It's really helpful, and conducive to cooperation. Though you may be right. I remember when I was walking down a hill, the moment the base of a tree was covered by the hill the entire thing vanished.
Very nice.Thanks!
About that building appearing a different size, is it possible it's a different building? I'm not sure what year that picture was taken, but the other picture of Chicago from New Buffalo was taken in quite a while ago, and I noticed some newer buildings when I was doing some identification of my own.
Also, was the building elevation above the lake factored into those diagrams? They average about 20 feet I think. Or probably not enough to matter?
Now if jroa could post his diagram.
(http://krfoto.zenfolio.com/img/s9/v17/p345511614-3.jpg)
I have a query about this picture.
Ok, as we can see, half the buildings disappear and this is attributed to the curvature of the earth, right?
Well here's my problem with this.
If that is the case, we also are told that the earth is round, so it should curve from right to left but that line looks perfectly straight, yet my real query is.
If the buildings disappear by half, over that distance, then they should also angle away from each other, for instance, they should be slightly leaning left and right from that distance, also.
They may be spirit level straight up when your a mile away or a bit more, but not from that distance. Any reason why they are bolt upright, if the earth is a globe?