AI4QJ wrote:
I think I see why it no longer surprizes me after going through the smith
chart.
The 100 ohm line (10 degrees) is open. The 600 ohm line has a load impedance
of -j567 ohms, it is not open. The fact that it is terminated with an
impedance (the 100 ohm line) adds degrees on the chart. We should expect the
reactance of the 100 ohm line to add phase angle at the termination similar
to a "discreet component". Hope this makes sense; the smith chart makes it
very clear.

---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open

The Smith Chart does make it clear what is happening.
Here is the math to go with it. The impedance at the
junction of the two lines is:

-j100*tan(90-10) = -j100*tan(80) = -j567 ohms
-j600*tan(43.4) = -j600*tan(43.4) = -j567 ohms

The phase shift at the junction of the two lines is:
80-43.4 = 36.6 degrees

Time permitting, I will work up the phasor diagrams of
the component voltages (or currents) at the junction
where rho = (600-100)/(600+100) = 0.7143
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
On Thu, 06 Dec 2007 21:22:09 GMT, Cecil Moore
wrote:

Based on your questions, an ordinary prudent man would
assume that you are just wasting my time. Next thing
is that you will be ragging on me for the number of
postings I had to make in answering your questions.

Well, you didn't answer them all did you?

I answered them until the total number of them
started approaching infinity when your motive
became clear.
--
73, Cecil http://www.w5dxp.com

So, what voltage magnitudes were presented to the inputs of your
scope?

On Dec 7, 11:21 am, Cecil Moore wrote:

---43.4 deg 600 ohm line---+---10 deg 100 ohm line---open

The Smith Chart does make it clear what is happening.
Here is the math to go with it. The impedance at the
junction of the two lines is:

-j100*tan(90-10) = -j100*tan(80) = -j567 ohms
-j600*tan(43.4) = -j600*tan(43.4) = -j567 ohms

The phase shift at the junction of the two lines is:
80-43.4 = 36.6 degrees

Time permitting, I will work up the phasor diagrams of
the component voltages (or currents) at the junction
where rho = (600-100)/(600+100) = 0.7143

If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

Would the phase shift at the junction still be
36.6 degrees?

....Keith

Tom Donaly wrote:
I think I see why it no longer surprizes me after going through the
smith chart.
The 100 ohm line (10 degrees) is open. The 600 ohm line has a load
impedance of -j567 ohms, it is not open. The fact that it is
terminated with an impedance (the 100 ohm line) adds degrees on the
chart. We should expect the reactance of the 100 ohm line to add phase
angle at the termination similar to a "discreet component". Hope this
makes sense; the smith chart makes it very clear.

Do you want to work that out mathematically?

For the stub to be electrically 1/4WL, the following
must hold where L1 and L2 are in degrees.

-jZ01*tan(L1) = -jZ02*cot(L2) = -jZ02*tan(90-L2)

-j600*tan(43.4) = -j100*cot(10) = -j567 ohms at the junction

When L1 = L2, the stub is half Z01 and half Z02.
Such a stub is very close to 1/2 the physical length
of a single-Z0 stub when Z01/Z02 = 6.
--
73, Cecil http://www.w5dxp.com

Ian White GM3SEK wrote:
The 4th edition does use degrees for the electrical
lengths of the plain unloaded sections (which is valid from everyone's
point of view); but it no longer implies that the loading coil
"replaces" any number of degrees.

"Replace" seems to mean different things to different
people so it is not a good word to use without a stated
definition. It would probably be better to say the loading
coil "occupies" a certain number of degrees in a loaded
antenna.

The number of degrees occupied by the coil varies but it
is in the tens of degrees for a 75m mobile loading coil.
Here is an EXCEL file that computes the Z0 and VF of a
loading coil assuming it meets the "less than 1" test
included in the computation. Of course, the results
are only approximate since some secondary effects, such
as wire diameter, are ignored.

http://www.w5dxp.com/CoilZ0VF.xls

The VF of a 75m Texas Bugcatcher coil is ~0.02 at 4 MHz.
Since it is ~7 inches long, it occupies ~43 degrees of
antenna. The stinger occupies ~10 degrees so the coil
indeed does not "replace" 80 degrees of antenna. It
*occupies* 43 degrees of the antenna. The rest of the
necessary phase shift, 90-43-10 = 37 degrees, occurs at
the coil to stinger impedance discontinuity where the
Z0 of the coil is ~4000 ohms and the Z0 of the stinger
is ~400 ohms. A 10/1 ratio of Z0s causes a considerable
phase shift in the traveling waves, not in the standing-
waves.

One side of the argument recognizes only the phase shift
through the coil. The other side of the argument recognizes
only the phase shift at the top of the coil. Both sides
are partially right and partially wrong. Interestingly,
the truth lies just about half way in between the two
rail arguments. About half of the "missing degrees" are
contributed by the part of the antenna *occupied* by the
coil while the rest is contributed by the impedance
discontinuity between the coil and the stinger.

I don't know the detailed history behind that change, but I do know one
thing: ON4UN is not a man to be swayed by "political" influence. The
change in the 4th edition would be because he was challenged to look
again at the *technical* issues, and then he made up his own mind.

If he changed his mind based on experiments using standing-
wave current measurements, he is still wrong. I have tried
to contact him using his ARRL email address, but got no
reply.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
You have done this before; postulating
explanations that only work in the complexity
of the "real" world, but fail when presented with
the simplicity of ideal test cases.

For Pete's sake, Keith, Ohm's law doesn't even
work when R=0.

Then, when the explanations fail on the simple
cases, claiming these cases are not of interest
because the real world is more complex.

I define the boundary conditions within which my
ideas work. Whether they work outside those defined
conditions is irrelevant. I believe they do work
for ideal conditions, but I don't have the need
to prove a "theory of everything".

Every model that we use has flaws. Asking me to
come up with a flawless "theory of everything"
model is an obvious, ridiculous diversion but
you already know that.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

This example would correspond to a larger coil and
a shorter stinger in a loaded mobile antenna.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

...
It gets the females turned on.

ROFLOL!

My gawd man, you must be a riot at a party!

Regards,
JS

On Dec 7, 1:00 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the 100 ohm line was only 5 degrees long, how
long would the 600 ohm line have to be to obtain
0 ohms at the input?

-jcot(5) = -j11.43 normalized to Z0=100 ohms

-j100(11.43) = -j1143 ohms at the junction

-j1143/600 = -j1.905 normalized to Z0=600 ohms

arctan(1.905) = 62.3 degrees of Z0=600 ohm line

Would the phase shift at the junction still be
36.6 degrees?

The new phase shift would be 90-5-62.3 = 22.7 deg.

62.3 + 5 + 22.7 = 90 degrees

This example would correspond to a larger coil and
a shorter stinger in a loaded mobile antenna.
--
73, Cecil http://www.w5dxp.com

The smith chart shows this well. If I go up only 5 degrees on the
outer circle of open-end (infinite impedance on the 100 ohm line),
lower values of electrical angle corresponds to higher reactance, in
this case -j1143 for the 5 degree line instead of -j567 for the 10
degree line at the junction. The new phase shift at the junction
should be, and is, now lower since the 100 ohm line has a higher
capacitive reactance at the junction. As the 100 ohm line is shortened
to 0 degrees, we have a 600 ohm transmission line that is open and now
the 600 ohm line must be lengthened to the full 90 degrees for 1/4W.
This would correspond to a coil with no stinger.