Jim Kelley wrote:
Apparently I should have added that distributed resistive losses in
series create distributed voltage drops with a common current through
all, but distributed resistive losses in shunt create a distribution of
Norton current sources, where the shunt current on the transmission line
or radiator at a given point is the sum of all equivalent current
sources between that point and the end of the transmission line/radiator.

Distributed shunt resistive losses would imply dielectric losses
which certainly exist but are minimum at HF. We could even assume
a worst case open-wire transmission line made from resistance wire
and located in the vacuum of free space. There doesn't seem to be
any valid way to justify asserting that shunt losses exactly equal
series losses in every possible transmission line at every possible
frequency under every possible conditions. Asserting such is just
an admission that one it trying to force reality to match the math
model rather than vice versa.

In a flat transmission line without reflections, if the E-field
drops, the characteristic impedance of the transmission line
forces energy to migrate from the H-field to the E-field, such
that the constant V/I ratio remains equal to Z0. Thus, the H-field
supplies energy to compensate for the losses in the E-field.
--
73, Cecil, W5DXP

Don't forget, as always happens, that when Zo is assumed to be purely
resistive, as is always done, it automatically forces wire resistance loss
to equal shunt conductance loss. You've lost a degree of freedom. Which
has a considerable bearing on your arguments.
----
Reg

Reg Edwards wrote:
Don't forget, as always happens, that when Zo is assumed to be purely
resistive, as is always done, it automatically forces wire resistance loss
to equal shunt conductance loss. You've lost a degree of freedom. Which
has a considerable bearing on your arguments.

Reg, seems I learned back in the dark ages that if Z0 is assumed to
be purely resistive, it forces wire resistance loss to equal shunt
conductance loss - AND BOTH OF THEM ARE EQUAL TO ZERO, i.e. the
line is lossless.

Actually, the argument is not about lossless lines, but about lines
with an attenuation factor term. I'm assuming that when Z0 is not
purely resistive, the ratio of voltage to current still equals a
constant Z0. If the E-field is attenuated by series I^2*R losses,
the H-field will supply energy to the E-field in an amount that will
maintain the Z0 constant ratio. If the H-field is attenuated by shunt
I^2*R losses, the E-field will supply energy to the H-field in an
amount that will maintain the Z0 constant ratio. That's why the
attenuation factors are identical - there's simply no other alternative.
--
73, Cecil, W5DXP

Cecil,

I am curious how you separate "reality" from the "math model". I have
never directly measured the properties of open-wire transmission lines
located in the vacuum of free space. Have you?

How do you know that "reality" is correct and the "math model" is wrong?

The second paragraph is even more curious. Do you have a reference for
this migration of energy from the H-field to supply the suffering
E-field? I must have missed that day in class.

73,
Gene
W4SZ

Cecil Moore wrote:


Distributed shunt resistive losses would imply dielectric losses
which certainly exist but are minimum at HF. We could even assume
a worst case open-wire transmission line made from resistance wire
and located in the vacuum of free space. There doesn't seem to be
any valid way to justify asserting that shunt losses exactly equal
series losses in every possible transmission line at every possible
frequency under every possible conditions. Asserting such is just
an admission that one it trying to force reality to match the math
model rather than vice versa.

In a flat transmission line without reflections, if the E-field
drops, the characteristic impedance of the transmission line
forces energy to migrate from the H-field to the E-field, such
that the constant V/I ratio remains equal to Z0. Thus, the H-field
supplies energy to compensate for the losses in the E-field.
--
73, Cecil, W5DXP

"Cecil Moore" wrote in message
...
Reg Edwards wrote:
Don't forget, as always happens, that when Zo is assumed to be purely
resistive, as is always done, it automatically forces wire resistance
loss
to equal shunt conductance loss. You've lost a degree of freedom.
Which
has a considerable bearing on your arguments.

Reg, seems I learned back in the dark ages that if Z0 is assumed to
be purely resistive, it forces wire resistance loss to equal shunt
conductance loss - AND BOTH OF THEM ARE EQUAL TO ZERO, i.e. the
line is lossless.

Actually, the argument is not about lossless lines, but about lines
with an attenuation factor term. I'm assuming that when Z0 is not
purely resistive, the ratio of voltage to current still equals a
constant Z0. If the E-field is attenuated by series I^2*R losses,
the H-field will supply energy to the E-field in an amount that will
maintain the Z0 constant ratio. If the H-field is attenuated by shunt
I^2*R losses, the E-field will supply energy to the H-field in an
amount that will maintain the Z0 constant ratio. That's why the
attenuation factors are identical - there's simply no other alternative.
--
73, Cecil, W5DXP

===============================

Dear Cec,

You ought to have more sense than try to argue with ME about transmission
lines. You had better return to the dark-ages before Heaviside and start
again from square one.

If I say, when Zo is made purely resistive, that Series Resistance and Shunt
Conductance losses automatically become equal to each other even when
NEITHER is zero, then I really do mean "When Zo is made purely resistive,
Series Resistance and Shunt Conductance losses automatically become equal to
each other even when NEITHER is zero."

So your argument, whatever it is, falls as flat as a pancake on Good Friday.
;o) ;o)
----
Reg

Cec, it's Worship of the Great Smith Chart which has led you astray. My
warnings have been disregarded. ;o)
---
Reg.

Gene Fuller wrote:
How do you know that "reality" is correct and the "math model" is wrong?

With a ridiculous question like that, I rest my case! Reality is
*always* correct. If the math model disagrees with reality, it is
simply wrong! Do you really think that what happens only in your
mind is reality? As my Mother once said, "If that's what you think,
think again!"

Example: The 19th century math model didn't explain the orbit of
Mercury. Do you think you can control the orbit of Mercury in
your mind? I suppose in your world, you really can do that.

The second paragraph is even more curious. Do you have a reference for
this migration of energy from the H-field to supply the suffering
E-field? I must have missed that day in class.

You were probably out with a cold that day in kindergarten. It's
called "conservation of energy". If the ratio of the E-field to
the H-field is a constant, any change in either one will result
in a change in the other. This is a well known and accepted fact
of physics in the field of optics. How did you miss it? Hint: What
happens to isotropic radiation in 3D space? Both the E-field and
H-field decrease in value per square meter and THEIR RATIO STAYS
EXACTLY THE SAME.

FYI, the characteristic impedance or index of refraction forces the
ratio of the E-field to the H-field to a constant value. When a
forward wave in a 50 ohm transmission line encounters a 50 ohm
to 300 ohm impedance discontinuity, the ratio of the voltage to
current in the forward wave changes from 50 to 300. I'm extremely
surprised that you don't know and don't accept that fact of physics.
--
73, Cecil, W5DXP

On Fri, 19 Nov 2004 17:59:01 -0600, Cecil Moore
wrote:
I suppose in your world, you really can do that.
21st Century math doesn't explain the orbit of pluto, but 28th century
math will....

Reg Edwards wrote:
So your argument, whatever it is, falls as flat as a pancake on Good Friday.

I am assuming lossy lines, where R G, as is typical of transmission
lines used at HF frequencies. So exactly where does my argument fall flat?

Actually, it seems that it is your argument that is falling flat as R is
rarely, if ever, equal to G at HF frequencies. Where would we ever obtain
such a terrible dielectric at HF that G would be equal to R? Maybe 9913
filled with water?
--
73, Cecil, W5DXP

Reg Edwards wrote:
Cec, it's Worship of the Great Smith Chart which has led you astray. My
warnings have been disregarded. ;o)

Reg, I can't conceive of a transmission line so terrible at
HF as to have R=G. Did you have 9913 filled with water in mind?
--
73, Cecil, W5DXP