On Dec 28, 12:36*pm, Richard Fry wrote:
"Art Unwin" wroteThe antenna compendium states that an assumption
is made with MoM programs that an assumption is made that current
in a radiator is sinusoidal where as we all know that the current
degrades in aplitude dependent on radiator length.

_____________

Art, the current distribution along even the shortest fractional
wavelength, constant OD radiator also is ~sinusoidal.

Current always is near zero at the open end of a linear radiator of
every
physical length. The shape of the current wave formed along a very
short
radiator appears to be very close to triangular. *But in fact that
"triangular" current distribution is just a very short section of a
sinusoidal waveform.

N.B. that MoM programs show exactly this for radiators that are very
short
in terms of electrical wavelengths. *This also is proven
mathematically in
the antenna engineering texts of Kraus, Balanis, Johnson & Jasik,
etc.

RF

O.K. have it your way. At the end of a radiator voltage is a maximum
as current is zero
ie the curves of current and current crosses each other. We can then
use the absolute standard
equatiion of E = I R. Using this formula for understanding conditions
at the end of a radiator
we can state that E, I and R equals zero ala a non closed circuit.
Kraus used four travelling waves in his analysis of the helical
antenna an analysis that was not corrobarated by
following examiners or the application of the NEC (MoM) programs where
disturbing differences was never resolved.
You introduce wavelength as if it was a standard without considering
the velocity factor and where a transmission line analogy
does not satisfy a helical antenna because of slow wave created in a
similar way to cavitation as explored by Bernoulle
or by the addition of sharp corners encountered by current flow
As far as what has been proven in text books they are only reflect the
conditions placed on the problem but also assumption of correct theory
applied.
This is why history shows the evolution of science is a series of
broken theories whose value is measured by their resistance to attack
over time.
I would remind you that the metric of time has NOT stopped. But as I
stated earlier you can have it your way without objection from me
Art

Art wrote:
"We can then use the absolute standard equation of E = I R."

For ac (RF) that`s not true. The formula is E=IZ, where Z includes
reactance and resistance in quadrature. I`m not piling on but some
readers may believe Art.

Best regards, Ricxhard Harrison, KB5WZI

On Dec 28, 4:12*pm, (Richard Harrison)
wrote:
Art wrote:

"We can then use the absolute standard equation of E = I R."

For ac (RF) that`s not true. The formula is E=IZ, where Z includes
reactance and resistance in quadrature. I`m not piling on but some
readers may believe Art. *

Best regards, Ricxhard Harrison, KB5WZI

Yes you are correct but the original equation was E=IR
which preceeds the implication of impedance which is a derivative
of my equation and came about with the addition of A.C. technology.
If the impedance is totally resistive then my statement is not untrue
Now to avoid the nitpicking are you saying that E=IZ cannot be used
for calculations at the end of an antenna and if so" WHY "
Art

Art wrote:
"Now to avoid nitpicking are you saying that E=IZ cannot be used for
calculations at the end of an antenna and if so "WHY"?"

It is complicated by multiple currents. Like an open-circuited
transmission line, electrical conduction stops at the end of the
conductor. Current then becomes a phasor problem.

Collapse of conduction current induces a voltage which combined with the
incident voltage almost doubles the total voltage at this spot in many
cases. This reverses the direction of current in the conductor. Due to
capacitance at his high-voltage spot with the iniverse, displacement
current flows into free space from open-circuited antenna ends. It is
usually smaller than the conduction current.

Best regards, Richard Harrison, KB5WZI

On Dec 28, 5:35*pm, (Richard Harrison)
wrote:
Art wrote:

"Now to avoid nitpicking are you saying that E=IZ cannot be used for
calculations at the end of an antenna and if so "WHY"?"

It is complicated by multiple currents. Like an open-circuited
transmission line, electrical conduction stops at the end of the
conductor. Current then becomes a phasor problem.

Collapse of conduction current induces a voltage which combined with the
incident voltage almost doubles the total voltage at this spot in many
cases. This reverses the direction of current in the conductor. Due to
capacitance at his high-voltage spot with the iniverse, displacement
current flows into free space from open-circuited antenna ends. It is
usually smaller than the conduction current.

Best regards, Richard Harrison, KB5WZI

You skated over the difference between an open circuit of the
transmission
line compared to the end of an antenna.
The analogy is flawed and will be shown when the resistance in the
center of a radiator is disclosed via the
computor programs. You never did supply the information needed to
justify the values of E,I and R when
the current value crosses the zero line on a graph. You can ofcourse,
declare that none of the given factors
can ever be equal to zero by jumping the datum line !!!!! By the way,
could you state a situation where the
displacement current is LARGER than the conduction current so I may
review it in the light of Newtonian laws?
Art
Art

Art wrote:
"could you state a situation where the displacement current is LARGER
than the conduction current so I may review it in the light of Newtonian
laws?"

I have difficulty in imagining current between the plates of a capacitor
exceeding the current through the capacitor`s leads.

Values of voltages and currents anywhere along an antenna primarily
depend on the impedance of the antenna at that point and then are
dictated by the phasors of the incident and reflected totals at the same
point. Arnold B. Bailey in Fig. 7-28 on page 368 of "TV And Other
Receiving Antennas" shows current distribution on a half-wave dipole
which smoothly varies from zero at its ends to maximum at its center.

Experience shows that a quarter-wave back from a maximum impedance
point, a minimum impedance point is created by incident and reflected
phasors.

Best regards, Richard Harrison, KB5WZI

Art Unwin wrote:
You never did supply the information needed to
justify the values of E,I and R when
the current value crosses the zero line on a graph.

In simple terms, when the standing-wave current has
a zero amplitude at a current node, none of the energy
is in the magnetic field and all of the energy is in
the electric field. That's why a voltage maximum appears
at a current minimum. When the current equals zero, the
virtual impedance, E/I, is infinite.

This is essentially what happens at the end of a dipole
or monopole or open-circuit stub. The characteristic
impedance of a #14 wire 30 feet above ground is very
close to 600 ohms. Given that Z0, we can treat a dipole
element as a lossy transmission line and calculate the
voltage at the end of the dipole element.

If we model a 1/4WL 600 ohm open-circuit stub with
EZNEC and adjust the resistivity to 0.0000021 ohm-m
to simulate the radiation resistance of a dipole
wire, the feedpoint impedance of the stub is 35 ohms
and conditions on the lossy stub are very close to
the conditions on a dipole element.
--
73, Cecil http://www.w5dxp.com

On Dec 29, 9:11*am, Cecil Moore wrote:
Art Unwin wrote:
You never did supply the information needed to
justify the values of E,I and R when
*the current value crosses the zero line on a graph.

In simple terms, when the standing-wave current has
a zero amplitude at a current node, none of the energy
is in the magnetic field and all of the energy is in
the electric field. That's why a voltage maximum appears
at a current minimum. When the current equals zero, the
virtual impedance, E/I, is infinite.

This is essentially what happens at the end of a dipole
or monopole or open-circuit stub. The characteristic
impedance of a #14 wire 30 feet above ground is very
close to 600 ohms. Given that Z0, we can treat a dipole
element as a lossy transmission line and calculate the
voltage at the end of the dipole element.

If we model a 1/4WL 600 ohm open-circuit stub with
EZNEC and adjust the resistivity to 0.0000021 ohm-m
to simulate the radiation resistance of a dipole
wire, the feedpoint impedance of the stub is 35 ohms
and conditions on the lossy stub are very close to
the conditions on a dipole element.
--
73, Cecil *http://www.w5dxp.com

At the end of the radiator you state the energy is transfered to the
field
so I would imagine there is zero skin effect at that point and the
chain of
skin effect is still present on the outside of the radiator, this
because a full period has not yet elapsed
This equates to a displacement current across the capacitance gap
(plates) between
the outside and the inside of the radiator which is the only current
route available
when the capacitor field expires. Note that this energy is released
prior to the end of the current flow period
because of the absence of the skin effect at that time.
Cecil I am examining all the holy cows that pervade the science of
radiation
as it is universally accepted that radiation is not fully understood,
thus the many hats!
At the moment I see no mechanism that supports the capacitor field to
expire in the direction of incoming current
prior to the completion of the forward period.
Regards
Art