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Snoopy
21st Sep 2003, 11:45
I seem to remember a formula for calculating the VHF range of a transmitter taking into account the curvature of the earth. I think that it was:

Square root (height in feet) x 1.something

I have forgotten what something is....does anyone know what something is and/or can actually confirm whether the formula is correct and whether the answer unit is statute miles or nautical miles?

Burger Thing
21st Sep 2003, 12:26
I believe it is 1.25 x SQR(Transmitter height in feet) + 1.25 x SQR(Receiver height in feet)

The answer should be in nautical miles

But that works of course only for the line of sight. Any big mountains in between will spoil the business.

donut king
21st Sep 2003, 12:59
1.25x (square root distance agl...in feet)

AGL= distance above transmitter.

Answer= xxxx n.m.


I think!!!!!!!!

Therefore... 5000 ft AGL equals signal reliability of 88.39 n.m. Sounds about right!

Anyone else please correct/add input if I'm wrong!

D.K.

P.S. DME is another story!!!

Ex Douglas Driver
21st Sep 2003, 13:05
d = square root (h^2 + 2hr)

where:
h = height above earth's surface
r = earth's radius = 6371km or 19418808ft

1000 FT = 32.4 NM
10000FT = 102.5 NM

This gives the visible horizon from one point. If you want to figure out the range between 2 elevated objects, then add together the 2 values of d.


So that would make it
d (in NM) = 1.025 x square root height in feet

Snoopy
21st Sep 2003, 13:21
Thanks for the input!
Much appreciated