I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of ( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place

On Nov 30, 8:10*pm, Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

*THE CHARGE WHEN ACCELERATED *CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in *the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of *( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place

hey there groups archivist, if there is such a thing... remove this
post in accordance with his own request that such garbage be removed.

Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways


snip crap, but left plenty

WOW!!!

Look everybody he's totally making up physics again!

Art - get back on your meds, you're nuts again.

tom
K0TAR

Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of ( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place
TROLL!

orfus wrote:
Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of ( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place
TROLL!

Nope. Local loony.

You, however, are a troll until proven otherwise.

tom
K0TAR

tom wrote in
. net:

orfus wrote:
Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

THE CHARGE WHEN ACCELERATED CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of ( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place
TROLL!

Nope. Local loony.

You, however, are a troll until proven otherwise.

tom
K0TAR


Ok, at the risk of stirring muddy water, I'm curious now, I'm new to this
group, and the subject as there clearly seems to be more to it than I knew. I
also don't know of those archives mentioned so I haven't seen the context.

So in simple terms (hopefully) what is the truth of it? As far as I knew, a
photon at RF with energy but no mass will produce a current that changes over
time in a metal that it hits, though I imagine that as metal has resistance
there must also be a voltage too. I've also heard of the 'skin effect' that
means that at high RF frequencies, current flow tends to stay on the surface,
so clearly the picture isn't as simple as DC and Ohm's law. I also know that
when photons in optical fibres meet boundaries between layers they don't
reflect simply on one side, within one region of specific refractive index,
there's apparently some more complex information exchange that amounts to the
photon crossing the border before returning. Which makes me suspect that
equally exotic action happens when RF photons hit metal sheilds. So what IS
correct? And even if there is more to it, does the aggregate of many photons,
and the wave analysis of their behaviour, reduce to a simple model that makes
the OP correct?

I'm asking this because calls of 'troll' and 'loony' aren't working for me.

On Nov 30, 9:05*pm, Lostgallifreyan wrote:
tom wrote e.net:



orfus wrote:
Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

*THE CHARGE WHEN ACCELERATED *CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in *the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of *( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place
TROLL!

Nope. Local loony.

You, however, are a troll until proven otherwise.

tom
K0TAR

Ok, at the risk of stirring muddy water, I'm curious now, I'm new to this
group, and the subject as there clearly seems to be more to it than I knew. I
also don't know of those archives mentioned so I haven't seen the context..

So in simple terms (hopefully) what is the truth of it? As far as I knew, a
photon at RF with energy but no mass will produce a current that changes over
time in a metal that it hits, though I imagine that as metal has resistance
there must also be a voltage too. I've also heard of the 'skin effect' that
means that at high RF frequencies, current flow tends to stay on the surface,
so clearly the picture isn't as simple as DC and Ohm's law. I also know that
when photons in optical fibres meet boundaries between layers they don't
reflect simply on one side, within one region of specific refractive index,
there's apparently some more complex information exchange that amounts to the
photon crossing the border before returning. Which makes me suspect that
equally exotic action happens when RF photons hit metal sheilds. So what IS
correct? And even if there is more to it, does the aggregate of many photons,
and the wave analysis of their behaviour, reduce to a simple model that makes
the OP correct?

I'm asking this because calls of 'troll' and 'loony' aren't working for me.

It's fairly straightforward, actually, if you believe in Faraday's law
of magnetic induction. That law says that for any closed loop
(through air, through a conductor, through anything), there is an
electromotive force (a voltage source, if you will) whose magnitude is
proportional to the rate of change of magnetic flux enclosed by the
loop. As there is no voltage drop along a perfect conductor, if your
closed loop follows the path of a perfect conductor, there is no
voltage drop around that loop, and therefore the rate of change of the
total magnetic flux enclosed by that loop must be zero. If the
perfect conductor is a closed box, then you can draw loops anywhere
through that conductor and you will never see a changing magnetic
field enclosed by that loop. Thus, the inside of the box and the
outside are magnetically independent; things happening on one side
(magnetically) are not sensed on the other side.

You can understand how this works if you realize that a changing
magnetic field outside the box that would penetrate the box if it
weren't there will induce currents in the conducting box (or even just
in a closed loop of wire). Those currents will (in a perfect
conductor) be exactly the right magnitude to cause a magnetic field
that cancels the external one everywhere inside the closed box (or the
net flux enclosed by a loop of wire). An example: if you short the
secondary of a mains transformer, the primary will draw lots of
current at its rated voltage: it's very difficult for the primary to
change the magnetic flux in the core.

Does the electric field shielding from a perfect conductor need any
explanation?

Of course, an imperfect conductor will be an imperfect magnetic
shield. But a perfect conductor won't let any change of field
through, no matter how slow (no matter how low an EMF it generates),
so a perfect conductor works as a shield all the way down to DC. A
box made with an imperfect conductor is essentially a perfect shield
if the box's wall thickness is at least many skin-depths thick at the
frequency of interest.

That's a quick beginning. You can find lots more about this in E&M
texts. There's even useful stuff about it on the web. ;-)

Cheers,
Tom

K7ITM wrote:
I'm asking this because calls of 'troll' and 'loony' aren't working for me.

It's fairly straightforward, actually, if you believe in Faraday's law
of magnetic induction. That law says that for any closed loop (through
air, through a conductor, through anything), there is an electromotive
force (a voltage source, if you will) whose magnitude is proportional
to the rate of change of magnetic flux enclosed by the loop. As there
is no voltage drop along a perfect conductor, if your closed loop
follows the path of a perfect conductor, there is no voltage drop
around that loop, and therefore the rate of change of the total
magnetic flux enclosed by that loop must be zero. If the perfect
conductor is a closed box, then you can draw loops anywhere through
that conductor and you will never see a changing magnetic field
enclosed by that loop. Thus, the inside of the box and the outside are
magnetically independent; things happening on one side (magnetically)
are not sensed on the other side.

You can understand how this works if you realize that a changing
magnetic field outside the box that would penetrate the box if it
weren't there will induce currents in the conducting box (or even just
in a closed loop of wire). Those currents will (in a perfect
conductor) be exactly the right magnitude to cause a magnetic field
that cancels the external one everywhere inside the closed box (or the
net flux enclosed by a loop of wire). An example: if you short the
secondary of a mains transformer, the primary will draw lots of current
at its rated voltage: it's very difficult for the primary to change the
magnetic flux in the core.

Does the electric field shielding from a perfect conductor need any
explanation?

Of course, an imperfect conductor will be an imperfect magnetic shield.
But a perfect conductor won't let any change of field through, no
matter how slow (no matter how low an EMF it generates), so a perfect
conductor works as a shield all the way down to DC. A box made with an
imperfect conductor is essentially a perfect shield if the box's wall
thickness is at least many skin-depths thick at the frequency of interest.

That's a quick beginning. You can find lots more about this in E&M
texts. There's even useful stuff about it on the web. ;-)


Here is a link to a generalized proof of the skin effect:

http://www.ifwtech.co.uk/g3sek/misc/skin.htm

This is exactly equivalent to Tom's explanation above. The detailed
proof is quite mathematical but it is solidly based in classical physics
- Faraday's Law and Ampere's theorem (both of which are embodied in
Maxwell's equations). This derivation produces the well-known equations
for current density as a function of depth, conductivity and
permeability.

The special feature of this particular proof is that it's much more
general than the ones you see in better-known textbooks - and therefore
much more powerful. It shows that if RF current is flowing in/on *any*
conducting surface, for *any* reason, then the skin effect will be
present.

The possible reasons why RF current may be flowing can be divided into
two main groups:

* "Circuit conditions" - the conductor is part of a circuit that makes
RF current flow.

* "Electromagnetic induction" - the conductor is intercepting an
incident electromagnetic wave which induces a current.

In either case, an RF current flows... and wherever that happens, there
you will also find the skin effect.



--

73 from Ian GM3SEK
http://www.ifwtech.co.uk/g3sek

K7ITM wrote in news:c52a1b1d-ef32-4d69-bf61-
:

It's fairly straightforward, actually, if you believe in Faraday's law
of magnetic induction. That law says that for any closed loop
(through air, through a conductor, through anything), there is an
electromotive force (a voltage source, if you will) whose magnitude is
proportional to the rate of change of magnetic flux enclosed by the
loop. As there is no voltage drop along a perfect conductor, if your
closed loop follows the path of a perfect conductor, there is no
voltage drop around that loop, and therefore the rate of change of the
total magnetic flux enclosed by that loop must be zero. If the
perfect conductor is a closed box, then you can draw loops anywhere
through that conductor and you will never see a changing magnetic
field enclosed by that loop. Thus, the inside of the box and the
outside are magnetically independent; things happening on one side
(magnetically) are not sensed on the other side.

You can understand how this works if you realize that a changing
magnetic field outside the box that would penetrate the box if it
weren't there will induce currents in the conducting box (or even just
in a closed loop of wire). Those currents will (in a perfect
conductor) be exactly the right magnitude to cause a magnetic field
that cancels the external one everywhere inside the closed box (or the
net flux enclosed by a loop of wire). An example: if you short the
secondary of a mains transformer, the primary will draw lots of
current at its rated voltage: it's very difficult for the primary to
change the magnetic flux in the core.

Does the electric field shielding from a perfect conductor need any
explanation?

Of course, an imperfect conductor will be an imperfect magnetic
shield. But a perfect conductor won't let any change of field
through, no matter how slow (no matter how low an EMF it generates),
so a perfect conductor works as a shield all the way down to DC. A
box made with an imperfect conductor is essentially a perfect shield
if the box's wall thickness is at least many skin-depths thick at the
frequency of interest.

That's a quick beginning. You can find lots more about this in E&M
texts. There's even useful stuff about it on the web. ;-)

Cheers,
Tom

Thanks, that helps, especially the paragraph about creating a magnetic field
in response that tends to cancel the original one, and the thickness of metal
with regard to frequency. The OP (Art Unwin) mentioned cancellation in more
complex terms, so I'm still not clear if this validates what he said or not.
It appears to but he mentions stuff I'm not likely to grasp in just an hour
or two of effort.. What I'm getting at is that I'm not sure if his calling
orthodoxy into question is all that drew the flak, or if there's something
obviously wrong in his post that I'm missing.

Also (though I'll likely find out about this when I look deeper), why is it
often ok for a Faraday cage to have holes in it? Braided screens, meshes,
perforated metal sheets, etc, I've seen many shields that are not a complete
'seal'... UHF TV cables especially seem to be very loosely shielded but they
work. Conversely, I found some nice coax in a skip once that had two heavy
braids amounting to almost complete coverage around a single fine stranded
core. (Found outside a telephone exchange, but I don't know what frequency
they were intended for, though I used some for an outdoor VHF receiving
quarter wave dipole with good results, and I suspect it will do for a SW
longwire once I get a matching transformer for it).

On Nov 30, 11:05*pm, Lostgallifreyan wrote:
tom wrote e.net:



orfus wrote:
Art Unwin wrote:
I have been reading the groups archives on shield antennas and Faraday
shields and the different auguments regarding how shielding or the
Faraday shield works. Frankly it is a total mess and should be removed
so that hams are not mislead.
Shielding is very simple.
A particle with a electromagnetic field strikes the outside of the
shield.
The magnetic field of same passes thru the shield some might say it is
coupled to the inside of the shield.
The magnetic vector component is out of phase with the electrical
field so it will be just a static particle at rest on the inside but
no inline with the electrical field vector which is now a staic
particle at rest on the outside
We now have a arbitrary boundary as discused by Gauss
For equilibrium all vectors impinging on the boundary must be aligned
such that they cancel.
To accomplish this the inner vector or charge MUST move sideways

*THE CHARGE WHEN ACCELERATED *CREATES A TIME VARYING CURRENT ALONE
WHILE THE OTHER FIELD VECTORS CANCEL OUT
( I believe that this was the object intended in *the cross field
antenna)

As with a applied varying current leaves a xmitter to create
radiation, so must the receiver obtain a time varying current.

Maxwells equations show equations with the electric field, the
magnetic field and a time varying current. When you have a electrical
field or vector of a static particle at rest outside the boundary
opposing the static vector on the inside of the boundary you have
nothing left EXCEPT a time varying current in the closed circuit.
For informative descriptions of how radiation occurs view the QRZ
forum of *( antenna construction and design ) threads (3) on the
double helix
antenna ( see you there)
Somebody some where should re write the above such that a definition
is left for those who follow and remove the garbage which is now in
place
TROLL!

Nope. Local loony.

You, however, are a troll until proven otherwise.

tom
K0TAR

Ok, at the risk of stirring muddy water, I'm curious now, I'm new to this
group, and the subject as there clearly seems to be more to it than I knew. I
also don't know of those archives mentioned so I haven't seen the context..

So in simple terms (hopefully) what is the truth of it? As far as I knew, a
photon at RF with energy but no mass will produce a current that changes over
time in a metal that it hits, though I imagine that as metal has resistance
there must also be a voltage too. I've also heard of the 'skin effect' that
means that at high RF frequencies, current flow tends to stay on the surface,
so clearly the picture isn't as simple as DC and Ohm's law. I also know that
when photons in optical fibres meet boundaries between layers they don't
reflect simply on one side, within one region of specific refractive index,
there's apparently some more complex information exchange that amounts to the
photon crossing the border before returning. Which makes me suspect that
equally exotic action happens when RF photons hit metal sheilds. So what IS
correct? And even if there is more to it, does the aggregate of many photons,
and the wave analysis of their behaviour, reduce to a simple model that makes
the OP correct?

I'm asking this because calls of 'troll' and 'loony' aren't working for me.

If you go back to the arbitary boundary of the Gaussian law of statics
and view it as a
Faraday shield it all becomes quite simple. If one adds a time varying
field you have the duplicate of Maxwells laws for radiation, where
the outside of the boundary is the radiator.
The Faraday shield supplies the transition from a static to a dynamic
field for xmission and
the reverse action for receiving.
Very basic my dear Watson, and a vindication that particles and not
waves create radiation
which puts it in line with deductions when other methods are applied.