What you are describing is totally different than the original claim
(and the subsequent discussion) that the earth is an RF sink. What was
the purpose in doing that?
Yes, you can make the earth act like a parallel path back to the
transmitter ... but what does that gain you?? You've taken some of the
common mode current and given it a different route, but it still gets
back to the transmitter ... AND THAT PORTION OF THE COMMON MODE CURRENT
BYPASSES ANY COMMON MODE CHOKES THAT YOU MAY HAVE PUT ON THE COAX! It's
self defeating.
I don't know why this is so difficult to understand. Lightning can be
shunted to ground because the potential field of the lightning is
generated between the clouds and the ground. There is no such inherent
potential field between the common mode current flowing on the shield of
the coax and the ground. Reference my early example of a portable rig
with no connection to ground anywhere. The potential with respect to
the common mode current is between the rig and the antenna. Yes, if the
rig is grounded you can create that alternate path for the common mode
current by grounding the coax at some point ... but like I said above
that doesn't help anything ... at least not as far as the rig is concerned.
People who believe that the earth is inherently a sink for RF keep
assuming that because the earth is large that it has some sort of
sinking capacity for everything. It doesn't. Look at it this way ...
if you had an infinitely large capacitance the size of a football (to
simulate the earth) and connected it to the shield of the coax at a
position where the common mode current had a maximum value, what would
you get? You wouldn't shunt that current aside, and you wouldn't
dissipate it. The current literally would have no "push" (E-field,
voltage, whatever) to enter that capacitor. I don't know why folks are
so willing to abandon raw physics it comes to RF.
Bleeding of static charge to ground is similar to lightning because that
static charge is created with respect to ground ... AND IT IS A DC
CURRENT THAT STAYS WHERE IT GOES! Common mode RF currents are AC ...
they just go back and forth.
At this point I'm tired of flogging this horse. Believe me or not, but
the idea of the earth being an infinite sink for RF is a fallacy that
just won't die in ham radio circles.
Dave AB7E
On 9/29/2025 11:11 PM, Michael Tope wrote:
Dave,
Say you have a dipole up 50ft and you have some common-mode current on
the coax shield (perhaps the choke at the feed point is not very good
or worse yet, there isn't one). If you bond the coax shield to an 8ft
ground rod at the point where the coax reaches the ground, you will
create a circuit path for RF current to flow that is in parallel with
the shield of the coax leading back toward the transmitter. Now an 8
ft ground rod (depending on soil properties) isn't a very good RF
ground, so the amount of RF current that is shunted there might not be
much, but it won't be zero. Moreover, if I start attaching radials to
that ground rod the RF impedance of that grounding point will drop and
more of that shield current will be diverted away from the shield
leading back to the transmitter.
Lightning protection is, at least in part, about shunting fast
rise-time currents either from direct hits or induced currents from
nearby strokes. The other part is about having everything in the
electrical system bonded to a common point so the voltage on
everything rises together during the stroke (potential difference is
the enemy). Bleeding of static charge is related, but not the same
thing (that may save your receiver front-end). If you look at the
spectrum of a lightning stroke there is a lot of energy in the MW and
HF bands. A robust RF ground should make a good lightning ground and
vice versa). Seems like there is also (at least with a direct hit) a
DC component since there is a net movement of charge. I've always
wondered about the insulated THHN in the radial system for my HF
vertical and the DC component. Perhaps that's where the ground rods
come in :-).
73, Mike W4EF..............
On 9/29/2025 3:40 PM, David Gilbert wrote:
The original postulation from Brian, K6STI, was that connecting the
shield of the coax to ground at a point where any common mode
current might exist would shunt it to ground. I've claimed that's a
fallacy, because ground is NOT an RF sink and RF is not a static
charge that can be just bled off somewhere. RF is AC, and any
connection to ground is bidirectional. At most, connecting the
shield to ground just adds another element (most likely capacitive)
to the overall network. The current doesn't just disappear into the
earth.
Dave AB7E
On Sun, 28 Sep 2025 08:33:30 -0500, Kelly Taylor via
TowerTalk<towertalk@contesting.com> wrote:
Question: If the Earth WAS an RF sink, why would you want it to be?
Any RF “sunk” into the ground is RF that’s not available to radiate.
Better to design a system to put as much RF as possible into the
air, no?
Silly me… ;-)
73, kelly, ve4xt
Sent from my iPhone
On Sep 28, 2025, at 06:30, Jim Lux wrote:
The field from the antenna (and from the feedline, if there's any
current in the shield, or it's unbalanced) certainly does interact
with the soil under the antenna (and houses, trees, etc.).
The question is "how much" (which NEC can answer, as long as you're
willing to accept the "uniform soil property" model).
And that depends mostly on "how close is it" - after all,
instruments called sounders fly on spacecraft and measure the EM
properties of the soil below at a distance of hundreds or thousands
of km: whether on Earth, Mars (MARSIS, SHARAD), or Europa (REASON).
They work at 9 MHz, and REASON also has a VHF mode.
We can even get a sort of worst case - There's a paper by Dave
Rutledge and Michael Muha that that has some simple equations for a
dipole laying on the ground. For very dry soil with epsilon 3 +
0.005j, the signal propagating into the soil is about 5-6 dB
greater than the signal propagating into space. It roughly goes as
n^3 (where n is the index of refraction - sqrt(epsilon), so
epsilon^1.5). George Hagn at SRI spent quite a while trying to
measure soil properties with dipoles at various heights above the
ground.
It's behind the IEEE Paywall, but it might be available elsewhere:
D. Rutledge and M. Muha, "Imaging antenna arrays," in IEEE
Transactions on Antennas and Propagation, vol. 30, no. 4, pp.
535-540, July 1982, doi: 10.1109/TAP.1982.1142856.
When it comes to "drive a rod" vs "radials" (or some form of
counterpoise), one way to look at it is that the radials make a
"higher conductivity" soil, and there's all kinds of interesting
trades about wires in the ground vs wires on ground vs wires above
ground, which lots of people have looked at: Rudy N6LF has done a
lot of experiment at frequencies of interest to hams; J.R. Wait has
published dozens of papers on the electromagnetics of wires close
to, or immersed in, a dielectric. As the phrase has it, this is a
"well studied problem".
Of some considerable interest is that the soil is not homogeneous
and the RF propagates quite a ways below the surface (hence the
effectiveness of sounders at doing subsurface imaging). So it's
very much a "build it and try it" (which is where the 120 radial
thing comes from: that's enough that empirically, it doesn't matter
what kind of soil is under that dense radial field)
On Sat, 27 Sep 2025 21:24:42 -0700, David Gilbert wrote:
I already did with the example of a floating portable setup. Current
requires an E-field to push it. You could connect a grounded wire to a
point on the coax shield and it wouldn't shunt any common mode current
to ground because there is no E-field (voltage) for it.
An earth ground is a grounding point for lightning strikes because the
current that flows in lightning is the result of charge buildup (an
E-field) between clouds and ground.
Earth ground affects transmitted RF because the radiated RF
impinges on
the earth and is absorbed and reflected, the ratio between the two
being
affected by the parameters of the earth (conductivity and
permittivity).
There is no such E-field between the coax and the earth due to the
common mode current on the coax shield. Even if you view the earth as
some sort of super large capacitor, it would require an E-field to
push
current into it.
The earth is NOT an RF sink.
Dave AB7E
On 9/27/2025 12:55 PM, Brian Beezley wrote:
"That's a fallacy. It simply isn't."
Dave, it would be helpful if you'd supply your reasoning.
In many ways I regard ground as just another conductor. However,
unlike a wire, it is normally without resonance effects. That's the
"current sink" aspect. Current will flow from a wire into ground if
you make a connection. If you're using a ground rod, the impedance at
the connection depends on the rod length, rod diameter, and the
characteristics of the soil. If the soil is uniform, reflections
don't
occur, unlike for a wire of finite length. The current dissipates as
it spreads within the ground, which acts like an infinitely long wire
with a traveling wave. However, when ground strata are distinct and
well defined, resonance can occur. An example shown for the
stratified
ground calculator described in the writeup below exhibits strong
resonance. A water table 200 feet below a desert surface magnifies
surface ground conductivity by a factor of 10, which is pretty
amazing. I think such situations are rare because I suspect most
variation in ground characteristics occurs gradually rather than as
distinct strata, which is necessary for resonance.
https://k6sti.neocities.org/sg
Brian
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