Using the Pythagoras theorem for an OBC rightangled triangle it is possible to deduce the formula to the distance to the horizon.
AB = h (height of eyes of the observer above sea level in undergrounds)
OA = OC = r (ray of the Earth)
BC = d (looked distance to the horizon of the observer)
BO = h + r
CO = r
so
BO^{2} = BC^{2} + CO^{2}
(h + r)^{2} = d^{2} + r^{2}
next because
We are placing the mean of the ray in front of the globe R = 6370 x 10^{3} m and we are sharing the result through 1852 m in order to get nautical miles.
The composite thickness is causing atmospheres that light rays are yielding to certain bending, so the distance to the horizon will be a little bit bigger. The pattern is longing for the refraction for the average this way:
A rough formula is often practiced, where instead of value 2,08 we are putting value 2. Cursory defining the height of the observer above sea level, changing visibility
and changes in the refraction are causing that one should treat the result with the big carefulness.
We can from the model calculate the distance to the horizon or to the object about the known height (for example lights of the lighthouse) using the moment which the light is hiding behind the horizon. To grab such a moment hold of best when the light is visible for the observer standing aboard and invisible for the sitting observer.
