What I am still not understanding, is . . . .


To clear away all misconceptions and confusion, try returning to square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V

----
Reg

Cecil,

If you prefer your voltages to flow and your currents to drop, have at
it. :-)

Here's a hint.

In business, politics, crime (sorry for the redundancy), and a bunch of
other stuff there is an old saw, "Follow the money."

In electronics the appropriate dictum is, "Follow the electrons."

There is no law regarding conservation of voltage. There is a
fundamental law about conservation of change, and an equally strong law
dealing with continuity of current.

Go ahead and label a change in current at two points on a wire a "drop"
if you like, but don't confuse this change with a drop in voltage. They
ain't the same thing.

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

I am quite familiar with standing waves, thank you. I have no
disagreements with Terman, Kraus, Balanis, or any other legitimate
experts.


What I am still not understanding, is since the exponential equations
for voltage and current in a transmission line are identical except
for the Z0 term, how can something happen to the current without
the same thing happening to the voltage at the same time? How can
something happen to the voltage without also happening to the current
at the same time? In a matched system, the voltage and current arrives
at the load at exactly the same time attenuated by exactly the same
amount. But that voltage didn't flow and that current didn't drop???
--
73, Cecil, W5DXP

Gene Fuller wrote:
Go ahead and label a change in current at two points on a wire a "drop"
if you like, but don't confuse this change with a drop in voltage. They
ain't the same thing.

In the matched exponential transmission line equations, the
attenuation factors for the voltage and current are identical,
i.e. they decrease by exactly the same percentage.
--
73, Cecil, W5DXP

"Frank" wrote in message
news:t4okd.90482$VA5.33610@clgrps13...

"Frank" wrote in message
news:H4hkd.141267$9b.112169@edtnps84...
Modeled #14 AWG, copper conductor, 32ft monopole, 29 radials of 25ft, and
base 6" above (nominal lambda/1000) Sommerfeld/Norton ground of Er = 13,
sigma = 0.013 S/m at 1.8 MHz. All segments 6".

NEC2 computes:
Zin = 2.87 - j1358 Efficiency 92%

RADIALS2 computes (with radials 1mm below ground):
Zin = 1.55 - j1310 Efficiency 23.5%

Not a large amount of difference, but thought I had gotten closer results
with a different monopole, but seem to have deleted the code (Not sure
why such a large difference in efficiency). NEC2 is supposed to provide
a reasonable approximation of a buried radial monopole when at about
lambda/1000 above ground. Be interested in any comments, and what NEC4
provides if anybody has it.

73,

Of course the higher efficiency is due to NEC calculating only the I^2R
losses, and not the TRP. TRP should be fairly easy to calculate since the
pattern is "phi" independent. Have not checked to see if there is a TRP
card.
Note that a 32 ft monopole mounted on a perfect ground has an input
impedance of 1.58 - j1311 Ohms. The efficiency is reduced to 86% due to
increased I^2R losses.

Frank

From the calculated field strength (as a function of Theta) the TRP for 100
W input, which includes copper and ground losses, shows 27.4 W, or 27.4%
efficient. In very close agreement with the RADIALS2 program. The only
noticeable discrepancy appears to be in the real part of Zin.

Frank

Frank, as you say, the height of the radials in the NEC2 model is only 6
inches above ground.

The radials are shallow-buried in the RADIALS2 model. It can't do elevated
radials.

The ground loss resistance as height decreases, as seen by the antenna,
increases very fast percentage-wise as the radials get within a few inches
of the ground. It is due to very close magnetic and electric coupling to
ground. Radials are transmission lines, insulated from but running very
close to a resistive slab of soil.

This would account for the computed higher input resistance of the radials
( 3.5 - j*3.3 ohms ) ( for 29, 25-feet long radials. Rg=77, K=13 ) in
program RADIALS2.

The calculated antenna input impedance in RADIALS2 is that of the antenna
alone. For feedpoint impedance add the input impedance of the radial system.
Presumably, NEC2 does not compute the input reactance of radials.

Efficiency is calculated in the usual way from the sum of antenna input
resistance and radials' input resistance.

If you contrive to change the radials input reactance without changing
frequency or the antenna, you will notice the loading coil tunes it out
along with antenna reactance.

Incidentally, when elevated radials are near the ground their velocity
factor decreases fast which makes a mess of the usual recommendation to
prune them to 1/4-wave free-space length.

When radials are actually lying on the ground surface the velocity factor
decreases to roughly 0.5 of the velocity of light. When buried the
underground VF can fall to as small as 0.15 depending on soil permittivity.
(or moisture content.)
----
Reg, G4FGQ

Reg, thanks for the info. I see I was making an error with RADIALS2. Did
not realize that you had to add the radial impedance to the antenna
impedance; I thought it was computed in the final result. I figured
something was weird since the input impedance was similar to the antenna
modeled over a perfect ground. NEC2 does compute the input impedance of the
complete structure, but as mentioned before it is limited, in that all wires
must be = lambda/1000 above the ground (at 1.8 MHz about 6"). I am just
starting to delve into computational electromagnetics, so do not know that
much about NEC. It uses the "Method of moments" in its computations. The
theory of operation manual is available for download at several web sites.

Anyway, with RADIALS2, I now get an input impedance of 5.1 - j1303, and with
NEC2 2.9 - j1358. I did try entering a negative number for the depth of
the radials, but RADIALS2 did not like it.

Your comments about the effect on the radials of being buried are also very
interesting, and obviously indicate the reason for our slightly different
results. The very low VF of buried radials indicates that the length is
less important.

As for efficiency, NEC2 computes a normalized far E-field at 1 meter. For
phi independent structures it becomes trivial to integrate the power density
over a hemispherical region to arrive at the true total radiated power. 100
Watts into the antenna radiates 27 Watts, again, very close to RADIALS2'
computed efficiency of 23.5%. Come to think of it, I guess I could have
estimated the losses -- as you do -- by comparing the input impedance of an
antenna over a perfect ground with the same antenna over a lossy ground.
Still I think it was more fun playing with Excel spread sheets and coming up
with a similar answer.

Regard,

Frank

"Reg Edwards" wrote in message
...
Frank, as you say, the height of the radials in the NEC2 model is only 6
inches above ground.

The radials are shallow-buried in the RADIALS2 model. It can't do elevated
radials.

The ground loss resistance as height decreases, as seen by the antenna,
increases very fast percentage-wise as the radials get within a few inches
of the ground. It is due to very close magnetic and electric coupling to
ground. Radials are transmission lines, insulated from but running very
close to a resistive slab of soil.

This would account for the computed higher input resistance of the radials
( 3.5 - j*3.3 ohms ) ( for 29, 25-feet long radials. Rg=77, K=13 ) in
program RADIALS2.

The calculated antenna input impedance in RADIALS2 is that of the antenna
alone. For feedpoint impedance add the input impedance of the radial
system.
Presumably, NEC2 does not compute the input reactance of radials.

Efficiency is calculated in the usual way from the sum of antenna input
resistance and radials' input resistance.

If you contrive to change the radials input reactance without changing
frequency or the antenna, you will notice the loading coil tunes it out
along with antenna reactance.

Incidentally, when elevated radials are near the ground their velocity
factor decreases fast which makes a mess of the usual recommendation to
prune them to 1/4-wave free-space length.

When radials are actually lying on the ground surface the velocity factor
decreases to roughly 0.5 of the velocity of light. When buried the
underground VF can fall to as small as 0.15 depending on soil
permittivity.
(or moisture content.)
----
Reg, G4FGQ

Reg Edwards wrote:
To clear away all misconceptions and confusion, try returning to square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V

Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?
--
73, Cecil, W5DXP

Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V


Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?

Depends on whether 'a' is in series or in shunt.

73, ac6xg

Jim Kelley wrote:

Cecil Moore wrote:
So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?

Depends on whether 'a' is in series or in shunt.

It would be too much of a coincidence for 'a', the attenuation
factor, to be the same whether in series or in shunt. Most of
the attenuation at HF is due to I^2*R losses, a series event.

Transmission lines are distributed networks involved with EM wave
energy transmission. I^2*R losses can cause a decrease in current
just as it can cause a decrease in voltage. The sequence of events
is obvious.

1. The RF voltage drops because of I^2*R losses.
2. The proportional E-field decreases because of the voltage drop.
3. Since the E-field to H-field ratio is fixed by Z0, the H-field
decreases as does the ExH power in the wave.
4. Since the RF current is proportional to the H-field, the current
decreases by the same percentage as the voltage.

The chain of cause and effect is obvious. The current decreases
because of I^2*R losses in the transmission line which is a
distributed network, not a circuit.
--
73, Cecil, W5DXP

Jim Kelley wrote:



Cecil Moore wrote:

Reg Edwards wrote:

To clear away all misconceptions and confusion, try returning to
square one
and begin again with a clean slate.

dV/dz = -(R+j*Omega*L)*I

dI/dz = -(G+j*Omega*C)*V



Those are essentially the differential equations used by Ramo and
Whinnery to develop the following exponential equations for load
matched transmission lines (no reflections):

V = V+(e^-az)(e^-jbz)

I = V+(e^-az)(e^-jbz)/Z0

where 'a' is the attenuation factor. So why does an attenuated
voltage drop while a current, attenuated by exactly the same
percentage, doesn't drop?


Depends on whether 'a' is in series or in shunt.

73, ac6xg

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.

73, Jim AC6XG

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

"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

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

In "Transmission Lines" by Chipman, he gives an example where
R = 0.1 ohm/m and G = 0.9 micromhos/m. For Z0 to be a purely
resistive 50 ohms, G would have to be 40 micromhos/m making
the transmission line considerably more lossy just to achieve
a purely resistive Z0. Real world transmission lines rarely
have a purely resistive characteristic impedance.

The formula for the attenuation factor is R/2*Z0 + G*Z0/2
That's 0.001 + 0.0000225, so you can see that G has negligible
effect on losses, i.e. virtually all losses in the above
example are series I^2*R losses.

The attenuation factor is 0.0010225 for both the voltage and
current so it's obvious that the current attenuation is caused
by the series I^2*R losses, the same thing that causes the
voltage attenuation.
--
73, Cecil, W5DXP

David Ryeburn wrote:

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?

Sorry, should have said, "... where R/Z0 G*Z0, as is typical of
transmission lines used at HF frequencies."
--
73, Cecil, W5DXP

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?

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

Its easy for Cecil. He doesn't have the slightest trouble.

On Fri, 19 Nov 2004 23:48:22 -0600, Cecil Moore
wrote:
I am assuming lossy lines, where R G, as is typical of transmission
how can you say one of them is much larger than the other?
Sorry, should have said, "... where R/Z0 G*Z0
just a matter of Z0² ... no big deal
perhaps it shoulda been R· Z0² G ;-)

Reg Edwards 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?

Its easy for Cecil. He doesn't have the slightest trouble.

Just forgot to render them both dimensionless with the Z0
term. I suspect you knew what I meant anyway. :-)
--
73, Cecil, W5DXP

Gene Fuller wrote:
As for the E- and H-fields, this just gets more amusing by the minute.
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?

Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.

Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP

Cecil,

Nice try.

I cannot find the word "migrate" in your reference. However, I do find
several examples of your own interpretations mixed in.

When in doubt, change the subject?

More rattlesnake physics?

73,
Gene
W4SZ

Cecil Moore wrote:
Gene Fuller wrote:

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

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?


Here's the reference: From "Fields and Waves" by Ramo and Whinnery.
Take a close look at the exponential transmission line equations
for flat lines (no reflections):

V = Vmax(e^-az)(e^wt-bz) (1)

I = Vmax(e^-az)(e^wt-bz)/Z0 (2)

'a' (alpha) is the attenuation factor. The two equations are identical
except for the Z0 term. If you divide equation (1) by equation (2),
you get Z0. In a flat transmission line (no reflections) the current
is ALWAYS equal to the voltage divided by the characteristic impedance
of the transmission line. The voltage and current are attenuated by
EXACTLY the same factor. If the voltage drops because of I^2*R losses,
the current must decrease by exactly the same percentage. (I have avoided
calling it a current drop so it wouldn't upset you.)

Since the attenuation factor is R/2*Z0 + G*Z0/2 and since, for most
transmission lines used on HF, R/2*Z0 G*Z0/2, the current attenuation
is caused by the series I^2*R drop in the voltage and the V/I=Z0 ratio
that must be maintained - pretty simple logic.

I must have missed that day in class.


Yep, you must have. But it's not too late to learn what you missed
that day. :-)
--
73, Cecil, W5DXP

Gene Fuller wrote:
I cannot find the word "migrate" in your reference. However, I do find
several examples of your own interpretations mixed in.

You expect me to remember the exact wording after 40 years?
Perhaps your prof, like mine, told you to skip sections 1.22-
1.28 in Ramo and Whinnery. I didn't take his advice - I read
them.

So Gene, please point out the error in my logic. If the
current attenuation is primarily from R, the series
resistance, what other explanation can there be than
energy is being supplied by the H-field to the sagging
E-field? Can you think of any other rational, logical,
non-emotional, technical explanation, given these
exponential transmission line equations?

V = Vmax(e^-az)*e^j(wt-bz)

I = Vmax(e^-az)*e^j(wt-bz)/Z0

V/I = Z0

(I think you are just trying to punish me for saying my
dog has a soul. :-)
--
73, Cecil, W5DXP

Cec, I was well aware of what Chipman was about to write years before he
wrote his most excellent, most reliable book on the subject of transmission
lines. There are are very few errors. All of those which I have found can be
attributed to the printer. But you don't find these unless you actually use
the book, fully understand the book, and do some practical sums.

There are far too many people who use books as bibles because they have
found them verbally convincing but who have never actually used them by
inserting practical engineering numbers. No-one unfamiliar with numbers can
call himself an engineer. More likely he is a plagiarising Old Wife.

I do not worship Chipman. I think he is still plodding around in his 90's. I
don't worship anybody.

Whilst on the subject of reliability, I have very recently had a serious
accident.

My corkscrew broke.

It can only have been due to metal fatigue and all the use its had.
Probably made in Taiwan or Korea.

For 5 days, for one reason or another, I have been confined to the house
without the opportunity to replace corkscrews. Yet there have been 4
unopened bottles of perfectly good wine in the cooler, with clean glasses.
Calamity! Absolute misery!

But my long mechanical engineering experience came to the rescue. I
discovered a 1/4" Philips screwdriver and a hammer. With lots of hammering I
eventually drove the cork (of a bottle of Premieres Cotes de Bordeaux,
sweet-white), inside the bottle.

The cork floats on the top of the wine and there's few problems with
pouring. The cork remains quite intact. No contaminating bits to spit out.

So everything is now back to normal. I shortly expect to obtain a new
corkscrew - this time with a spare.

Hic!
----
Reg.



"Cecil Moore" wrote in message
...
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

Reg Edwards wrote:
There are far too many people who use books as bibles ...

Ain't that the truth? I post "1+1=2" and somebody wants a reference.
I just bought "Mathematics From the Birth of Numbers" by Jan Gullberg.
In addition to information on virtually all branches of mathematics,
it gives the history of the branches. It's really interesting.

With lots of hammering I
eventually drove the cork (of a bottle of Premieres Cotes de Bordeaux,
sweet-white), inside the bottle.

At the beer/wine busts at Texas A&M during the 50's, nobody
could ever remember to bring a corkscrew so that's the way
we did it. Sometimes we forgot a bottle opener and used the
bumper of my old '49 Chevvy to open the beer.
--
73, Cecil, http://www.qsl.net/w5dxp