Dave Oldridge wrote:
My question is one of Physics.

Lest anyone think I have invented something new, I'm sure
Walter Maxwell was doing these sorts of Smith Chart
calculations while I was in high school.

Given the following 1/4WL resonant stub with Z01 = 600
ohms and Z02 = 4000 ohms:

Source----Z01A----x----Z02----y----Z01B----open

Boundary conditions a The feedlines are lossless. The
source sees zero ohms. The Z01A length is 45 degrees. The
Z02 length is 45 degrees. How long is the Z01B length?

This is a problem easily solved by using the outside circles
of two Smith Charts, one normalized for 600 ohms and the
other normalized for 4000 ohms. The procedure is:

Assuming the source is seeing zero ohms, calculate the
impedance at 'x'. Plot that impedance on a Z02 Smith Chart.
Calculate the impedance at point 'y' of the Z02 length. Plot
that impedance on the first Smith Chart and see how long
Z01B needs to be to reach to infinity.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Dave wrote:

Cecil --- you are begging the question!!


Of course I am, nobody knows *why* things are the way they
are. There is no 'why' built into quantum mechanics. There
is only probability. Why does one photon wind up in an
inner interference ring and one wind up in an outer
interference ring? Nobody knows.


Methinks you have left the world of deterministic Physics and moved into Quantum
Physics. Such comments do not contribute to the answer. You are not using
Quantum mechanics to try to explain to me what is happening. You are using
transmission line models. Stick to the transmission line model.

I thought antenna EM Physics was deterministic.

- - - -

your model.

You stated:

"Anything done below the coil affects the number of degrees subtracted
from the antenna by the bottom element to coil interface. Anything
done to the stinger affects the number of degrees added to the
antenna by the coil to stinger interface. For a given element delta
length, the number-of-degrees-added effect is greater than the number-
of-degrees-subtracted effect."

The 4 inch spring is approximately 0.646 degrees long at 5.3 MHz.

Original: Source-----Z01A-----+-----Z02-----+-----Z01B-----open

Case #1: Source-----Z01A-----+-----Z02-----+-0.646-+-----Z01B-----open

Case #2: Source-----Z01A-----+-0.646-+-----Z02-----+-----Z01B-----open

Z01A has not changed. Z02 has not changed. Z01B has not changed. [The L/D ratios
for these components have remained the same and the Zo of each element is
therefore the same.] The physical characteristics of 0.646 have not changed. The
Capacitance to the coil is the same in either case. Only the location has changed.

Do you claim the complex impedance of 0.646[case 1] is different from the
complex impedance of 0.646[case 2]. Why?

Or, do you claim the net Z01B + 0.646[Case 1} is now electrically longer, in
degrees, than Z01A + 0.646[case 2]. Why?

Dave wrote:
Or, do you claim the net Z01B + 0.646[Case 1} is now electrically
longer, in degrees, than Z01A + 0.646[case 2]. Why?

I already explained how to prove to yourself why those
two cases are different. Please perform the Smith Chart
gymnastics and then report your findings.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Dave wrote:
Or, do you claim the net Z01B + 0.646[Case 1} is now electrically
longer, in degrees, than Z01A + 0.646[case 2]. Why?

I already explained how to prove to yourself why those
two cases are different. Please perform the Smith Chart
gymnastics and then report your findings.

If one takes one foot of stinger and transfers it to the
base element section, the antenna length remains the same
and the resonant frequency increases. I assume you already
knew that. So of course, 0.646 is electrically longer when
it is on top of the coil than when it is on the bottom of
the coil. The Smith Chart exercise will show why that is
true.
--
73, Cecil http://www.qsl.net/w5dxp