Roy Lewallen wrote:
I maintain that the authors of that book know more than you do about the
topic by at least an order of magnitude -- more likely about three.

That doesn't prove you understand what they said.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"We can deduce the ratio of the voltage to current from the feedpoint
impedance."

What`s the difference between a 1/4-wave of transmission line and a
1/2-wave dipole?

The transmission line is designed for low radiation and the antenna is
designed for high radiation.

Current will be turned around by the open-circuit at the end of the
transmission line. At the end of the antenna, displacement current will
flow between oppositely polarized parts of the antenna, the earth, and
other surrounding objects, and some conduction current will be
redirected in the opposite direction on the antenna.

Impedance of the transmission linee is uniform in its presentation to
either one of the two traveling waves traversing the line, forward or
reflected..

SWR changes along an antenna due to radiation, inefficiencies,
irregularities, etc. An average characteristic (surge) impedance is used
for antenna calculations.

SWR for the usual open-circuit 1/2-wave dipole:
SWR = Avg. Zo of the antenna / radiation resistance.
Avg. Zo = 276 log 1/periphery of antenna conductor
log is the base 10 unit
periphery is measured in the same units as length, usually in
wavelengths.

Loaded antennas tend to be high Q resonant circuits and this limits
their bandwidths. Low resistance allows a large current as reactance is
cancelled at resonance. The large current causes large voltage drops
across the antenna inductance and capacitance. I`ve seen mobile antennas
which were ready to fire up coronas at their tips at the slightest
excuse. This demonstrates high reactance (energy storage) and low
resistance (radiation and loss).

There are several variables which determine the voltage and current at
the end of the dipole. The current does not drop to zero simply because
the conduction in the forward direction stops. Displacement current
continues the flow to some extent. The dipole has a Q, and the wave
traveling on the dipole generates a voltage to current ratio related to
the size and configuration of the antenna.

You can predict that forward voltage on a transmission line will double
at its open-circuit end, and you can predict the forward voltage. I
think prediction of voltage at the open-circuit end of an antenna is
harder.

Best regards, Richard Harrison, KB5WZI

Roy Lewallen wrote:
I maintain that the authors of that book know more than you do about the
topic by at least an order of magnitude -- more likely about three. If
you really want to know the answer to your silly question, you should
study what they've written and try to understand it, rather than posting
it as a question to me on this newsgroup.

Hmmmmmm, perhaps you and your authors would understand the subject better
by reading and understanding a good book on quantum electrodymanics. You
will not fully understand fields until you accept the fact that photons
can have four, not two, polarizations. I maintain that the developers of
quantum electro-dynamics know more than you and your authors combined.
You have earlier rejected the latest physics theories. I doubt that you
will understand fields until you fully understand virtual photons.

(See, two can play your silly argumentum ad verecundiam game.) :-)
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Harrison wrote:
The current does not drop to zero simply because
the conduction in the forward direction stops. Displacement current
continues the flow to some extent.

Many electrons do not flow through air unless the air is ionized.
"Displacement current" is not understood very well and is a sort
of virtual current. The actual electron flow is taking place in
the rest of the circuit. When a lot of electrons flow between the
plates of a capacitor, for instance, the voltage capability of the
capacitor has been exceeded and it is in failure mode.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil, W5DXP wrote:
"When a lot of electrons flow between the plates of a capacitor, for
instance, the voltage capability of the capacitor has been exceeded and
it is in a failure mode."

Yes. Conduction current flows when electrons penetrate the dielectric.
Capacitors consist of conductors separated by insulation. This
dielectric bars conduction up to some voltage limit where the insulation
breaks down. This insulation can be a high vacuum, but at some voltage,
a spark can jump the gap even in a vacuum.

We have vacuum capacitors, both fixed and variable. We have vacuum tubes
which have measured interelectrode capacitances. We have radio waves
which propagate as the result of alternating magnetic and electric
fields. These propagate in the high vacuum of deep space where
corpuscular particles may be speeding through like infinitesimal
bullets, but evidence is that no matter is required to allow influence
by electric and magnetic fields. An alternating electric field, even in
a vacuum where no electrons are in motion, produces an alternating
magnetic field which generates an alternating electric field, and so on
and so forth ad infinitum, ad nauseam.

We know this works as we use radio signals in deep space and view light
from the sun and other cosmic sources. It was J.C. Maxwell who
speculated that displacement current produces a magnetic field the same
as conduction current does, to explain how radiation occurs.

Best regards, Richard Harrison, KB5WZI

Cecil, W5DXP wrote:
"We can deduce the ratio of voltage to current from the feedpoint
impedance."

Kraus says:
"The current I1 at a distance from the nearest current maximum, as shown
in Fig. 6-11 is given by

I1 = Io cos beta x

where I1 = terminal current,

Io = maximum current

When x=0, R1 = Ro; but when x=lambda/4, R1 = infinity if Ro is not zero.
However, the radiation resistance measured at a current minimum
(x=lambda/4) is not infinite as would be the calculated from:

R1 = Ro/ cos squared (beta x), since an actual antenna is not
infinitesimally thin and the current at a minimum point is not zero.
Nevertheless, the radiation resistance at a current minimum may in
practice be very large, i.e., thousands of ohms.

I think the voltage may be measured at the end of a dipole using a diode
probe spaced from the dipole by an insulator. The ground return can be
provided by the exterior of a 1/2-wavelength coax which is well
grounded.

For calibration, the easily determined voltage at the dipole feedpoint
can be indicated by the diode probe.

The coupling can be kept small if there is sufficient antenna power. The
antenna does not need to be reconfigured into a halo to measure its
voltage at its open-circuit end.

Best regards, Richard Harrison, KB5WZI

You do realize, of course, that the reading you get will depend on how
you've oriented the coax. Which orientation would provide the "correct"
answer? And why?

Roy Lewallen, W7EL

Richard Harrison wrote:
. . .
I think the voltage may be measured at the end of a dipole using a diode
probe spaced from the dipole by an insulator. The ground return can be
provided by the exterior of a 1/2-wavelength coax which is well
grounded.
. . .

Roy, W7EL wrote:
"Which orientation would provide the "correct" answer?"

Maybe none.

We know what the correct answer is at the feedpoint. By trial and error
we can find the most stable and least critical placement for our test
lead. We can calibrate our voltage indicator with a sample at the
feedpoint under the conditions that prevail.

A test lead perpendicular to the balanced antenna suffers the least
interference. It`s subject to imperfections and it`s subject to
improvements to overcome the imperfections.

In AM broadcasting we get away with hanging a sampling loop on each
tower of a directional array to monitor both tower current and phase. It
works well enough.

Do you have a better idea than a diode probe to sample voltage at the
end of a dipole?

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
Do you have a better idea than a diode probe to sample voltage at the
end of a dipole?

I use a 4th dimension ground.
--
73, Cecil http://www.qsl.net/w5dxp



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Do you have a better idea than a diode probe to sample voltage at the
end of a dipole?

===============================

The only people person who needs to know the volts at the ends of an antenna
conductor such as a 1/2-wave dipole is the antenna designer. He has to
concern himself with insulators, breakdown voltages, the effects of corona
discharge and other such mundane (to other people) things.

Antenna designers, of which extremely few are needed in this small world,
being practical, sensible people who are obliged by professional
self-discipline to keep economy in time and materials foremost in their
minds, do not waste valuable resources researching the 'end-voltage'
question, purchasing latest space-age technology equipment just for a
one-off job, and employing teams of incompetent but highly-paid assistants
and workmen to make one solitary measurement.

Of course they don't !

In a few seconds they just do a little school arithmetic : -

Volts at end of dipole = Q * Vin / 2

Where Vin is centre-fed dipole feedline volts and Q is the dipole's resonant
Q factor.

Q is dipole inductive reactance divided by radiation resistance.

According to the ARRL handbook, Heaviside, Terman, me and countless others,
Q = Omega*L/R

Richard, you are familiar with these elementary notions. Do you not feel a
little saddened, like me, that amateurs, even future professional engineers,
are handicapped by having unnecessarily complicated, incorrect even, ideas
knocked into their heads by do-gooders on this newsgroup, trying to appear
knowledgeable, who should know better ?

I hasten to distinguish between accumulated PRACTICAL EXPERIENCE and
TECHNICAL BAFFLEGAB. The latter is easily recognisable.
===
Reg, G4FGQ