Note that, with a resonant Q of 11, should an open-circuit fault occur
at
a distance of 1/4-wavelength the voltage at the fault can rise to a
million
volts or more.

How do you compute this ?

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It is computed in exactly the same way at 60 Hz as it is at 60 MHz.

A 1/4-wavelength line behaves as any other tuned circuit. The voltage at
the open end rises to Q times the voltage applied at the input end.

With a short-circuit line the current at the short circuit rises to Q times
the current at the input end.

The Q of a tuned circuit is the reactance of the inductance divided by the
wire resistance. Q = Omega*L / R.

The Q of a transmission line is the reactance of the line's overall
inductance divided by the line's overall resistance. And again, Q = Omega*L
/ R.

Since both L and R are directly proportional to line length, for a given
line, Q is a constant and is independent of length.

Omega = 2*Pi*Freq.

There are other more complcated ways of calculating the resonant rise in
open-circuit voltage from the line's transmission and propagation
properties. But they all give the same answer of course. Such calculations
provide a means of checking for program software bugs.

The hardest part of the exercise is calculating the inductance and
resistance from the line's physical dimensions and operating frequency.
Which are needed anyway to calculate all the many other output quantities
from the program. Q is just a spin-off.
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Reg, G4FGQ