At what FL should you see the curvature of the earth?
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At what FL should you see the curvature of the earth?
Got my pictures back from a recent trip and one picture caught my attention:
Blue sky, almost black at the very top of the picture, no clouds and I think I can make out the curvature of the earth.
The picture was taken at FL410 or 430 at about 45N. The lens was set to about 100mm.
Question is: Is the curvature visable at FL430 or is it just the effect of my lens? At what altitudes can one make it out?
7 7 7 7
Blue sky, almost black at the very top of the picture, no clouds and I think I can make out the curvature of the earth.
The picture was taken at FL410 or 430 at about 45N. The lens was set to about 100mm.
Question is: Is the curvature visable at FL430 or is it just the effect of my lens? At what altitudes can one make it out?
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FL0.01 (approx)
To answer the camera related part first:
Since you had the lens "zoomed" it is unlikely that there was a fisheye effect due to the focal length of the lens. In fact, if anything one might expect the lens to lessen any apparent curvature at that setting.
If you are concerned about distortion in the lens there is a simple test, if you can be bothered. Find a conveniently large man made object which you know (or assume) is rectilinear - a tall office block is a useful one. From a fair distance back, such that it does not appear distorted to the naked eye, take a few photos of it. Rotate the camera through a few random angles and repeat. If the developed pictures look fairly consistent you can probably be reasonably confident that the lens is good. (It's hardly a very scientific method, but probably good enough)
Regarding the FL question, and my rather flippant answer above. In theory the curvature of the earth should be visible from any altitude - the problem is that irregularity in the earth's surface masks the effect at lower altitudes, and the observer introduces a perception issue.
Essentially, from any altitude the observer forms a cone, with his location at the tip and the horizon being the base of the cone. Looking along the surface of the cone - which is what you do when you look at the 'horizon', for that is simply the cone's surface, the base appears curved. The higher you go, the more pronounced the curve. But at any height above that of a worm, it's theoretically visible.
Oh, and to keep any flat-earthers and Discworld fans happy, you aren't seeing the curvature of the surface - just the endge of the surface. We know the earth is a sphere now, and interpret it accordingly. But a flat disc (with optional elephants and turtle) would look the same.
For the case you quoted, at that FL you'd have a visual range to the horizon of ~270nm. If you consider the cone I mentioned before, it has a height of about 18nm (the base of the cone runs well below the earth under you) and a radius of some 270nm also. A very shallow cone, you're not that high. If we consider your camera lens has about a 30 deg field of view, and was looking along the cone, then the circular base of the cone would pass out of view 15 degs either side of the centre of the picture, or about 70 nm each side of the centre of the picture. In 70 nm the earth curves by about 1 deg, and you have 1 deg either side. So you should have about 2 degs of curvature in your picture, which ought to be discernable.
Stirring things up is the possibility of irregularity of the horizon. You'd really need no significant variation in the altitude of the horizon (whether clouds, sea, whatever) to make sure that what you were seeing was really the curvature.
Since you had the lens "zoomed" it is unlikely that there was a fisheye effect due to the focal length of the lens. In fact, if anything one might expect the lens to lessen any apparent curvature at that setting.
If you are concerned about distortion in the lens there is a simple test, if you can be bothered. Find a conveniently large man made object which you know (or assume) is rectilinear - a tall office block is a useful one. From a fair distance back, such that it does not appear distorted to the naked eye, take a few photos of it. Rotate the camera through a few random angles and repeat. If the developed pictures look fairly consistent you can probably be reasonably confident that the lens is good. (It's hardly a very scientific method, but probably good enough)
Regarding the FL question, and my rather flippant answer above. In theory the curvature of the earth should be visible from any altitude - the problem is that irregularity in the earth's surface masks the effect at lower altitudes, and the observer introduces a perception issue.
Essentially, from any altitude the observer forms a cone, with his location at the tip and the horizon being the base of the cone. Looking along the surface of the cone - which is what you do when you look at the 'horizon', for that is simply the cone's surface, the base appears curved. The higher you go, the more pronounced the curve. But at any height above that of a worm, it's theoretically visible.
Oh, and to keep any flat-earthers and Discworld fans happy, you aren't seeing the curvature of the surface - just the endge of the surface. We know the earth is a sphere now, and interpret it accordingly. But a flat disc (with optional elephants and turtle) would look the same.
For the case you quoted, at that FL you'd have a visual range to the horizon of ~270nm. If you consider the cone I mentioned before, it has a height of about 18nm (the base of the cone runs well below the earth under you) and a radius of some 270nm also. A very shallow cone, you're not that high. If we consider your camera lens has about a 30 deg field of view, and was looking along the cone, then the circular base of the cone would pass out of view 15 degs either side of the centre of the picture, or about 70 nm each side of the centre of the picture. In 70 nm the earth curves by about 1 deg, and you have 1 deg either side. So you should have about 2 degs of curvature in your picture, which ought to be discernable.
Stirring things up is the possibility of irregularity of the horizon. You'd really need no significant variation in the altitude of the horizon (whether clouds, sea, whatever) to make sure that what you were seeing was really the curvature.
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I would have thought the effect of earth curvature was mainly due to the fact that the image was distorted because the camera wasn't pointed exactly at the horizon. If it was pointed slightly above the horizon, you would expect to see apparent curvature as seen. Look at photographs of tall buildings taken horizontally from the ground- they look as if they are leaning over. I think there would be a slight effect of curvature if the camera was pointed exactly at the horizon because the horizon is slightly curved, though in the field of view of a camera through an aircraft window, not enough to be identifiable.
I have quite clearly identied curvature by eye at cruise altitude. You need a clear horizon, and high altitude shows this best, but on a really clear day I would have thought it would be apparent from 'medium' altitudes upwards.
I have quite clearly identied curvature by eye at cruise altitude. You need a clear horizon, and high altitude shows this best, but on a really clear day I would have thought it would be apparent from 'medium' altitudes upwards.
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Terry Pratchett notwithstanding, call me odd ( and many do ) and it may be my eyesight, but I've always noticed it at ground level - the horizon.
However, how far away is the horizon, given an uninterrupted view, say from a beach? I seem to remember my old geography teacher saying it was 11 miles.
However, how far away is the horizon, given an uninterrupted view, say from a beach? I seem to remember my old geography teacher saying it was 11 miles.
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Earth Curves
Well. I sat and thought about this for a long time, then I read Mad Flt Sc`s answer. Nope, cant follow that!!
And BLK 33, Ive heard differing answers on this, I think it depends on ambient temperature and light refraction etc, I heard on a clear day it was 26 or 27 miles, but again, Id love to hear something a bit more concrete.
Have a nice day all,
Speedy
And BLK 33, Ive heard differing answers on this, I think it depends on ambient temperature and light refraction etc, I heard on a clear day it was 26 or 27 miles, but again, Id love to hear something a bit more concrete.
Have a nice day all,
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It all depends how high you are
The distance to the horizon is incredibly dependent upon how high you are, and how high the object is (or how tall). So both the 11 miles and 26-27 mile answers can be correct.
Hence the spotting tops on battleships etc.
10-11 miles is a reasonable estimate for two small ships to mutually see each other. Battleships and the like would have a reasonable chance of seeing each other in the 20-25 miles range. But of course although the ships are that distance apart, the horizon is a little closer, it's as if you were looking over a hill.
That's one reason why you can see France from Kent (and vice versa) - as long as you are up on the cliffs. Trying to look from beach to beach is probably not possible (Straits of Dover are about 25 miles wide, from what I recall)
Speedbird, if I could work out how to draw a picture that I could post it would be a much simpler explanation. Doesn't seem to be a useful picture facility here (which is odd, some other vBulletin sites do have that feature, I guess it's disabled to conserve bandwidth)
Hence the spotting tops on battleships etc.
10-11 miles is a reasonable estimate for two small ships to mutually see each other. Battleships and the like would have a reasonable chance of seeing each other in the 20-25 miles range. But of course although the ships are that distance apart, the horizon is a little closer, it's as if you were looking over a hill.
That's one reason why you can see France from Kent (and vice versa) - as long as you are up on the cliffs. Trying to look from beach to beach is probably not possible (Straits of Dover are about 25 miles wide, from what I recall)
Speedbird, if I could work out how to draw a picture that I could post it would be a much simpler explanation. Doesn't seem to be a useful picture facility here (which is odd, some other vBulletin sites do have that feature, I guess it's disabled to conserve bandwidth)
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I may have this completely cockeyed, but I seem to recall that a good formula for working out how far away the horizon is (and I got this from Pprune, I'll have you know) is
d = 1.17 x sqrt(h)
where (d) is the distance in nautical miles and (h) is the height in feet.
Assuming you're at sea level and about 6 feet tall, you'd be able to see the grand distance of 2.6NM. Someone standing on our two ships would be what, 40, 50 feet high? That would give you about 8 and a half nautical miles.
41,000 feet gives you 237NM, give or take.
Someone tell me I've got this right??
d = 1.17 x sqrt(h)
where (d) is the distance in nautical miles and (h) is the height in feet.
Assuming you're at sea level and about 6 feet tall, you'd be able to see the grand distance of 2.6NM. Someone standing on our two ships would be what, 40, 50 feet high? That would give you about 8 and a half nautical miles.
41,000 feet gives you 237NM, give or take.
Someone tell me I've got this right??
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Yeah, your answers come out about right.
If you want to use any unit, this formula is more useful
d=sqr of (h^2 + 2rh) (just draw some triangles to get it)
Where
d=distance to horizon
h=height of viewpoint
r=radius of earth = 6371km
If you want to use any unit, this formula is more useful
d=sqr of (h^2 + 2rh) (just draw some triangles to get it)
Where
d=distance to horizon
h=height of viewpoint
r=radius of earth = 6371km
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Those formulae are approximate, and I doubt they would give accurate figures at such small heights, ie 6 ft.
Similarly at extreme heights your horizon will never be greater than half the diameter of the globe, taken from a point immediately below you.
Similarly at extreme heights your horizon will never be greater than half the diameter of the globe, taken from a point immediately below you.
