Cecil Moore wrote:
Richard Clark wrote:

I will agree in this respect, but not all Toms are Rauchs. The term
"Current Drop" is abhorrent to some (a pollution of technical
language), an irritant to others, and inconsequential to many who
simply enjoy the cat fight.


Heh, heh, so you don't believe there is a current drop between the
current maximum point and current minimum point on a transmission
line with reflections? Seems to me going from 2 amps at a current
maximum to 0.1 amps at a current minimum is a measurable drop in
total current.

Would you please provide a proof that going from 2 amps to 0.1 amps
is NOT a drop in total current?

Just one more example of trying to use lumped circuit analysis methods
on distributed network problems. Are you guys ever going to learn?

Hint: With distributed networks involving an appreciable percentage of
a wavelength, there are definitely current drops in the series loop.
This certainly applies to 75m Bugcatcher coils used on standing-wave
mobile antennas.
--
73, Cecil http://www.qsl.net/w5dxp


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Next, Cecil, you're going to be talking about a "current gradient"
and a "scalar current field." Here's a question for you, Cecil, and
Richard Harrison, and Yuri, too: how do you take the gradient of
the current at a point on a transmission line, and, if were possible
to do so, what is the physical significance of the result?
73,
Tom Donaly, KA6RUH

Tom Donaly, KA6RUH wrote:
": how do you take the gradient of the current at a point on a
transmission line?"

Not sure I understand the question. Gradient is the rate of change and
that`s the derivative of the current at a given point. Over a certain
path it is the difference between the path ends and can be averaged for
the path.

For convenience, Kraus has collected transmission line formulas. I`m not
a typist so I`ll just say they are near the end of the new edition, page
890. In the 1950 edition they can be fornd on pages 506 and 507, also
near the end of the book.

Work out your own example.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Tom Donaly, KA6RUH wrote:
": how do you take the gradient of the current at a point on a
transmission line?"

Not sure I understand the question. Gradient is the rate of change and
that`s the derivative of the current at a given point. Over a certain
path it is the difference between the path ends and can be averaged for
the path.

For convenience, Kraus has collected transmission line formulas. I`m not
a typist so I`ll just say they are near the end of the new edition, page
890. In the 1950 edition they can be fornd on pages 506 and 507, also
near the end of the book.

Work out your own example.

Best regards, Richard Harrison, KB5WZI


Thank you, Richard, you just made my point.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:

Cecil Moore wrote:
Heh, heh, so you don't believe there is a current drop between the
current maximum point and current minimum point on a transmission
line with reflections? Seems to me going from 2 amps at a current
maximum to 0.1 amps at a current minimum is a measurable drop in
total current.

Next, Cecil, you're going to be talking about a "current gradient"
and a "scalar current field." Here's a question for you, Cecil, and
Richard Harrison, and Yuri, too: how do you take the gradient of
the current at a point on a transmission line, and, if were possible
to do so, what is the physical significance of the result?

A total current gradient obviously exists on a transmission line
with current minimums and maximums. You can locate those points
with a simple pickup loop. The current gradient is caused
by the superposition of forward and reflected current waves as
described in any transmission line textbook.

"Taking the gradient" seems to me to be unnecessary and just a
logical diversion away from the qualitative conceptual discussion.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:

Tom Donaly wrote:

Cecil Moore wrote:

Heh, heh, so you don't believe there is a current drop between the
current maximum point and current minimum point on a transmission
line with reflections? Seems to me going from 2 amps at a current
maximum to 0.1 amps at a current minimum is a measurable drop in
total current.


Next, Cecil, you're going to be talking about a "current gradient"
and a "scalar current field." Here's a question for you, Cecil, and
Richard Harrison, and Yuri, too: how do you take the gradient of
the current at a point on a transmission line, and, if were possible
to do so, what is the physical significance of the result?


A total current gradient obviously exists on a transmission line
with current minimums and maximums. You can locate those points
with a simple pickup loop. The current gradient is caused
by the superposition of forward and reflected current waves as
described in any transmission line textbook.

"Taking the gradient" seems to me to be unnecessary and just a
logical diversion away from the qualitative conceptual discussion.
--
73, Cecil http://www.qsl.net/w5dxp


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You and Richard need a refresher course in electromagnetics. I hope
Yuri doesn't fall into the same trap.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
You and Richard need a refresher course in electromagnetics. I hope
Yuri doesn't fall into the same trap.

As usual, zero technical content from you.

Tom, here's a "circuit" for you.

+--------------------------------+
| |
source Load
| |
+--------------------------------+

The source is delivering 200 watts in the form of V=100V and
I=2A in phase. The load is 4050 ohms. Using your circuit model,
you assert that the current through the source is equal to the
current through the load since it is a series circuit. Yet, if
the current through the 4050 ohm resistor were actually 2.0A,
the power to the load would be about 16,000 watts, thus violating
the conservation of energy principle.

Is there a current drop from the source to the load? Of course!
Does this violate Kirchhoff's laws? Of course not!

Why doesn't your circuit model work? Because the wires
between the source and the load are 1/4WL of 450 ohm ladder-line
thus rendering the circuit model invalid for the application.

YOUR CIRCUIT ANALYSIS MODEL DOES *NOT* WORK ON DISTRIBUTED NETWORK
PROBLEMS!!!
--
73, Cecil http://www.qsl.net/w5dxp


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On Wed, 20 Oct 2004 00:16:49 -0500, Cecil Moore
wrote:
Does this violate Kirchhoff's laws?
Of course it does, several times.

Richard Clark wrote:

Cecil Moore wrote:
Does this violate Kirchhoff's laws?

Of course it does, several times.

Got news for you, Richard. The current drop from a current
loop to a current node is NOT a violation of Kirchhoff's laws.
It is a characteristic of distributed networks.
--
73, Cecil http://www.qsl.net/w5dxp


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On Wed, 20 Oct 2004 09:17:57 -0500, Cecil Moore
wrote:
a violation of Kirchhoff's laws.
that's right

Cecil Moore wrote:
Tom Donaly wrote:

You and Richard need a refresher course in electromagnetics. I hope
Yuri doesn't fall into the same trap.


As usual, zero technical content from you.

Tom, here's a "circuit" for you.

+--------------------------------+
| |
source Load
| |
+--------------------------------+

The source is delivering 200 watts in the form of V=100V and
I=2A in phase. The load is 4050 ohms. Using your circuit model,
you assert that the current through the source is equal to the
current through the load since it is a series circuit. Yet, if
the current through the 4050 ohm resistor were actually 2.0A,
the power to the load would be about 16,000 watts, thus violating
the conservation of energy principle.

Is there a current drop from the source to the load? Of course!
Does this violate Kirchhoff's laws? Of course not!

Why doesn't your circuit model work? Because the wires
between the source and the load are 1/4WL of 450 ohm ladder-line
thus rendering the circuit model invalid for the application.

YOUR CIRCUIT ANALYSIS MODEL DOES *NOT* WORK ON DISTRIBUTED NETWORK
PROBLEMS!!!
--
73, Cecil http://www.qsl.net/w5dxp


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You still don't get it, Cecil, which is o.k. since I didn't expect you
to. I don't doubt that Yuri can find a coil that exhibits a different
current at one end than at the other; I have an antenna that exhibits
the same behavior, and I made it that way on purpose. However, the
term "current drop" as used by Yuri was wrong. There is no place for
it in electromagnetic theory, and if you had known enough theory to
understand that, you wouldn't have answered as you did.
73,
Tom Donaly, KA6RUH