"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

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

Cec, I assume you know the simple formula for Zo from R,L,G,C.

Write it down. It will then be obvious, to make the angle of Zo equal to
zero it is necessary only that the angle of R+j*Omega*L be made equal to the
angle of G+j*Omega*C.

Or even more simple, for Zo to be purely resistive, G = C*R/L

If R and G exist, as they always do, then the line cannot be lossless.
----
Regards, Reg.

Richard Clark wrote:
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....

We have all the math we need to handle Pluto's orbit, and have had for a
century. We do not, however, know all the bits out there that may
perturb it. That said, we can still predict it quite well enough for
anyone's current needs.

tom
K0TAR

Reg Edwards wrote:
Cec, I assume you know the simple formula for Zo from R,L,G,C.

Write it down. It will then be obvious, to make the angle of Zo equal to
zero it is necessary only that the angle of R+j*Omega*L be made equal to the
angle of G+j*Omega*C.

Or even more simple, for Zo to be purely resistive, G = C*R/L

If R and G exist, as they always do, then the line cannot be lossless.

No argument, Reg. Nothing you said contradicts anything I said.
--
73, Cecil, W5DXP

On Fri, 19 Nov 2004 21:04:45 -0600, Tom Ring
wrote:
That said, we can still predict it quite well enough for
anyone's current needs.

Hi Tom,

Exactly, but you were responding to a dry comment on the nature of
"reality vs. models" which was absurd from the get-go.

To say that reality drives models or vice-versa (especially from
armchair theorists) is naive in the extreme. Anyone's "knowledge" of
reality is simply its own corrupt model and not any particular state
of enlightenment. The naive part is in not knowing the errors
bounding this "knowledge."

Barring revolution, anarchy, social upset or the rest, futurity will
tend generally to more resolution in the answer and a finer reduction
of error, but it will never be fully resolved because the answer could
not be encompassed by the mind.

73's
Richard Clark, KB7QHC

Cecil,

You nicely ducked the question, again.

This is not about philosophy, existentialism, the meaning of life,
planetary orbits, or any other such fluff.

Again, how do you know the "reality" for your transmission line in free
space?

This is a straightforward question. Do you have data? Do you have a
reference that you consider reality? Are you basing your reality on a
model? Is it just a matter of blind faith? Something else?

You have many times made similar remarks about us "stupid folks" who get
misled by math models. How do you get your information?


As for the E- and H-fields, this just gets more amusing by the minute.

73,
Gene
W4SZ

Cecil Moore wrote:
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

Richard Clark wrote:

On Fri, 19 Nov 2004 21:04:45 -0600, Tom Ring
wrote:

That said, we can still predict it quite well enough for
anyone's current needs.


Hi Tom,

Exactly, but you were responding to a dry comment on the nature of
"reality vs. models" which was absurd from the get-go.

To say that reality drives models or vice-versa (especially from
armchair theorists) is naive in the extreme. Anyone's "knowledge" of
reality is simply its own corrupt model and not any particular state
of enlightenment. The naive part is in not knowing the errors
bounding this "knowledge."

Barring revolution, anarchy, social upset or the rest, futurity will
tend generally to more resolution in the answer and a finer reduction
of error, but it will never be fully resolved because the answer could
not be encompassed by the mind.

73's
Richard Clark, KB7QHC

Including revolution, anarchy, social upset or the rest, the minds will
encompass everything to the finest resolution, because they will be the
ones we built to surpass us. Unless we destroy ourselves first, it is
inevitable.

tom
K0TAR

Reg Edwards wrote:
Write it down. It will then be obvious, to make the angle of Zo equal to
zero it is necessary only that the angle of R+j*Omega*L be made equal to the
angle of G+j*Omega*C.

Reg, there is no contradiction between what I have said and what you have said.

Continuing the discussion - For average transmission lines used on HF
frequencies, "... the value of G ... is likely to be too small to affect the
attenuation factor ..." Quote from "Transmission Lines" by Chipman, page 94.

Some of Chipman's calculations indicate that, for a typical 10 MHz example,
R is about 0.1 ohm/meter while G is about 0.9 micromhos/meter. That's about
a 100,000:1 ratio making G negligible as far as attenuation factor goes.

The attenuation factor depends almost entirely on R, the series resistance
parameter. G, the parallel conductance parameter, has a negligible effect
on the attenuation factor at HF.

Since, at HF, the attenuation factor consists almost entirely of series
resistance, and since the attenuation factor is identical for voltage and
current, it logically follows that the series resistance is primarily
responsible for the attenuation of the current.

Or even more simple, for Zo to be purely resistive, G = C*R/L

Actually, that is only an approximation for low-loss lines.
--
73, Cecil, W5DXP

In article , Cecil Moore
wrote:

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?

If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

David

--
David Ryeburn

To send e-mail, use "ca" instead of "caz".

Gene Fuller wrote:
Again, how do you know the "reality" for your transmission line in free
space?

Simple, we know the characteristics of copper wire and we know the
dielectric properties of a vacuum. Examples abound in textbooks.

This is a straightforward question. Do you have data?

Of course, don't you? If not, maybe you had better get some.

You have many times made similar remarks about us "stupid folks" who get
misled by math models. How do you get your information?

The distributed network model was developed because the circuit model
failed in systems that are an appreciable portion of a wavelength. I'm
surprised that you don't have that information.

As for the E- and H-fields, this just gets more amusing by the minute.

The attenuation factor for the average transmission line at HF is almost
entirely due to the resistance in the conductors as the shunt conductance
is usually negligible. Resistance in the conductors causes a voltage drop.
Yet, we know that the current is attenuated by exactly the same percentage
as the voltage. Since G is negligible, R must be responsible for the
decrease in current. What laws of physics can account for that fact?

Z0 determines the ratio of HF voltage to HF current. Therefore, if the
voltage drops and the V/I ratio stays constant, energy must be transferred
from the H-field to the E-field. That's pretty simple stuff, Gene, maybe
sophomore level.
--
73, Cecil, W5DXP

David Ryeburn wrote:
If R is measured in ohms and G is measured in siemans (or mhos) how can
you say one of them is much larger than the other?

Sorry, I should have said (effect of R) (effect of G)
on the losses. Or 0.1 ohms/meter in series has a greater
effect than 1/(0.9 micromhos/meter) in parallel. That's
about 7 magnitudes difference based on an example in
"Transmission Lines" by Chipman.
--
73, Cecil, W5DXP