"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to
the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open
or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt.

Hank, EM reflections are covered on this web page.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

In particular: "Clearly, if the wavelength of the incident light and the
thickness of the film are such that a phase difference exists between
reflections of p, then reflected wavefronts interfere destructively, and
overall reflected intensity is a minimum. If the two reflections are of
equal amplitude, then this amplitude (and hence intensity) minimum will
be zero."

"In the absence of absorption or scatter, the principle of conservation of
energy indicates all "lost" reflected intensity will appear as enhanced
intensity in the transmitted beam. The sum of the reflected and
transmitted
beam intensities is always equal to the incident intensity. This important
fact has been confirmed experimentally."

In order for (rearward-traveling) "reflected intensity" to "appear as
(forward-traveling) enhanced intensity in the transmitted beam", the
momentum of that (rearward-traveling) intensity must change directions.
Thus, it appears that total destructive interference between two rearward-
traveling reflected waves in a transmission line will reverse the
direction
of momentum of the energy in those canceled reflected waves.

We need to change a few of your statements:

Any power not dissipated or radiated by an antenna is reflected back.
"Dissipation" means EM energy transformed into heat, according to
the IEEE Dictionary.

The transmitter/tuner end will not re-reflect 100% of the reflected energy
unless there exists a short, open, pure reactance, or "total destructive
interference" as explained in _Optics_, by Hecht.

Besides a short or an open, a purely reactive impedance will cause
100% energy reflection. Apparently, so will "total destructive
interference". Quoting from _Microwave_Transmission_, by J. C. Slater:

"The method of eliminating reflections is based on the interference
between waves. ... The fundamental principle behind the elimination
of reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase."

That's pretty clear. We get one set of rearward-traveling reflections
at the match point. We get another set of rearward-traveling reflections
at the antenna. If these two sets of reflections are equal in magnitude
and opposite in phase at the match point, they cancel each other and the
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.
--
73, Cecil http://www.qsl.net/w5dxp
Cecil,

I am not quite sure what you are saying. But, I ran a SPICE simulation of
the following:
1V 1MHz source with resistor R0 feeding a 50 Ohm 250 ns transmission line
shorted at the far end. Independent of R0, in steady state the voltage at
the input end of the transmission line will be 1V. The effect of R0 is to
limit how long it takes to reach steady state. For R0 = 50 Ohms, it is one
cycle; for R0 = 500 Ohms, it is about 8 cycles, as eyeballed off the
waveform display.