Cecil Moore wrote:
Gene Fuller wrote:
But you already knew that . .

Of course I did. "Through the coil" does NOT mean "through
the coil wire". It means "through the coil". You still uttered
a falsehood but I doubt that you will ever admit it.

Cecil,

You got me. I omitted the word "wire".

Of course your time delay "impossibility" complaint makes absolutely no
sense at all if you accept that W8JI was not talking about a wave
traveling through 50 feet of wire in 3 ns. He was talking about a wave
traveling 10 inches in 3 ns. I mistakenly assumed that no one in this
discussion even remotely considered superluminal wave propagation.

Please accept my humble apology for an utter lack of fine craftsmanship
in word games.

73,
Gene
W4SZ

Cecil Moore wrote:
Tom Donaly wrote:
O.k., Cecil, you said it, now prove it. There's no requirement
for a 90 degree phase shift when you do the math.

Are you using the equation for forward and reflected
current? If not, you need to do so. The phase shift
is not in the standing-wave current. Standing-wave
current phase is fixed with respect to the source.
Absolutely *nothing* happens to the standing-wave
current at the impedance discontinuity.

The reflected current is known to be in phase with
the forward current at the feedpoint. The forward
current is reflected at the tip of the antenna and
undergoes a 180 degree phase shift. Something must
account for the other 180 degrees or else the
feedpoint impedance would not be resistive. I am
working on a graphic that illustrates what happens
at the impedance discontinuity.

Please enlighten us on how the reflected current
gets back in phase with the forward current without
undergoing a phase shift of 180 degrees in its
round-trip path. It is my understanding that the
forward phasor rotates in one direction while the
reflected phasor rotates in the opposite direction.
The key concept there is that a phasor is always
rotating.

I have waded through the math before but I cannot
locate my notes after moving. If you can figure
out a reasonable answer to the above question, I
will certainly consider it.

How about answering the other part of my post, Cecil.
I didn't use reflection mechanics to reach my conclusion.
I did use two port ABCD parameters (the hard way) which can
be derived from reflection mechanics. Anyway, you're being too
simple. If you're going to use reflection mechanics, you have to
account for all the reflections, and you have to explain yourself
each step of the way. Anyway, destroy a few brain cells
thinking about the other part of my post and get back to me.
73,
Tom Donaly, KA6RUH

Cecil Moore wrote:
John Smith wrote:
By the way, one week in the hospital cost: Hospital $80,000,
emergency room $1,900, ambulance $1

I'm sorry that happened. Get well quick.

Off topic, I know you have a harley though ...

Had putts since I was 16, never had an accident out of the dirt, nothing
ever serious, etc.

Be careful on that monster of yours--wishing ya all the luck--and, oh
yeah, TAKE CARE on that road! They ARE out to get ya ... ;-)

Regards,
JS

Gene Fuller wrote:
Cecil Moore wrote:
"Through the coil" does NOT mean "through
the coil wire". It means "through the coil". You still uttered
a falsehood but I doubt that you will ever admit it.

You got me. I omitted the word "wire".

In that case, I apologize for jumping on you. The VF of
the coil is approximately double what it would be if
the current followed the wire. That's because of the
inter-winding interaction. But the magnetic fields
from coil#1 do not magically jump 10 inches to coil#100
as W8JI implies they do. His measurements are off by
a magnitude because he used standing-wave current for
his measurements. I suspect he actually measured less
than 3 ns "delay" but knew he couldn't afford to post
a faster-than-light measurement.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
If you're going to use reflection mechanics, you have to
account for all the reflections, and you have to explain yourself
each step of the way.

I can do that, Tom, but I am standing by for an emergency
trip to New York which involves my daughter's life.

In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Tom Donaly wrote:
If you're going to use reflection mechanics, you have to
account for all the reflections, and you have to explain yourself
each step of the way.

I can do that, Tom, but I am standing by for an emergency
trip to New York which involves my daughter's life.

In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)

Actually, I have. They're built into the formulas for transmission line
voltages and currents on my calculator. My calculator is a Cecil
disciple.
I hope your daughter comes through in good shape. I understand the
worry, believe me.
73,
Tom Donaly, KA6RUH

Cecil Moore wrote:
... I am standing by for an emergency
trip to New York which involves my daughter's life.
...

Prayers and best wishes ... God protect and speed.

Warmest regards,
JS

Tom Donaly wrote:
Cecil Moore wrote:
In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)

Actually, I have. They're built into the formulas for transmission line
voltages and currents on my calculator.

So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Tom Donaly wrote:
Cecil Moore wrote:
In the meantime, let's see if we can agree if 43.4 degrees
of 600 ohm line is terminated in -j567 ohms, the forward
current will be in phase with the reflected current, i.e.
the impedance looking into the line will be zero ohms. Have
you used any current reflection coefficients recently? :-)

Actually, I have. They're built into the formulas for transmission line
voltages and currents on my calculator.

So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?

No, but I agree with myself that whatever voltage is applied to
the input will be canceled by all the reflections in the circuit
adding up to a voltage at said input that will cancel the input voltage.
It's like finding the zeros of a network equation.
73,
Tom Donaly, KA6RUH

On Dec 6, 9:48 pm, Cecil Moore wrote:
So are we agreed that a 43.4 degree stub terminated in
0-j567 ohms impedance is electrically 1/4WL, i.e. 90
degrees long?

There are many ways to get the 0 input impedance:
- 43.4 degrees of 600 ohm line terminated in a lumped 0-j567 impedance
(assuming I recall the problem corrrectly and you did the math
correctly)
- 43.4 degrees of 600 ohm line followed by 46.6 degrees of 600 ohm
line, open at the end
- 43.4 degrees of 600 ohm line followed by 10 degrees (IIRC) of 100
ohm line, open at the end
- a short
- 180 degrees of any impedance line shorted at the end
- and many, many more

Are you claiming that all of these are electrically 1/4WL ?
Even the 180 degree line? And the short? Seems like a stretch.

And when looked at in detail (think time domain for moment), they each
behave quite differently.

....Keith