JB, Roger, LF,
It isn't quite that simple. The standard analysis of a long thin ground rod
(as per David Gibson's article) is to estimate the radius of the buried
hemisphere which would give the same capacitance. Knowing the resistivity of
the ground and the estimated capacitance you can work out the equivalent
resistance. I don't have the analysis to hand but I seem to recall it models
the rod as an oblate spheroid (as long as the length/diameter ratio is not too
great which it isn't in the caving situation).
I seem to recall that the effective radius of a thin rod is about one fifth of
its length. To get the benefits of multiple ground rods you need to space them
by at least twice their length.
73
John F5VLF
On 27 May 2010, at 15:13CEST, John Bruce McCreath wrote:
>
> Hello LFers,
>
> While laying in bed this morning pondering the inside of my eyelids I got to
> thinking about the electrodes used
> for "through the earth" communications. What is more important, the depth of
> the electrode or its surface area?
> A typical 3/4 inch diameter by 4 foot long ground rod has a surface area of
> 113 square inches, while a metal
> plate 1 foot square has a surface area of 288 square inches. If depth is the
> key, then obviously the rods have
> the advantage, but if it's surface area, the plates win hands down. Even in
> stoney soil, it's relatively easy to
> make a slit-like hole into which could be slipped a sheet of galvanized metal
> with a lead attached. To make
> a good connection to the surrounding soil, pour some "kitty litter" into the
> slit and moisten it with water so as
> to improve the contact between the plate and the surrounding soil. Am I onto
> something here or have I
> overlooked some important detail?
>
> 73, J.B., VE3EAR
>
> LowFER Beacon "EAR"
> 188.830 kHz. QRSS30
>
>
>
>
>
>
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