Cecil Moore wrote:
Tom Donaly wrote:
Because I don't have to prove you wrong, Cecil, you have to prove
yourself right.

Sorry, I don't, Tom. Hecht, Born and Wolf have already
proven those fundamentals of physics to be true. You
have to prove me, Hecht, and Born and Wolf wrong.

And of course, you will mount every diversion known to
man to avoid facing the technical facts as explained by
Hecht, Born & Wolf, Melles-Groit, and the FSU web page.

Sorry Cecil, quoting sources you can't possibly understand, yourself,
won't prove anything. Let me know if you ever plan on doing it right.
73,
Tom Donaly, KA6RUH

On Sat, 17 Nov 2007 21:46:30 -0500, "Stefan Wolfe"
wrote:


"Richard Clark" wrote in message
.. .
On Sat, 17 Nov 2007 20:30:54 -0500, "Stefan Wolfe"
wrote:

Well, for one powers do not add, just energies.

Well, power is a vector quanity subject to the rules of vector math.
But
look at it in terms of the energy domain:

The total energy in the system over any time interval is equal to the
sum
of
the each power integrated over each segment of the transmission line
during
the same time interval.

This isn't getting any better.

Since Energy is the integral of power over time,

Now from bad to worse.

You are going to have to unwind a lot of your own problems before you
can get to Cecil's tricks of subterfuge.

What are the units you use to express the integral of watts over time?

Watt-Hours?

OK, fine. I prefer joules. Stay calm, Richard.

Joules is the integral of watts over time? Beyond superlatives!

Now consider the equation

Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)

For energy, integrate Ptotal for 1 second:

= (50 - 0) + (50 - 0) - 2SQRT P1*P2 sine(A)
= 100 - 2*SQRT*P1*P2(sine(0 - n*2*pi)) = 100- 0 = 100

The integral of one full sine or cosine cycle totals zero assuming zero
reference.
It isn't in integral form.

No ENERGY lost, even though at certain times the power was greater than 100W
while at other times it was less.
that's not the meaning of integration

Let's rewind up that list of charades to revisit:
Well, power is a vector quanity subject to the rules of vector math.
and ponder the implication of a negative power, for simplicity:
P1 = 50W @ 90deg
P2 = 50W @ 270deg
what does the math reveal?

73's
Richard Clark, KB7QHC

Gene Fuller wrote:
Assuming one does not make an error in the setup of the
problem (perhaps a poor assumption) or in the math, the energy will
always come out correctly. It is not an independent consideration.

Energy has not been a consideration at all for many
people such as yourself. That's why you guys have
missed the boat as far as energy is concerned. You
have only given lip service to the conservation of
energy principle. When you are pressed for details,
you whole argument falls apart.

You got it wrong. Try again.

Please point out in detail what is wrong with it.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
Sorry Cecil, quoting sources you can't possibly understand, yourself,
won't prove anything. Let me know if you ever plan on doing it right.

The technical content of your objection is noted, Tom.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
and ponder the implication of a negative power

Power to the load equals forward power minus reflected power.

The negative sign for power in that case is associated with
direction of the Poynting vectors, positive for forward and
negative for reflected.

Constructive interference power minus destructive interference
power equals zero.

The negative sign for power is associated with destructive
interference.

Both of those cases are simply mathematical conventions.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
and ponder the implication of a negative power, for simplicity:
P1 = 50W @ 90deg
P2 = 50W @ 270deg
what does the math reveal?

Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Ptotal = 50 + 50 + 2*SQRT(2500)cos(180)

Ptotal = 50w + 50w - 100w = 0

There's 100 watts of destructive interference.
The two waves are canceled in their original direction
of flow. The energy in the two waves is redistributed
to regions that permit 100 watts of constructive
interference to occur. If the above happens in a
transmission line, the "lost" energy in the canceled
waves will simply be redistributed in the only other
direction available.

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/waveinteractions/index.html

"... when two waves of equal amplitude and wavelength that are
180-degrees ... out of phase with each other meet, they are not actually
annihilated, ... All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction of
light."
--
73, Cecil http://www.w5dxp.com

On Sun, 18 Nov 2007 00:16:47 -0500, "Stefan Wolfe"
wrote:

Let's rewind up that list of charades to revisit:
Well, power is a vector quanity subject to the rules of vector math.
and ponder the implication of a negative power, for simplicity:
P1 = 50W @ 90deg
P2 = 50W @ 270deg
what does the math reveal?

I am not trying to define power as negative with respect to zero
Negative with respect to zero? What about with respect to positive?
Or even a smaller negative! How about half negative (only 90 degrees
shift instead of 180)?
What happened to:
Well, power is a vector quanity subject to the rules of vector math.

I was so wanting to see your solution, much less how:
Ptotal = P1 + P2 + 2*SQRT(P1*P2)cos(A)
migrated into the negated sine in:
= (50 - 0) + (50 - 0) - 2SQRT P1*P2 sine(A)

I never expected such a trove of mistakes cascading through successive
replies. This is a troll, right?

No wonder you embrace anonymity, this would be embarrassing,
otherwise.

73's
Richard Clark, KB7QHC

On Nov 17, 4:03 pm, Cecil Moore wrote:
K7ITM wrote:
Of course it was obvious from the base posting in this thread that the
"waves" must be on a TEM transmission line. But you still haven't
said anything about HOW you launched two distinct waves but got them
to combine into one.

The waves are launched by the external reflection from a
Z0-match and the internal reflection from the load.
--
73, Cecil http://www.w5dxp.com

So the waves are going opposite directions along the line??

Exactly. Misdirection is the primary tool used by magicians (illusionists) to
distract us into thinking something is occurring which really isn't. Its
utility hasn't been lost on those wanting to divert our attention from the
flaws in their arguments.

Anyway, re-thinking on the case being here discussed, we must admit that the
issue could have been readily solved considering the basic antenna theory.

We know that two stacked dipoles yield a gain of up to 3 dB with respect to a
single dipole. In other words, two dipoles each fed with power P/2 produce, in
some regions of the space, a higher field than that produced by a single dipole
fed with power P, although the total transmit power has not changed.

The extra power measured at the receiver is obviously "created" at the expense
of power taken away from other regions of the space (according to the transmit
antenna pattern).

Too fundamental to deserve further discussions!

73

Tony I0JX

What are the units you use to express the integral of watts over time?
Joules


Frank GM0CSZ / KN6WH