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 *P _{1}* located at a distance

*l*from

_{1}*H*and with a height

*h*above the sea level. This point

_{1}*P*represents the ship with a height

_{1}*h*. To the right side of

_{1}*H*there is point

*P*localted at a distance

_{2}*l*of

_{2}*H*and with a height

*h*. This point

_{2}*P*represents the position of the observer. The triangle

_{2}*M*,

*P*,

_{1}*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 *R ^{2 }*on both sides, we get:

After applying the square root on both side, we obtain the equation for determining the distance from the horizon *l _{1}*:

For the entire distance *l* from *P _{1}* tor

*P*we obtain finally:

_{2}The distance *l* thus corresponds to the maximum viewing distance of a vessel with a height *h _{1}* and an observer with an elevation

*h*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

_{2}*l*

_{1},*l*and

_{2}*l*. Of course, this calculation needs to know the value for the earth's radius. This value is approximately 6370 km.