Reg,

The K3LC study is probably the most definitive look at radials since BL&E.
The following URL takes you to the *.pdf file wherein they evaluate radials
under various soil conditions, on 160/80/40 meters. They used NEC-4 for
their study. This is the study that resulted in the "radial optimization"
formula that several of us have referred to.

http://www.ncjweb.com/k3lcmaxgainradials.pdf

Since this is all NEC-4...it should compliment what you are having Frank do.

Here is the formula info:

(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. 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: (wire length in meters)

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

"Reg Edwards" wrote in message
...
"Frank's" wrote
What is interesting, in my preliminary results, is that there is
only a 2% improvement in sky wave total radiated power with
120 radials over 36
=====================================
Frank,

If what you are saying is that efficiency is the same for both 36 and
120 radials, then, at least at 8 MHz, B,L&E's findings for LF do not
apply at HF.

Amateurs do not use LF. They use HF.
----
Reg.

"hasan schiers" wrote in message
...
Reg,

The K3LC study is probably the most definitive look at radials
since BL&E.
The following URL takes you to the *.pdf file wherein they evaluate
radials
under various soil conditions, on 160/80/40 meters. They used NEC-4
for
their study. This is the study that resulted in the "radial
optimization"
formula that several of us have referred to.

http://www.ncjweb.com/k3lcmaxgainradials.pdf

Since this is all NEC-4...it should compliment what you are having
Frank do.

Here is the formula info:

(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. 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: (wire length in meters)

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
========================================
Hasan,

Thanks very much for the formula of which I was entirely unaware. I
will study it.

I notice that it disregards resistivity and permittivity of the ground
under under the antenna which, obviously, ought be taken into account
even when only crudely known.

There's a great difference between a soil resistivity of 25 and 5000
ohm-metres.

This is similar to BL&E who simply state that if the length and number
of radials are greater than certain amounts then soil characteristics
don't matter.
----
Reg.

Hi Reg,

Notice that the purpose of the formula is to get the maximum performance
from a finite length of available wire. So it doesn't make much difference
what the ground characteristics are..the goal is to put down enough and the
right length so that the ground characteristics don't matter any more.

For more detail, go the the url I listed, as it shows the effects of various
ground characteristics:

http://www.ncjweb.com/k3lcmaxgainradials.pdf

With this data, the generic formula and your own work, something synthetic
could result that is even better than your most recent efforts. It is a VERY
interesting subject to those of us using ground mounted verticals.

73,

....hasan, N0AN

"Reg Edwards" wrote in message
...

========================================
Hasan,

Thanks very much for the formula of which I was entirely unaware. I
will study it.

I notice that it disregards resistivity and permittivity of the ground
under under the antenna which, obviously, ought be taken into account
even when only crudely known.

There's a great difference between a soil resistivity of 25 and 5000
ohm-metres.

This is similar to BL&E who simply state that if the length and number
of radials are greater than certain amounts then soil characteristics
don't matter.
----
Reg.