Willis Tower

Quote from: Saros on February 24, 2015, 12:13:57 PM

Quote from: Lemmiwinks on February 24, 2015, 09:42:46 AM

Quote from: Saros on February 24, 2015, 04:45:08 AM

I am not sure what SKEP tries to imply, but it is totally possible to see the Willis tower from 47 miles away. That is even possible within the RE in mind. Just use a regular horizon distance calculator and bear in mind the Willis tower has a height of 442 m without the antenna(which is not small at all). Just punch in the numbers in a calculator and the result is that from a height of 2 meters you could see something 442 m tall from a distance of 49 miles. So, in fact that observation doesn't prove anything as it is consistent with RE as well since even the horizon distance calculators allow it. Check out http://members.home.nl/7seas/radcalc.htm
For h1 enter 2 m for h2 enter 442 m and you will see how far is the visible horizon. Bear in mind that the photo was taken from a higher position than 2 m, so that explains why you see so much of the tower.

It is ridiculous how people who believe in round Earth start denying something is possible simply because it was suggested by a FE'er when in fact it is possible even within their theory... This just suggests they are not so bright after all. Anyone can get confused!

It appears Skep is a FE'r, so your last paragraph doesn't make much sense to me.

I don't care what he believes in. I just wanted to point out that it is absolutely possible to see the Willis tower from that distance. He suggested otherwise. What do you believe in?

But you obviously did, because you assumed he was a RE'r and only didn't believe it could be seen because a FE'r said it could.  Here, I'll quote the relevant part:

Quote

It is ridiculous how people who believe in round Earth start denying something is possible simply because it was suggested by a FE'er when in fact it is possible even within their theory...

Of course you can see the tower from that distance, and in that picture you can see the curvature of the earth as well, its a rather great piece of Round Earth proof.

I assumed he is a RE'er because he probably is. Why would a FE'er try to debunk an argument intended to prove flat Earth?

I don't see any curvature of the Earth though. That is not an example of curvature, but an example of how objects gradually vanish in the distance starting from the bottom.

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

I don't see any curvature of the Earth though. That is not an example of curvature, but an example of how objects gradually vanish in the distance starting from the bottom.

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

That sounds like it may be a better-developed description of what Mr. Bloomington was struggling to say but couldn't manage to get into coherent sentences.

If I understand what you're saying, the water surface is actually flat, but it looks like it's rising as distance from you increases. This happens because light from something above water level in the distance gets "mixed with" light reflecting from the water, and the further away the distant object is, the higher the "mixing" will take place, making the bottom get lost in light from the surface of the water, and the further away it is, more of it gets lost from the bottom up. Is that it, or at least close?

Neat idea, but I don't think it explains what we see. Think of a lighthouse on a small island some miles from you with a high mountain range many more miles beyond it. As you look at the lighthouse, the bottom, say, 20 feet are hidden by the water. meanwhile, the water continues to appear to rise behind it as it obscures, let's say the bottom one thousand feet of the more distant mountains - that is, the water appears to rise behind the lighthouse until it's 1000 feet up the mountains. This isn't what we see, though. We see the horizon cut off the lower thousand feet of the distant mountains, the bottom of the lighthouse, and the mountains again, all in a single unbroken line regardless the distance to each of the objects, as long as they all are past the horizon.

Clearly the best idea yet! Keep 'em coming!

On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all.

Wrong. Scattering of light making things blue, like the daytime sky, occurs because it is scattered downwards. Things in the distance should dissipate to redness not blueness, just like the setting sun.

Quote from: Saros on February 24, 2015, 01:26:18 PM

On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all.

Wrong. Scattering of light making things blue, like the daytime sky, occurs because it is scattered downwards. Things in the distance should dissipate to redness not blueness, just like the setting sun.

Actually, what I said is not wrong but comes directly from the official science books. Try again.

Quote from: Saros on February 24, 2015, 01:26:18 PM

I don't see any curvature of the Earth though. That is not an example of curvature, but an example of how objects gradually vanish in the distance starting from the bottom.

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

That sounds like it may be a better-developed description of what Mr. Bloomington was struggling to say but couldn't manage to get into coherent sentences.

If I understand what you're saying, the water surface is actually flat, but it looks like it's rising as distance from you increases. This happens because light from something above water level in the distance gets "mixed with" light reflecting from the water, and the further away the distant object is, the higher the "mixing" will take place, making the bottom get lost in light from the surface of the water, and the further away it is, more of it gets lost from the bottom up. Is that it, or at least close?

Neat idea, but I don't think it explains what we see. Think of a lighthouse on a small island some miles from you with a high mountain range many more miles beyond it. As you look at the lighthouse, the bottom, say, 20 feet are hidden by the water. meanwhile, the water continues to appear to rise behind it as it obscures, let's say the bottom one thousand feet of the more distant mountains - that is, the water appears to rise behind the lighthouse until it's 1000 feet up the mountains. This isn't what we see, though. We see the horizon cut off the lower thousand feet of the distant mountains, the bottom of the lighthouse, and the mountains again, all in a single unbroken line regardless the distance to each of the objects, as long as they all are past the horizon.

Clearly the best idea yet! Keep 'em coming!

I have never seen your lighthouse example in real life. Could you please provide at least a photo? But nevertheless, I can explain it away too since your logic appears wrong. When I have a bit more time will write you a message here.

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

Apart from the fact none of what you said here has any science whatsoever to back it up. Why does this peculiar effect create a horizon at exactly the same distance you would expect to see one if you were standing on a sphere with a radius of 4000 miles. Is it just coincidence?

Quote from: Saros on February 24, 2015, 01:26:18 PM

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

Apart from the fact none of what you said here has any science whatsoever to back it up. Why does this peculiar effect create a horizon at exactly the same distance you would expect to see one if you were standing on a sphere with a radius of 4000 miles. Is it just coincidence?

It is absolutely scientific. You can't see anything unless it reflects light. Therefore the water in front of the object you're trying to see also reflects light which eventually reaches your pupil. Over distance, however, this prevents us from seeing stuff. The things vanish in the distance starting from the bottom. It is common sense, since near the surface the light blends into one before reaching your eyes. The farther away, the stronger the effect till you cannot see anything but blue. By the way, this effect is most distinct over water for obvious reasons.

Quote from: herewegoround on February 25, 2015, 12:46:31 AM

Quote from: Saros on February 24, 2015, 01:26:18 PM

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

Apart from the fact none of what you said here has any science whatsoever to back it up. Why does this peculiar effect create a horizon at exactly the same distance you would expect to see one if you were standing on a sphere with a radius of 4000 miles. Is it just coincidence?

It is absolutely scientific. You can't see anything unless it reflects light. Therefore the water in front of the object you're trying to see also reflects light which eventually reaches your pupil. Over distance, however, this prevents us from seeing stuff. The things vanish in the distance starting from the bottom. It is common sense, since near the surface the light blends into one before reaching your eyes. The farther away, the stronger the effect till you cannot see anything but blue. By the way, this effect is most distinct over water for obvious reasons.

If there is a direct line from your eye to an object which is large enough to be seen, there is no reason why it shouldn't be visible. What you said doesn't mean anything. Why would the presence of water underneath something prevent you from seeing it from a distance? I have studied optics I don't know of any reason why this should be. You also haven't explained why this creates a horizon at the exact distance you would expect there to be one on a sphere with the radius of the Earth.

Quote from: Saros on February 25, 2015, 06:33:39 AM

Quote from: herewegoround on February 25, 2015, 12:46:31 AM

Quote from: Saros on February 24, 2015, 01:26:18 PM

It is due to the simple fact that air/water molecules deflect the light path and the light we observe reflected from an object mixes with the light coming from the objects/environment on its path, thus in this case it blends with the light coming from the water right in front of it, so you cannot see the bottom part of the building, but instead you see water. The surrounding incoming light is eating away the object observed in it. On top of that molecules scatter blue light more efficiently than red light, so it is only natural that everything would eventually dissipate in blueness, but it doesn't mean there is a curvature at all. That is also the reason why when you go up you see more of the object, and the higher you go you see more and more. Think about it.

Apart from the fact none of what you said here has any science whatsoever to back it up. Why does this peculiar effect create a horizon at exactly the same distance you would expect to see one if you were standing on a sphere with a radius of 4000 miles. Is it just coincidence?

It is absolutely scientific. You can't see anything unless it reflects light. Therefore the water in front of the object you're trying to see also reflects light which eventually reaches your pupil. Over distance, however, this prevents us from seeing stuff. The things vanish in the distance starting from the bottom. It is common sense, since near the surface the light blends into one before reaching your eyes. The farther away, the stronger the effect till you cannot see anything but blue. By the way, this effect is most distinct over water for obvious reasons.

If there is a direct line from your eye to an object which is large enough to be seen, there is no reason why it shouldn't be visible. What you said doesn't mean anything. Why would the presence of water underneath something prevent you from seeing it from a distance? I have studied optics I don't know of any reason why this should be. You also haven't explained why this creates a horizon at the exact distance you would expect there to be one on a sphere with the radius of the Earth.

There is no direct line to the object after a certain distance. Didn't you read what I posted? What I said makes perfect sense only if you try to understand it. Light gets scattered on its path to your eye and it doesn't travel straight forever, hence there is no direct line of sight over a big distance. That is why when you go higher you start seeing the object again, but from a certain fixed position the horizon is always limited(except in cases of anomalous refraction). Nothing to do with curvature. You're basically trying to suggest that light will have to reach your pupil regardless of dispersion and scattering. That is impossible. The dispersion is greater near the surface if you're near the surface, of course, that is just relative to your position and it is a matter of perspective. The radius has nothing to do with that, in fact, it is indeed a rough coincidence that the formula seemingly produces correct results. The key is the light itself and its properties while travelling through the atmosphere and through other media combined with the reflective properties of the observed object and the position you're observing it from. That is it.

So light only scatters upward?

So light only scatters upward?

Is that really a question. Do you expect the light to scatter continuously forward near the surface and why would it be the case when the environment in which it is travelling is highly dispersive? Additionally, no one said it scatters only upward, but it does scatter upward too, and that is why you eventually lose sight of the object.

There is no direct line to the object after a certain distance. Didn't you read what I posted? What I said makes perfect sense only if you try to understand it. Light gets scattered on its path to your eye and it doesn't travel straight forever, hence there is no direct line of sight over a big distance. That is why when you go higher you start seeing the object again, but from a certain fixed position the horizon is always limited(except in cases of anomalous refraction). Nothing to do with curvature. You're basically trying to suggest that light will have to reach your pupil regardless of dispersion and scattering. That is impossible. The dispersion is greater near the surface if you're near the surface, of course, that is just relative to your position and it is a matter of perspective. The radius has nothing to do with that, in fact, it is indeed a rough coincidence that the formula seemingly produces correct results. The key is the light itself and its properties while travelling through the atmosphere and through other media combined with the reflective properties of the observed object and the position you're observing it from. That is it.

I don't mean to sound rude because you are being polite at least but that is scientifically illiterate nonsense. There is no evidence at all to back it up, in fact everything that is known about light contradicts it.

Quote from: Saros on February 25, 2015, 11:50:02 AM

There is no direct line to the object after a certain distance. Didn't you read what I posted? What I said makes perfect sense only if you try to understand it. Light gets scattered on its path to your eye and it doesn't travel straight forever, hence there is no direct line of sight over a big distance. That is why when you go higher you start seeing the object again, but from a certain fixed position the horizon is always limited(except in cases of anomalous refraction). Nothing to do with curvature. You're basically trying to suggest that light will have to reach your pupil regardless of dispersion and scattering. That is impossible. The dispersion is greater near the surface if you're near the surface, of course, that is just relative to your position and it is a matter of perspective. The radius has nothing to do with that, in fact, it is indeed a rough coincidence that the formula seemingly produces correct results. The key is the light itself and its properties while travelling through the atmosphere and through other media combined with the reflective properties of the observed object and the position you're observing it from. That is it.

I don't mean to sound rude because you are being polite at least but that is scientifically illiterate nonsense. There is no evidence at all to back it up, in fact everything that is known about light contradicts it.

Could you please provide some scientific source which clearly states otherwise? I mean, have you read any scientific journals or books which contradict to what I shared here? It is totally scientific to make the claim that light gets scattered when travelling through air or water and that you cannot observe the source of light forever if it is located near the surface since it will quickly disperse over distance. Anyway, I might have explained it in layman's terms, but it is not nonsense.

Quote from: herewegoround on February 26, 2015, 12:41:48 AM

Quote from: Saros on February 25, 2015, 11:50:02 AM

There is no direct line to the object after a certain distance. Didn't you read what I posted? What I said makes perfect sense only if you try to understand it. Light gets scattered on its path to your eye and it doesn't travel straight forever, hence there is no direct line of sight over a big distance. That is why when you go higher you start seeing the object again, but from a certain fixed position the horizon is always limited(except in cases of anomalous refraction). Nothing to do with curvature. You're basically trying to suggest that light will have to reach your pupil regardless of dispersion and scattering. That is impossible. The dispersion is greater near the surface if you're near the surface, of course, that is just relative to your position and it is a matter of perspective. The radius has nothing to do with that, in fact, it is indeed a rough coincidence that the formula seemingly produces correct results. The key is the light itself and its properties while travelling through the atmosphere and through other media combined with the reflective properties of the observed object and the position you're observing it from. That is it.

I don't mean to sound rude because you are being polite at least but that is scientifically illiterate nonsense. There is no evidence at all to back it up, in fact everything that is known about light contradicts it.

Could you please provide some scientific source which clearly states otherwise? I mean, have you read any scientific journals or books which contradict to what I shared here? It is totally scientific to make the claim that light gets scattered when travelling through air or water and that you cannot observe the source of light forever if it is located near the surface since it will quickly disperse over distance. Anyway, I might have explained it in layman's terms, but it is not nonsense.

Provide the scientific source which says that light would be scattered like this. I studied both geometric and physical optics as part of my degree in physics and there was nothing we learned that would suggest that light would be scattered if it was travelling close to a surface. Not even anything similar. Why would it be? Even if there was such a phenomenon it wouldn't be effective over such a tiny distance. Light takes about 4 x 10^-9 seconds to travel that far. It's no distance at all. If you honestly think this is a valid theory you need to provide some mathematics, a proper explanation. It just sounds like another ad hoc idea thrown together to overcome the horizon problem.

Quote from: Saros on February 26, 2015, 08:05:07 AM

Quote from: herewegoround on February 26, 2015, 12:41:48 AM

Quote from: Saros on February 25, 2015, 11:50:02 AM

There is no direct line to the object after a certain distance. Didn't you read what I posted? What I said makes perfect sense only if you try to understand it. Light gets scattered on its path to your eye and it doesn't travel straight forever, hence there is no direct line of sight over a big distance. That is why when you go higher you start seeing the object again, but from a certain fixed position the horizon is always limited(except in cases of anomalous refraction). Nothing to do with curvature. You're basically trying to suggest that light will have to reach your pupil regardless of dispersion and scattering. That is impossible. The dispersion is greater near the surface if you're near the surface, of course, that is just relative to your position and it is a matter of perspective. The radius has nothing to do with that, in fact, it is indeed a rough coincidence that the formula seemingly produces correct results. The key is the light itself and its properties while travelling through the atmosphere and through other media combined with the reflective properties of the observed object and the position you're observing it from. That is it.

I don't mean to sound rude because you are being polite at least but that is scientifically illiterate nonsense. There is no evidence at all to back it up, in fact everything that is known about light contradicts it.

Could you please provide some scientific source which clearly states otherwise? I mean, have you read any scientific journals or books which contradict to what I shared here? It is totally scientific to make the claim that light gets scattered when travelling through air or water and that you cannot observe the source of light forever if it is located near the surface since it will quickly disperse over distance. Anyway, I might have explained it in layman's terms, but it is not nonsense.

Provide the scientific source which says that light would be scattered like this. I studied both geometric and physical optics as part of my degree in physics and there was nothing we learned that would suggest that light would be scattered if it was travelling close to a surface. Not even anything similar. Why would it be? Even if there was such a phenomenon it wouldn't be effective over such a tiny distance. Light takes about 4 x 10^-9 seconds to travel that far. It's no distance at all. If you honestly think this is a valid theory you need to provide some mathematics, a proper explanation. It just sounds like another ad hoc idea thrown together to overcome the horizon problem.

Looking for a reply to this.

Also looking for cikljamas to explain why he intentionally doctored a photo to prove his point. Also looking for him to draw some more lines on the unedited photos.

You can't see anything unless it reflects light.

Erroneous.  As usual LOL.

A tungsten filament, when subject to an applied current, emits photons.  That's how an incandescent light bulb works.  Zero "reflection" involved.

You're welcome.   

Quote from: Saros on February 25, 2015, 06:33:39 AM

You can't see anything unless it reflects light.

Erroneous.  As usual LOL.

A tungsten filament, when subject to an applied current, emits photons.  That's how an incandescent light bulb works.  Zero "reflection" involved.

You're welcome.   

LOL

I am obviously talking about objects which DON'T emit their own light. Nice try trolling again.

Quote from: ausGeoff on February 28, 2015, 12:03:28 AM

Quote from: Saros on February 25, 2015, 06:33:39 AM

You can't see anything unless it reflects light.

Erroneous.  As usual LOL.

A tungsten filament, when subject to an applied current, emits photons.  That's how an incandescent light bulb works.  Zero "reflection" involved.

You're welcome.   

I am obviously talking about objects which DON'T emit their own light. Nice try trolling again.

So now you're qualifying your argument after it's been pointed out (by me) that one of your contentions was incorrect?  You really need to double check your claims before you post them.  One faulty claim can lead to others.  Tsk, tsk...
 

Quote from: Saros on February 28, 2015, 01:06:34 AM

Quote from: ausGeoff on February 28, 2015, 12:03:28 AM

Quote from: Saros on February 25, 2015, 06:33:39 AM

You can't see anything unless it reflects light.

Erroneous.  As usual LOL.

A tungsten filament, when subject to an applied current, emits photons.  That's how an incandescent light bulb works.  Zero "reflection" involved.

You're welcome.   

I am obviously talking about objects which DON'T emit their own light. Nice try trolling again.

So now you're qualifying your argument after it's been pointed out (by me) that one of your contentions was incorrect?  You really need to double check your claims before you post them.  One faulty claim can lead to others.  Tsk, tsk...

As much as it pains me to agree with Saros, his statement was quite clear in the context of this thread ausGeoff. Your comment was wholly unnecessary and needlessly combative.

I don't have much to add in this discussion besides a photo. This zoomed in photo was taken ~60 miles away from Kenosha, Wisconsin, looking south at Chicago. The buildings visible from left to right are the John Hancock Center, Aon Center, Trump International Hotel and Tower and the Sears Tower. Have a good one.

I don't have much to add in this discussion besides a photo. This zoomed in photo was taken ~60 miles away from Kenosha, Wisconsin, looking south at Chicago. The buildings visible from left to right are the John Hancock Center, Aon Center, Trump International Hotel and Tower and the Sears Tower. Have a good one.

http://i.imgur.com/JYrQ3OB.jpg

That's a cool picture! Do you know the height above lake level it was taken from?

I don't have much to add in this discussion besides a photo. This zoomed in photo was taken ~60 miles away from Kenosha, Wisconsin, looking south at Chicago. The buildings visible from left to right are the John Hancock Center, Aon Center, Trump International Hotel and Tower and the Sears Tower. Have a good one.

The height of the Willis tower to the tip is 527m, the architectural height is 442m. So, on the round earth, even if an observer was 5 meters above water level he couldn't have seen at that distance (60 miles) any part of Willis tower (let alone other buildings) except antenna.

Even if an observer was 30 meters above water level he couldn't have seen at that distance (60 miles) any part of Willis tower (let alone other buildings) except antennas.

Thus the question about the height the picture was taken from.

The first time T4 posted this picture he gave the distance as 50 miles. Kenosha's PR literature gives the distance to Chicago as 55 miles. Measuring the distance to the Willis Tower from the Kenosha waterfront in Google Earth may give a better idea of the distance.

The obvious follow-up question would be "is this unusual?" Your horizon calculator's refraction model has to be based on average conditions since they don't ask for things like temperature and air pressure, both of which are significant. Kenosha's civic literature describes a couple of 50-foot lighthouses on the lakefront (and there likely are some taller buildings; 30m doesn't seem out of the question for a city that size), but I don't see anything about seeing Chicago in the distance, which suggests this is not common.

Setting that aside for the moment, how do you explain this example of the buildings, again, "disappearing from the bottom" if the Earth is flat? Since this photo is taken from north of Chicago, the Hancock Tower is closer than the Willis Tower. Why are both of them cut off by the same, straight, horizon line? Shouldn't the Willis be cut off higher, since it's further away but sitting with their bases at the same level? This would be necessary if Saros' idea were correct.

Quote from: t94 on March 01, 2015, 02:00:51 PM

I don't have much to add in this discussion besides a photo. This zoomed in photo was taken ~60 miles away from Kenosha, Wisconsin, looking south at Chicago. The buildings visible from left to right are the John Hancock Center, Aon Center, Trump International Hotel and Tower and the Sears Tower. Have a good one.

The height of the Willis tower to the tip is 527m, the architectural height is 442m. So, on the round earth, even if an observer was 5 meters above water level he couldn't have seen at that distance (60 miles) any part of Willis tower (let alone other buildings) except antenna.

The height of the tower to the tip is 527m, the architectural height is 442m. The difference between the two is 85m which is about 1/6 of the total. I would estimate then that only the top 1/3 of the entire tower is visible in this picture. So the lowest part of the tower showing is about 350m high.

You didn't state the height the picture was taken at but it's quite clearly not at sea level, I am going to estimate it is 5m above sea level. That's about a 2 storey house.

The distance to the horizon is calculated using √((R+h)2-R2), where R is the radius of the Earth, 6400 km, and h is the height of your eyes above sea level.

The distance to the horizon from the lowest part of the building is √((6400+0.35)2-64002)=67km.

The distance to the horizon if your eyes are 5m above sea level is √((6400+0.005)2-64002)=8km.

Therefore, if the picture was taken 5m above sea level and just managed to see a point on the tower 350m above sea level then the distance from the point the picture was taken to the tower was 67km + 8km = 75km or 47 miles.

At any time when the sea is calm and the weather clear, the light of the Eddystone may be seen from an elevation of five feet above the water level; and according to the Admiralty directions, it "maybe seen thirteen nautical (or fifteen statute), miles," 1 or one mile further away than the position of the observers on the above-named occasion; yet, on that occasion, and at a distance of only fourteen statute miles, notwithstanding that it was a very fine autumn day, and a clear background existed, not only was the lantern, which is 80 feet high, not visible, but the top of the vane, which is 100 feet above the foundation, was, as stated in the report "entirely out of sight." There was, however, a considerable "swell" in the sea beyond the breakwater.

That vessels, lighthouses, light-ships, buoys, signals, and other known and fixed objects are sometimes more distinctly seen than at other times, and are often, from the same common elevation, entirely out of sight when the sea is rough, cannot be denied or doubted by any one of experience in nautical matters.

The conclusion which such observations necessitate and force upon us is, that the law of perspective, which is everywhere visible on land, is modified when observed in connection with objects on or near the sea. But how modified? If the water were frozen and at perfect rest, any object on its surface would be seen again and again as often as it disappeared and as far as telescopic or magnifying power could be brought to bear upon it. But because this is not the case--because the water is always more or less in motion, not only of progression but of fluctuation and undulation, the "swells" and waves into which the surface is broken, operate to prevent the line of sight from passing absolutely parallel to the horizontal water line.

If it is argued that "there are times when the surface of the sea is perfectly calm, and that at such times at least, if the earth is a plane, the telescope ought to restore the hull of a ship, irrespective of distance, providing its power is great enough to magnify it," the reply is that practical experiments have proved that during what is called a "dead calm," the undulations or waves in the water amount to more than 20 inches, as will be seen from the following extracts:--

ON THE DIMENSIONS OF OCEAN WAVES.

"This interesting subject was very fully entered into at a recent meeting of the Academy of Sciences, by Admiral Coupvent de Bois:

"It is not easy to ascertain the height of the waves of the ocean; nevertheless, the method adopted for the purpose is capable of affording sufficiently exact results. The point in the shrouds corresponding with a tangent to the tops of the highest waves is ascertained by gradually ascending them, and making observations until it is reached. That point being determined, the known dimensions of the ship give the height of the waves above the line of flotation, which corresponds with the horizon of the sea, in the trough of the wave. In this way the following results were obtained:--



"The lengths of the waves have also been measured, and it has been found that, for example, waves of 27 feet in height, are about 1640 feet in length."

It is well known that even on lakes of small dimensions and also on canals, when high winds prevail for some time in the same direction, the ordinary ripple is converted into comparatively large waves. On the "Bedford Canal," during the windy season, the water is raised into undulations so high, that through a powerful telescope at an elevation of 8 inches, a boat two or three miles away will be invisible; but at other times, through the same telescope the same kind of boat may be seen at a distance of six or eight miles.

During very fine weather when the water has been calm for some days and become as it were settled down, persons are often able to see with the naked eye from Dover the coast of France, and a steamer has been traced all the way across the channel. At other times when the winds are very high, and a heavy swell prevails, the coast is invisible, and the steamers cannot be traced the whole distance from the same altitude, even with a good telescope.

Instances could be greatly multiplied, but already more evidence has been given than the subject really requires, to prove that when a telescope does not restore the hull of a distant vessel it is owing to a purely special and local cause...Read more : http://www.theflatearthsociety.org/forum/index.php?topic=62885.msg1661871#msg1661871

When the water is frozen then we have different result (we can see the very bottom of a distant mountains) :

At any time when the sea is calm and the weather clear, the light of the Eddystone may be seen from an elevation of five feet above the water level; and according to the Admiralty directions, it "maybe seen thirteen nautical (or fifteen statute), miles," 1 or one mile further away than the position of the observers on the above-named occasion; yet, on that occasion, and at a distance of only fourteen statute miles, notwithstanding that it was a very fine autumn day, and a clear background existed, not only was the lantern, which is 80 feet high, not visible, but the top of the vane, which is 100 feet above the foundation, was, as stated in the report "entirely out of sight." There was, however, a considerable "swell" in the sea beyond the breakwater.

That vessels, lighthouses, light-ships, buoys, signals, and other known and fixed objects are sometimes more distinctly seen than at other times, and are often, from the same common elevation, entirely out of sight when the sea is rough, cannot be denied or doubted by any one of experience in nautical matters.

The conclusion which such observations necessitate and force upon us is, that the law of perspective, which is everywhere visible on land, is modified when observed in connection with objects on or near the sea. But how modified? If the water were frozen and at perfect rest, any object on its surface would be seen again and again as often as it disappeared and as far as telescopic or magnifying power could be brought to bear upon it. But because this is not the case--because the water is always more or less in motion, not only of progression but of fluctuation and undulation, the "swells" and waves into which the surface is broken, operate to prevent the line of sight from passing absolutely parallel to the horizontal water line.

If it is argued that "there are times when the surface of the sea is perfectly calm, and that at such times at least, if the earth is a plane, the telescope ought to restore the hull of a ship, irrespective of distance, providing its power is great enough to magnify it," the reply is that practical experiments have proved that during what is called a "dead calm," the undulations or waves in the water amount to more than 20 inches, as will be seen from the following extracts:--

ON THE DIMENSIONS OF OCEAN WAVES.

"This interesting subject was very fully entered into at a recent meeting of the Academy of Sciences, by Admiral Coupvent de Bois:

"It is not easy to ascertain the height of the waves of the ocean; nevertheless, the method adopted for the purpose is capable of affording sufficiently exact results. The point in the shrouds corresponding with a tangent to the tops of the highest waves is ascertained by gradually ascending them, and making observations until it is reached. That point being determined, the known dimensions of the ship give the height of the waves above the line of flotation, which corresponds with the horizon of the sea, in the trough of the wave. In this way the following results were obtained:--



"The lengths of the waves have also been measured, and it has been found that, for example, waves of 27 feet in height, are about 1640 feet in length."

It is well known that even on lakes of small dimensions and also on canals, when high winds prevail for some time in the same direction, the ordinary ripple is converted into comparatively large waves. On the "Bedford Canal," during the windy season, the water is raised into undulations so high, that through a powerful telescope at an elevation of 8 inches, a boat two or three miles away will be invisible; but at other times, through the same telescope the same kind of boat may be seen at a distance of six or eight miles.

During very fine weather when the water has been calm for some days and become as it were settled down, persons are often able to see with the naked eye from Dover the coast of France, and a steamer has been traced all the way across the channel. At other times when the winds are very high, and a heavy swell prevails, the coast is invisible, and the steamers cannot be traced the whole distance from the same altitude, even with a good telescope.

Instances could be greatly multiplied, but already more evidence has been given than the subject really requires, to prove that when a telescope does not restore the hull of a distant vessel it is owing to a purely special and local cause...Read more : http://www.theflatearthsociety.org/forum/index.php?topic=62885.msg1661871#msg1661871

When the water is frozen then we have different result (we can see the very bottom of a distant mountains) :

If there was even a fraction credibility in these claims what we would see would be very different. The distance at which ships start to disappear would vary enormously. As it is, it doesn't. It is always exactly as would be expected if you were standing on a sphere with a radius of 4000 miles.

It's slightly tragic that flat Earth proponents bring up the Bedford experiment; a poorly conducted 18th century fiasco with anomalous atypical results. When it was repeated with proper controls the results were perfectly in line with round Earth predictions.

How far across water you can see land depends on how far away the land is and how far above sea level it rises. The shortest distance across the English Channel is 20 miles or 33 km.

The distance to the horizon from a point 100m above sea level is √(6400.1²-6400²)=35km. If the land in France rises just 100m within 2km of the shore line it would be perfectly possible to see it from England.

When the water is frozen then we have different result (we can see the very bottom of a distant mountains) :

How distant are those mountains? How do you know you're seeing the very bottom of them?

This is simply idle speculation without a lot more details.

Quote from: cikljamas on March 02, 2015, 08:15:39 AM

When the water is frozen then we have different result (we can see the very bottom of a distant mountains) :

How distant are those mountains? How do you know you're seeing the very bottom of them?

This is simply idle speculation without a lot more details.

Of course, it is a speculation. Cikljamas would expect us to believe it is true just because he says so. We know nothing about the photos and about the mountains. All this conversation would be meaningful if someone really took the time to investigate the issue by going out and taking photos and doing measurements himself and then coming back here to present the results instead of speculating and interpreting pictures.

Quote from: Alpha2Omega on March 02, 2015, 09:37:15 AM

Quote from: cikljamas on March 02, 2015, 08:15:39 AM

When the water is frozen then we have different result (we can see the very bottom of a distant mountains) :

How distant are those mountains? How do you know you're seeing the very bottom of them?

This is simply idle speculation without a lot more details.

Of course, it is a speculation. Cikljamas would expect us to believe it is true just because he says so. We know nothing about the photos and about the mountains. All this conversation would be meaningful if someone really took the time to investigate the issue by going out and taking photos and doing measurements himself and then coming back here to present the results instead of speculating and interpreting pictures.

 Sorry, his credibility is forever broken here. Purposefully and deliberately altered photos to prove a point, which when he got called out on made him disappear from the forums for 3 days and to not respond to the doctored photos, despite him posting in the same thread after the fact.


Edit: put my statement in someone else's quote