Editing Line-of-sight propagation (section)

=== Vacuum distance to horizon ===
{{main|Horizon distance}}
[[File:RadioHorizont_h_d.jpg|thumb|318x318px|''R'' is the radius of the Earth, ''h'' is the height of the transmitter (exaggerated), ''d'' is the line of sight distance]]
Assuming a perfect sphere with no terrain irregularity, the distance to the horizon from a high altitude [[Transmitter station|transmitter]] (i.e., line of sight) can readily be calculated.

Let ''R'' be the radius of the Earth and ''h'' be the altitude of a telecommunication station. The line of sight distance ''d'' of this station is given by the [[Pythagorean theorem]];

: <math>d^2=(R+h)^{2}-R^2= 2\cdot R \cdot h +h^2</math>

The altitude of the station ''h'' is much smaller than the radius of the Earth ''R.'' Therefore, <math>h^2</math> can be neglected compared with <math> 2\cdot R \cdot h</math>. 

Thus:

: <math>d \approx \sqrt{ 2\cdot R \cdot h}</math>

If the height ''h'' is given in metres, and distance ''d'' in kilometres,<ref>Mean radius of the Earth is ≈ 6.37×10<sup>6</sup> metres = 6370 km. See [[Earth radius]]</ref>
: <math>d \approx 3.57 \cdot \sqrt{h}</math>

If the height ''h'' is given in feet, and the distance ''d'' in statute miles,

: <math>d \approx 1.23 \cdot \sqrt{h}</math>

[[File:RadioHorizont h H.jpg|thumb|318x318px|''R'' is the radius of the Earth, ''h'' is the height of the ground station, ''H'' is the height of the air station ''d'' is the line of sight distance]]
In the case, when there are two stations involve, e.g. a transmit station on ground with a station height ''h'' and a receive station in the air with a station height ''H'', the line of sight distance can be calculated as follows:
 <math>d \thickapprox \sqrt{2 R} \, \left( \sqrt{h} + \sqrt{H}\right) </math>