On Tue, 04 May 2004 12:46:02 -0700, Jack Twilley
I recognize that salt water is far more conductive than fresh water,

Hi Jack,

Unfortunately, this is a persistent illusion that begs real comparison
to real conductors. Salt water is miserable as a conductor, and its
special place in the pantheon of noble applications has little to do
with "conductivity."

but fresh water's still superior to sand and the like.

Even this is arguable, sand has less loss (as sand is one of the most
common precursors to making glass, it is hard to suggest it presents
issues of conduction or loss).

That being
said, I am wondering about using a pond as a ground screen and
mounting the antenna itself on an island (or a raft) in the middle of
the pond.

No doubt you will get a raft of anecdotal support, but not much data.

What I don't know is just how large a pond do I need in order for
something like this to work?

I will do something dangerous and make a presumption. To work, as you
suggest it through the example of the AM stations, you need to be
operating in the 160M band. OK, so that was a caprice of guesswork,
the remainder of your post offers other opportunity to suggest you are
building your house on sand.

For those who absolutely require less variables in their equations,
imagine a standard dipole tuned for 20m strung roughly 45 feet above
ground level between two trees, one on either side of a fresh water
pond.

Well, this is where you are in over your head (water metaphors are
abundant in this topic). This, again, requires presumptions insofar
as the original observation was driven by the AM example. However, at
this point we will depart from the low frequency mandate to examine
another mandate: polarization and your presumption of conductivity.

A horizontally polarized antenna seeing a horizontally conducting
surface is a scenario that describes a self-short-circuit.
Horizontally polarized waves meeting the earth (a conductive one)
immediately snuff themselves (how long would your car battery last
with a screwdriver held across its poles?).

On the other hand, vertical antennas do not suffer this fate - and for
the same reason: it is a current wave (or at least the magnetic
component inducing such a current, in a conductive earth) that spans
earth making a perfectly reasonable relationship to continued
propagation.

How wide does the pond have to be at that point (and others)
for it to work right? Even answers like "the pond will have to be
wider than the dipole is long" or "there will be no noticeable impact
on performance" are fine if they're based in reality, and ideally in
math and physics I can understand.

Well, once you divorce yourself of the notion of using a horizontal
antenna, it becomes a matter of "ray tracing" from the vertical
polarized source, out to the reflecting surface. If you want a
radiation lobe to bounce away at an angle of 45°; then this surface
has to be as far away as the origin of the ray's source is above
ground. If you want a radiation lobe to bounce away at an angle of
5°; then this surface has to be 10 or 20 wavelengths away (which is
why large bodies of water are so attractive).

OK now to drop the other shoe.

The reason why this all works can be described through conduction, but
that is messy and far from intuitive (why is a poor conductor better
than a poorer conductor that is sometimes better than the poor
conductor?).

The whole matter of near earth conductivity (through the application
of trig) hardly matters a whit in regard to DX angles. DX angles are
a property of the earth in the far region (at least 5 wavelengths
away, which describes the almost certain frustration of planting long
enough radials that will make any difference in that regard).

The trick is to simply abandon the thought of the antenna except as a
point representing the source of the ray to be traced to ground.
Think of the wave propagating along that ray. It has left the antenna
far behind and is in its native media of 377 Ohms. It then happens to
intersect another media - salt water. Salt water's characteristic Z
presents that wave with a 10:1 SWR mismatch, and as we all know
exceedingly little power passes through that interface, and nearly all
of it is reflected (again, in the same angle of reflection already
described by the ray we started with).

THIS is the power of salt water, marsh water, and other so-called
conductive surfaces. The conductivity is for s**t and matters just as
much - it is mismatch that does our work. Even average earth presents
a mismatch, but not nearly so effective; thus power passes through the
interface at its critical angle (another optics concept that is
intimately tied into this ray-tracing). We call this (as does the
optics community) the Brewster Angle. Optics has a simple method of
forecasting the angle, and it is the same math that gives us SWR. a
ratio.

Well, there is more that could be said, but this is enough to part the
waves.

73's
Richard Clark, KB7QHC