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originally posted by: WaxingGibbons
a reply to: samara11278
Since the side windows sit at an ~90 degree angle to the round one, giving the cupola a "cup" shape... if you were to look out of those side windows, even though you could not see the full hemisphere of Earth from this distance, you could see black space on the outside edge of the area visible to you.
Not in a perfect sphere around a portion of Earth......
Elementary geometry tells us that, because the angle between the dashed lines at G is a right angle, the distance OG from the observer (O) to the horizon (G) is related to the radius R and the observer's height h by the Pythagorean Theorem:
(R + h)2= R2 + OG2
or
OG2 = (R + h)2 − R2 .
But if we expand the term (R + h)2 = R2 + 2 R h + h2, the R2 terms cancel, and we find
OG = sqrt ( 2 R h + h2 ).
It's customary to use the fact that h R at this point, so that we can neglect the second term. Then
OG ≈ sqrt ( 2 R h )
is the distance to the horizon, neglecting refraction.
The solid arc OH now represents the curved line of sight; H is the (refracted) apparent horizon. Notice that refraction lets us see a little farther, if the ray is concave toward the Earth, as shown here.
If we can assume a constant lapse rate in the air between the eye and the Earth's surface, and if the observer's height h is small compared to the 8-km height of the homogeneous atmosphere, then we can assume the curved ray is an arc of a circle. This assumption makes things easy, because the relative curvature of the ray and the Earth's surface is all that matters. In effect, we can use the previous result, but just use an effective radius of curvature for the Earth that is bigger than the real one.
This assumption is made so often that it's conventional in surveying and geodesy to use a “refraction constant” that's just the ratio of the two curvatures. A typical value of the ratio is about 1/7; that is, the ray curves about 1/7 as much as the Earth does (or, equivalently, the radius of curvature of the ray is about 7 times that of the Earth's surface).
Using this “typical” value means we should just use the formula given above, but use a value R′ instead of R for the effective radius of the Earth, where
1/R′ = 1/R − 1/(7R) = 6/(7R) ,
so that
R′ = R × 7/6 .
originally posted by: WaxingGibbons
a reply to: samara11278
No, from 450 km you can only see about 4700 km of the surface, so there is no way you would see space around the edges of this area, at all, in fact, since space is not next to this area, there is more surface next to this area. The Earth would simply fill up the whole cupola up to the frame, especially with the pics I posted which are taken from deeper inside the cupola, about 2 m from the windows.
Again, from 450 km you simply cannot see more than around 4700 km of the surface......
originally posted by: EartOccupant
With an Fish I lens I CAN make the Shot
With and Without glass in the windows
But now there is distortion of the edges of the windows, they round up.