A half wavelength in a transmission line with velocity factor of 66% is
approximately 0.66 * 492/f(MHz).

Here's why. Wavelength = v / f where v is the velocity of
electromagnetic waves and f is the frequency in Hz. v = c * vf where c
is the velocity of light in free space and vf is the velocity factor.
The velocity of light in free space (c) is approximately 984,000,000
feet/second. So one wavelength, in feet, in any medium is approximately
984,000,000 * vf / f(Hz), or 984 * vf / f(MHz). A half wavelength is
then half of this, or about 492 * vf / f(MHz) feet.

A resonant "half wave" antenna is always shorter than a free-space half
wavelength for a variety of reasons. Its actual length depends on wire
diameter, end insulators, the effect of ground, and so forth. 468/f is a
useful approximation that's usually in the ballpark but seldom exactly
correct.

Roy Lewallen, W7EL

Tac wrote:
I know free space half wave dipole length in feet is 492/f mhz.
I also know 468/f mhz is for "antenna cross section...stray
capacitance...called end effect because the ends of the antenna are
made farther apart electrically than they are physically..." as laid
out in the "Practical Antenna Handbook 3rd edition p.138".

This is suspiciously close to the 95% velocity propagation of parallel
line.
My question is, for a given coax with 66% velocity propagation, is the
formula 66% of 492 (325/f) or
66% of 468 (309/f)?

Thanks in advance