(This formula is from both QST and Low Band Dx'ers Handbook by Devoldre)
Obviously a ton of short radials does not equal a ton of long radials, but
it can get you really close. See the articles for limitations of the
formula. For most circumstances (N > 32 and average soil), the formula is
fine. For very poor soil, make sure you use a lot more wire than enough for
only 16 radials, for example. In other words, don't use the formula like
this: Gee, I only have 200 feet for radial wire. It will give you the right
answer....but...when you only use so little total wire, your losses will be
quite a bit worse than the 0.5 to 1 dB that was the goal of the studies.
The formula goes like this:
N = ((2*PI*W)^0.5)/1.2
N equals the square root of the quantity (2*PI*WireLength) divided by 1.2
Where N = number of radials
Where W = length in meters of available wire to make the radials
Length of radials = W/N
and the constant 1.2 is the tip separation in meters to produce the proper
density on 80m ..this would be twice the density one needs for 160m and half
what is needed on 40m. The value for minimum tip separation is simply .015
wavelength. So if you calculate a full wavelength for the freq in use,
multiply it
by .015 and that gives you the value for tip separation in the formula
above. For 80m it is 1.2 meters
Example:
You have 500 meters (about 1640 feet) of radial wire available for your 80m
vertical. How many and how long should the radials be:
46 radials, 10.8 meters (35.6 feet) will produce the lowest possible loss
for this amount of available wire.
This is a simple formula for how many radials to put down if you have
"only so much wire". These days with copper prices through the roof, it pays
to be economical and still stay within 0.5 to 1 dB of "what's best".
73,
...hasan, N0AN
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