Michael Tope wrote:
I am a bit behind on ARRL Handbooks, Richard, but from
what you describe, this is the same figure that appears in my
1992 edition (chapter 38, figure 2). In any case, what is
shown in the figure agrees with my understanding of "broken",
although admittedly when I made my previous post, I was
thinking of the case where the shield is broken on the side of
the loop opposite the feedpoint. For the purposes of this
discussion, however, it doesn't matter whether the break is
at the top (opposite the feed) or at the bottom (adjacent
to the feed). In either case, current induced on the inside
of the shield by current flowing on the center conductor loop
has a continuous back to ground via the outside surface of the
shield. IOW, the gap doesn't suppress the eddy current, rather
it forces it to flow on the outside surface of the shield, thereby
causing the loop to radiate.

Absolutely nothing, neither electic nor magnetic, couplesthrough the
wall of a conductor more than several skin depths thick. This isn't
anything that can be debated, it is simply how it works. It is very
easy to demonstrate, it takes only a few minutes and a minimum of test
equipment.

It is something very basic in physics and underlies how coaxial cables
and things with shields of all types work.

The gap is the feedpoint no matter where the gap is placed. The
radiation and coupling of any time-varying field, magnetic or electric,
occurs on a frequency where the shield is more than a few skin depths
thick comes by the gap.

This is such a very basic thing it is important everyone understand it.


I think the handbook has it right.


Yes, I agree it does. If you connect the shield at both ends, the
loop can't radiate because the eddy current caused by current
flowing on the inner conductor loop will confined to the inside of
the shield.

Absolutely. When the gap is closed there is no potential difference
across the gap the outside of the shield is not connected to the inside
of the shield via the potential developed across the gap. The outer
wall is not coupled to the inner wall, the feedpoint is shorted.

When the gap is opened, the outside of the shield IS the antenna. Not
the inside or anything inside the inside.

Likewise, eddy currents induced on the outside of the
shield by EM waves passing the antenna will be confined to the
outside of the shield if there is no gap (reciprocity holds - the
antenna won't receive with no gap).

Again true. This is a very basic thing we must understand if we are to
understand how shields, walls, or conductors of any kind or form work
with HF currents, voltages, or fields of any type.

There isn't any way to change this effect.

73 Tom

wrote in message
ups.com...

Michael Tope wrote:
I am a bit behind on ARRL Handbooks, Richard, but from
what you describe, this is the same figure that appears in my
1992 edition (chapter 38, figure 2). In any case, what is
shown in the figure agrees with my understanding of "broken",
although admittedly when I made my previous post, I was
thinking of the case where the shield is broken on the side of
the loop opposite the feedpoint. For the purposes of this
discussion, however, it doesn't matter whether the break is
at the top (opposite the feed) or at the bottom (adjacent
to the feed). In either case, current induced on the inside
of the shield by current flowing on the center conductor loop
has a continuous back to ground via the outside surface of the
shield. IOW, the gap doesn't suppress the eddy current, rather
it forces it to flow on the outside surface of the shield, thereby
causing the loop to radiate.

Absolutely nothing, neither electic nor magnetic, couplesthrough the
wall of a conductor more than several skin depths thick. This isn't
anything that can be debated, it is simply how it works. It is very
easy to demonstrate, it takes only a few minutes and a minimum of test
equipment.


I don't think we disagree on that point, Tom. Perhaps I should
have chosen my words more carefully. I didn't mean to imply
that gap somehow forces the current on the inside of the shield
to pass through shield. When I said that the gap forces the current
to flow on the outside surface of the shield, I meant that in the
sense that the eddy current flows on the inside of the shield until
it reaches the break in the shield at which point the current flow
wraps around the edge of the shield and onto the outside surface
(thereby reversing direction relative to the direction of the eddy
current on the inside of the shield). The skin effect in effect
separates the shield into two distinct conductors, the inner surface
being one conductor and the outer surface of the shield being the
other. The gap is the circuit node where these two independent
conductors are connected. The eddy current flows out of one
conductor (the inner surface of the shield ) and into the other
conductor (the outer surface of the shield).

73, Mike W4EF.............................................. ...........

wrote:
Absolutely nothing, neither electic nor magnetic, couplesthrough the
wall of a conductor more than several skin depths thick.

Do 60 Hz magnetic fields penetrate RF coax since
the conductor is not more than several skin depths
thick?
--
73, Cecil http://www.qsl.net/w5dxp

Tom, W8JI wrote:
"Absolutely nothing, neither electric nor magnetic, couples through the
wall of a conductor several skin depths thick."

That`s wrong for a "Faraday screen".

Terman is right. At the bottom of page 38 of his 1955 edition he writes:
"It is possible to shield electrostatic flux without simultaneously
affecting the magnetic field by surrounding the space to be shielded
with a conducting cage that is made in such a way as to provide no
low-resistance path for the flow of eddy currents, while at the same
time offering a metallic terminal upon which electrostatic flux lines
can terminate."

An example exists in the AM broadcast stations I`ve worked in. Every
tower was coupled to its transmission line through a 1:1 air-core
traansformer. Two identical single-layer solenoids sharing the same
axis. Between the coils was a metal picket fence. One end of the pickets
was firmly grounded to the coupling cabinet. The other end of all
pickets was an open circuit. Electric lines of force were intercepted by
the pickets and directly shorted to ground. However, the fences had no
effect on the magnetic coupling between them because the open circuit at
the ends of the pickets prevented circulating currents which would have
opposed magnetic coupling according to Lenz`s law.

Voila! Magnetic coupling but no electrostatic coupling between coils of
a transformer.

It`s time for W8JI to turn-off his misinformation machine.

Best regards, Richard Harrison, KB5WZI

Richard,
"this can't be" because "gurus" know otherwise.
Why do you hate Tom? You don't like anything he says on his "myth
overturning" web pages. He describes in a such detail and explains that
"shield is an antenna" - why don't you get it? :-))))
According to Tom, RF gets induced on the outside "wire" of the shield, then
it crolls to the "inside" wire of the shield around the edge of the tubing
and sees another wire and jumps over, and then to coax.
If tubing or shield was the antenna, then it would receive DX and near field
signals the same way. The fact that shield is shielding the near field
signals should make any guru wonder.
There was ZS1 on TopBand reflector reporting that he used shielded loop and
other loop antennas, and shielded loop was the only one that suppressed the
local TV birdies. Tom "explained" to him "how things work" and he apologized
that he did not mean to have this as an example of what I was saying.

There are other examples where shield "doesn't shield" - like link coupling
made of coax with end shield open and center conductor soldered to the
shield. As I mentioned I have magnetothermia machine that produces about
200W from single shielded loop, according to Tom, it should be frying the
coax in the gap, with all that RF power trying to make the corner :-)

There is more nonsense on his web site.

73 Yuri, K3BU


"Richard Harrison" wrote in message
...
Tom, W8JI wrote:
"Absolutely nothing, neither electric nor magnetic, couples through the
wall of a conductor several skin depths thick."

That`s wrong for a "Faraday screen".

Terman is right. At the bottom of page 38 of his 1955 edition he writes:
"It is possible to shield electrostatic flux without simultaneously
affecting the magnetic field by surrounding the space to be shielded
with a conducting cage that is made in such a way as to provide no
low-resistance path for the flow of eddy currents, while at the same
time offering a metallic terminal upon which electrostatic flux lines
can terminate."

An example exists in the AM broadcast stations I`ve worked in. Every
tower was coupled to its transmission line through a 1:1 air-core
traansformer. Two identical single-layer solenoids sharing the same
axis. Between the coils was a metal picket fence. One end of the pickets
was firmly grounded to the coupling cabinet. The other end of all
pickets was an open circuit. Electric lines of force were intercepted by
the pickets and directly shorted to ground. However, the fences had no
effect on the magnetic coupling between them because the open circuit at
the ends of the pickets prevented circulating currents which would have
opposed magnetic coupling according to Lenz`s law.

Voila! Magnetic coupling but no electrostatic coupling between coils of
a transformer.

It`s time for W8JI to turn-off his misinformation machine.

Best regards, Richard Harrison, KB5WZI

"Yuri Blanarovich" wrote in message
...

There are other examples where shield "doesn't shield" - like link
coupling made of coax with end shield open and center conductor soldered
to the shield. As I mentioned I have magnetothermia machine that produces
about 200W from single shielded loop, according to Tom, it should be
frying the coax in the gap, with all that RF power trying to make the
corner :-)


Yuri, think about how the "link coupling" magnetic loop you describe
above works. When the loop is energized where does the RF current
leaving the center conductor go? It has to flow onto the outside of the
shield. Where else could it go?

RF current "makes the corner" around to the outside surface of the
shield in coax all the time. If it didn't we wouldn't need choke
balun's.

73, Mike
W4EF.............................................. ...................

"Michael Tope" wrote in message
. ..

"Yuri Blanarovich" wrote in message
...

There are other examples where shield "doesn't shield" - like link
coupling made of coax with end shield open and center conductor soldered
to the shield. As I mentioned I have magnetothermia machine that produces
about 200W from single shielded loop, according to Tom, it should be
frying the coax in the gap, with all that RF power trying to make the
corner :-)


Yuri, think about how the "link coupling" magnetic loop you describe
above works. When the loop is energized where does the RF current
leaving the center conductor go? It has to flow onto the outside of the
shield. Where else could it go?

RF current "makes the corner" around to the outside surface of the
shield in coax all the time. If it didn't we wouldn't need choke
balun's.

We need RF chokes and baluns to supress curents induced on the shield from
the unbalance at the antenna feedpoint.
Sooo, according to W8JI "teachings", RF current gets induced onto the
outside surface of tubing, then crolls around the edges and goes inside the
tubing?
Sooo, we should cork the elements, or the current will get confused inside
of dark tubing elements, Eh?
Any formulas to calculate the resonance of such "antenna"??

73, Mike
W4EF.............................................. ...................

--
Yuri Blanarovich, K3BU, VE3BMV

On Mon, 22 May 2006 10:42:40 -0400, "Yuri Blanarovich"
wrote:


"Michael Tope" wrote in message
...

"Yuri Blanarovich" wrote in message
...

There are other examples where shield "doesn't shield" - like link
coupling made of coax with end shield open and center conductor soldered
to the shield. As I mentioned I have magnetothermia machine that produces
about 200W from single shielded loop, according to Tom, it should be
frying the coax in the gap, with all that RF power trying to make the
corner :-)


Yuri, think about how the "link coupling" magnetic loop you describe
above works. When the loop is energized where does the RF current
leaving the center conductor go? It has to flow onto the outside of the
shield. Where else could it go?

RF current "makes the corner" around to the outside surface of the
shield in coax all the time. If it didn't we wouldn't need choke
balun's.

We need RF chokes and baluns to supress curents induced on the shield from
the unbalance at the antenna feedpoint.
Sooo, according to W8JI "teachings", RF current gets induced onto the
outside surface of tubing, then crolls around the edges and goes inside the
tubing?
Sooo, we should cork the elements, or the current will get confused inside
of dark tubing elements, Eh?
Any formulas to calculate the resonance of such "antenna"??

73, Mike
W4EF.............................................. ...................

Yuri,

It is true that current will not flow on the inside of a tube from
current on the outside. The "waveguide beyond cutoff" effect keeps it
from doing so. The currents quickly cancel a short distance inside the
tube.
However, if you put a conductor inside that tube (wire) now it acts
like a coax cable and the energy on the center conductor couples to
the inside wall of the tube. At the end of the tube the current is
free to wrap around to the outside.

73
Gary K4FMX

"Yuri Blanarovich" wrote in message
...

RF current "makes the corner" around to the outside surface of the
shield in coax all the time. If it didn't we wouldn't need choke
balun's.

We need RF chokes and baluns to supress curents induced on the shield from
the unbalance at the antenna feedpoint.

Actually what oftentimes happens with a coax feed is that the RF
current leaving the inside of the feedline shield can flow in two
directions. It can flow down the antenna element half connected to the
shield (desired path), or it can flow down the outside of the shield
(undesired path). The electrons are dumb, all they are looking for is
the path of least resistance. They can't tell that the metal surface
on the outside of the coax isn't supposed to be part of the antenna.
The only way to keep current from flowing down the shield is make
the antenna element-half connected to the shield look like a lower
impedance than the outside of the shield. If you place ferrite beads
around the outside of the shield, this will raise the impedance of the
shield path, thereby diverting the bulk of the RF current into the
element-half and off of the shield's outside surface.

Sooo, according to W8JI "teachings", RF current gets induced onto the
outside surface of tubing, then crolls around the edges and goes inside
the tubing?

As per K4FMX's comments, this can only happen if there is a
center conductor inside the tubing, or if the tubing diameter is greater
than ~1/2 wavelength in diameter, otherwise the inside of the tubing
looks like a circular waveguide beyond cutoff. This is why coax
of a given diameter becomes useless above a certain upper frequency
limit. Once the I.D. of the coax becomes a significant fraction of a
wavelength in diameter, the coax will start to support propagation of
waveguide modes (e.g. non-TEM modes). At HF frequencies, even
large diameter tubing is well beyond waveguide cutoff, so there is no
concern about "corking" open tubing with no center conductor (it
corks itself).

73, Mike W4EF.............................................. .......

Sooo, we should cork the elements, or the current will get confused inside
of dark tubing elements, Eh?
Any formulas to calculate the resonance of such "antenna"??

Michael Tope wrote:
The electrons are dumb, all they are looking for is
the path of least resistance.

Hmmmm, electrons that know ohm's law sound pretty
smart to me. :-)
--
73, Cecil http://www.qsl.net/w5dxp