chuck wrote:

If the skin depth at 14 MHz is about 2.4 inches, can we roughly assume
that the RF resistance of that path is no less than 52.8 ohms (2.4*22
ohms)? This assumes most of the RF current would occur in the top one
inch (attenuation at one inch would be about 15 dB), and that the
resistance at 14 MHz is equal to the DC resistance.

What you'd need to do is look at the I^2 * R loss for every little pie
slice of water the current flows through. It's greatest near the antenna
(assuming a vertical) where the current density is greatest. In that
region, the current density is greatest and R is also greatest, so
that's where the majority of loss occurs. (Which is why a radial wire
field is useful for land installations -- its resistance is least near
the antenna.) So you can't just calculate a single value of R or I based
on the current and cross section at some point -- the entire area over
which the current is flowing must be taken into account. The modeling
program does just that.

Don't get too worried about the skin depth. Shallower skin depth is an
indication of a better conductor. The skin depth in metal is extremely
thin, yet it's a better conductor yet.

A path one inch deep by 16 feet long (1/4 wavelength at 14 MHz) would
then have no less than 26.4 ohms resistance at 14 MHz.

True but irrelevant. The current at the far end is much less than the
current at the near end.

Now imagine a system of multiple one foot wide by 16 feet long copper
radials on the ground with 26.4 ohm resistance distributed uniformly in
each radial. Obviously such a system will be lossy, with an average
radial resistance of 13.2 ohms.

While the analogy is a stretch, it illustrates the difficulty I am
having in understanding how seawater can be considered more efficient
than even a single slightly elevated radial, which is reported to be
less than 1 dB worse than 120 quarter wavelength buried radials
(ignoring slight pattern distortion). So even if seawater does
constitute a less lossy ground plane than a single radial (yeah, apples
and oranges, but we can weigh their juices I think) it would be better
by less than 1 dB. .

The problem is that the analogy is too much of a stretch. Too many
incorrect assumptions were made, resulting in an invalid conclusion.

Then there is the issue of the one foot long "grounding rod" immersed in
the sea. If the above back-of-the-envelope analysis is valid, it would
seem that a even one inch long rod would be more than sufficient.

As it turns out, a one inch rod is nearly as good, even though it
doesn't extend to the entire depth where significant current is flowing.
Half the total current is below about 1.7 inches deep. To connect
directly with essentially all the current requires at least several skin
depths. Here's the relative current on a foot long wire directly below a
quarter wave vertical at 14 MHz:

Depth (in.) I
0.5 0.81
1.5 0.53
2.5 0.35
3.5 0.23
4.5 0.15
5.5 0.10
6.5 0.07
.. . .
10.5 0.01
11.5 0.006

If we
were dealing with a pool of liquid mercury or silver, this would have
considerable intuitive appeal for me. But the seawater model is
troubling. I imagine seawater to be a lot like earth, except more
homogeneous and with orders of magnitude higher conductivity. And I
imagine a perfect ground plane to have conductivity orders of magnitude
higher than seawater. I imagine even a modest system of copper radials
to appear more like liquid mercury than seawater does.

At RF, taking skin depth into account, there's about 5 orders of
magnitude difference between the conductivities of copper and average
soil. Sea water is 30 times more conductive (at RF) than average soil,
so it's still far short of copper. But Suppose we had a conductor which
was 10 orders of magnitude more conductive than copper -- would it make
any difference if our ground plane was made out of that or out of
copper? How about 3 orders of magnitude less conductive? The fact is
that in this application, 30 times better than soil is adequate for the
water to behave a lot more like copper than like soil.

A modest system of radials in soil looks like very, very small cross
sections of copper (remember the skin depth in copper!) separated by
very large regions of soil.

Out of curiosity, I altered the conductivity of the water in my computer
model. Dropping it by a factor of 10 at DC (about 3 at RF) results in a
reduction of about one dB in field strength, or about 25% in efficiency
when using a single ground wire. So salt water has just about the
minimum conductivity you can get by with if you want really good
efficiency with a single ground wire.

Where am I going astray?

In oversimplifying the problem and using analogies which aren't quite right.

Roy Lewallen, W7EL