Richard Harrison wrote:

Keith wrote:
"Yes, indeed. But there is no power."

Actually I wrote much more than that and the stuff which preceded was
much more important.

Keith`s perception is flawed. He is addressing r-f with a d-c mindset.
R-F power flow does not stop at an SWR zero-voltage point. Like "Old Man
River" it just keeps rolling along.

P(t) = V(t) * I(t) is much more general than DC or sinusoidal RF. It
works
for any signal shape you choose, even non-repetitive ones.

It certainly works for my electic company who integrate P(t) and
regularly send me a bill.

The zero-voltage point exists because the phase relationship between two
oppositely traveling waves is fixed, and the zero-volts point is where
the vectors cancel.

You must simultaneously sense both waves to find a zero. Sensing either
wave alone finds no dip in voltage. Carramba!

From this response it is clear that you are not yet in group b).

But is it:
group a) P(t) is not always equal to V(t) * I(t); or
group c) energy can flow when P(t) is a constant 0
?

Or maybe my list is incomplete and there is a 4th option which I missed.

May I respectfully suggest that if you are having difficulty selecting
which option applies to your thinking (and yet can't produce a fourth
option), that you try to discover the source of your discomfort. From
there, enlightenment may arise.

For myself, I recognized the "double think" and so rejected c).
P(t) = V(t) * I(t) seemed so fundamental that it had to be accepted
thus I was forced to reject a).
This left only b). Although it rejects some of the often stated and
accepted explanations about transmission lines, it was the lesser
of the evils. Some thinking revealed the weaknesses in the 'often
stated and accepted' explanations and now all is consistent.

This process took some time, prompted and assisted by the never
ending arguments which go on in this group. So I do thank those
mis-guided souls who argue endlessly and the patient answerers
who respond mostly for the benefit of the lurker.

But be that as it may, do try and figure out whether option a),
b) or c) best describes your thinking. Only an increased
understanding can arise.

....Keith

Richard Harrison wrote:

Keith wrote:
"Yes, indeed. But there is no power."

Keith`s perception is flawed. He is addressing r-f with a d-c mindset.
R-F power flow does not stop at an SWR zero-voltage point. Like "Old Man
River" it just keeps rolling along.

The zero-voltage point exists because the phase relationship between two
oppositely traveling waves is fixed, and the zero-volts point is where
the vectors cancel.

Using Keith's "logic", he would also be forced to assert that in a bright
ring-dark ring light interference pattern, the energy in the bright rings
is trapped between the dark rings and just sits there and circulates.
That's one reason why the field of optics is so far ahead of RF transmission
lines. Everyone in optics understands interference patterns which is exactly
what RF standing waves are.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
In reality there is not zero volts in the incident wave or in the
reflected wave. There`s full voltage coming and going. The volts just
happen to be out-of-phase at this point.

Yes, indeed. But there is no power.

Power is the same as irradiance in optics. When total V=0, it is simply
the result of destructive interference. Perhaps this quote from _Optics_,
by Hecht, will enlighten you. "The principle of conservation of energy
makes it clear that if there is constructive interference at one point,
the "extra" energy at that location must have come from elsewhere. There
must therefore be destructive interference somewhere else."

The voltage goes to zero because two voltage waves are engaged in destructive
interference. The current goes to maximum because two current waves are engaged
in constructive interference. The momentum in the voltage waves simply transfers
to the current waves and they just keep on rolling along. There is no mechanism
of physics existing at that point to change the momentum of the waves. Believing
that no energy crosses a superposed V=0 boundary is just a wet dream. The current
is at an absolute maximum point so plenty of charge carriers are crossing that
boundary.
--
73, Cecil http://www.qsl.net/w5dxp
"One thing I have learned in a long life: that all our science, measured
against reality, is primitive and childlike ..." Albert Einstein



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On Wed, 20 Aug 2003 11:36:57 -0500, W5DXP
wrote:

That's one reason why the field of optics is so far ahead of RF transmission
lines. Everyone in optics understands interference patterns which is exactly
what RF standing waves are.

Hi Cecil,

And you have yet to confirm it (or refute it) at the bench. Is your
interest more rhetorical than actual?

We engaged in a rather lengthy interchange to this topic, up to the
point of how you could evidence this for yourself. My methods and
data support your statement's sense. I offered only one proviso of
your rig being able to withstand the demands of such mismatch and you
offered your SGC could do that easily. As is stands left at that, the
remaining task should occupy no more than 10 Minutes of 16 readings
and note taking.

73's
Richard Clark, KB7QHC

W5DXP wrote:

wrote:
In reality there is not zero volts in the incident wave or in the
reflected wave. There`s full voltage coming and going. The volts just
happen to be out-of-phase at this point.

Yes, indeed. But there is no power.

Power is the same as irradiance in optics. When total V=0, it is simply
the result of destructive interference. Perhaps this quote from _Optics_,
by Hecht, will enlighten you. "The principle of conservation of energy
makes it clear that if there is constructive interference at one point,
the "extra" energy at that location must have come from elsewhere. There
must therefore be destructive interference somewhere else."

My knowledge of optics is insufficient to comment on any analogies you
choose to draw. Fortunately, a knowledge of optics is unnecessary to
understand circuits and transmission lines.

The voltage goes to zero because two voltage waves are engaged in destructive
interference. The current goes to maximum because two current waves are engaged
in constructive interference. The momentum in the voltage waves simply transfers
to the current waves and they just keep on rolling along. There is no mechanism
of physics existing at that point to change the momentum of the waves. Believing
that no energy crosses a superposed V=0 boundary is just a wet dream.

This puts you in
group a) P(t) is not always equal to V(t) * I(t); or
group c) "double think".

Care to think about which and comment?

The current
is at an absolute maximum point so plenty of charge carriers are crossing that
boundary.

Yes indeed, but current by itself is not energy. Remember
P(t) = V(t) * I(t) [unless you choose option a)]
Both volts and amps are simultaneously necessary for power.

....Keith

W5DXP wrote:
It happens all the time in optics and no optics engineer
would be silly enough to assert that the bright ring energy is trapped
and circulating between the dark rings.

Nor would he be silly enough to assert that energy first goes to the
dark ring and then turns around and goes to bright ring.

ac6xg

Richard Clark wrote:

W5DXP wrote:
That's one reason why the field of optics is so far ahead of RF transmission
lines. Everyone in optics understands interference patterns which is exactly
what RF standing waves are.

And you have yet to confirm it (or refute it) at the bench. Is your
interest more rhetorical than actual?

Actually, I confirmed it in my back yard some 15 years ago.
It wasn't rocket science.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:
My knowledge of optics is insufficient to comment on any analogies you
choose to draw. Fortunately, a knowledge of optics is unnecessary to
understand ... transmission lines.

Equally unfortunately, that's just a delusion of yours.

Care to think about which and comment?

I have no idea what you are talking about.

Yes indeed, but current by itself is not energy.

Hmmmmm, I^2*Z0 is not power? (Somebody get the net).
--
73, Cecil http://www.qsl.net/w5dxp



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On Wed, 20 Aug 2003 18:25:17 -0500, W5DXP
wrote:

Richard Clark wrote:

W5DXP wrote:
That's one reason why the field of optics is so far ahead of RF transmission
lines. Everyone in optics understands interference patterns which is exactly
what RF standing waves are.

And you have yet to confirm it (or refute it) at the bench. Is your
interest more rhetorical than actual?

Actually, I confirmed it in my back yard some 15 years ago.
It wasn't rocket science.

Hi Cecil,

I see you haven't got a clue what I wrote. Never mind.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
W5DXP wrote:

Richard Clark wrote:
And you have yet to confirm it (or refute it) at the bench. Is your
interest more rhetorical than actual?

Actually, I confirmed it in my back yard some 15 years ago.
It wasn't rocket science.

I see you haven't got a clue what I wrote. Never mind.

On the contrary, my bench was located in my back yard in Queen
Creek, AZ at the time - during the dry season, of course.
--
73, Cecil http://www.qsl.net/w5dxp



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