Loopfan wrote:
. . .
I have been cutting my coax loops to less than 1/10th of a wavelength
and also taking the velocity factor of the cable into account. After
your help with analyzing transmission lines, and proving for myself that
the outside shield is the antenna by way way of being able to attenuate
it with an RF choke, I am now wondering if I should *NOT* take the
velocity factor into account and make my loops with disregard to the
velocity factor? Or does the jacket contribute to the velocity factor
of the outer-surface of the cable?

The transmission line velocity factor applies only to the inside of the
coax, where the fields are entirely within the dielectric material.
Waves on the outside of the shield are propagating at nearly the speed
of light. A jacket will slow propagation by somewhere around 2 - 3%.

It's not apparent why you need to worry about the exact electrical size
of the loop anyway, unless you're trying to predict with good accuracy
how much tuning C you'll need.

Question 2: Have we come to any conclusions about how the current gets
from the outer-skin surface of the shield to the inner-skin surface of
the shield?

Sure. It flows from the outside to inside at the gap, around the cut
edges of the shield.

I don't want to rehash an old topic, so I'll be just as
happy to say that it merely *does*. I'm wondering if there is a field
set up on the outer skin edge that encompasses the inner-skin edge and
transfers current that way, or can I view the inner and outer skins as
more or less the same conductive skin surface that has a 180 degree bend
in it so that its analogous to the inside and outside being the same
sort of "outer" skin? Looks like I'm confusing myself...

Actually, you're pretty close. The current simply flows around the edge.
It might help to visualize the shield as being hollow, with a separate
inner and outer "shell", connected only at the cut end of the gap.
That's what it looks like to the RF. The field on the outside of the
shield doesn't penetrate the shield, nor does the field on the inside.
So the two have no effect on each other -- except at the gap. The
current gets from one surface to the other purely by conduction.

. . .

Roy Lewallen, W7EL