Keith Dysart wrote:
But a key realization is that the behaviour on the line
can not depend on arcane details about the construction of the
generator.

Agreed, but the behavior on the line also doesn't depend
upon where the forward and reverse traveling wave came from.
All that is important is their existence. The reverse
traveling wave might come from a separate source and the
line doesn't know the difference.

So when such statements are made, it is instructive
to explore the situation with a different generator. If the outcome
is different, then there is something wrong with the theory.

No, if the outcome is different, something different is
happening while abiding by the wave reflection model rules.
Why are you surprised that changes occur when something is
changed?

Note
that the standard explanations only require that the impedance
of the source be known.

If the "standard explanations" were correct, this would
have been put to bed a long time ago.

But how does it know, since the output (source) impedances are
the same for both cases?

Conclusion: The output source impedance is NOT the impedance
encountered by the reflected wave. That's essentially what
Walter Maxwell's paper says.

I think you have made this mistakte before. In my expression P=VI,
V and I are simultaneous instantaneous values and P is the
instantaneous power. If you want average you need to integrate
and divide.

No, if you want to average sinusoidals , use RMS values, which
is the result of said integrating and dividing. Somebody else
already did all the work.

Your expression applies only to sine waves and V and I are peak
voltages.

I is a peak voltage??????? You are more confused than
I ever realized. :-)

Interestingly, the expression P=V*I*cos(theta) for sine
waves is always derived by starting with Pinst=Vinst*Iinst,

Interestingly, you forgot to say you were talking about
instantaneous power until now.

True. But there IS energy flowing at every point that is not a voltage
minimum or maximum. At these points there is NEVER any energy
flowing.

Every half cycle in every EM traveling wave, the instantaneous
power is zero. Would you therefore argue that traveling waves
cannot transfer any energy? Please get real.

Given two parallel water pipes carrying 100 gallons/minute
in opposite directions. I can see you arguing that there
is no NET flow of water. But I cannot see you arguing that
there is no flow of water at all when the pressure of just
one stream can knock you off your feet.

Well that does leave you with a bit of a conundrum since on
an open ended line, zero energy is flowing at every quarter
wavelength point back from the load.

The conundrum is all yours. Every half cycle, every EM
traveling wave is at a zero instantaneous power point.
If EM traveling waves can transfer energy while the
instantaneous power is zero every half cycle, superposing
two of them doesn't present a conundrum at all.

At such a point a real instantaneous wattmeter would always
indicate zero.

A real instantaneous wattmeter would always indicate zero
at the zero crossing of every EM traveling wave in the
universe so exactly how does the light from Alpha Centauri
ever make it to Earth when the instantaneous power is zero
every half cycle? Please get real.
--
73, Cecil http://www.w5dxp.com