Cecil Moore wrote:

Jim Kelley wrote:
Electromagnetic waves can cancel, but rocks can't.

However, two EM waves have to exist before they can cancel.

Yes, but what about the rock? :-)

If they exist, they posses both energy and momentum.

So if I "possess" an American Express card, do I "possess" money? No.
I simply "possess" the potential to purchase something with it at a
point of sale.

73, Jim AC6XG

Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:
However, two EM waves have to exist before they can cancel.

And that makes rocks like waves?

That makes real waves tangible like real rocks. The wave particles
are just smaller. OTOH, "People who live in glass houses shouldn't
throw stones." is an intangible.

If they exist, they posses both energy and momentum.

Bet ya can't prove it without first transfering it to something.

_Optics_, by Hecht is good enough for me. "It is possible
to compute the resulting (momentum) force via Electromagnetic
Theory, whereupon Newton's Second Law suggests that the *wave
itself carries momentum*. (all emphasis his, not mine) ... As

Excellent, now try and understand what is intended by _ALL_ of the words
in the sentence.

Maxwell showed, the *radiation pressure* equals the energy density
of the EM wave. ... When the surface under illumination is perfectly
reflecting, the beam that entered with a velocity of +c will emerge
with a velocity of -c. This corresponds to twice the change in
momentum that occurs on absorption, ..."

Yes, I'm familiar with the subject. I've been familiar with it for a
long time.
However it is incorrect to infer that interactions between waves would
be the same as interactions between waves and matter!

It's obvious that the energy in the TV ghosting wave makes a round-
trip to the match-point and back to the RCVR.

It's obvious that the signal has taken multiple paths, at least.

That's obviously a
change in the direction of momentum of the reflected wave.

Perhaps. You're describing back scattering, which should exhibit a
Compton effect wavelength shift if true. But again, you're describing
an interaction with matter. Photons don't interact with each other in
the same way they interact with matter.

73, Jim AC6XG

"Cecil Moore" wrote in message
...
Tam/WB2TT wrote:

"Cecil Moore" wrote:
Richard, you know you are going against the conventional wisdom on
this newsgroup. Ghosting cannot exist during steady-state so if
ghosting exists it simply means that you are still in the transient
state and the steady-state doesn't exist (yet).

Looking for the smiley face.

Is there a smiley face that means, "sad but true"? Many otherwise
intelligent, knowledgeable, educated engineers have attempted to force
their metaphysical "steady-state" agenda on uninitiated and unsuspecting
victims. One is saddened by such an event and one wonders why. Does
a steady-state religion or creed exist within amateur radio? If so,
what are its purpose and goals?
--
73, Cecil http://www.qsl.net/w5dxp


From this I infer that Cecil believes that once the transient portion of
the response has concluded you are in the steady state and no more
reflections are occurring. Is this a correct re-statement of your belief,
Cecil?
Steve N.

"Tam/WB2TT" wrote in message
...

"Henry Kolesnik" wrote in message
.. .
Richard
...I got
out the book REFLECTIONS II by Walt Maxwell W2DU. I'm typing verbatim
from
page 2-2 and 23-1

I paraphrase / quote the quote:
Power reflected from a mismatch back into the PA is not absorbed there.
He writes:
"when the pi-network tank is tuned to resonance, a virtual
short circuit to rearward traveling waves is created at the input
of the network. Consequently, instead of the reflected power
reaching the tubes of the amplifier, it is totally re-reflected
toward the load by the virtual short "

Hank, I'm sorry, but I believe this is not correct as stated. The word
"totally", I believe is misleading and contrary to what I think all will
agree. Namely that there MUST be SOME real part (although the true amount
is disputed here) to the Zout of the PA and therefore SOME power MUST be
absorbed there. I believe the point of contention truely is just how much
is absorbed and how much is reflected...

...Also what happens in a transistor final with no pi?

AGAIN. Be careful that you DO NOT keep assuming that the full power is the
incident power. Incident meaning forward power or that which the
transmitter is sending toward the load. I have a real corker below. Be
careful. Do not try this at home, I am a professional. (yea, I know this'll
spawn all kinds of grief)


Tam sez:
Henry,
Here is an example of what you just said. Take a sine wave source, and
connect it to a 1/4 wave section of shorted transmission line through a
series resistor R. The reflected wave will reach this resistor 1/2 cycle
later, and will be in phase with the source. For a lossless transmission
line, there will be *0 Volts across the resistor*. There will be 0 current
through the resistor, and the reflected wave will be re reflected for all
values of R, including R=Z0, because the reflected wave will not "know"
what
R is. You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.
Tam/WB2TT

Tam,
Although I believe you have digressed somewhat, I will follow this path
since it contains a closely related concept. You obviously have a pretty
good handle on much of this (as others do, up to a point, and struggle to
make their mental models fit the reality - or to make other's mental models
fit theirs)...However, there is a paradox here which appears to be the root
of this disagreement and your example hit it right on the head. You have
some implicit assumptions here.

1) Zero volts across the resistor = (not said, but implied) the t-line acts
like an open at the input end, therefore there is no current into the line
which results in the zero V across R. V=I^2 * R. I think everybody will
agree with this. (shorted 1/4 wave acts like an open and an open 1/4 wave
looks like a short - mantra of all, no?).

2) Resistor current = zero therefore reverse traveling wave gets reflected
toward the load end. The implication clearly is that with zero resistor
current, no energy can be flowing out that way, so it must be reflected.

HOWEVER...
Been a while since I went to this depth and interesting to do so, though
unnecessary, I will anyway...

If the input to this stub acts like an open, there can be NO current,
thus no power entering, therefore there can be no forward wave, no reflected
wave and no summation of waves to make the open in the first place and no
need to re-reflect the reflected wave from the resistor who (or is it whom)
has no current. This appears to be the root cause of this problem.
Now, we can say that (and I think it has been said) that it makes no
difference how large these two waves (which cancel each other to form the
open circuit at the stub input) can be any absolute magnitude. 1 amp 10
amps 100 amps doesn't matter - they cancel. So what are they really?
There are two things to consider as you work out how you will resolve
this paradox in your mental model.
1) I believe the most important -- If the Z looking in to the stub is high,
how can you send a large amount of power down the line. If you say it is an
infinite Z then you have a really big problem explaining how it got there in
the first place.
2) There is a tendency to assume that the forward power is the same power as
when the load is not a short, but Zo.

Interesting puzzle, but I don;t need to go further. At some point you must
'believe' something and I can comfortably stop there...for now.

Oh well...
--
Steve N, K,9;d, c. i My email has no u's.

"Henry Kolesnik" wrote in message
. ..
Before it came down my 20 meter antenna had a SWR approaching 20:1 and I
was
running a Collins 30S-1 and no tuner!. I did that for several years and
would be still doing if the roofers hadn't destroyed ~130 ft. dipole. One
of these days after my back improves I'll put another one up.


Henry,
I don't have a problem with the 20:1, obviously you were matching enough
to get out, but I keep coming back to: Why did you keep doing it if it
burned up your tubes?
Steve N.

"Henry Kolesnik" wrote in message
.. .


...a cheap SWR meter.


OOPS!
Steve

Cecil, W5DXP wrote:
"Anything about dissipationless resistances or negative resistances (in
"Transmission Lines, Antennas, and Wave Guides")?"

On page 73, the characteristic resistance of free-space is defined as
the sq rt of the permeability of space devided by the dielectric
constant of space. The units are henries/m and farads/m. The solution is
a voltage to current ratio of 376.7 ohms, or 120 pi ohms.

As free-space is a perfect (lossless) medium for radio waves, it is a
dissipationless resistance.

On page 13, the characteristic resistance (Ro) of a transmission line is
given in formula (14.3) as the sq rt of L/C, but this is an
approximation for low-loss lines as there are no perfect lines.

Negative resistance is a gain instead of a loss. The authors of
"Transmission Lines, Antennas, and Wave Guides" were writing for WW-2
officers being trained at Harvard University in radio and radar. The
phone system was then using some "negative resistance repeaters" but
neither these nor "active antennas" were big at that time. You can only
make up the loss in a two-wire phone loop with amplification. Any more
gain and the loop "sings" (breaks into oscillation).

Best regards, Richard Harrison, KB5WZI

"Cecil Moore" wrote in message
...
Richard Clark wrote:

wrote:
You asked - I answered.

Did you? What was the original question?

I don't know. I wasn't around 7 billion years ago.


Chuckle, chuckle. CUTE!
Steve N.

OH! NO! Vortex vs. Bernoulli
Steve N.

--
Steve N, K,9;d, c. i My email has no u's.
"alhearn" wrote in message
om...
This is absolutely the best thread I've seen in years. It's
educational, thought provoking, entertaining, and revealing about the
human psyche.

The thread reminds of some of those that I monitor from time to time
regarding aeronautics and fluid dynamics on the topic of how an
airplane wing generates lift. Believe or not, there is still no
consensus after 100+ years. It's interesting in that it parallels this
thread in many ways -- attempting to interpret various abstract
mathmatical definitions of a physical process in a way that it "makes
sense." We're fortunate in that such things don't actually have to
make sense to work.

Al

"Jim Kelley" wrote in message
...
Cecil Moore wrote:

Tam/WB2TT wrote:
You can get the same answer from knowing that the impedance looking
into a 1/4 wave section of shorted transmission line is infinite.

Ever measure the forward and reflected currents halfway into a
shorted 1/4WL stub? How can currents be flowing unimpeded into
and out of an infinite impedance?

The question is a little misleading because the direction of the flow of
current changes every half cycle and is transverse, or orthogonal to the
direction of wave propagation. In a transmission line, the current
flows through Z0, ostensibly, which is essentially the impedance from
one conductor to the other at every point along the transmission line.
Other than that, superposed forward and reflected waves behave just as
you described, naturally.

73, Jim AC6XG
There is no current in the steady state. The steady state voltage is
independent of source impedance, which affect only how long it takes to
reach that. I ran a simulation on this, and you can see that as the voltage
builds up, the current decreases

Tam