Roy Lewallen wrote:
But Cecil (and, I'm afraid,
others) also see waves of average power and sometimes energy, which seem
to follow different rules.

Power and energy are scalars, Roy. Of course, scalars follow
different rules. Maybe the problem is that you are trying
to use phase math on power.
--
73, Cecil http://www.w5dxp.com

Dr. Honeydew wrote:
Cecil Moore wrote:
A Bird wattmeter reads 100 watts forward and 100w reflected. The
current in the source is zero. The source is not only not sourcing any
forward power, it is also not sinking any reflected power.

What complete and utter Texas-size bullsh*t. It's obvious that the
source is sourcing the forward voltage wave, and it's sucking up
entire reverse voltage wave from the line.

And doing it while magically expending zero energy.
Perpetual motion is possible, after all.

If zero power is being dissipated in the source, it cannot
be sourcing the forward voltage wave and it cannot be
sucking up the reverse voltage wave.
--
73, Cecil http://www.w5dxp.com

Roy Lewallen wrote:
Unlike voltage waves, which are very well
known and subject to over a century of analysis using well established
mathematics and physical principles, the waves of average power follow
rules which constantly change to suit the needs of the moment.

The rules of waves of average power come from the
field of optics and they haven't changed in many
decades. You see, average power is all that light
physicists can measure. They call it intensity or
irradiance. To deny the body of physics embodied
in optics is ignorant and ridiculous.
--
73, Cecil http://www.w5dxp.com

Tom Ring wrote:
Roy Lewallen wrote:
rules which constantly change to suit the needs of the moment.
Watterson fans will recognize the rules for propagating power waves as
closely resembling those of Calvinball.

Hopefully some of those here will get that. I would be surprised if
it's more than one in four.

The rules for propagating waves of EM energy have been
nailed down for generations. Optical physicists don't
have the luxury of measuring voltage and current. They
must necessarily measure average power density. They
are quite good at it and their average power density
equations are quite accurate and mature. They obviously
know a lot more about EM waves than most of the posters
here.
--
73, Cecil http://www.w5dxp.com

Tom Ring wrote:
Roy Lewallen wrote:

rules which constantly change to suit the needs of the moment.
Watterson fans will recognize the rules for propagating power waves as
closely resembling those of Calvinball.

Roy Lewallen, W7EL

Hopefully some of those here will get that. I would be surprised if
it's more than one in four.

Three out of four readers haven't yet figured out how to use Google?
That's pretty grim. "Calvinball" brought about 70,000 hits, or so the
results screen said.

Roy Lewallen, W7EL

On Apr 18, 8:20 pm, Cecil Moore wrote:
Jim Kelley wrote:
Since you insist that waves can have an effect on other waves, then
you should at least be able to detail either mathematically or
phenomenalogically the effect y has on x, and x has on y as well as
provide some natural process that would cause this effect. Please
elaborate. Thanks.

In the s-parameter equation, b1 = s11(a1) + s12(a2) = 0,
the interaction of s11(a1) and s12(a2) results in wave
cancellation. The effect of each wave on the other is
to reverse the direction and momentum of both waves.

That is what happens at a Z0-match in a transmission
line. That is what happens at the surface of thin-
film when reflections are being canceled.

Again, the redistribution of the wave energy is certainly
an interaction that wouldn't exist with either wave alone.
--
73, Cecil http://www.w5dxp.com

Redistribution is an interaction....interesting. But, you were
telling us about how waves interact with other waves. I'm interested
to know what effect x has on y, and vice versa? We have x + y making
z. So after that, tell us how have x and y changed as a result of
their "interaction"?

73, Jim AC6XG

On Thu, 19 Apr 2007 01:25:29 GMT, Gene Fuller
wrote:

On 18 Apr 2007 17:03:40 -0700, Keith Dysart wrote:
.=2E.Keith

Hi Richard,

Very interesting. I also use Agent for reading newsgroups. I copied and
used Keith's long message without the slightest difficulty. Everything
worked as he said it would.

Wonder what is different?

Hi Gene,

Good question. It was just a couple of ripples in the time-space
continuum perhaps. I can't put my finger on any commonality,
especially for the quote above when it is distinctly outside of any
association with an equal symbol.

Having patched up the text files, the modeler works quite nicely. Nice
piece of work.

It puts the popular generic title of "____ for Dummies" to the test;
because if you fill in the blank with Analog, then there is still one
dummy beneath the waves at low tide.

73's
Richard Clark, KB7QHC

Cecil Moore wrote:

Tom Ring wrote:

Roy Lewallen wrote:

rules which constantly change to suit the needs of the moment.
Watterson fans will recognize the rules for propagating power waves
as closely resembling those of Calvinball.


Hopefully some of those here will get that. I would be surprised if
it's more than one in four.


The rules for propagating waves of EM energy have been
nailed down for generations. Optical physicists don't
have the luxury of measuring voltage and current. They
must necessarily measure average power density. They
are quite good at it and their average power density
equations are quite accurate and mature. They obviously
know a lot more about EM waves than most of the posters
here.

Well, that's one who doesn't. Too bad, since Calvinball was a wondrous
game. I miss it.

tom
K0TAR

Jim Kelley wrote:
Redistribution is an interaction....interesting. But, you were
telling us about how waves interact with other waves. I'm interested
to know what effect x has on y, and vice versa? We have x + y making
z. So after that, tell us how have x and y changed as a result of
their "interaction"?

In a transmision line, when z=0, x and y are permanently
changed. Their energy components combine into one re-reflected
wave. The separate identities of x and y disappear at the
instant that z becomes zero.

In order to measure s11 and s12, a2 is turned off. The result is:

a1----|
|----s21(a2)
s11(a1)----|

Note that s11(a1) has already reflected from the impedance
discontinuity and there are no other impedance discontinuities
between it and the source. Should be smooth sailing.

In order to measure s21 and s22, a1 is turned off. The result is:

|----s22(a2)
s12(a2)----|
|----a2

Note that s12(a2) has already passed through the impedance
discontinuity and there are no other impedance discontinuities
between it and the source. Should be smooth sailing.

s11(a1) and s12(a2) are your two waves. They exist and are
so measurable that their measurements results in knowing
the value of s11 and s12.

For b1 = s11(a1) + s12(a2) = 0, s11(a1) and s12(a2) must
be of equal magnitude and opposite phase. That's exactly
what happens at a Z0-match.

s11(a1) and s12(a2) *never* encounter an impedance discontinuity.
They are effects of a1 and a2 encountering an impedance discontinuity.
The only thing s11(a1) and s12(a2) encounter are each other and that
interaction completely changes those two waves. The two waves cancel
and their energy components are re-distributed in the opposite
direction. s11(a1) and s12(a2) never encounter an impedance
discontinuity.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Jim Kelley wrote:
Since you insist that waves can have an effect on other waves, then
you should at least be able to detail either mathematically or
phenomenalogically the effect y has on x, and x has on y as well as
provide some natural process that would cause this effect. Please
elaborate. Thanks.

In the s-parameter equation, b1 = s11(a1) + s12(a2) = 0,
the interaction of s11(a1) and s12(a2) results in wave
cancellation. The effect of each wave on the other is
to reverse the direction and momentum of both waves.

That is what happens at a Z0-match in a transmission
line. That is what happens at the surface of thin-
film when reflections are being canceled.

Again, the redistribution of the wave energy is certainly
an interaction that wouldn't exist with either wave alone.

Cecil,

Do you see the common factor in your response about "wave interaction"?
In all of your examples there is an interface or some sort of
discontinuity. Nobody argues that waves are forever unchanging. However,
those changes take place only through interaction with interfaces or
other discontinuities.

73,
Gene
W4SZ