K7ITM wrote:
(Yawn) So, I have this system where there's a wave in each direction
and they are identical amplitudes so that there is zero loss to
radiation or thermal dissipation. And in this system there is a series
coil through which the waves pass, and the current at each end of the
coil is different amplitude. That means that the coulombs/second
passing a point at one end of the coil is different than the
coulombs/second passing the other end of the coil. The currents can be
in phase or counter-phase. In fact, if the phases of the currents at
the two ends of the coil were not the same, then even equal-amplitude
currents at each end would imply that, except at certain instants of
time, there are differing coulombs/second passing the points at either
end of the coil.

What happens to that imbalance in charge? Where does it go? What do
we call something that behaves that way? What's so freakin' special
about that?

The charge briefly piling up and then being sucked out of such an
inductor is the same place charge piles up and is sucked out of parts
of a transmission lines with standing waves on them. That is the
shunt capacitance to the rest of the universe from each part of the
coil or transmission line that momentarily stores this charge.

So, I guess the word you are trying to get me to say is "capacitance".
Nobody says it is "freakin' special", though. Its common as dirt.

What do I win?

John Popelish wrote:

K7ITM wrote:
What happens to that imbalance in charge? Where does it go? What do
we call something that behaves that way? What's so freakin' special
about that?

The charge briefly piling up and then being sucked out of such an
inductor is the same place charge piles up and is sucked out of parts of
a transmission lines with standing waves on them.

Seems you got sucked in by a myth, John. The forward current is equal
at both ends of the coil. The reflected current is equal at both ends
of the coil. That takes care of any question of charge imbalance. There
simply isn't any.

Assume the coil is 90 degrees long and that the forward current is one
amp and the reflected current is one amp.

At one end of the coil, the forward and reflected currents are 180
degrees out of phase. The standing wave current is zero.

At the other end of the coil, the forward and reflected currents
are in phase. The standing wave current is 2 amps.

Now do you see why standing wave current is considered not to be flowing?
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
John Popelish wrote:

K7ITM wrote:

What happens to that imbalance in charge? Where does it go? What do
we call something that behaves that way? What's so freakin' special
about that?


The charge briefly piling up and then being sucked out of such an
inductor is the same place charge piles up and is sucked out of parts
of a transmission lines with standing waves on them.


Seems you got sucked in by a myth, John. The forward current is equal
at both ends of the coil.

Now, cut that out! Standing waves have sinusoidal current swings that
vary in amplitude with location. Location includes the two ends of a
coil.

The reflected current is equal at both ends
of the coil.

Smile when you say that.

That takes care of any question of charge imbalance. There
simply isn't any.

Oh poo. At current nodes charge piles up and spreads out, on
alternating half cycles. For one half cycle, the pile is positive,
and for the next it is negative. This is a basic transmission line
concept. If transmission lines had no shunt capacitance, there would
be no place to put this charge. But there is, so it is no problem.
Whether the transmission line is coax, twin line or a slow wave helix
makes little difference. The process is similar. Isn't this what you
have been arguing?

Assume the coil is 90 degrees long and that the forward current is one
amp and the reflected current is one amp.

At one end of the coil, the forward and reflected currents are 180
degrees out of phase. The standing wave current is zero.

At the other end of the coil, the forward and reflected currents
are in phase. The standing wave current is 2 amps.

Okay.

Now do you see why standing wave current is considered not to be flowing?

I see how no current is considered to be flowing. Current is charge
flowing. AC current is charge flowing back and forth.

But I see how two waves going in opposite directions create a standing
wave where the magnitude of the sinusoidal current at different points
along the standing wave have different magnitudes. And that between
the nodes where the amplitude is zero, the phase of the current
variation is constant.

John Popelish wrote:
Oh poo. At current nodes charge piles up and spreads out, on
alternating half cycles. For one half cycle, the pile is positive, and
for the next it is negative. This is a basic transmission line
concept. If transmission lines had no shunt capacitance, there would be
no place to put this charge. But there is, so it is no problem. Whether
the transmission line is coax, twin line or a slow wave helix makes
little difference. The process is similar. Isn't this what you have
been arguing?

If the forward traveling wave is equal in magnitude at both ends of the
coil, there is no net storage of energy due to the forward traveling wave.

If the reflected traveling wave is equal in magnitude at both ends of the
coil, there is no net storage of energy due to the reflected traveling
wave.

Superposing those two waves still results in no net storage of energy.
Sorry, got to hit the road.
--
73, Cecil http://www.qsl.net/w5dxp

Cecil Moore wrote:
John Popelish wrote:

Oh poo. At current nodes charge piles up and spreads out, on
alternating half cycles. For one half cycle, the pile is positive,
and for the next it is negative. This is a basic transmission line
concept. If transmission lines had no shunt capacitance, there would
be no place to put this charge. But there is, so it is no problem.
Whether the transmission line is coax, twin line or a slow wave helix
makes little difference. The process is similar. Isn't this what you
have been arguing?


If the forward traveling wave is equal in magnitude at both ends of the
coil, there is no net storage of energy due to the forward traveling wave.

Over a complete cycle, I agree, Within a single cycle, standing waves
slosh charge back and forth between adjacent current nodes, piling up
positive charge at one and negative charge at the next. This is the
reason that the voltage peaks at the tip of a quarter wave antenna.
It is a current node (because current has no place to go from there),
so charge piles up and produces voltage. But over a complete cycle,
the net charge movement is zero (the positive piles are he same size
as the negative piles).

If the reflected traveling wave is equal in magnitude at both ends of the
coil, there is no net storage of energy due to the reflected traveling
wave.

Same thing I said last paragraph.

Superposing those two waves still results in no net storage of energy.
Sorry, got to hit the road.

I'll put this on hold till you get back. Have fun.

Cecil wrote,
"The forward current is equal at both ends of the coil. The reflected
current is equal at both ends of the coil."

If that's really true, then the net current is precisely equal at both
ends of the coil. I thought you had been claiming that the current is
different at each end. Which way is it going to be? If they are
different phases, then they are NOT equal. If they are different
phases, where does the phase shift COME FROM? If I allow a wave in one
direction ONLY and the currents at the two ends are DIFFERENT in phase,
WHAT HAPPENS inside the coil to make them different? Where does the
extra charge come from, or go to?

It's all very simple. Yawn.

Hint: Replace the coil with a piece of coaxial transmission line,
formed into a loop so the input and output ends are adjacent. Short
the outer conductors together and notice that nothing changes in terms
of the voltages across each end of the line and currents in the center
conductors at each end. Note the difference in current at the two ends
of the line, and note the current in the single outer conductor
terminal of this three-terminal system. Notice that the sum of all
three currents at every instant in time is essentially zero (current
direction taken as positive going into each terminal). Got it yet? Do
you understand WHAT it is, besides the inductance, that allows a coil
to look like a transmission line? Do you understand that the mode is
not quite TEM, so some of the usual TEM transmission line behaviour is
not going to hold?

Cheers,
Tom

K7ITM wrote:

Cecil wrote,
"The forward current is equal at both ends of the coil. The reflected
current is equal at both ends of the coil."

If that's really true, then the net current is precisely equal at both
ends of the coil.

I was speaking above about the magnitudes only, not the phases.
It was clear from the rest of my posting that was the assumption.
The fact that you attempted to change the meaning by trimming is
noted.

So to be perfectly clear, here is my statement re-worded using
a 45 degree phase shift through the coil.

The forward current magnitude is equal at both ends of the coil.
The reflected current magnitude is equal at both ends of the coil.

At the bottom of the coil, the forward current is 1 amp at zero deg.
At the bottom of the coil, the reflected current is 1 amp at zero deg.
At the bottom of the coil, the standing wave current is 2 amps at
zero deg.

At the top of the coil, the forward current is 1 amp at -45 deg.
At the top of the coil, the reflected current is 1 amp at +45 deg.
At the bottom of the coil, the standing wave current is 1.4 amp at
zero deg.

I asked if you knew how to do phasor math but you trimmed out that
phasor math part of my posting.
--
73, Cecil http://www.qsl.net/w5dxp

No, Cecil, I did not try to change the meaning by trimming. I was
simply pointing out a basic flaw in your whole development. You use
the differing phase to establish that a travelling wave in each
direction results in a difference in the standing wave current at each
end, but then you try to use amplitude only to show no net current into
the coil. Now use the SAME phase difference you used to develop the
standing wave, and use it to determine the net AC current into the
coil, AT SOME PHASE. Now use the same phase difference in the other
direction to see that it also results in a net AC current AT SOME
PHASE. AND for the case where there is a standing-wave current
difference between the two ends of the coil, the net coil current is
EXACTLY as predicted by the vector sum of the two travelling wave net
currents.

Now you decide. Can I do phasor math? Do you need a specific example
with numbers, or can YOU work that out yourself? Suggest you use the
example from your previous posting. If that causes any difficulty, try
it with 180 degrees phase shift through the component. I've done it,
and it keeps giving me precisely the same answer as a full cycle of
instantaneous currents.

It appears that Cecil is back with many postings, but he seems to be
ignoring answering my question. Perhaps he's unable to do so. Just so
the lurkers understand that indeed it is possible to work through the
phasor math, here goes. Here's exactly the scenario Cecil set up,
quoted from his posting:

==========

"So to be perfectly clear, here is my statement re-worded using
a 45 degree phase shift through the coil.

The forward current magnitude is equal at both ends of the coil.
The reflected current magnitude is equal at both ends of the coil.

At the bottom of the coil, the forward current is 1 amp at zero deg.
At the bottom of the coil, the reflected current is 1 amp at zero deg.
At the bottom of the coil, the standing wave current is 2 amps at
zero deg.

At the top of the coil, the forward current is 1 amp at -45 deg.
At the top of the coil, the reflected current is 1 amp at +45 deg.
At the bottom of the coil, the standing wave current is 1.4 amp at
zero deg."

==========

OK, so the difference in "FORWARD" current from the bottom to the top
is:
fwd.bottom.current - fwd.top.current
= 1A at 0 degrees - 1 amp at -45 degrees
= 1+j0 - sqrt(.5)-j*sqrt(.5)
= 1-sqrt(.5) + j*sqrt(.5)
(about 0.765 at 67.5 degrees)


The difference in "REFLECTED" current from the bottom to the top is:
refl.bottom.current - refl.top.current
= 1A at 0 degrees - 1 amp at +45 degrees
= 1+j0 - sqrt(.5)-j*sqrt(.5)
= 1-sqrt(.5) - j*sqrt(.5)

The SUM of these two differences is:
[1-sqrt(.5) + j*sqrt(.5)] + [1-sqrt(.5) - j*sqrt(.5)]
= 2 - 2*sqrt(.5) + j0
= 2 - sqrt(2) + j0
= 2 - sqrt(2) at zero degrees

The standing wave current at the bottom of the coil is 2 amps just as
Cecil suggests at one point: It's the sum of the "forward" and
"reflected":
net current at the bottom = sw.bottom.current
= 1+j0 + 1+j0
= 2+j0
= 2 at zero degrees

Presumably Cecil meant that the standing wave current at the TOP
(not the BOTTOM) of the coil is 1.4 amps at 0 degrees. That's close,
but more exactly, it's
net current at the top = sw.top.current
= sqrt(.5)-j*sqrt(.5) = sqrt(.5)+j*sqrt(.5)
= 2*sqrt(.5)
= sqrt(2)
= sqrt(2) at zero degrees.

So the difference in net current (that is, the difference in the
standing wave current) between the top and the bottom of the coil
in this example is exactly:

sw.bottom.current - sw.top.current
= 2 at zero degrees - sqrt(2) at zero degrees
= 2 - sqrt(2) at zero degrees

So, we see that the difference in current between the bottom and
the top is exactly the same, independent of whether we just use
the standing-wave currents, or the currents in the "forward"
travelling wave plus the currents in the "reflected" wave. That it's
also
exactly the same answer you get by looking at a full cycle of
instantaneous currents is left as an exercise (fairly simple) for the
reader.

Either way, there is a difference, and that current must go
somewhere. It should be pretty easy to account for it. In
fact, it's not even very hard to predict fairly accurately in
the case of a loading coil in an antenna perpendicular to a ground
plane or equivalently in a symmetrical doublet.

Cheers,
Tom