# Distance between two antennas

I am trying to find out the formula to calculate how high antennas need to be for Line of Sight (LoS) propagation.

I found:

d = 3.57sqrt(h)


also

d = 3.57sqrt(Kh)


$d$ can also be worked out using

d = 3.57( sqrt(K[h1]) + sqrt(K[h2]) )


Where $d$ is the distance between an antenna and the horizon (or between two antennas) in kilometers, $h$ is the height of the antenna(s) in meters, and $K$ is used to account for the curvature of the earth (which is usually $\frac{4}{3}$).

The problem with this equation is it is making the antennas ridiculously high for the distance I am trying to calculate. The question I am trying to answer is: “Two antennae are used for line of sight propagation. The antennae are spaced $150$ km apart. Determine the required antennae heights.”

My calculations:

d=3.57sqrt(4/3)(h)
d=3.57(sqrt[4/3])(sqrt[h])
d=3.57(1.1547)(sqrt[h])
150=(4.1223)sqrt(h)
150/4.1223=sqrt(h)
sqr(36.3875)=sqr(sqrt(h))
1324.05=h   1324meters = h1 + h2  -> each antenna needs to be 762meters high


Is this the correct method? Or have I chose the wrong equation totally?

• I have to ask what is this for? – EhBabay May 8 '13 at 22:20
• It will be more likely to get a good answer if you ask a pure mathematical question. – user59671 May 8 '13 at 22:32

The exact distance, $d$, from the top of the tower of height $h$ to the horizon is $$d=\sqrt{2rh+h^2}$$ where $r$ is the radius of the Earth. However, if we assume the height of the tower is insignificant in comparison to the radius of the Earth, we can make the following distance approximation: $$d=\sqrt{2rh}$$ Since $r=6371000\text{ m}$, This gives $$d=3570\sqrt{h}$$ where $d$ and $h$ are measured in meters.

As I mentioned in a comment to Abel, if you have two towers of height $h_1$ and $h_2$, the maximum distance between them is the sum of each of their distances to the horizon; that is, $$d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$$ $\hspace{4.5cm}$

If the towers are of equal height $h$, then the maximum distance between them is $2\sqrt{2rh}$ $$d=7139\sqrt{h}$$

• Thank you, that is a great explanation. I appreciated the detail given! Also, thank you to the other guys who answered too. Very much appreciated guys. This question was bothering me, all because I didn't think to halve 'd'. – Aaron Doyle May 9 '13 at 16:21
• what about refraction ? – vasilakisfil Mar 22 '18 at 17:40
• @vasilakisfil: no mention was made regarding refraction, neither atmospheric nor gravitational. If the problem is stated without parameters for refraction (wavelength of light, density and refractive index of media, etc.), the best we can do is an approximation ignoring refraction. Besides, the tags are only geometry and trigonometry. – robjohn Mar 23 '18 at 1:56

A: the first formula is the correct one. It is based on the curvature of a spherical earth. The reason for the variable $K$ is unclear.

B: Picture yourself seated at ground level, exactly between the two antennas (75 km from each) Each antenna must be tall enough to just reach you at that antenna's horizon. IOW, use 75 km for the distance in the first equation, to give the height of each antenna.

I thought I'd just show you where the right equation comes from.

Your situation is shown in the following picture. The earth is depicted green, the LOS is depicted red and the radius of the earth as well as the antennae are depicted black.

From this picture we see that Pythagoras' Theorem applies and we have $(R+h)^2=R^2+\left(\frac{d}{2}\right)^2$. Expanding $(R+h)^2$ now gives us $2Rh+h^2 = \left(\frac{d}{2}\right)^2$ and since $h<<R$ we may neglect the $h^2$ to obtain $2Rh = \left(\frac{d}{2}\right)^2$ and finally $$h = \frac{d^2}{8R}.$$

In your case this gives $h = \frac{\left(150\,\mathrm{km}\right)^2}{8\cdot 6378.1\,\mathrm{km}} \approx 441\,\mathrm{m}$.

• In the problem stated above, there are two towers of different heights. The distance of the tower with height $h_1$ to the horizon is $d_1=\sqrt{2rh_1}$. The distance of the tower with height $h_2$ to the horizon is $d_2=\sqrt{2rh_2}$. Thus, the distance between the towers is $d_1+d_2=\sqrt{2rh_1}+\sqrt{2rh_2}$. If $h_1=h_2=h$, then this reduces to $2\sqrt{2rh}=\sqrt{8rh}$. – robjohn May 9 '13 at 1:52