On Mon, 25 Aug 2003 21:19:24 -0500, W5DXP
wrote:
Two waves are flowing with unchanging energies and producing an absolutely
black ring, i.e. zero irradiance equals zero power.

Hi Cecil,

Only when you violate conservation of energy, i.e. you have not summed
over the complete interval, only that part that satisfies your
argument (which allows anyone to perform similar miracles).

73's
Richard Clark, KB7QHC

Richard Clark wrote:
W5DXP wrote:
Two waves are flowing with unchanging energies and producing an absolutely
black ring, i.e. zero irradiance equals zero power.

Only when you violate conservation of energy, ...

No violation at all, Richard. The power (irradiance) that is lost
in the black rings is gained in the bright rings as explained in
_Optics_, by Hecht where he says constructive interference is
always equal to destructive interference.
--
73, Cecil http://www.qsl.net/w5dxp



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wrote:

Richard Harrison wrote:
They are contradictory. That`s a fatal flaw.

This may be, but there seems to be some reluctance to point out WHICH
step is contradictory.

The reluctance seems to be your unwillingness to read the replies
which point out exactly that.

I'll accept that, but no energy flows past the short or open where the
voltage or current is zero which is consistent with p(t) = v(t) * i(t).

No NET energy flows past such a point. Plenty of component energy flows
past the point in both directions as explained in Ramo, Whinnery, & Van
Duzer.

I do understand the derivation from this separation of the going and
coming signals. But I am unable to rationalize this explanation with
the very basic and widely accepted p(t) = v(t) * i(t) and the observed
zero voltages and currents.

You are talking about NET energy and NET power. Drop down one level to
component energy and power and all will become clear - especially if
you take a look at light wave interference patterns.

Do think of that voltage or current 0 at the shorted or open end of the
line. In your model, this voltage or current 0 produces a reflection
while other voltage or current zeroes do not. Is this consistent?

Yes, perfectly consistent. The short or open end of the line is a physical
impedance discontinuity. The other voltage and current zeros are NOT
physical impedance discontinuities. They are a result, not a cause of
anything. You are simply confusing cause and effect.

Consider adding another quarter wave section to the end of the line.
The voltage or current 0 remains and to the left of the original
end of the line nothing changes, but by your model, suddenly there
is no longer a reflection at the original end of the line, but it
has moved to the new end. But the voltage and current distributions
have remained the same. There is no observable difference.

Are you telling us that you cannot locate the physical short or open
at the end of the line and don't observe any difference between a physical
short or open and a piece of continuous transmission line? All you have
to do is open your eyes.
--
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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wrote:
W5DXP wrote:
Actually, not. You continue to resist pointing out which step is wrong.

You obviously have not read all of my replies.

Is it step 2)?
"2) At quarter wave points along this line, voltages and currents
which are always 0 can be observed -- the standing wave"

This seems to be generally accepted.

This is where your confusion starts. There are points where the
NET voltage and NET current are zero. Those are merely the points
where the forward Poynting vector and reflected Poynting vector
are out of phase.

Is it step 3)?
"3) Power is the rate at which energy flows"

This is the definition of power.

So please apply it to the forward Poynting vector and the reflected
Poynting vector as described in Ramo & Whinnery.

Is it step 4)?
"4) The power (rate of energy flowing) at time t can be computed
using p(t) = v(t) * i(t)"

This is the well know expression of power in terms of volts and
amps.

This is NET power. Your sin is not a sin of commission. It is a
sin of omission.

Is it step 5)?
5) Substituting a voltage or current which is always 0 into the
expression above will result in a power which is also always 0

Just normal subsitution of actual values into an equation.

Again, you are dealing only with NET power when you need to be dealing
with component energies.

Is it step 6)?
"6) From 2) and 5), the power (rate of energy flowing) at quarter
wave points will be 0"

The result obtained after substitution.

Again, your assertions apply ONLY to NET power which is not what
is being discussed.

Is it step 7)?
"7) From 6), the energy crossing quarter wave points is 0"

If steps 1) to 6) are not in error, then step 7) follows.

The NET energy is zero. The component energies are not zero.

Is it step 8)?
"8) From 7), energy can not be flowing down and up the line
crossing quarter wave points"

Simply follows from 7).

And completely false for component energies.

You have not yet pointed to any error in the derivation.

Yes, I have, numerous times. Let me give you an analogy. To prove
my point I say: "My pickup is white. Please prove otherwise." My
statement is absolutely true and absolutely irrelevant. So are
yours.

If your extensive study of optics leads you to believe that
the conclusion is incorrect, and, if the conclusion is incorrect,
there must be an error in the derivation. What is it?

Your error is one of omission. Your assertions are simply not
relevant to the discussion.

But the derivation is quite simple.

Yes, too simple and completely irrelevant.

Since you have not yet pointed to an error in the derivation
(which would be the obvious way to close the question), I
conclude that you have been unable to locate such an error.
Seems reasonable, does it not?

Since you have not proven that my pickup is not white, you lose
the argument. See, I can do the same thing you are attempting.
--
73, Cecil http://www.qsl.net/w5dxp



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On Tue, 26 Aug 2003 09:17:33 -0500, W5DXP
wrote:

Richard Clark wrote:
W5DXP wrote:
Two waves are flowing with unchanging energies and producing an absolutely
black ring, i.e. zero irradiance equals zero power.

Only when you violate conservation of energy, ...

No violation at all, Richard. The power (irradiance) that is lost
in the black rings is gained in the bright rings as explained in
_Optics_, by Hecht where he says constructive interference is
always equal to destructive interference.

Hi Cecil,

So you contradict yourself. Two waves do not produce a black ring,
they produce two rings (one dark, one bright) over an interval of one
wavelength (the proper bounds for an energy solution) and the sum of
them are the same as the applied powers - not zero. Choosing lesser
bounds to craft your "proof" constitutes an invalid proof.

73's
Richard Clark, KB7QHC

Richard Clark wrote:

W5DXP wrote:
No violation at all, Richard. The power (irradiance) that is lost
in the black rings is gained in the bright rings as explained in
_Optics_, by Hecht where he says constructive interference is
always equal to destructive interference.

So you contradict yourself. Two waves do not produce a black ring,
they produce two rings (one dark, one bright) ...

Don't know what planet you live on, Richard, but on this one if there
are two rings, one dark and one bright, there exists a dark ring. No
contradiction at all. Please sober up and try to do a better job next
time.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Clark wrote:

W5DXP wrote:
That's the flaw in your logic and this is about the 5th time I have
explained it to you.

And what makes this remarkable?

.... his ability to ignore the great body of scientific facts contained
in the field of physics from which I have previously quoted. Most of what
we are discussing has been known for about 300 years. If he (and you) choose
to ignore the known facts of physics, there is absolutely nothing I can do
about it. Understanding interference patterns is centuries old but my daughter
still cannot understand them. Probably only a small percentage of the human
population understand interference patterns - which is a pity.
--
73, Cecil http://www.qsl.net/w5dxp



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On Mon, 25 Aug 2003 21:54:51 -0400, wrote:

Consider again, Assertion A, in more detail this time:
1) Consider a shorted ideal transmission line excited by a single
frequency sinusoidal signal which has reached steady state
2) At quarter wave points along this line, voltages and currents
which are always 0 can be observed -- the standing wave
3) Power is the rate at which energy flows
4) The power (rate of energy flowing) at time t can be computed
using p(t) = v(t) * i(t)
5) Substituting a voltage or current which is always 0 into the
expression above will result in a power which is also always 0
6) From 2) and 5), the power (rate of energy flowing) at quarter
wave points will be 0
7) From 6), the energy crossing quarter wave points is 0
8) From 7), energy can not be flowing down and up the line
crossing quarter wave points

Assume for the moment that all this optical talk has convinced me
that conclusion 8) from Assertion A is false. For Assertion A
to be false, one of the steps in Assertion A must be false.
All the steps in Assertion A seem correct to me.
Can you help me find the false step which makes Assertion A false?

Hi Keith,

One obvious and glaring mistake that you share with everyone is found
in your step (2).

To observe these variables necessarily involves extracting energy and
consuming power. Hence the Assertion is rendered false in its own
confirmation.

So what now? Step back and retract (2)? Then where's your proof?
Offer that the "assumption" that replaces the reality of this step
renders the reality unnecessary?

These issues are then argument in faith between those who have none.
This is because other methods that do employ reality exist to offer
"proof" in equal measure, and yet no one seems prepared to go there,
or show any awareness of those paths.

You need only recite the virtues of the TR/ATR tube in RADAR systems
to offer Richard, KB5WZI, a model he can accept, and one that is
solidly based in reality. You could posit the complete solution to
the interference model so expounded upon by Cecil, except he seems to
have a patent on its application to his uniquely crafted solutions.

73's
Richard Clark, KB7QHC

On Tue, 26 Aug 2003 11:13:45 -0500, W5DXP
wrote:
Richard Clark wrote:

W5DXP wrote:
That's the flaw in your logic and this is about the 5th time I have
explained it to you.

And what makes this remarkable?

there is absolutely nothing I can do about it.

Hi Cecil,

And yet you gust on in indifference. Well, I am going to step inside
out of the elements and watch the leaves dance. ;-)

73's
Richard Clark, KB7QHC

On Tue, 26 Aug 2003 11:04:48 -0500, W5DXP
wrote:

Richard Clark wrote:

W5DXP wrote:
No violation at all, Richard. The power (irradiance) that is lost
in the black rings is gained in the bright rings as explained in
_Optics_, by Hecht where he says constructive interference is
always equal to destructive interference.

So you contradict yourself. Two waves do not produce a black ring,
they produce two rings (one dark, one bright) ...

Don't know what planet you live on, Richard, but on this one if there
are two rings, one dark and one bright, there exists a dark ring. No
contradiction at all. Please sober up and try to do a better job next
time.

Hi Cecil,

I live on the blue marble your planet is now approaching as close as
it has been in 70,000 years. The sobriety you speak of is a matter of
having more oxygen than you. As for jobs? Write when you find work,
buckaroo. We have a dead rover somewhere near you that could stand
fixing. ;-)

73's
Richard Clark, KB7QHC