Steve Nosko wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

Assuming yes... I think you are saying that we have a 450 ohm line and a 50
ohm load and therefore there is a reflection. Is that where you are?

Yes, there are reflections on the 450 ohm ladder line and none on the 50
ohm coax.

There
is no reflected energy on the 50 ohm coax and therefore no momentum in
the reflected waves on the 50 ohm coax.

I think you contradicted yourself here. Are you saying that there IS or
IS NOT reflected energy/waves on the 50 ohm section (to the left of the
"--x--"??

No contradiction. There's no reflected energy on the 50 ohm coax because
those two reflections are cancelled at the Z0-match point 'x'. They are
equal in magnitude and opposite in phase.

So what changes the direction
and momentum of the energy wave reflected from the load?

You lost me here. Perhaps you are asking; "If there is a wave
reflected at the end with the 50 ohm, back toward the left, how does this
power then get absorbed IN that load?" Is this the question?

No, there is 178 joules/sec rejected by the load on the 450
ohm line. That energy possesses direction and momentum toward the source.
What reverses the direction and momentum of that reflected energy?

The wave, in the 450 section, reflected from the load is NOT the power being
delivered TO the load, so it does not have to be "changed' to be delivered
there.

There is 100 joules/sec traveling to the right and none to the left on the 50 ohm
coax. There is 278 joules/sec traveling to the right and 178 joules/sec to the left
on the 450 ohm ladder-line. How does the rearward traveling 178 joules/sec, rejected
by the load, get turned around and join the forward traveling 100 source joules/sec
in order to add up to 278 joules/sec of forward power on the ladder line that is
incident upon the load?

It's a simple question: How does energy rejected by the load become energy incident
upon the load?
--
73, Cecil http://www.qsl.net/w5dxp



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Interesting question and actually BEFORE I read R. Clark's post I decided to
say that I will begg off at this point. I may be overcome with curiosity on
this question in the near future (like I did for the QST WATTMETER article
calculations), but for now I prefer not to take the challenge. Joules are
much more than I want to consider. Job, Wife, Mother, Mother-in-law,
sister-in-law all needing care and a job that is full of crap. I use this
for enjoyable discussions that I can contribute to and this a way more brain
power that I want to devote. I suspect you, Cecil, can explain it. I know
that a 1/2 wave repeats the load Z, smith Chart and all that and can work
with that.

you win. 73

--
Steve N, K,9;d, c. i My email has no u's.
"Cecil Moore" wrote in message
...
Steve Nosko wrote:
100W XMTR---50 ohm coax---x---1/2 WL 450 ohm ladder-line---50 ohm load

Assuming yes... I think you are saying that we have a 450 ohm line and a
50
ohm load and therefore there is a reflection. Is that where you are?

Yes, there are reflections on the 450 ohm ladder line and none on the 50
ohm coax.

There
is no reflected energy on the 50 ohm coax and therefore no momentum in
the reflected waves on the 50 ohm coax.

I think you contradicted yourself here. Are you saying that there
IS or
IS NOT reflected energy/waves on the 50 ohm section (to the left of the
"--x--"??

No contradiction. There's no reflected energy on the 50 ohm coax because
those two reflections are cancelled at the Z0-match point 'x'. They are
equal in magnitude and opposite in phase.

So what changes the direction
and momentum of the energy wave reflected from the load?

You lost me here. Perhaps you are asking; "If there is a wave
reflected at the end with the 50 ohm, back toward the left, how does
this
power then get absorbed IN that load?" Is this the question?

No, there is 178 joules/sec rejected by the load on the 450
ohm line. That energy possesses direction and momentum toward the source.
What reverses the direction and momentum of that reflected energy?

The wave, in the 450 section, reflected from the load is NOT the power
being
delivered TO the load, so it does not have to be "changed' to be
delivered
there.

There is 100 joules/sec traveling to the right and none to the left on the
50 ohm
coax. There is 278 joules/sec traveling to the right and 178 joules/sec to
the left
on the 450 ohm ladder-line. How does the rearward traveling 178
joules/sec, rejected
by the load, get turned around and join the forward traveling 100 source
joules/sec
in order to add up to 278 joules/sec of forward power on the ladder line
that is
incident upon the load?

It's a simple question: How does energy rejected by the load become energy
incident
upon the load?
--
73, Cecil http://www.qsl.net/w5dxp



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Steve Nosko wrote:
you win. 73

It's not a contest, Steve, it is a search for the truth. I thought
maybe you could offer something compelling. The following two
questions are very similar:

1. How does RF power (joules/sec) rejected by the load wind up being
incident upon that same load some time later in a Z0-matched system?

2. How does irradiance (joules/sec) rejected by a pane of glass wind
up being incident upon that same glass pane some time later in a non-
glare air/thin-film/glass system?

Question #2 was answered decades ago by optics physicists. Question #1
still remains unanswered by RF physicists and engineers even though it
has virtually the same answer as Question #2.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil,

With reference to question #2:

Who says such a silly thing? Not Melles-Griot, who appear to be your
favorite optical reference source. Not Born and Wolf in "Principles of
Optics", which is the ultimate optics reference book. Not any
professional optics experts I have ever encountered.

If you go ahead and solve the antireflective glass problem using
standard Maxwell's equations (sorry, Reg and Peter) with standard
boundary conditions for E and H fields you will find there is not the
slightest bit of confusion. This analysis is shown in many optics and
E&M textbooks.

In the perfectly antireflective case all of the waves keep moving in the
same direction, from air to thin film to glass. If one postulates the
existence of a wave in the reverse direction it will be discovered that
the amplitude of that wave is zero, meaning it does not exist. There is
no "bouncing back and forth" of confused energy not knowing where and
how to turn around.

The interference model is useful and intuitive for what it is meant to
explain. However, don't expect this sort of simple handwaving model to
be extendable to all sorts of silliness about energy and momentum transfer.

73,
Gene
W4SZ

Cecil Moore wrote:
Steve Nosko wrote:

you win. 73


It's not a contest, Steve, it is a search for the truth. I thought
maybe you could offer something compelling. The following two
questions are very similar:

1. How does RF power (joules/sec) rejected by the load wind up being
incident upon that same load some time later in a Z0-matched system?

2. How does irradiance (joules/sec) rejected by a pane of glass wind
up being incident upon that same glass pane some time later in a non-
glare air/thin-film/glass system?

Question #2 was answered decades ago by optics physicists. Question #1
still remains unanswered by RF physicists and engineers even though it
has virtually the same answer as Question #2.

On Wed, 10 Mar 2004 19:53:20 GMT, Gene Fuller
wrote:

However, don't expect this sort of simple handwaving model to
be extendable to all sorts of silliness about energy and momentum transfer.

It all goes to the measure of Standard handWaving Ridiculousness.

Gene Fuller wrote:
In the perfectly antireflective case all of the waves keep moving in the
same direction, from air to thin film to glass.

To prove that to be a true statement you must prove that the transistion
point between materials of different indices of refraction results in zero
reflections. Good luck on that one.

For instance, one can change the thin-film thickness from 1/4WL to 1/2WL
and cause exactly the opposite effect, i.e. extreme glare.

If you are using the quantum electrodynamics model, please let us know.
Most of the rest of us are using the EM wave reflection model.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil,

I haven't the foggiest idea what model you might be using. I am using
the classical model that is found in virtually any textbook that deals
with plane waves in non-conducting media.

I don't have to "prove" anything. Just set up the standard wave
equations with the standard boundary conditions and the problem
practically solves itself. The non-zero remaining waves are all moving
in the same direction. I forgot to ask them if they realize that Cecil
doesn't approve of such behavior.

I suppose this is an prime example of being seduced by "math models",
but I believe that is a lesser fault than being seduced by Cecil's
imaginary models.

73,
Gene
W4SZ

Cecil Moore wrote:

Gene Fuller wrote:

In the perfectly antireflective case all of the waves keep moving in
the same direction, from air to thin film to glass.


To prove that to be a true statement you must prove that the transistion
point between materials of different indices of refraction results in zero
reflections. Good luck on that one.

For instance, one can change the thin-film thickness from 1/4WL to 1/2WL
and cause exactly the opposite effect, i.e. extreme glare.

If you are using the quantum electrodynamics model, please let us know.
Most of the rest of us are using the EM wave reflection model.

Gene Fuller wrote:
I don't have to "prove" anything. Just set up the standard wave
equations with the standard boundary conditions and the problem
practically solves itself. The non-zero remaining waves are all moving
in the same direction. I forgot to ask them if they realize that Cecil
doesn't approve of such behavior.

You should have warned us that you were talking about NET waves and
NET energy transfer. I'm not discussing that at all. I am talking about
component waves and component energy transfer without which standing
waves cannot exist. Or maybe you can offer an example of standing waves
in the absence of at least two waves traveling in opposite directions.
If you can do that, I will admit defeat.

I suppose this is an prime example of being seduced by "math models",
but I believe that is a lesser fault than being seduced by Cecil's
imaginary models.

It is indeed an example of being seduced by the NET math model. Please
transfer over to the component math model and rejoin the discussion.
Lots of interesting things are happening below the threshold of the
NET math model. The NET math model doesn't explain anything except
the NET results. If your bank account balance doesn't change from one
month to another, do you also assume that you have written no checks
and have no income for that month? Literally speaking, please get real!
--
73, Cecil http://www.qsl.net/w5dxp



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Gene Fuller wrote:
In the perfectly antireflective case all of the waves keep moving in the
same direction, from air to thin film to glass.

I just realized what you are saying. Your above statement is wrong about
"all of the waves". Your above statement is correct about "net irradiance".
I'm not talking about "net" anything. I am talking about the component
forward and reflected waves which are easily proven to exist.
--
73, Cecil http://www.qsl.net/w5dxp



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