Here's a question overflowing from eHam.net and it's
not a trick question.

Assume that the radiating portion of a 40m vertical
is made out of 33 feet of RG-213 and the braid is
the radiator. The center conductor of the coax is
left floating at both ends. How much RF voltage
and/or current will be induced in that center wire
when using the outside braid as the radiator for
100 watt operation? How much of an EM field can
exist inside the coax braid?
--
73, Cecil http://www.qsl.net/w5dxp

In article ,
Cecil Moore wrote:

Here's a question overflowing from eHam.net and it's
not a trick question.

Assume that the radiating portion of a 40m vertical
is made out of 33 feet of RG-213 and the braid is
the radiator. The center conductor of the coax is
left floating at both ends. How much RF voltage
and/or current will be induced in that center wire
when using the outside braid as the radiator for
100 watt operation? How much of an EM field can
exist inside the coax braid?

No RF Current, as the cenbter conductor is "Floating"......
There maybe some voltage buildup , but it will not have current flow
untill the center conductor makes a circuit with something......

You wrote:
No RF Current, as the cenbter conductor is "Floating"......
There maybe some voltage buildup , but it will not have current flow
untill the center conductor makes a circuit with something......

Well, that's essentially what I assumed. But W8JI disagrees
so now I am not sure my assumption was correct.
--
73, Cecil http://www.qsl.net/w5dxp

The center conductor isn't entirely floating.

If the shield were closed at both ends, there would be no fields in the
coax, but the coax center conductor at each end can have current
induced on it from the outside world.

I think you'd go about describing the coupling as being via "fringing
fields" at the ends if you were to think of a free end of coax with
fields in it radiating into space, I'm thinking of the reciprocal
behavior...

But this is a behavior where I'm imagining a differential-mode current
existing to start with...

I think the answer is highly influenced by the exact details of the end
of the coax and where it is with respect to other objects... the
coupling is very light and mostly to the shield at each end... and I
think this makes it not entirely unlike just putting the shield wire
and the center conductor wire in parallel in space with the ends tied
together.

I think current flows in the center conductor in phase with the current
in the shield and there's very little differential mode current if any.

Certainly I'm wrong, I'm going to think on it some more and try to
figure out how wrong...

What current flows in the center conductor if you short it to the
shield with one wire at each end? What about if you do it with a solid
metal cap ?

73,
Dan

Cecil Moore wrote:

Here's a question overflowing from eHam.net and it's
not a trick question.

Assume that the radiating portion of a 40m vertical
is made out of 33 feet of RG-213 and the braid is
the radiator. The center conductor of the coax is
left floating at both ends. How much RF voltage
and/or current will be induced in that center wire
when using the outside braid as the radiator for
100 watt operation? How much of an EM field can
exist inside the coax braid?

From my EM311 days:

If I integrate the fields inside a short conducting cylinder, since there is no
charge within the surface integral, there is no field within the cylinder.

So the volume integral, ?E.dL = 0 = No voltage. But, this isn't the case!

Now, if the cylinder is a 1/4 wavelength with distributed L and C ... ???

Since the cylinder is long compared to a wavelength, the distributed capacitance
will couple a voltage to the inner conductor. The terminal impedance is open
circuited [High Z] so no current flows.

Conclusion: a standing wave exists on the inner conductor. It is caused by the
distributed capacitance and the magnitude of the standing wave on the cylinder.

Been away from EM for almost 50 years. I've probably forgotten too much.

"Dave" wrote in message
...
Cecil Moore wrote:

Here's a question overflowing from eHam.net and it's
not a trick question.

Assume that the radiating portion of a 40m vertical
is made out of 33 feet of RG-213 and the braid is
the radiator. The center conductor of the coax is
left floating at both ends. How much RF voltage
and/or current will be induced in that center wire
when using the outside braid as the radiator for
100 watt operation? How much of an EM field can
exist inside the coax braid?

From my EM311 days:

If I integrate the fields inside a short conducting cylinder, since there
is no charge within the surface integral, there is no field within the
cylinder.

So the volume integral, ?E.dL = 0 = No voltage. But, this isn't the case!

Now, if the cylinder is a 1/4 wavelength with distributed L and C ... ???

Since the cylinder is long compared to a wavelength, the distributed
capacitance will couple a voltage to the inner conductor. The terminal
impedance is open circuited [High Z] so no current flows.

ah, but it isn't! remember, 1/4 wave long (more or less) makes the open at
one end look like a short at the other end. there is also that distributed
capacitance all along the length between the inner side of the shield and
the center conductor... and also, the driving voltage at the ends are about
90 degrees out of phase, so there could be some non-trivial currents in the
center conductor.



Conclusion: a standing wave exists on the inner conductor. It is caused by
the distributed capacitance and the magnitude of the standing wave on the
cylinder.

Been away from EM for almost 50 years. I've probably forgotten too much.

Dave wrote:
remember, 1/4 wave long (more or less) makes the open at
one end look like a short at the other end.

There is an open at both ends. Does that make it look like
a short at both ends?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Dave wrote:

remember, 1/4 wave long (more or less) makes the open at one end look
like a short at the other end.


There is an open at both ends. Does that make it look like
a short at both ends?

If it looked like a short at both ends, it would look like an open at both ends.

Built a skeleton piece of giant coax in EZNEC, been playing with it as
a monopole over MININEC ground.

I'm not going to claim much since it's probably wildly inapplicable,
but I will claim that the answer to this question ain't trivial and
depends on the details of the ends.

I find a current in the center conductor, It's very much bigger when
the center conductor is slightly extended past the ends of the "shield"
than when it's slightly inside. The shield is eight wires arranged in
a regular octogon with the tops and bottoms tied together. It shows
more or less normal fat monopole behavior when used alone. (33 foot
element resonant around 6.4 MHz)

A fatter wire used for the center conductor has greater current than a
thinner one. A very thin wire exhibits almost no current.

The peak current in the center conductor is about 3% of what's flowing
in the shield for a 1 foot "diameter" (the circumscribed circle around
the octagon) shield and a 6 inch diameter inner conductor. The current
is very low at either end of the center conductor and peaks about 2/3
from the top.

This dumb model is full of holes of many kinds, I'm sure, but it does
seem to show nontrivial effects of changing things a little bit near
the ends of the coax (slight extension of center conductor outside
of/retraction of center conductor into the shield)

Dan

Cecil Moore wrote:
Here's a question overflowing from eHam.net and it's
not a trick question.

Assume that the radiating portion of a 40m vertical
is made out of 33 feet of RG-213 and the braid is
the radiator. The center conductor of the coax is
left floating at both ends. How much RF voltage
and/or current will be induced in that center wire
when using the outside braid as the radiator for
100 watt operation? How much of an EM field can
exist inside the coax braid?

I know this will shock regular group users, but he has taken things WAY
out of context in the question above!

The actual problem is this:

A fellow placed a relay at the top of a half square antenna to change
directions by switching from one flat top and drop wire to another.
This is a VOLTAGE fed antenna at the ground. The vertical wires at the
antenna ends have to be an electrical 1/4 wl long on the OUTSIDE for
the system to work properly.

Cecil suggested he simply run the relay wires up inside a "shield" to
the relay, and the shield would prevent the relay control wires from
affecting the very high feed impedance at the base. The shield could be
used as the actual vertical antenna lead.

Now I know to many people the problem is obvious. The problem is the
IMPEDANCE of the open stub formed at the bottom of the vertical sleeve
by the inner wire that has to go to a control system of some type and
the outer sleeve.

That impedance has to be many ten's of kilo ohms so the shunting
impedance is high compared to the impedance of the sleeve.

Full RF voltage of the feedpoint is also across the gap where the
center wires leave the shield.

In order for the shield to have some meaningful effect on the system
other than simply running the wires down in parallel with the fed wire,
the impedance between the inner wire and shield must be VERY high at
the bottom. It can of course be a SHORT at the top, since the relay
just sits up there in the air with only the contacts making a
connection, so the top is easy to handle with some bypass caps.

What Cecil totally misses is he formed what is in effect the electrical
equivalent of a sleeve balun. The velocity factor of the transmission
line forming this stub has to be the SAME as the outside of the sleeve
so the INSIDE is 1/4 wl long electrical, and the the loss has to be
very low. Otherwise the common mode impedance of the relay wires
exiting the shield will not be several times higher than the antenna
feed impedance, which is several k-ohms.

I've seen antenna manufacturers make the same mistake Cecil just made,
and assume that running a cable down the center of a "hot" mast that is
part of an antenna means the wires have zero current and zero effect
since they are inside the shield, but anyone with any understanding of
how the system works would catch the flaws in this idea right away.

The flaw is the differential IMPEDANCE between the shield and the shell
forming an antenna has to be several times the common mode impedance of
the shell or the system won't be worth a flip. Without that high
impedance, the inner wire might as well just run down the outside of
the sleeve and a couple good HV high impedance RF chokes be used to
supply relay control voltage.

As a matter of fact at AM BC stations, when using two way or RPU
antennas on the hot base insulated towers, I never bothered with
running the cables INSIDE the tower. I went up 1/4 wl above the base,
and bonded the cables to the tower. I spaced the cables a foot or so
off the tower face on large insulators, so it formed an open 1/4 wl
very low loss stub. This made the differential mode impedance of the
open stub end at the ground very high, and allowed the cables to be
brought away from the hot tower base without interacting a large amount
with the tower base impedance.

This is a very simple common system that is often used in antennas
(often in BC systems) , and once in a while used incorrectly by Hams
and Ham antenna manufacturers (like Gap and MFJ and a few manufacturers
of Ham log periodics).

Cecil will catch on with help I'm sure, I just don't have time to walk
him through the problem step by step.

73 Tom