Cecil Moore wrote:
Tom Donaly wrote:
The same thing can be accomplished, above, using an open stub and
measuring the voltage at both ends. All this is just theoretical,
though, because line loss will skew the results. Besides, why try to
measure current or voltage when all you have to do is measure length
and frequency?

Tom, you are not going to understand what I am saying until
you perform the stub exercise I provided. Just do one at a
time. Assume ideal lossless conditions with VF=1.0.

---600 ohm line---+---10 deg 100 ohm line---open

How many degrees of 600 ohm line does it take to make the
above stub look like 1/4 wavelength, i.e. 90 degrees?

Until you perform the exercise, you are just creating
diversions and avoiding the technical truth.

If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.
73,
Tom Donaly, KA6RUH

Cecil Moore wrote:
Keith Dysart wrote:
On Dec 4, 11:13 pm, Cecil Moore wrote:
Tom Donaly wrote:
50 ohm Shorted line 8.5655 meters long. Frequency = 7 Mhz. 2 volt 50
ohm
generator. Current at input = 12.361 milliamps. Current at short =
40 milliamps. Divide the current at the imput of the line by the
current
at the short and take the arc sine (in radian mode) of the result. This
is 1.2566. Now take Beta = .14671 and multiply it by the length 8.5655.
This also equals 1.2566, which is the angular length of the shorted
line. Will someone explain how this works to Cecil?
You won't understand what I am talking about until you perform
the stub experiments that I previously posted.

---600 ohm line---+---10 deg, 100 ohm line---open-circuit

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

--5 deg, 100 ohm line--+--600 ohm line--+--5 deg, 100 ohm line--open

How many degrees of 600 ohm line does it take to resonate
that stub to an electrical 1/4WL?

You have said multiple times that the electrical length of a
quarter wave stub must be 90 electrical degress, so the
computation is too easy...

1) x + 10 = 90
x = 80 degrees for the 600 Ohm line
2) 5 + x + 5 = 90
x = 80 degrees for the 600 Ohm line

although I suspect others will disagree with your solution.

I have not yet provided a solution. Your's is *wrong*.

The 90 degree physical solution is *wrong* because it results
in more than 90 electrical degrees. Please try again.

No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?
73,
Tom Donaly, KA6RUH

Cecil Moore wrote:
Keith Dysart wrote:
I thought that when you specified 5 and 10 degrees in your
problem statement, you meant electrical degrees. That is, the
phase shift encountered by the forward travelling wave.

That specification is the same for physical and electrical
degrees because we are dealing with a single Z0 piece of
transmission line. The 100 ohm line is indeed 10 degrees
long both physically and electrically.

Certainly, the answer was in terms of electrical degrees. That is,
the phase shift encountered by the forward travelling wave.

You, and others, are going to be surprised to find out the
600 ohm section is only 43 degrees of physical length. How
can 43 degrees of 600 ohm line add to 10 degrees of 100 ohm
line to equal 90 electrical degrees of stub? Hint: Like I
told Roy and Tom years ago, there's a 37 degree phase shift
at the impedance discontinuity between the 600 ohm line and
the 100 ohm line. 43+37+10 = 90 electrical degrees.

Understand that simple stub example and you will understand
loaded mobile antennas. Most of the "experts" here are just
full of you-know-what.

Actually, you got it wrong, Cecil. It's -43 degrees, which means
if you wanted to make one it would be about 317 degrees. (Actually,
I got -46.613 degrees.) Maybe you can make a length using negative
degrees, but it's tough for me to do.
73,
Tom Donaly, KA6RUH

Keith Dysart wrote:
But again, given that the key message is "that the system phase shift
is NOT equal to the sum of the phase shifts of the components.",
why is the question of delay through the coil important?

Is it just to have a "debate" with some called experts?

That's as good a reason as any. :-)

Or does it offer some advancement in the solution of antenna
problems? Having computed (or measured) the delay through
the coil, how would this alter the design of the antenna?

I don't know.
--
73, Cecil http://www.w5dxp.com

Yuri Blanarovich wrote:
Now I also understand the small "bump" increase in the current at the bottom
of the coil due to some loses that reflected wave encounter on the way
"there and back" to the tip of the radiator from the bottom of the coil (?)

I suspect that current bump is caused by magnetic linkage
between the central windings while the end windings would
not experience it to such a degree. It is the thing that
increases the velocity factor over the purely helical
path.
--
73, Cecil http://www.w5dxp.com

Gene Fuller wrote:
No one has ever said that there is a 3 ns delay *through* the coil.

That's simply a false statement. Here's W8JI exact words:

"On 80-meters ... time delay is about 3nS. How does the
current travel through the inductor so fast?

Gene, "through the inductor" is through the coil. W8JI
said there is a 3 ns delay *through* the coil. W7EL
seems to agree.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
If you know how to do it, Cecil, don't be coy about it. Just state
your case and be done with it. Since you already stated that
the total electrical length is 90 degrees, you're just asking me
to prove your point. Do it yourself, and then I'll tell you
whether I agree with you or not.

Does that mean you don't know how to do it?

I already did it on another thread, Tom. Adding 43 degrees
of Z0=600 ohm feedline to the 10 degrees of Z0=100 ohm
feedline will turn the stub into an electrical 1/4
wavelength (90 degree) open stub. And that's exactly
how base-loaded mobile antennas work.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
No, Cecil, it's your theory. You have to provide the method and
then everyone else will decide whether or not they agree with you.
You're not chicken are you?

Actually, I wanted to see if anyone besides me could
solve the problem - no one else has.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
Actually, you got it wrong, Cecil. It's -43 degrees, which means
if you wanted to make one it would be about 317 degrees.

Tom, you are just showing your ignorance. 43 degrees
of 600 ohm line attached to 10 degrees of 100 ohm
line makes a dandy 1/4WL stub. Why are you so totally
ignorant of that fact of physics?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
No one has ever said that there is a 3 ns delay *through* the coil.

That's simply a false statement. Here's W8JI exact words:

"On 80-meters ... time delay is about 3nS. How does the
current travel through the inductor so fast?

Gene, "through the inductor" is through the coil. W8JI
said there is a 3 ns delay *through* the coil. W7EL
seems to agree.

Cecil,

You're losing your touch. That isn't exactly a subtle selective quote.

Here are the rest of the words following the question, "How does the
current travel through the inductor so fast?"

"At first this seems impossible, but the answer is actually quite
obvious. Time-varying current gives rise to time-varying magnetic flux.
This magnetic flux, since conductor spacing is close and the distance
very small, links the starting turn very tightly to the next turn. The
rapidly changing magnetic flux causes charges to move in the next
conductor, and the changing magnetic field couples through all the close
spaced turns with very little time delay. It is this magnetic flux
coupling that provides the primary mechanism for energy transfer through
the inductor, and the path is much shorter than the circuitous and much
longer path along the conductor."


But you already knew that . .

73,
Gene
W4SZ