Pete Smith wrote:
> I wonder if anyone knows how much ground conductivity varies within the
> broad contours shown on the USGS map? For example, my area of West
> Virginia is listed in the 2 millisiemen countour, probably because much of
> the area is karst geology with only a narrow skin of soil over
> limestone. However, the area where my tower is footed is an ancient river
> bed, with 6-7 feet of usually moist soil. It seems (from experience only)
> to be rather better than the general expectation of poor ground. I presume
> that the depth of soil needed depends in some large part on the skin depth
> at the frequency of interest, but how far from a vertical antenna does soil
> conductivity continue to affect field strength?
>
There are two aspects to the answer to your question..
1) Near field ohmic losses.. the soil right under the antenna has higher
currents, tapering off as you move away. Get 4 or 5 antenna heights
away, and I imagine that the current is pretty small. (the radials fix
this problem..)
2) reflection loss, where the elevation angle will affect the location
of the soil which affects the field strength at that radiation
direction. (a radial field,unless it's gargantuan, or, you only care
about radiation at very high angles, isn't going to make a huge
difference here. Making it rain would possibly help)
As for the impact of a change in conductivity...
For the former, it's pretty direct.. double the conductivity, halve the
loss resistance. (if you fed from a constant current or constant voltage
source, that would also halve the power loss, but changing the loss
resistance changes the impedance, etc...)
for the latter, you have to look at the equations for reflection of a
plane wave aka the Fresnel equations..
http://en.wikipedia.org/wiki/Fresnel_equations
Rp= {[n1*cos(thetat)-n2*cos(thetai)]/[n1*cos(thetat) +
n2*cos(thetai)]}^2
thetai is the angle of incidence (perpendicular = 0)
tehtat is the angle of transmission (perpendicular = 0)
n1*sin(thetai) = n2*sin(thetat)
n is the index of the medium = sqrt(mu*epsilon)
OR...
As given in the ARRL antenna book:
A = (k' * sin(elev)-sqrt(k' - cos(elev)^2))/
(k' * sin(elev)+sqrt(k' - cos(elev)^2))
where
k' = epsilonr - j*[ 1.8E4 *sigma/fMHz]
where epsilonr is the relative permittivity (e.g. 5 to 20)
and sigma is the conductivity (in S/m)
You then have to look at the relative path lengths and figure out if the
phase of the reflection affects it at your particular elevation.
(This is why programs like HFTA don't deal with vertical polarization...
H pol is a lot easier.. almost total reflection at low angles and the
phase term is small)
You can probably assume mu=1, so epsilon is your important number here,
and the big factor in epsilon is the water content.
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