Cecil Moore wrote:
John Popelish wrote:

To the center conductor, carrying the standing wave, the shield is the
outside world. If there is no shield, the outside world is the
outside world, as far as displacement current goes. Do you imagine
this current changes in some way other than magnitude and wave
velocity when you wrap a shield around a wire carrying a standing wave?


No, that is your point, not mine. My point is that displacement
current to real ground is non-existent outside of a coax shield
(unless common mode current exists)

With you, so far..

and that it is usually a
secondary effect if the coax shield doesn't exist.

And then we part ways.

The primary
reason for the variation in standing wave current along the line
is the phasor sum of the forward and reflected wave phasors that
are rotating in opposite directions. Do you understand phasor
addition? 1 at zero + 1 at 180 deg = zero at a standing wave
node? Displacement current to real ground doesn't cause that.

I am making the point that if the displacement currents were
insignificant, outside a coax, then the speed of light for waves out
there would be infinite. And they are not, therefore those
displacement currents cannot be assumed to be insignificant.

I am explaining distributed network theory to you.


:-) How? By denying the existence of the individual H-fields
in forward and reflected EM waves? Now, that's really funny.

Exactly the opposite. I am explaining the distributed effect of the E
field along the wave.

http://www.qsl.net/w5dxp/travstnd.GIF


And I have agreed with that. Why do you keep bringing it up?


Because that's the whole point of this discussion. If you
agree with that, there is no reason to continue. I just
don't care about instantaneous current, Brownian motion, or
the exact location and velocity of every electron carrier.
There's too much uncertainty involved.

You are avoiding the very facts that would allow you to make an air
tight argument for your beliefs about "the whole point of the
discussion". You somehow picture current as a continuous thing from
one end of a conductor to the other, when it carries a traveling
energy wave. This is a misconception.

You appear to accept that current is a localized kind of thing (parts
of the conductor carry current, but those parts are separated by
nodes) when two traveling waves going in opposite directions
superpose, but have no concept that explains how this happens, only a
mathematical function that quantifies it.

What you don't get is, that the currents that each of those traveling
waves would have generated were localized, to begin with. Local
current cycles and voltage cycles are the water the energy waves ride
on over arbitrarily long distances along conductors and transmission
lines.

I know you don't care about this factoid, but understanding it would
allow you to think about "the entire reason for this discussion" much
more clearly.

John Popelish wrote:
I am making the point that if the displacement currents were
insignificant, outside a coax, then the speed of light for waves out
there would be infinite. And they are not, therefore those displacement
currents cannot be assumed to be insignificant.

But I am not talking about displacement currents within the
transmission line, as exists in free space. I am talking *solely*
about the displacement currents to *earth ground*. I contend that
those are often secondary effects as proven by the coax example.
Just how much displacement current to "earth ground" is there for
a coil located halfway between here and Alpha Centauri?

Exactly the opposite. I am explaining the distributed effect of the E
field along the wave.

And completely ignoring the H-field? In the treatment of those fields,
the only variation is Z0. For EM fields, there is no "across" and no
"through". The difference between voltage and current essentially
disappears except for their Z0 ratio. The equation for current in
a transmission line is identical to the equation for voltage except
for the Z0 term. Current "drops" are commonplace in lossy transmission
lines.

For instance, what is the current at the end of 200 feet of RG-58
terminated by a 50 ohm antenna used on 446 MHz when the source
current is 2 amps?

You are avoiding the very facts that would allow you to make an air
tight argument for your beliefs about "the whole point of the
discussion". You somehow picture current as a continuous thing from one
end of a conductor to the other, when it carries a traveling energy
wave. This is a misconception.

Maybe in the field of physics - not in the field of RF engineering.
For any two current points, I can calculate a point in between.
Sorry, but that's a characteristic of a *continuous* single-valued
function and can be proven mathematically. I admit to being a EE/math
major. I didn't take many pure physics courses so I am missing your
point about me being able to prove anything additional. Maybe it
will dawn on me after awhile.

What you don't get is, that the currents that each of those traveling
waves would have generated were localized, to begin with.

I realize that is the physicist talking and it agrees with my
earlier assertion that standing wave current doesn't flow. I
guess I'm so dense that I need help in proving what you think
I can prove with that information. Right now, I am apparently
missing something, maybe because of too much California Merlot.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
John Popelish wrote:

I am making the point that if the displacement currents were
insignificant, outside a coax, then the speed of light for waves out
there would be infinite. And they are not, therefore those
displacement currents cannot be assumed to be insignificant.


But I am not talking about displacement currents within the
transmission line, as exists in free space. I am talking *solely*
about the displacement currents to *earth ground*. I contend that
those are often secondary effects as proven by the coax example.
Just how much displacement current to "earth ground" is there for
a coil located halfway between here and Alpha Centauri?

Almost exactly as much as there is between a coil that is a half
wavelength from a grounded surface.

Exactly the opposite. I am explaining the distributed effect of the E
field along the wave.

And completely ignoring the H-field? In the treatment of those fields,
the only variation is Z0. For EM fields, there is no "across" and no
"through". The difference between voltage and current essentially
disappears except for their Z0 ratio. The equation for current in
a transmission line is identical to the equation for voltage except
for the Z0 term. Current "drops" are commonplace in lossy transmission
lines.

For instance, what is the current at the end of 200 feet of RG-58
terminated by a 50 ohm antenna used on 446 MHz when the source
current is 2 amps?

Somewhat less then 2 amps. Loss certainly occurs along that length,
at that frequency? So what? Are you thinking that this is the
predominate mechanism that is altering the current magnitude through
your coil? It is part of the answer, but not the whole answer.

You are avoiding the very facts that would allow you to make an air
tight argument for your beliefs about "the whole point of the
discussion". You somehow picture current as a continuous thing from
one end of a conductor to the other, when it carries a traveling
energy wave. This is a misconception.

Maybe in the field of physics - not in the field of RF engineering.
For any two current points, I can calculate a point in between.
Sorry, but that's a characteristic of a *continuous* single-valued
function and can be proven mathematically. I admit to being a EE/math
major. I didn't take many pure physics courses so I am missing your
point about me being able to prove anything additional. Maybe it
will dawn on me after awhile.

That is the best outcome I can hope for.

What you don't get is, that the currents that each of those traveling
waves would have generated were localized, to begin with.

I realize that is the physicist talking and it agrees with my
earlier assertion that standing wave current doesn't flow.

No current flows. Charges flow (move) the magnitude of that movement
past any point is current. I am picking nits, here, but the
distinction is important if you want to build on these simple
concepts. By the way, I am an EE, not a physicist, but I have to
think physics to do engineering.

I guess I'm so dense that I need help in proving what you think
I can prove with that information. Right now, I am apparently
missing something, maybe because of too much California Merlot.

Sounds like something I might do, this afternoon.

John Popelish wrote:

Cecil Moore wrote:
For instance, what is the current at the end of 200 feet of RG-58
terminated by a 50 ohm antenna used on 446 MHz when the source
current is 2 amps?

Somewhat less then 2 amps.

Does "somewhat" cover 24 dB of losses? :-) The point is that the
current "drops" by exactly the same amount as the voltage. That's
a characteristic of distributed networks as opposed to lumped
circuits. In a Z0 RF environment, the current has to "drop" by exactly
the same amount as the voltage to maintain the Z0 ratio. There are
really no "across" and "through" concepts as exist in DC circuitry.

I guess I'm so dense that I need help in proving what you think
I can prove with that information. Right now, I am apparently
missing something, maybe because of too much California Merlot.

Sounds like something I might do, this afternoon.

Which, helping or imbibing? :-)
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
John Popelish wrote:

Cecil Moore wrote:

For instance, what is the current at the end of 200 feet of RG-58
terminated by a 50 ohm antenna used on 446 MHz when the source
current is 2 amps?


Somewhat less then 2 amps.


Does "somewhat" cover 24 dB of losses? :-) The point is that the
current "drops" by exactly the same amount as the voltage. That's
a characteristic of distributed networks as opposed to lumped
circuits. In a Z0 RF environment, the current has to "drop" by exactly
the same amount as the voltage to maintain the Z0 ratio. There are
really no "across" and "through" concepts as exist in DC circuitry.

I guess I'm so dense that I need help in proving what you think
I can prove with that information. Right now, I am apparently
missing something, maybe because of too much California Merlot.


Sounds like something I might do, this afternoon.


Which, helping or imbibing? :-)

Merlot is what we Californians ship to out of state Republicans
in hopes of poisoning them into not voting in the next election.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Merlot is what we Californians ship to out of state Republicans
in hopes of poisoning them into not voting in the next election.

What do you ship out to Libertarians?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
Tom Donaly wrote:

Merlot is what we Californians ship to out of state Republicans
in hopes of poisoning them into not voting in the next election.


What do you ship out to Libertarians?

Libertarians don't believe in handouts.
73,
Tom Donaly, KA6RUH