Loopfan wrote in message arthlink.net...
I have a few more questions, but first I'd like to thank everyone for
their feedback. After doing some usenet searching, it seems that I am
retracing the footsteps of those from about 1995 onwards. I'd like to
thank W7EL, W8JI and others for making my head hurt. grin I feel like
that character in Close Encounters making a mash-potato mountain in his
living room....

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?

Since you're making the loop small compared with a wavelength anyway,
just how does it matter? I would think the difference in feedpoint
impedance (the gap) between jacketed and unjacketed line would be very
small indeed.

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? 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...

So one way to make the loop, assuming the gap is at the top, is to
make one side out of coax (which becomes the feedline) and the other
side out of solid rod the same OD as the feedline. Attach the center
conductor of the feedline where it comes out at the gap across the
gap to the center of the solid face of the other side. Equivalently,
make both sides out of coax, but put a shorting disc from outer to
center where the center re-enters the coax after the gap. Now it's
easy to see that the voltage across the gap is simply the voltage
delivered to the feedline.

It's true that whatever current is on the outside of the outer
conductor right at the gap must be balanced by a current on the inside
of the outer conductor on the coax side. Otherwise, charge would pile
up at the gap. And on the side with the shorting disc or solid
conductor, the current on the outside of the conductor must be
balanced by a current in the disk and practically by the same current
delivered to the inner conductor of the coax where it attaches to the
disk. Again, if it didn't, charge would pile up there. It's OK to
pile up a little charge if there's a capacitance to put it on, but in
a very small volume, there can't be much capacitance.

Remember, too, that the current is there (in a receiving antenna)
because we put a load across the gap by means of the coax feedline.
Otherwise, any current would only be there to charge the capacitance
of the gap (to a voltage corresponding to the EMF given by Faraday's
law, less a tiny I*R drop in the conductor).

Cheers,
Tom