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Originally Posted by nattydread
Hmmm. Does anyone know how far a typical merchant would have to be before it disapears below the horizon? How far before its smoke would also be below the horizon?
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Hi!
It all depends upon the height of the observer and the height of the object being observed. For a rough geometric solution, the square of the distance to the horizon from your eye is equal to the height of the eye above the water times (2 x the radius of the Earth); the Earth's radius may be averaged as ~6372 km). The distance from the observed object to its horizon works on the same principle. Note this approach does not take into account the effects of atmospheric refraction, which "bends" light around the earth's curvature, thus extending the range at which objects can be seen; nor does it take into account the curvature of the earth, which has only a small effect unless you're almost in space.
For example, assume the observer's eye is 5 meters above sea level (maybe he is standing next to the UZO), and he is trying to observe a column of smoke just over 50 meters tall, under which chugs a coastal steamer.
- distance (U-boat to horizon) = sqrt(0.005 km x 2 x 6372 km) = roughly 7.98 km
distance (horizon to top of smoke) = sqrt(0.05 km x 2 x 6372 km) = roughly 25.24 km
total distance = (U-boat to horizon) + (horizon to top of smoke) = 7.98 km + 25.24 km = about 33.2 km
Hope this helps!
Pablo