"Ian White said -

Looking edge-on at the line, we have two conductors carrying equal
and
opposite currents, but one is slightly farther away than the other
so
their transverse radiated fields do not quite cancel out.

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

Ian, Oh yes they do.

Next to each half wavelength of line there is another half wavelength
of line in which the current is in antiphase with it. And so, in the
far field, the fields from adjacent half-wavelengths of line cancel
each other out.

Now you'll say my logic falls down when the line length is an odd
number of half wavelengths. But you must not consider half
wavelengths of line to be behaving independently of each other.
----
Reg, G4FGQ

Reg Edwards wrote:

Ian, Oh yes they do.

Next to each half wavelength of line there is another half wavelength
of line in which the current is in antiphase with it. And so, in the
far field, the fields from adjacent half-wavelengths of line cancel
each other out.
. . .

No, they don't. They cancel only in two directions, directly normal to
the plane containing the wires. Radiation occurs in all other
directions, because the fields don't add in antiphase. An example of an
antenna which uses two closely spaced elements carrying equal
out-of-phase currents is the W8JK.

Roy Lewallen, W7EL

"Roy Lewallen" bravely wrote to "All" (08 Apr 05 15:05:11)
--- on the heady topic of " VF, low-loss line, high-impedence line -
relationship"

RL From: Roy Lewallen
RL Xref: aeinews rec.radio.amateur.antenna:28146

RL Asimov wrote:
Is there any electron current in the conductor or not?

RL In a perfect conductor, no. In a real but good conductor like copper,
RL it's confined to a very thin layer at the surface. Look up "skin
RL effect" in any electromagnetics or basic text on RF, or google it.

If I put an RF voltmeter/ampmeter at any point along the perfect
conductor I will get a zero reading is that right?

So then if the conductor was perfect then I wouldn't need a conductor
at all in the first place? Seems a shame to waste perfect unobtainium
just to have the last word. ;-)


RL First of all, a mismatch doesn't cause loss.

An impedance mismatch in any medium causes a scattering of the energy.

RL If I connect a 50 ohm source to a one wavelength, 300 ohm transmission
RL line and connect the other end of that line to a 50 ohm resistor,
RL there's a 6:1 mismatch at both ends of the line.
[,,,]
RL No loss is caused by the mismatch. No "scattering of energy occurs".
RL All of the energy from the source arrives at the load, where it was
RL intended.

So why bother with high priced cable when a piece of clothes hanger
will do just as well, encantations? I think your are still in April 1
mode, Mr.Roy. We are way past diatribe here, thank you.

A*s*i*m*o*v

.... Pandora's Law: Never open a box you didn't close yourself

I hope that most readers will see from the examples I've given that the
assertions you've made are false. And perhaps some will be spurred to do
a little reading and studying in order to better understand the topic. I
don't think anything else I'm likely to say will change the minds of
those who aren't interested in either looking at the evidence or
learning more about the subject, so my time is better spent at other
pursuits. I'll leave you with your notions undisturbed and intact.

Roy Lewallen, W7EL

Asimov wrote:
. . .
So why bother with high priced cable when a piece of clothes hanger
will do just as well, encantations? I think your are still in April 1
mode, Mr.Roy. We are way past diatribe here, thank you.

A*s*i*m*o*v

... Pandora's Law: Never open a box you didn't close yourself

On Friday, 08 Apr 2005 22:59:00 -500, In a bit of not so clever
editing "Asimov"
wrote in a response to a perfectly clear post by Roy:


RL If I connect a 50 ohm source to a one wavelength, 300 ohm transmission
RL line and connect the other end of that line to a 50 ohm resistor,
RL there's a 6:1 mismatch at both ends of the line.
[,,,]
RL No loss is caused by the mismatch. No "scattering of energy occurs".
RL All of the energy from the source arrives at the load, where it was
RL intended.

So why bother with high priced cable when a piece of clothes hanger
will do just as well, encantations? I think your are still in April 1
mode, Mr.Roy. We are way past diatribe here, thank you.

What Roy actually said was:

"If I connect a 50 ohm source to a one wavelength, 300 ohm
transmission line and connect the other end of that line to a 50 ohm
resistor, there's a 6:1 mismatch at both ends of the line. The power
supplied by the source and the power delivered to the load are exactly
the same as if I had used a 50 ohm line instead. This is, of course,
overlooking resitive loss in the line. If you consider the resistive
loss, it can be greater in one line than the other (the 300 ohm line
might be less lossy), depending on the physical construction of the
line."

So you see, our shyster poster left out the part where Roy considers
that resistive loss plays a part, and by inference makes copper wire a
better choice than clothes hangers.

On Sat, 09 Apr 2005 01:56:47 -0700, Roy Lewallen
wrote:

I hope that most readers will see from the examples I've given that the
assertions you've made are false. And perhaps some will be spurred to do
a little reading and studying in order to better understand the topic. I
don't think anything else I'm likely to say will change the minds of
those who aren't interested in either looking at the evidence or
learning more about the subject, so my time is better spent at other
pursuits. I'll leave you with your notions undisturbed and intact.


And it is to be hoped, expressed in another forum.

"Roy Lewallen" wrote
Reg Edwards wrote:

Ian, Oh yes they do.

Next to each half wavelength of line there is another half
wavelength
of line in which the current is in antiphase with it. And so, in
the
far field, the fields from adjacent half-wavelengths of line
cancel
each other out.
. . .

No, they don't. They cancel only in two directions, directly normal
to
the plane containing the wires. Radiation occurs in all other
directions, because the fields don't add in antiphase. An example of
an
antenna which uses two closely spaced elements carrying equal
out-of-phase currents is the W8JK.

Roy Lewallen, W7EL

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

Roy, I've never head of a W8JK. You are confusing the issue.

The problem is concerned with a LONG balanced transmission line and
its terminations which form part of the whole radiating system. And as
we can agree it is incorrect to consider parts of the system in
isolation.

To simplify the questions, wthout loss of rigor, it is best to
consider the line itself as being lossless with matched terminations.

I have stated that power radiated from the system is independent of
line length and nobody has disagreed. Indeed, a radiating power
calculating formula from reputable authors (of which I was unaware)
has confirmed this.

The power radiated from the system is identical to that radiated from
a monopole or short dipole, of length equal to wire spacing, with a
current equal to the current which flows in the terminations (ie., the
load). The terminations actually exist.

Radiated power = Load current-squared times calculated radiation
resistance.

That is obviously true down even to zero line length.

The implication is that radiation occurs only from the termination(s)
and that no radiation occus from the line. But, I repeat, we must NOT
consider the parts in isolation as do old wives.

You have stated that radiation from the line itself (in isolation)
must exist in the plane of the wires because of the finite spacing
between the line wires.

But we must consider only the far field. Not that in the immediate
vicinity of the line and its termination.

I suspect that the radiation pattern of a LONG-line SYSTEM converges
towards that from a monopole located in the position of the load.

Many of us are curious to acquire an idea of what the radiation
pattern looks like.

You are familiar with programs which produce far-field radiation
patterns. Do you know of a program which accurately produces the
radiation pattern of a very long close-spaced, zero resistance, pair
of wires terminated with a wire of length equal to wire spacing and
including a load resistance equal to Zo.

Patterns, of course, will change with frequency. It will be necessary
to statistically analyse results. Or just look at them from a common
sense point of view.

From a practical engineering viewpoint it is quite sufficient to know
what I innocently stated in the first place - the minute amount of
power lost is the load's radiation resistance times load current
squared and is independent of line length.
----
Reg, G4FGQ

Reg Edwards wrote:

"Roy Lewallen" wrote
Reg Edwards wrote:

Ian, Oh yes they do.

Next to each half wavelength of line there is another half
wavelength
of line in which the current is in antiphase with it. And so, in
the
far field, the fields from adjacent half-wavelengths of line
cancel
each other out.
. . .

No, they don't. They cancel only in two directions, directly normal
to
the plane containing the wires. Radiation occurs in all other
directions, because the fields don't add in antiphase. An example of
an
antenna which uses two closely spaced elements carrying equal
out-of-phase currents is the W8JK.

Roy Lewallen, W7EL

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

Roy, I've never head of a W8JK. You are confusing the issue.

The problem is concerned with a LONG balanced transmission line and its
terminations which form part of the whole radiating system. And as we
can agree it is incorrect to consider parts of the system in isolation.

To simplify the questions, wthout loss of rigor, it is best to consider
the line itself as being lossless with matched terminations.

I have stated that power radiated from the system is independent of
line length and nobody has disagreed.

Here on the back row, there's always been a hand raised in disagreement
on that point.

No argument that it's very, very small. But exactly zero - definitely
not.

Indeed, a radiating power calculating formula from reputable authors
(of which I was unaware) has confirmed this.

If you mean the Sterba reference, then please re-read it. All the
endorsements of the formula that you quote are peppered with caveats
such as "an approximation" and "providing that operations are confined
to wavelengths other than those within the ultra-short-wave region."

This is for the very good reason that some small amount of transverse
radiation does exist. Transverse radiation in the plane of the line is
small because the vector components of radiation from the two parallel
lines are equal in magnitude and almost exactly opposite in phase - but
never exactly opposite.

I am probably the only person in this discussion who has actually USED
parallel-wire lines "within the ultra-short-wave region". If you can
maintain good balance, the losses due to transverse radiation are
negligibly small for engineering purposes.

But to claim they are exactly zero is a physical absurdity... and I'll
always disagree with those.

(Sorry, I'll have to be out of this discussion again for a while.)

--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek