On 8/3/2015 5:10 AM, Jeff wrote:

This has been explained previously. A dipole is not balanced when it is
connected to the coax. The shield outer surface presents a third
element which makes the shield side of the dipole different from the
center conductor side. In the case of the resistor the current flowing
in one side must flow out the other, so it is balanced no matter what.
The dipole has no such requirement. If you restrict the current running
into one side and not the other it can do nothing about it.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

Ok, don't call the shield an element.

A balanced antenna does not imply balanced current. The load of the
antenna element on the coax will be equal, but the coax also has a
parallel load from the shield outer surface. The two loads in parallel
result in a different voltage on the end of the coax shield than on the
end of the coax inner conductor. This different voltage causes the
different current flow in the antenna element.

A good illustration showing this can be seen near the top of the web
page linked below.


It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

I can't say for myself what happens in practice, but others here and on
the web say there *is* severe distortion in the antenna pattern. See
this link about halfway down the page. This is a EZNEC+ simulation. ymmv

http://www.tomthompson.com/radio/EHa...ommonMode.html

--

Rick

On 8/3/2015 4:49 AM, Jeff wrote:

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.

If there are high different potentials between different parts of the
car it is not a Faraday Cage.

You are applying some rule you learned without understanding it. By
ohms law there will be potential differences around the Faraday cage.


Also assuming that it is a Faraday cage then touching it whist the cage
is connected to a live conductor will not cause a shock, as everything
is at the same potential.

Only true if the conductor is perfect with no resistance.

--

Rick

rickman wrote:
On 8/3/2015 4:49 AM, Jeff wrote:

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.

If there are high different potentials between different parts of the
car it is not a Faraday Cage.

You are applying some rule you learned without understanding it. By
ohms law there will be potential differences around the Faraday cage.


Also assuming that it is a Faraday cage then touching it whist the cage
is connected to a live conductor will not cause a shock, as everything
is at the same potential.

Only true if the conductor is perfect with no resistance.

Pedantically true but practically irrelevant; for a real world cage the
voltage differences will be small fractions of a volt.


--
Jim Pennino

Jeff wrote:

This has been explained previously. A dipole is not balanced when it is
connected to the coax. The shield outer surface presents a third
element which makes the shield side of the dipole different from the
center conductor side. In the case of the resistor the current flowing
in one side must flow out the other, so it is balanced no matter what.
The dipole has no such requirement. If you restrict the current running
into one side and not the other it can do nothing about it.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

Jeff

The coax shield does create a 3rd element but the effect of it highly
depends on on the length of the coax and whether or not the shield is
grounded somewhere along the way.

The effect can be anywhere from negligable and barely measurable to
extremely significant and can be seen with an antenna modeling program.


--
Jim Pennino

Jeff wrote:

So if what you are saying is correct then terminating a piece of coax
with a 50 ohm resistor will not cause a perfect match and the load will
not be 50 ohms due to the shunt impedance to earth of the coax braid.

I think not.

Maybe you can draw a diagram and show yourself where the current flows.
The earth connection on the shield outside is in parallel with the
earth connection on the shield inside. You can't have any current flow
on the outside because any current flowing on the shield inside balances
with the current flowing in the center conductor. The resistor requires
the two currents are equal with none left to travel down the shield
outside.


Indeed, as you state there is no current flow on the outer of the coax;
the same situation exists for a perfectly balanced dipole attached
directly to the coax, as far as the coax is concerned the situation is
no different, all it sees is a perfectly matched load, just like the
resistor, all of the power is dissipated in the antenna and none flows
anywhere else including the coax outer.

Jeff

That current flows on the outside of the shield is observable and
measurable.

How much flows is totally dependant on how long the shield is and
whether or not it is grounded somewhere along the way.


--
Jim Pennino

On 8/3/2015 1:39 PM, wrote:
rickman wrote:
On 8/3/2015 4:49 AM, Jeff wrote:

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.

If there are high different potentials between different parts of the
car it is not a Faraday Cage.

You are applying some rule you learned without understanding it. By
ohms law there will be potential differences around the Faraday cage.


Also assuming that it is a Faraday cage then touching it whist the cage
is connected to a live conductor will not cause a shock, as everything
is at the same potential.

Only true if the conductor is perfect with no resistance.

Pedantically true but practically irrelevant; for a real world cage the
voltage differences will be small fractions of a volt.

"Your car is largely a Faraday cage but you can still be electrocuted if
a live wire is in contact with the chassis and you touch it while inside."

Why don't we conduct an experiment. You are saying it will only be a
small fraction of a volt, we'll put you in the car. lol

--

Rick

On 8/3/2015 1:44 PM, wrote:
Jeff wrote:

This has been explained previously. A dipole is not balanced when it is
connected to the coax. The shield outer surface presents a third
element which makes the shield side of the dipole different from the
center conductor side. In the case of the resistor the current flowing
in one side must flow out the other, so it is balanced no matter what.
The dipole has no such requirement. If you restrict the current running
into one side and not the other it can do nothing about it.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

Jeff

The coax shield does create a 3rd element but the effect of it highly
depends on on the length of the coax and whether or not the shield is
grounded somewhere along the way.

The effect can be anywhere from negligable and barely measurable to
extremely significant and can be seen with an antenna modeling program.

Let me summarize...

a) No current on shield outer surface

b) Current on shield outer surface

C) Any conceivable combination of the above

Your answer is c). Thanks for playing!

--

Rick

rickman wrote:
On 8/3/2015 1:39 PM, wrote:
rickman wrote:
On 8/3/2015 4:49 AM, Jeff wrote:

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.

If there are high different potentials between different parts of the
car it is not a Faraday Cage.

You are applying some rule you learned without understanding it. By
ohms law there will be potential differences around the Faraday cage.


Also assuming that it is a Faraday cage then touching it whist the cage
is connected to a live conductor will not cause a shock, as everything
is at the same potential.

Only true if the conductor is perfect with no resistance.

Pedantically true but practically irrelevant; for a real world cage the
voltage differences will be small fractions of a volt.

"Your car is largely a Faraday cage but you can still be electrocuted if
a live wire is in contact with the chassis and you touch it while inside."

Why don't we conduct an experiment. You are saying it will only be a
small fraction of a volt, we'll put you in the car. lol

I notice you did not address what I actually wrote.

Yet another red herring from the red herring master; I would never call
a car a Faraday cage for a number of reasons.


--
Jim Pennino

rickman wrote:
On 8/3/2015 1:44 PM, wrote:
Jeff wrote:

This has been explained previously. A dipole is not balanced when it is
connected to the coax. The shield outer surface presents a third
element which makes the shield side of the dipole different from the
center conductor side. In the case of the resistor the current flowing
in one side must flow out the other, so it is balanced no matter what.
The dipole has no such requirement. If you restrict the current running
into one side and not the other it can do nothing about it.


No, a dipole is still balanced, the coax outer does not create a 3rd
element other than by coupling. If the coax is taken off at right angles
the coupling will be low as similar to both elements of the dipole. The
dipole will never be perfectly balanced but can be a very close
approximation.

It your contention about the coax acting as a 3rd element were true then
there would be severe distortion of both the impedance of the dipole and
to it radiation pattern. This is not seen in practice and it is also
demonstrable that there is little current flow on the coax outer when
the dipole is well matched to the coax.

Jeff

The coax shield does create a 3rd element but the effect of it highly
depends on on the length of the coax and whether or not the shield is
grounded somewhere along the way.

The effect can be anywhere from negligable and barely measurable to
extremely significant and can be seen with an antenna modeling program.

Let me summarize...

a) No current on shield outer surface

b) Current on shield outer surface

C) Any conceivable combination of the above

Your answer is c). Thanks for playing!

Nope, just yet another knee jerk response.

If you had actually read and understood what I wrote, the only possible
answer is b.

The only thing in question is the magnitude of the current; that there
is a current is a given.


--
Jim Pennino

On 8/3/2015 2:25 PM, wrote:
rickman wrote:
On 8/3/2015 1:39 PM, wrote:
rickman wrote:
On 8/3/2015 4:49 AM, Jeff wrote:

This is one of those cases where you have forgotten the details and
underlying premises. A Faraday cage can't stop currents from flowing
through the "cage". It stops fields from penetrating the cage by action
of the resulting currents in the cage. Your car is largely a Faraday
cage but you can still be electrocuted if a live wire is in contact with
the chassis and you touch it while inside. There can be a high
potential across different parts of the car from the current and that
current can pass through you if you touch the cage.

If there are high different potentials between different parts of the
car it is not a Faraday Cage.

You are applying some rule you learned without understanding it. By
ohms law there will be potential differences around the Faraday cage.


Also assuming that it is a Faraday cage then touching it whist the cage
is connected to a live conductor will not cause a shock, as everything
is at the same potential.

Only true if the conductor is perfect with no resistance.

Pedantically true but practically irrelevant; for a real world cage the
voltage differences will be small fractions of a volt.

"Your car is largely a Faraday cage but you can still be electrocuted if
a live wire is in contact with the chassis and you touch it while inside."

Why don't we conduct an experiment. You are saying it will only be a
small fraction of a volt, we'll put you in the car. lol

I notice you did not address what I actually wrote.

Yet another red herring from the red herring master; I would never call
a car a Faraday cage for a number of reasons.

Neither is a coax shield.

--

Rick