I'm using a line of site software to find an optimum tower height for my
location to receive UHF broadcasts centered around 600MHz. This software
has an atmospheric correction factor for the frequency you are trying to
receive. They give a value of 1.333 for microwave, but don't give an exact
frequency. Does anyone know how to find this factor for a specific
frequency? The range for this factor is 0.1 - 5. 0.1 would correspond to a
high frequency (microwave (light maybe)), and 5 would be a low frequency
(vlf?).

John Doe wrote:
"They gave a value of 1.333 for microwave, but don`t give an actual
frequency."

It`s true that there is a slight difference in the distances to the
optical and radio horizons. It`s not important because the effective
distance is inexact and varies with changes in atmospheric conditions.
For radio horizon, the geometric model is usually an earth 4/3 tha
actual. That`s where the 1.333 comes from. It would mean the earth
appears under normal conditions to be a little flatter than it is. Radio
range is increased a little over estimates based on a true geometric
model. Since a still atmosphere can cause layering of air temperatures,
refraction occurs sometimes in the early morning which bends radio waves
away from the earth. To account for these misfortunes, designers of
sensitive radio services sometimes use an earth model which is only 2/3
the size of the actual earth so contact is kept under lousy
line-of-sight conditions.

Using the customary 4/3 smooth earth estimate, an easily remembered
formula emerges:
D = sq.rt. 2H

D is the distance to the horizon in miles.
H is the height of the antenna in feet.

You have an antenna at 200 feet.
The sq rt of 400 is 20 miles.

20 miles is the distance over a bare landscape that you can communicate
with an antenna at about ground level. If both antennas were at 200
feet, you might be able to communicate 40 miles at line-of-sight
frequencies. The estimate is usually very good.

Best regards, Richard Harrison,KB5WZI