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

Art Unwin wrote:
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.

The "capacitive" field is the *electric* field which is
at a *maximum* amplitude at the tip of a dipole. It is
the magnetic (inductive) field that is close to zero
at the tip of a dipole.
--
73, Cecil http://www.w5dxp.com

On Dec 29, 10:26*am, Cecil Moore wrote:
Art Unwin wrote:
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.

The "capacitive" field is the *electric* field which is
at a *maximum* amplitude at the tip of a dipole. It is
the magnetic (inductive) field that is close to zero
at the tip of a dipole.
--
73, Cecil *http://www.w5dxp.com
Cecil
I still am looking for an explanation that prevents current flow thru
the center.
I recognise that the common thinking is to accept reflection but I
fail to see
how that can happen so I can follow up with the numbers.
The capacitor is limited with respect to the energy that it can retain
so what happens when that
limit is reached and the forward period has not come to an end? Yes,
the common thinking
is that the current changes direction to oppose the forward moving
current as with a reflection
where the eddy current in the reverse direction cancels the eddy
current moving in the other direction.
It is here that I am looking for a mechanism that justifies this
reasoning of reflection so I can begin to dispel the
closed circuit aproach as seen with a full wave radiator in
equilibrium
Best regards
Art
Art
but I am looking for actual proof

Art wrote:
"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 es still present on the outside of the radiator,
this is because the full period has not elapsed."

It will elapse as it does in every period so that the higher resistance
at RF will take its toll as predicted.

Because of skin effect, hollow conducting pipes exhibit almost as much
conducting ability as solid rods of the same material. This creates a
market for aluminum tubes as radiating elements and Copperweld wire in
antennas. High-powered transmitters use silver-plated coiled pipes to
carry distilled water to cool their final amplifiers. The space within
the coiled pipes carries no significant RF. RF energy does not flow out
on the surface of a conductor and return in the conducting material
within the conductor in ordinary circumstances, although phase lag
within a conductor increases as penetration depth increases.

Best regards, Richard Harrison, KB5WZI