Cecil Moore wrote:
Roger wrote:
Cecil Moore wrote:
Assume a constant power source and you will get the results
that Roy is talking about.

No, because you would find two waves of equal voltage and current
traveling in opposite directions, always arriving at exactly out of
phase, at the source.

No, because a *constant power source* is pumping joules/second
into the system no matter what voltage or current it requires
to move those joules/second into the system. It's like the
power source in "Forbidden Planet".

Now the question here is "Do the waves bounce off one another?"

Waves do NOT "bounce" off one another. At a physical
impedance discontinuity, the component waves can
superpose in such a way as to redistribute their energy
contents in a different direction. (Redistribution of energy
in a different direction in a transmission line implies
reflections.) In the absence of a physical impedance
discontinuity, waves just pass through each other.

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

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

There is a third possibility. The interaction of the two waves can
establish a very high resistance, so high that no current flows-zero.

Does this place us at a logical impasse, with current reversing and
voltage doubling at in one argument (at the open end), but not doubling
at the source end? No, the voltage will double at the source end when
stability is reached after one full cycle (in the 1/2 wave example).

Logically then, we must recognize that our source voltage WILL NOT
remain constant following the arrival of the reflected wave. Certainly
this is what we find when we retune our transmitters after changing
frequency.

What would be the logic of insisting that the input voltage be held
constant to the 1/2 wave example after it is shown that the reflected
wave must interact with the incoming wave give a very high impedance at
the source?

73, Roger, W7WKB