Proof earth isn't flat! (All you troglodytes read)

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.

Quote from: hoppy on July 29, 2013, 08:04:26 PM

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.

Quote from: Pyrolizard on July 29, 2013, 08:10:36 PM

Quote from: hoppy on July 29, 2013, 08:04:26 PM

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.

Put the math up, friend.  Just saying you've factored it in doesn't mean it's right, or prove that you even have.  Or that you've factored in refraction, which you seem persistent on ignoring.

Quote from: Pyrolizard on July 29, 2013, 08:10:36 PM

Quote from: hoppy on July 29, 2013, 08:04:26 PM

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.

As has been explained the difference in height isn't that simple.

hoppy...lets assume for just a fraction of a second that your math is correct (and I've run the numbers myself and pyrolizard and alex are right about everything) You or anyone else have yet to address any of the other concerns that assassin brought up at the beginning of this topic

Quote from: 29silhouette on July 29, 2013, 10:53:01 AM

Quote from: hoppy on July 28, 2013, 03:16:39 PM

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 expected drop over 48.5 miles is 1398'.

If the camera and buildings were all at water level, but they aren't.

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.

Quote from: hoppy on July 29, 2013, 08:04:26 PM

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...

Bendy light?! The irony and hubris is amazing...

No, not 'electromagnetic acceleration' (aka 'bendy light'), refraction, entirely different phenomena in so many ways.

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.

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.

Am I correct in saying you do not believe in refraction? Are you aware that your eye has a lens in it, which uses refraction to focus images onto your retina? And that binoculars, refractive telescopes, and microscopes use refraction to magnify images?

Refraction is real, and can be measured, predicted, and observed. 'Bendy light' cannot. Forgive me for not believing in an unobserved and unproven phenomena, and for accepting a known, understood, and easily reproduced one.

Quote from: Pyrolizard on July 29, 2013, 08:10:36 PM

Quote from: hoppy on July 29, 2013, 08:04:26 PM

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...

Hey, Ski, if you feel like disproving refraction then that's for another thread.  Until then, it's a very well documented phenomenon which should be accounted for.  It's the same thing that makes a spoon handle appear severed in a glass of water, so you don't need to be a scientist to test 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.

Atmospheric refraction is a fact that can not be ignored just because you're sick of hearing about it.

Why are FEers conflating refraction with electromagnetic acceleration?  Especially someone as smart as Ski, it does not make any sense.

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.

Ski, for debating whether refraction exists, please see here: http://www.theflatearthsociety.org/forum/index.php/topic,59200.msg1515779.html#msg1515779

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.

Every turn?  That is a bit dramatic.  I see it invoked in circumstances such as this where you are taking in to account all factors that may affect a view at a distance.  Please keep your criticism accurate.

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.

Quote from: hoppy on July 29, 2013, 12:12:53 PM

Quote from: 29silhouette on July 29, 2013, 10:53:01 AM

Quote from: hoppy on July 28, 2013, 03:16:39 PM

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 expected drop over 48.5 miles is 1398'.

If the camera and buildings were all at water level, but they aren't.

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 earth
 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.

Quote from: 29silhouette on July 29, 2013, 10:08:32 PM

Quote from: hoppy on July 29, 2013, 12:12:53 PM

Quote from: 29silhouette on July 29, 2013, 10:53:01 AM

Quote from: hoppy on July 28, 2013, 03:16:39 PM

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 expected drop over 48.5 miles is 1398'.

If the camera and buildings were all at water level, but they aren't.

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 earth
 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.

hoppy, I was going to suggest that I can make a diagram on AutoCAD.  It should not lie.  However, I do believe that, even if the results come out in favor of FET, the RE'ers will just blame refraction or something.  It is a losing battle with this bunch. 

Yes it is a losing battle. 29silohette is a good guy, so i responded to him. If you want to add anything go ahead.

Refraction is not necessary to continually calculate.  If you look at the image and it consistently shows the same appearance it is not a mirage. Because weather patterns constantly fluctuate, no mirage is ever consistent.  If I recall correctly, Rowbotham worked at the Bedford Canal for approximately nine months.  This proves that his experiments stood the test of criticism due to their frequent performance.  There existed no recorded change...  Thus, no mirage..

This knowledge holds fast onto these pictures as well.  A mirage every time fails to occur every single day.  Remember please that there are seasons and alterations of temperature and air within these seasons.  You will inevitably fail when you say," every single image possesses some form of a mirage." 

Superior mirages usually occur in arctic climates.  Truthfully, there is nothing like the arctic in these locations.  An inferior mirage or inferior refraction would be more common in these locations than a superior one.  For those who do not understand what a superior mirage entails, the image appears above the line of sight.  However, inferior mirages result in a sinking effect...

To sum it all up, do the experiment every day continuously through different season.  This eliminates the probability of the image being from refraction.

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. 

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...

Quote from: 29silhouette on July 29, 2013, 10:08:32 PM

Quote from: hoppy on July 29, 2013, 12:12:53 PM

Quote from: 29silhouette on July 29, 2013, 10:53:01 AM

Quote from: hoppy on July 28, 2013, 03:16:39 PM

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 expected drop over 48.5 miles is 1398'.

If the camera and buildings were all at water level, but they aren't.

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 earth
 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.

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.

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...

How can gravity be so powerful as to bend light enough to mimic refraction but not powerful enough to bend earth into a sphere (the most gravitationally efficient shape)? Additionally I also have deep knowledge of an experiment to prove refraction. Take a straw. Put it in a glass of water. Observe.

Quote from: hoppy on July 30, 2013, 10:17:42 AM

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, please. But to make this image properly, please take into consideration the following image and explanation:



The orientation of that pier in relation to the Willis Tower indicates this image was taken around 41^o 48' 04.98" N 86^o 44' 49.45 W at an elevation of at least 11 meters[nb]The ground at that area is 11 meters above the lake, but there appears to be a walkway that might be a few meters off the ground. Still, 11 is probably a good number to use.[/nb] above the surface of Lake Michigan.

The four most easily-identifiable buildings are, left-to-right:

Willis Tower, 74,282 meters away, 6 meters above the lake
Aon Center, 73,195 meters away, 17 meters above the lake
Trump Tower, 73,659 meters away, 12 meters above the lake
John Hancock Building, 73,528 meters away, 5 meters above the lake

I could try to identify the smaller buildings for you, if you wish, but that'll take time. Here's what I think the other buildings are, from left to right:

311 South Wacker Drive, 74,261 meters away, 6 meters above the lake
Franklin Center, 74,220 meters away, 6 meters above the lake
Legacy Tower, 73,489 meters away, 6 meters above the lake
Aqua, 73,099 meters away, 6 meters above the lake
Olympia Center, 73,504 meters away, 5 meters above the lake
Water Tower Place, 73,443 meters away, 5 meters above the lake
900 North Michigan, 73,676 meters away, 4 meters above the lake (this building is the rightmost two blocks)

EDIT:

Should also point out, for clarification, the Earth's supposed radius is 6,371.0 kilometers.

Quote from: 29silhouette on July 29, 2013, 10:53:01 AM

Quote from: hoppy on July 28, 2013, 03:16:39 PM

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.

http://www.sacred-texts.com/earth/za/za05.htm

The expected drop over 48.5 miles is 1398'.




 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

I hope this helps.

Hoppy. It was a nice thread you brought up, but you fail to miss one point. I can see where lolflatdisc made a mistake in his calculations and you perhaps see that as a victory for the flat earth theory... but still....just have a look at the photo!! The fact that the bottom parts of the buildings are missing, means these are hidden behind the horizon, which could only corresponds with a spherical earth. Tall buildings, several hundred feet tall and you only see the very top. It is proof a spherical earth and you do not need to understand any calculations. Simple but important point, proofing the earth being a sphere.

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.

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.

There is still the problem of the bottom portion of the buildings being obscured by the horizon. At 20' elevation for both the photographer and the buildings' bases, it's not very likely that waves are getting in the way, they'd have to be 20' high just to obscure the ground beneath the buildings! To have a significant portion of the buildings themselves actually obscured, there needs to be another explanation, and FET just doesn't have one.

Quote from: 29silhouette on July 29, 2013, 10:08:32 PM

Quote from: hoppy on July 29, 2013, 12:12:53 PM

Quote from: 29silhouette on July 29, 2013, 10:53:01 AM

Quote from: hoppy on July 28, 2013, 03:16:39 PM

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 expected drop over 48.5 miles is 1398'.

If the camera and buildings were all at water level, but they aren't.

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 earth
 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.

Keep repeating that and it might one day be true. If you're talking about RE drop, then you have to use RE maths which is a bit more complicated than the simple trigonometry involved in FE maths. Take a look at this diagram:



The orange line is the line-of-sight of something directly on the water of Lake Michigan (distances and curvatures greatly exaggerated, by the way). As you can see, it completely fails to see the target building (the long line)

A distance above the lake, it is able to see the building because it can now look down to see the horizon. The blue/purple line intersects the building a lot farther down than h₁ units lower than the orange line.

But notice how the blue-purple line is tangent to the Earth? If we placed a camera where it touches, then it would have the same view as the original (thus, the purple line). That's why it's easiest to calculate 'drop' from a camera on the surface of the lake at the horizon of the original camera than it is to calculate the drop from the original camera itself.