Richard Clark wrote:
We are dealing with component energies. The NET
energy is zero. The component energies are not zero. This would seem
to conflict with much of your wave mechanics for the last 6 months.

If you think that, it's obvious you didn't understand what I have been
saying for the last 6 months, which is what I had assumed. The wave
reflection model is perfectly consistent.

Why do you bring up bizillion volts? What is the conflict you see?
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:

I think the fundamental problem here is that the concept of waves of
average power necessarily assumes nice, smooth, non-varying "waves" of
DC (average) power flowing along the line. When two of these "waves"
occupy the same space, they necessarily combine to form another smooth
combination net "wave". There's no way to account for the reality that
power and energy are *not* flowing in smooth, DC waves, but rather as a
time-varying function. Power is flowing back and forth each cycle, and
energy is moving into and out of storage each cycle. The time-varying
information is thrown away as soon as the average is taken, and it can't
be retrieved.

Who needs it? Of what use is it? Don't phasors contain all the important
information? What problem are you trying to solve by resorting to
instantaneous energy and/or power? You will never understand
instantaneous energy until you understand virtual photons. That's a
reasonable statement based on the latest quantum electrodynamics. Do
you understand virtual photons? Are you going to try to understand
virtual photons in order to really understand instantaneous energy?
If not, why not? Methinks you are the pot calling the kettle black.

So if all you can envision is waves of average power, it's
impossible to understand what's really happening with respect to power
and energy along the line. You simply don't have the information
necessary to envision or understand it.

The optics physicists have done just fine understanding everything while
dealing with irradiance (average power). Instantaneous power is simply
not a very useful concept except in some esoteric applications.
--
73, Cecil http://www.qsl.net/w5dxp



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With all due respect, Cecil, I believe you've gone a bit 'round the bend.

Roy Lewallen, W7EL

W5DXP wrote:

Who needs it? Of what use is it? Don't phasors contain all the important
information? What problem are you trying to solve by resorting to
instantaneous energy and/or power? You will never understand
instantaneous energy until you understand virtual photons. . .

On Wed, 27 Aug 2003 19:49:14 -0500, W5DXP
wrote:

The optics physicists have done just fine understanding everything while
dealing with irradiance (average power). Instantaneous power is simply
not a very useful concept except in some esoteric applications.

Hi Cecil,

To date you haven't fielded many practical problems of "optics" as you
describe it. And the admission above is a poor excuse in place of
simple applications, never mind the esoteric ones you stumble over in
the dark bands.

A simple example which might offer you 50% chance of being right using
simple irradiance (average power, so average that their steady
illumination doesn't change during the entire time you ponder):

Which is brighter?
1.) 629 lux
2.) 5.0 millicandela/cm²

Which is visible in sunlight? (mark with an "x")
1.) _
2.) _
Sunlight being defined as the irradiance observed beneath the
equatorial Sun on the equinox at local noon on a cloudless day.

"If" you got any multiple guess right, a tougher question that
involves more skill and relates to the source power for each:

What is the source total intensity for each value above with either
source at 10 centimeters remote? (either source being isotropic)
1.) _____
2.) _____
expressed in Lumens total flux (or make it simple on yourself, Watts
total radiation)?

73's
Richard Clark, KB7QHC

Roy Lewallen wrote:
With all due respect, Cecil, I believe you've gone a bit 'round the bend.

Can't answer the questions, huh? :-) I don't see the usefulness of
instantaneous power. You don't see the usefulness of virtual photons.
If I have a problem, you have a similar problem. FYI, if fields don't
really exist, as they don't in particle physics, then virtual photons
are necessary to accomplish the function of the fields.

Who needs it? Of what use is it? Don't phasors contain all the important
information? What problem are you trying to solve by resorting to
instantaneous energy and/or power? You will never understand
instantaneous energy until you understand virtual photons. . .
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Clark wrote:
It's obvious you didn't understand what you have been
saying for the last 6 months.

Couldn't possibly be a lack of understanding on your part, eh?
Which is more likely? That I don't understand what I have been
saying or that you don't understand what I have been saying?
Exactly what is it that you think I have said that is wrong?

I have always said, and I will continue to say, if you understood
what I was saying, you would agree with me. You have said nothing
that has contradicted anything I have said.
--
73, Cecil http://www.qsl.net/w5dxp



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Richard Clark wrote:
Which is brighter?
Which is visible in sunlight? (mark with an "x")
What is the source total intensity for each value above with either
source at 10 centimeters remote? (either source being isotropic)
expressed in Lumens total flux (or make it simple on yourself, Watts
total radiation)?

Cute, but why is the above relevant?
--
73, Cecil http://www.qsl.net/w5dxp



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W5DXP wrote:
Roy Lewallen wrote:

With all due respect, Cecil, I believe you've gone a bit 'round the bend.


Can't answer the questions, huh? :-)

No, actually, I've answered it again and again and again. But it makes
no difference -- you continue asking.

Roy Lewallen, W7EL

On Wed, 27 Aug 2003 22:22:34 -0500, W5DXP
wrote:

Richard Clark wrote:
It's obvious you didn't understand what you have been
saying for the last 6 months.

Couldn't possibly be a lack of understanding on your part, eh?
Which is more likely? That I don't understand what I have been
saying or that you don't understand what I have been saying?
Exactly what is it that you think I have said that is wrong?

I have always said, and I will continue to say, if you understood
what I was saying, you would agree with me. You have said nothing
that has contradicted anything I have said.

Hi Cecil,

I didn't post a thing that you didn't already say. :-)

That it confuses you is par for the course, even with the prospect of
a 50% discount. Frankly, I couldn't imagine how you could possibly
continue to disagree - with yourself.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
You answered the question perfectly, you see no relevance with
standard optics to your "optics."

You are still confused, Richard. "My optics" is standard optics.
--
73, Cecil http://www.qsl.net/w5dxp



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