A common issue for watersports refers to the distance to the horizon. Often the question is asked, how far you can see a moving ship just beyond the horizon. We do suspect that this distance will depend on the vertical size of the ship and the observer's elevation above the sea surface. With the help of some simple mathematical considerations, applying not more than the Pythagorean theorem, we are able to answer this question. First of all, we should sketch the situation: It shows the circular cross-section of the earth with its radius R and center M. We attach to this circle a tangent through the point H. Left on this line there is point P1 located at a distance l1 from H and with a height h1 above the sea level. This point P1 represents the ship with a height h1. To the right side of H there is point P2 localted at a distance l2 of H and with a height h2. This point P2 represents the position of the observer. The triangle M, P1, H can be described by the Pythagorean theorem, applying the following relationship: By multiplying (binomial formula), we can write down the right side of the equation as: Substracting the term R2 on both sides, we get: After applying the square root on both side, we obtain the equation for determining the distance from the horizon l1: For the entire distance l from P1 tor P2 we obtain finally: The distance l thus corresponds to the maximum viewing distance of a vessel with a height h1 and an observer with an elevation h2 above the sea's surface. For reasons of symmetry, we can swap the roles of the ship and its observer. The following input form is calculating these three distances l1,l2 and l. Of course, this calculation needs to know the value for the earth's radius. This value is approximately 6370 km.

 h1 [m] h2 [m] l1 [km] l2 [km] l [km]

Due to the pressure gradient of the atmosphere, light beams are propagating in a curved manner, when beeing nearly parallel to the earth's surface. From this effect the horizon appears at a larger distance than it should by pure geometrical calculation. Nevertheless, in order to work with the above derived formula, the refraction effect can be taken into account by simply assuming a virtually enlarged radius of the earth. For the propagation of light (wavelength between 250 and 700 nanometers) this value is about 7680 km. Additonal to this refraction effect, radio waves show diffraction from the earth's surface. This effect is increasing with increasing wavelength of the radio signals. For VHF Radio (wavelength 2 m) the apparent earth radius can be assumed with about 8470 km. These different virtual earth radii can be defined below. The above calculation results are changed respectively.