Formula for VHF range anyone?
Thread Starter
Joined: Mar 2003
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From: Zürich, Switzerland
Formula for VHF range anyone?
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?
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?
Joined: Apr 2001
Posts: 142
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From: Around the World
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.
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.
Joined: Apr 2002
Posts: 243
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From: canada
just math!!!
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!!!
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!!!
Joined: Jul 2001
Posts: 287
Likes: 0
From: Hong Kong
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
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
Last edited by Ex Douglas Driver; 21st September 2003 at 12:19.
