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We've heard that ideas of the curvature of the earth come up when a ship is seen slowly disappearing down the horizon. How can we calculate the distance of the ship?
Answer 1:

I wondered about a simple formula for this as well. If you assume the earth is a perfect sphere, then a tower of height k meters will have a line of sight tangent to the ocean surface at a distance of:
Sqrt(R^2 - (R-k)^2) meters, neglecting the tilt of the tower relative to the tangent point. (i.e. we assume that this angle is small) This can be further approximated by noting that R (the earth Radius) is very much larger than the typical tower height. We get:
Sqrt(stuff above) = Sqrt(2*R*k) - Sqrt(k^3/8R) - Sqrt(k^5/512R^3) ...
Dropping the very small corrections, we get the neat formula:
Sqrt(2*R*k)
which is proportional to the geometric mean of the tower height and the earth radius. Plugging in the radius of the earth (6360km) and a 2meter tall man, we get: 5 km until the ocean is tangent to the man. So a 2 meter man can see a 2 meter man's head (standing on the surface) at about 10km.

From a 40meter (133') bluff -- the tangent point is 22.5km out. Thus two 40meter tall crow's nests can spot each other at 45km...
Without the crow's nest, you can only see the other ship at best at 22km...

Answer 2:

Here's an answer that's well written and illustrated:

http://boatsafe.com/nauticalknowhow/distance.htm

I got it by putting "height above horizon distance calculation" into the hotbot.com search engine.




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