Roy Lewallen wrote:
chuck wrote:
. . .
2. What would the ohmic losses be over a one square foot by 33 foot
path through salt water?

Let's see, salt water conductivity is about 5 S/m, which is 1.524 S/ft.
So the *DC* resistance of that piece of sea water would be 1.524 * 33 /
(1 * 1) ~ 50 ohms. . .

Oops. The DC resistance would be 33 / (1 * 1) / 1.524 ~ 22 ohms.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Roy Lewallen wrote:
chuck wrote:
. . .
2. What would the ohmic losses be over a one square foot by 33 foot
path through salt water?

Let's see, salt water conductivity is about 5 S/m, which is 1.524
S/ft. So the *DC* resistance of that piece of sea water would be 1.524
* 33 / (1 * 1) ~ 50 ohms. . .

Oops. The DC resistance would be 33 / (1 * 1) / 1.524 ~ 22 ohms.

Roy Lewallen, W7EL

Thank you for the detailed response, Roy.

A couple of issues still trouble me however.

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.

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.

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. .

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. 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.

Where am I going astray?

73,

Chuck
NT3G

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

The permittivity, K, of water is about 80.

The relative velocity of propagation along a wire immersed in water is
about VF = 1/Sqrt( K ) = 0.11

At a frequency of 7.5 MHz, a 1/4-wavelength of wire immersed in water
is only 1.1 metres = 43 inches long.

Furthermore, in salt sea water, considering a wire as a transmission
line, dielectric loss is so high there is little or no current flowing
at the end of a quarterwave radial wire. Longer wires can be
disregarded because they carry no current.

So, at 7.5 MHz, there is no point in considering a system which has
more than a radius of 1.1 metres. At higher frequencies the radius is
even less.

A copper coin, 1" in diameter, immersed in a large volume of salt
water, has an impedance low enough to be used as an efficient ground
for a 1/4-wave HF vertical antenna. It is limited by its power
handling capacity.

I have made measurements years ago but have no records as I didn't
attach any importance to them at the time. And still don't.

Unpolluted, clean, fresh pond water, is a different kettle of fish.
Permittivity is still about 80 but the resistivity is very much
greater. About 1000 ohm-metres is a reasonable value.
----
Reg.

Reg Edwards wrote:
The permittivity, K, of water is about 80.

The relative velocity of propagation along a wire immersed in water is
about VF = 1/Sqrt( K ) = 0.11

At a frequency of 7.5 MHz, a 1/4-wavelength of wire immersed in water
is only 1.1 metres = 43 inches long.

Furthermore, in salt sea water, considering a wire as a transmission
line, dielectric loss is so high there is little or no current flowing
at the end of a quarterwave radial wire. Longer wires can be
disregarded because they carry no current.

So, at 7.5 MHz, there is no point in considering a system which has
more than a radius of 1.1 metres. At higher frequencies the radius is
even less.

A copper coin, 1" in diameter, immersed in a large volume of salt
water, has an impedance low enough to be used as an efficient ground
for a 1/4-wave HF vertical antenna. It is limited by its power
handling capacity.

I have made measurements years ago but have no records as I didn't
attach any importance to them at the time. And still don't.

Unpolluted, clean, fresh pond water, is a different kettle of fish.
Permittivity is still about 80 but the resistivity is very much
greater. About 1000 ohm-metres is a reasonable value.
----
Reg.



Interesting info, Reg.

I also made some kitchen table-top sal****er measurements about a year
ago, but at much lower frequencies than you discuss. My measurements are
not handy at the moment, but they don't comport with yours. I utilized a
variety of electrode geometries: concentric, 4 pole, parallel plate,
etc. Measurements of electric field strength, conductivity, path
conductance, etc. are not difficult but interpretation of the data
stumped me.

As you remember, the conductance of a sal****er path is a direct
function of the path's cross-sectional area. A penny doesn't produce
much of a cross-sectional area at its end of the path. Maybe your
pennies are better than ours, Certainly worth more.

73.

Chuck

chuck wrote:
Reg Edwards wrote:
The permittivity, K, of water is about 80.

The relative velocity of propagation along a wire immersed in water is
about VF = 1/Sqrt( K ) = 0.11

At a frequency of 7.5 MHz, a 1/4-wavelength of wire immersed in water
is only 1.1 metres = 43 inches long.

Furthermore, in salt sea water, considering a wire as a transmission
line, dielectric loss is so high there is little or no current flowing
at the end of a quarterwave radial wire. Longer wires can be
disregarded because they carry no current.

So, at 7.5 MHz, there is no point in considering a system which has
more than a radius of 1.1 metres. At higher frequencies the radius is
even less.

A copper coin, 1" in diameter, immersed in a large volume of salt
water, has an impedance low enough to be used as an efficient ground
for a 1/4-wave HF vertical antenna. It is limited by its power
handling capacity.

I have made measurements years ago but have no records as I didn't
attach any importance to them at the time. And still don't.

Unpolluted, clean, fresh pond water, is a different kettle of fish.
Permittivity is still about 80 but the resistivity is very much
greater. About 1000 ohm-metres is a reasonable value.
----
Reg.



Interesting info, Reg.

I also made some kitchen table-top sal****er measurements about a year
ago, but at much lower frequencies than you discuss. My measurements are
not handy at the moment, but they don't comport with yours. I utilized a
variety of electrode geometries: concentric, 4 pole, parallel plate,
etc. Measurements of electric field strength, conductivity, path
conductance, etc. are not difficult but interpretation of the data
stumped me.

As you remember, the conductance of a sal****er path is a direct
function of the path's cross-sectional area. A penny doesn't produce
much of a cross-sectional area at its end of the path. Maybe your
pennies are better than ours, Certainly worth more.

73.

Chuck
Hi Chuck

So what would be the best size cross sectional area to achieve a close
to perfect RF ground from 1 to 30 mhz over sea water? Considering things
like corrosion, fowling, growth on the plate over time and any other
factors that would deteriorate the effectiveness of this connection. You
would want adequate safety margin when using this kind of simple direct
contact.

Bob

bob wrote:
chuck wrote:
Reg Edwards wrote:
The permittivity, K, of water is about 80.

The relative velocity of propagation along a wire immersed in water is
about VF = 1/Sqrt( K ) = 0.11

At a frequency of 7.5 MHz, a 1/4-wavelength of wire immersed in water
is only 1.1 metres = 43 inches long.

Furthermore, in salt sea water, considering a wire as a transmission
line, dielectric loss is so high there is little or no current flowing
at the end of a quarterwave radial wire. Longer wires can be
disregarded because they carry no current.

So, at 7.5 MHz, there is no point in considering a system which has
more than a radius of 1.1 metres. At higher frequencies the radius is
even less.

A copper coin, 1" in diameter, immersed in a large volume of salt
water, has an impedance low enough to be used as an efficient ground
for a 1/4-wave HF vertical antenna. It is limited by its power
handling capacity.

I have made measurements years ago but have no records as I didn't
attach any importance to them at the time. And still don't.

Unpolluted, clean, fresh pond water, is a different kettle of fish.
Permittivity is still about 80 but the resistivity is very much
greater. About 1000 ohm-metres is a reasonable value.
----
Reg.



Interesting info, Reg.

I also made some kitchen table-top sal****er measurements about a year
ago, but at much lower frequencies than you discuss. My measurements
are not handy at the moment, but they don't comport with yours. I
utilized a variety of electrode geometries: concentric, 4 pole,
parallel plate, etc. Measurements of electric field strength,
conductivity, path conductance, etc. are not difficult but
interpretation of the data stumped me.

As you remember, the conductance of a sal****er path is a direct
function of the path's cross-sectional area. A penny doesn't produce
much of a cross-sectional area at its end of the path. Maybe your
pennies are better than ours, Certainly worth more.

73.

Chuck
Hi Chuck

So what would be the best size cross sectional area to achieve a close
to perfect RF ground from 1 to 30 mhz over sea water? Considering things
like corrosion, fowling, growth on the plate over time and any other
factors that would deteriorate the effectiveness of this connection. You
would want adequate safety margin when using this kind of simple direct
contact.

Bob

Hello Bob,

Sorry, but I'm not able to answer your question as I'm still struggling
to find an appropriate mental construct. For the moment, I'm
suspending disbelief, as they say.

Roy, W7EL, has reported model results showing that a wire (probably a
few millimeters in diameter) only one foot long will produce
near-perfect (my words) results.

The greater the cross-sectional area, the better, of course, but it
would seem not to be a critical factor based on what both Roy and Reg
reported, A one inch diameter copper pipe would probably give you some
margin based on those reports. Make it a couple of feet long and slip it
through six inches or so of one of those foam "noodles" the kids use
when swimming. That will keep it afloat, ensure it is visible, and
protect the hull from damage when it collides. Remember that you will
have to figure out how to attach this pipe to your tuner. In a lot of
installations, that will mean six feet or more of wire (from tuner to
pipe) hanging over the gunwale. That wire is effectively part of your
antenna, and it will radiate. For convenience, it would make sense to
let the pipe float away from the hull by six feet or so, but that makes
the connecting wire even longer. If you hang something like that over
the side you'll doubtless want to secure it with some kind of UV
resistant line to take the strain off the wire, especially when under
way (ugh!).

On some boats, using the stainless rudder shaft could be a better
solution if you can attach to it. It is often near the surface of the
water. On other boats, the rudder shaft exits the hull well below sea
level and that probably wouldn't work.

Experiment by all means, and if you go the copper pipe route, just
remember that your zinc will be protecting whatever copper you immerse.
If you use a lot of copper (like a 1 foot diameter by 1 foot long
cylinder made of copper flashing with lots of holes drilled in it),
expect accelerated depletion of zinc.

Good luck.

Chuck

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To illustrate the order of magnitude of the effects :

An ideal shape of ground electrode is a hemisphere, of diameter D
metres, pressed into the soil, flush with the soil surface..

Its resistance to the soil is easily proved and calculated :

R = S / Pi / D ohms,

where S is the soil's resistivity in ohm-metres and D is the diameter
in metres. Pi = 3.14. The metric system is by far the most simple.

The resistivity of salt sea water is 0.22 ohm-metres, constant
wherever you may sample and test. Unaffected by the melting of the
glaciers.

So with a diameter of 0.22 metres = 9 inches, the electrode resistance
= 1 ohm. Low enough?

If the ground electrode is a ball with diameter = 9 inches, immersed
in sea water at a sensible depth, then the electrode resistance will
be halved.

At radio frequencies the impedance of the connection to ground will be
that of the connecting wire only, even before the resistance of the
connecting wire to the water is taken into account. The high
permittivity of water will also tend to decrease impedance at RF.

Another illustration, following Lord Kelvin :

The resistance of a ground rod to soil is given by :

R = S / 2 / Pi / L * ArcSinh( 2 * L / D ) ohms,

where S = soil resistivity, L = rod length in metres and D is rod
diameter. ArcSinh is the inverse hyperbolic Sine function you will
find on your pocket scientific calculators.

So in sea water, at low frequencies, a rod 12 inches in length and a
diameter of 1 inch will have a resistance of 1.2 ohms. At HF, because
of the very low propagation velocity in water, propagation effects
predominate and the rod must be considered as a very lossy
transmission line. But its impedance to ground is still very low
because Zo is very small.

So the hull of a metal boat makes an excellent ground. Just connect
to it with an alligator clip at the end of a length of wire and stop
worrying about it.

By the way, the practical units of resistivity in ohm-metres should be
much preferred to the academic units of milli-Siemens. When dealing
with milli-Siemens I find I have to stand on my head and look
backwards.

1 milli-Siemens = 1000 ohm-metres.

The clock tells me it's 7.30 in the morning in Birmingham, the idle,
depressed ex-industrial city of the Midlands, where there used to be
10,000 factories, now superceded by the hardworking Chinese, and I'm
already half way down a bottle of Spanish Campaneo red. Hic!
----
Reg, G4FGQ.

Reg Edwards wrote:
The permittivity, K, of water is about 80.

The relative velocity of propagation along a wire immersed in water is
about VF = 1/Sqrt( K ) = 0.11
. . .

When the material is conductive, like salt water, you also have to
consider the conductivity in determining velocity factor. The velocity
factor in salt water is 0.0128 at 7.5 MHz, 0.0175 at 14 MHz (based on
conductivity of 5 S/m and dielectric constant of 81).

Incidentally, you can get this information directly from EZNEC,
including the demo program. Select a real ground type, then find the
velocity factor, skin depth, and other information in Utilities/Ground Info.

Roy Lewallen, W7EL