Roy Lewallen, W7EL wrote:
"If you`ll read what I`ve written, you`ll hopefully see that my only
point of contention is with your claim that waves reflect from a
"virtual short". They do not."

Seems to me they do.

If you are lucky enough to have a copy of Terman`s 1955 opus, we can
reason together.
On page 91 is found Fig. 4-3 Vector (phasor) diagrams showing manner in
which incident and reflected waves combined to produce a voltage
distribution on the transmission line.

At an open circuit, the voltage phasors are in-phase.

E2, the reflected phasor, rotates clockwise as it travels back toward
the source.

E1, the incident phasor, rotates counter-clockwise as we look back
toward the source.

Looking 1/4-wavelength back from the open-circuit, E2 and E1, each
having rotated 90-degrees, but in opposite directions, are now
180-degrees out-of-phase.

On page 92, Fig. 4-4 shows the current, which summed to zero at the open
circuit, has risen to its maximum value at 1/4-wavelength back from the
open-circuit while the voltage dropped to its minimum, nearly zero,
maybe close enough to declare a "virtual short-circuit", 1/4-wavelength
back from the open-circuit.

What`s a short-circuit? Little voltage and much current.

What`s the difference between a physical short and the virtual short?
Nothing except the shunting conductor.

Is there current flowing at the open-circuit end of the 1/4-wave line
segment? No, the open-circuit won`t support current.

If a high-impedance generator of the same frequency were connected to
the virtual short point on the line, would it also be shorted? Yes.
Where? At the virtual short, not the open-circuit at the end of the
line.

Best regards, Richard Harrison, KB5WZI