How Far is the Horizon?

I'm not sure why this has become an interest for me in recent weeks. Partly, I think, because I've been (re-)reading Patrick O'Brian's Aubrey/Maturin series of historical novels about the English Navy during the first part of the 19^th century.

Anyway, it's fairly easy to work out with some basic maths if you know the radius of the earth.

Assume the earth is a sphere, with centre C; and imagine that you are looking at the horizon, H, from point E, which is a distance h above the earth:



The distance to the horizon is d and the radius of the earth is r – which NASA says is on average 6,371 km, approximately 3,959 miles.. Since ∠EHC is a right angle, we know from Pythagoras' Theorem that:


... which can be rearranged to find d:


That's all very well and good, but is not very useful. Let's define d_m to be the distance to the horizon in miles, and d_f, h_f and r_f to be the measurements in feet, respectively, of the distance to the horizon, our height above the earth and the radius of the earth. Then, because there are 5,280 (3 x 1,760) feet in a mile:


Substituting in the actual radius of the earth (3,959 x 5,280 feet) and rearranging the formula a bit, we get:


Now, it so happens that 3,959 is almost exactly three-quarters of 5,280, so:


And, because the first term is so small compared with the second, we can approximate even further, and say:


So, if you're 6 ft tall, the distance to the horizon when you're standing on a beach as the water's edge is about 3 miles. If you were on the deck of Captain Aubrey's Surprise, say 18 ft above sea level, you'd be able to see over 5 miles; but at the top of the mast, which was probably 80 ft above sea level, you'd be able to see over twice as far.

In fact, this approximate calculation is remarkable accurate. At altitudes up to about 20,000 ft the difference between the "exact" distance given by equation (A) above and the approximate equation (B) is only about 0.01% - just a few yards.