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?

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...

I'll be happy to do more usenet message searches if that will prevent a
total flareup of the same old topic.

Thanks and 73
Brian

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

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

Hello Roy!

Roy Lewallen wrote:

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%.

Great. Makes sense now.

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.

Well, I was thinking about trying to maximize the area surrounded by the
loop up to about 1/10th of a wavelength or so. What I found was that
the distributed capacitance of my cable at this length was too much for
my tuner to handle. So when I multiplied this length by my cable's
velocity factor, 0.66, suddenly the tuner could handle it and the loop
came alive! THIS is what got me on the wrong path thinking that the
center conductor was the antenna, and that I had discovered some
overlooked secret about the velocity factor. I was obviously mistaken.

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.

Right. Kind of like a hollow knife blade. You've got two physical
sides that meet at a cutting edge, but if you examine it closely, you'll
see that we are only really dealing with one outer conductive surface
"folded" so to speak back on itself. The edges of the exposed gap wires
represent the cutting edge - a small distance to be sure, but it is a
length of conductive material making a seamless join. Two sides, but
one surface.

Thanks again for the help.

73
Brian

Hello Tom!

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.

I was just trying to follow the mantra of trying to enclose as much area
as possible. I found that in the end it wasn't as critical as I
thought, ie changing a 10-foot circumference loop to 7-feet. As long as
I could get my tuner to handle it, it seemed that the Q of the circuit
made up for the small difference in the loss of the capture area. So in
the end, you are right, the difference in feedpoint impedance wasn't too
critical.

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

snip
Yup! Once I had accepted the fact that the shield is the antenna, all
the other loops I made with one half of the loop being solid wire,
shorted coax pieces, cross-connects etc, all made sense. They all
worked nearly the same except for some very slight inbalances in the
directional pattern, and some small differences in the depth of the
nulls. The light bulb turned on in my head, and now I could stop
cutting up so much coax. grin

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).

Aha! As in putting say a gapped director or reflector element built out
of copper tubing in front or behind of my coax-built rx loop. Now that
would be interesting, although most likely a mechanical nightmare.

Good info!

73
Brian