I only wish I had contributed B, L, and E`s findings on antenna radials.
My information has always come 2nd hand from its reprinting by Ed Laport
in "Radio Antenna Engineering. Ed has formulas to use in choosing your
ground system. I can`t find my copy of Ed`s excellent book at the
moment. Laport, like Walter, W2DU, is an RCA alumnus and has associated
with the famous pioneers. K6JHE did us a favor by posting the original
data.

Common sense says that earth closest to a vertical tower gets most of
the capacitive current between the tower and the earth. It is important
that density of the ground radials be high close to the tower to reduce
current in the lossy soil.

I think there is more to it.

The area of a circle around a tower is (pi)(r)(r), where r=distance from
the tower. Area grows as the square of the distance from the tower.
Assume a unit depth for the earth crust, and cross-section becomes equal
to the surface area.

The resistance of a conductor is its resistive coefficient times its
length divided by its cross-sectional area.

Total resistance seen by a ground wave traveling away from a tower is an
inverse function of the distance from the tower`s highly conducting
ground system. The farther from the tower you get, the more
cross-aection there is, so the less resistance there is in the earth.
This must be in textbooks, but I don`t recall seeing it. I once asked
what a-c resistance to use for the earth at 60 Hz, 50 some years ago, in
a student problem and was told to use 25 ohms, no matter what the
distance through earth was. That`s when I noodled out the above
explanation for the earth`s resistance. It should work at r-f too except
for skin effect which if I recall causes an increase in resistance
proportional to the square root of the frequency. The skin thickness is
proportional to the reciprocal of the square root of the frequency.

The point is that high conductivity is only needed very close to the
tower for the ground wave. For a sky wave, you need high conductivity at
the reflection point for a vertically polarized wave.

Best regards, Richard Harrison, KB5WZI