David or Jo Anne Ryeburn wrote:

wrote:
It might help if the language was made more understandable. Isn't the
problem in getting an absolute ground reference point at the end of
the dipole?

Nope, that's not the problem. The language I am using is standard.

For you, maybe, but not for the average amateur radio operator. Instead
of trying to demonstrate how smart you are, how about trying to explain the
concept in words that the average amateur radio operator can understand?
Feynman took that approach in his book, _QED_.

Again, the problem goes away if you use a 4 dimensional model and define
the 4th dimensional path distance to ground as zero.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
In four dimensions, the ground is not "artificial" and is
at the same potential as the Earth.

What is that supposed to mean?

73, Jim AC6XG

Cecil Moore wrote in message ...
David or Jo Anne Ryeburn wrote:

wrote:
It might help if the language was made more understandable. Isn't the
problem in getting an absolute ground reference point at the end of
the dipole?

Nope, that's not the problem. The language I am using is standard.

For you, maybe, but not for the average amateur radio operator. Instead
of trying to demonstrate how smart you are, how about trying to explain the
concept in words that the average amateur radio operator can understand?

How about:
1. Kirchoff's Voltage Law: The sum of the EMFs around a closed loop
equals the sum of the voltage drops around a loop. (It's commonly
MISstated as the sum of voltage drops around a loop is zero; it's much
better to distinguish between EMFs and voltage drops, as there are
EMFs which CANNOT be localized to one point in the loop.)
2. Faraday's Law of Magnetic Induction: For any closed loop, there
is an EMF proportional to the rate of change of magnetic flux enclosed
by the loop. (This is one EMF which cannot be localized--it just
drives the whole loop.)
3. Ohm's Law: Voltage drop is proportional to current times
resistance (so in a good conductor with modest current, there is very
little voltage drop).

From those, it should be obvious that the voltage you measure between
any two points depends on the path you take, in the presence of a
time-varying magnetic field (and perhaps at other times as well).
Didn't you previously post that you understood that?

As David has suggested, that set of three laws isn't the only way to
think about it, but I think they should serve the typical ham fairly
well; no integrals needed.

Cheers,
Tom

Jim Kelley wrote:

Cecil Moore wrote:
In four dimensions, the ground is not "artificial" and is
at the same potential as the Earth.

What is that supposed to mean?

Solve the problem using a 4D math model. The 3D variable
path problem goes away.
--
73, Cecil, W5DXP

Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:
In four dimensions, the ground is not "artificial" and is
at the same potential as the Earth.

What is that supposed to mean?

Solve the problem using a 4D math model. The 3D variable
path problem goes away.

If you've solved it, you should post the solution. (Otherwise, it's
just sounds like you're trying to impress somebody.)

73, Jim AC6XG

On Thu, 23 Oct 2003 10:45:21 -0700, Jim Kelley
wrote:
If you've solved it, you should post the solution.
The promise of that (following submission to a spectrum of vanity
publishers) could last two years alone!

73's
Richard Clark, KB7QHC

Roy Lewallen, W7EL, posted:

No, that's not the problem. The problem is
that you're looking for a single voltage
between two points separated in space. There
is no single value for that voltage. If you
made some kind of "artificial ground" close
to the antenna, then there are an infinite
number of possible voltages between it and
the Earth.

Well, I chose one particular configuration
and one particular integration path because
I was curious about the original question -
something about how much voltage would the
end insulator have to handle for 100 watts
of radiated power.

I chose a vertical, half-wave monopole fed
against perfect ground and looked at the
driving source data with EZNEC. The feed-
point impedance was 2188 +j66 ohms and a
driving current of .213 amps produced a
radiated power of 100 watts and a feedpoint
voltage of 466 volts. 1500 watts scales
that up to 1805 volts. Symmetry about the
ground would increase that to 3600 volts
for the free-space case. That is what I
would adopt as my design-to target for
end insulators.

I know it's crude, but I was just looking
for a ballpark figure.

Jim, K7JEB

Roy Lewallen, W7EL, posted:

No, that's not the problem. The problem is
that you're looking for a single voltage
between two points separated in space. There
is no single value for that voltage. If you
made some kind of "artificial ground" close
to the antenna, then there are an infinite
number of possible voltages between it and
the Earth.

As another case study, I analyzed a half-wave,
inverted-V antenna over perfect ground with the
ends of the antenna close to the ground. I
placed high-resistance (1e12-ohm) loads between
these open ends and ground as voltmeters. The
angle of the inverted-V was 90 degrees, placing
the apex of the antenna 0.35 wavelength above
ground.

The feedpoint impedance was trimmed up to be
19 + j0.4 ohms. Going right for the maximum,
1500 watts into this impedance requires an
8.9 amp current source.

The voltage to ground across the load on the
antenna endpoint is 5300 volts for 1500 watts
of radiated power.

This scales back to 1370 volts for 100 watts.

Again, this is just a special case of the general
problem. But it has a configuration that is easy to
implement in the EZNEC program and is quite
relevant to typical ham-radio, low-band dipole
installations.

If anyone wants to go over this data, or that for
the vertical, half-wave monopole, drop me an e-mail
at k7jeb (at) qsl.net and I will send over the .EZ
files.

Jim, K7JEB

Roy Lewallen wrote:
"The problem is that you`re looking for a single voltage between two
points in space. There is no single value for that voltage."

Keith Henney in his 1950 "Radio Engineering Handbook on page 638 wrote:
"A thin-wire dipole gives an end potential of about 3,900 volts for 1000
watts antenna input for a height of 1/4-wave. It will be higher for
smaller heights, and falls to a minimum of about 1700 volts as height
increases to 3/4-wave; beyond this point it settles down to the
free-space value of about 3000 volts. Potentials vary as the square root
of the power ratio and as the inverse square root of the capacitance per
unit length. For a potential of 3900 volts on a wire 0.101 in. in
diameter (No.10 B&S), the radial gradient is of the order of 31 kv per
cm."

At frequencies where the antenna has high reactance and low resistance,
a few watts of power produce very high potentials. At high altitudes,
high potentials can easily produce corona and flashover. Apply more
watts and produce corona and flashover at sea level.

One way to measure the voltage at the end of a dipole might be to
increase input to the dipole until orona cccurs and then scale that to
the power of interest.

Best regards, Richard Harrison, KB5WZI

K7JEB wrote:
I know it's crude, but I was just looking
for a ballpark figure.

Hi Jim, all of the estimates have been in the same
ballpark except for the ones that indicate the voltage
doesn't appreciably change from the center feedpoint
of a 1/2WL dipole to the open ends.

Looks like antennas and transmission lines obey Maxwell's
equations after all. :-)
--
73, Cecil, W5DXP