On Thu, 11 Aug 2005 12:39:56 -0500, (Richard
Harrison) wrote:

No potential exists to evoke a current flow.

Hi Richard,

The potentials (energy) abound, you even describe them. The sum of
all voltages (energy) equal zero at balance (absolutely no power into
the load of the meter), but that does not mean there has been any
cancellation of energy - that is glaringly obvious. Remove the load,
and you can set the bridge adjustments ANYWHERE and there would still
be no interaction of energies.

At absolutely every segment of the circuit, there is a potential
difference, and all must exist to present the null (which is evidence
of a potential difference of zero, not evidence of no potentials).
This is Kirchoff's law in the raw.

Contrary wise, removing the source of energy will give an exactly
identical reading on the current meter, and to say that no energy is
the same as balanced energy is a shuffle on par with the old shell
game - King Lear muttering "Nothing begets nothing."

For cancellation of energy to exist, your analogy proves there must be
an identical energy present to offset it. This explicitly admits
energy remains in the face of narrowly inspecting an isolated instance
of balance (or cancellation - all the same, only a matter of
topology).

73's
Richard Clark, KB7QHC

Richard Harrison wrote:
Given a perfect transmission line with a complete reflrction, a length
can be found which produces a reflection with with the same phase and
magnitude as that of the generator. With equal and same phase volts on
either side of the generator/line junction, current does not flow.

*Net* current doesn't flow. But a circulator and load will
separate the forward component from the reflected component.
For instance, at a point on a transmission line where the
net current is zero, the forward current may be 100 amps and
the reflected current may be 100 amps, just out of phase with
each other.

In your above example, if the source is a signal generator
equipped with a circulator and load resistor, the net current
at the signal generator terminals can be measured to be zero
while the circulator resistor is smoking. Thus the difference
between distributed networks and lumped circuits.
--
73, Cecil, http://www.qsl.net/w5dxp

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Richard Clark wrote:
The deletion was deliberate because energy does not move ...

The Sun's energy is necessary for life on earth. If the
Sun's energy doesn't move, how does it get from the sun
to the earth?
--
73, Cecil, http://www.qsl.net/w5dxp

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Richard Clark wrote:
For cancellation of energy to exist, your analogy proves there must be
an identical energy present to offset it.

One possibility is that the source simply delivers less energy.
One possibility is constructive interference.
One possibility is additional dissipation.
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 11 Aug 2005 13:18:53 -0500, Cecil Moore
wrote:
The Sun's energy is necessary for life on earth. If the
Sun's energy doesn't move, how does it get from the sun
to the earth?

If you wish the crutch of illusion, the photon moves, the energy is in
its wavelength. You may analyze the photon as a wave, or as a
particle, but none of the energy is ever destroyed - short of
collision with an anti-photon. Reflection, interference,
what-have-you never presents us with opportunity to indulge in nuclear
pyrotechnics. If the sun were suddenly replaced with its equivalence
in Dark Energy, I don't think we would freeze.

On Thu, 11 Aug 2005 13:23:35 -0500, Cecil Moore
wrote:
One possibility is that the source simply delivers less energy.
Turning off the source of the bridge will balance the meter too?
Amusing.
One possibility is constructive interference.
That is one whacked out lack of balance.
One possibility is additional dissipation.
The meter suddenly becomes lossy?

Richard Clark wrote:
If you wish the crutch of illusion, the photon moves, the energy is in
its wavelength. You may analyze the photon as a wave, or as a
particle, but none of the energy is ever destroyed ...

That's nice but we weren't talking about energy being
destroyed. We were talking about energy moving. You said
energy doesn't move.

The deletion was deliberate because energy does not move.

So what is the origin of the energy that turns my arms brown?
--
73, Cecil, http://www.qsl.net/w5dxp

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On Thu, 11 Aug 2005 14:35:25 -0500, Cecil Moore
wrote:

Richard Clark wrote:
If you wish the crutch of illusion, the photon moves, the energy is in
its wavelength. You may analyze the photon as a wave, or as a
particle, but none of the energy is ever destroyed ...

That's nice but we weren't talking about energy being
destroyed. We were talking about energy moving. You said
energy doesn't move.
As indicated above.

The deletion was deliberate because energy does not move.

So what is the origin of the energy that turns my arms brown?
Were you ancestors slave holders? The old in-and-out, I suppose.
Sweat and Texas dust? Presuming the sweat comes from work.
The power of dissipated wavelengths from photonic sources delivering
them? Did the thermometer pop out yet?
Rug burn?
Too much caffeine sipping?

Odd how many personal problems get posted here. Try
play.doctor.dermatology and add more details for your symptoms

Cecil, W5DXP wrote:
"Net current doesn`t flow."

I`m inclined to agree, but at first current flows with a volts to amps
ratio rqual to the Zo of the line until a reflection returns to the
connection point of the stub. Then the total phase rotation within the
stub has reached 360-degrees. The complete reflection supplies a
reflected voltage wqual to the incident voltage. There is no difference
of potential or phase to evoke current. It`s the equivalent of a very
high impedance. Almost no more current is motivated to flow, once the
steady-state condition is reached.

Alexander Wing wrote on page 29 of "Transmission Lines, Antennas, and
Wave Guides":
"Suppression of Even Harmonics.- An application of a short-circuited
quarter-wavelength line is to suppress any unwanted even harmonics in
the output of a radio transmitter. A short-circuited one-quarter
wavelength long at the desired output frequency may be connected across
the output terminals or across the antenna feeder at any point without
placing much load on the transmitter at the fundamental or desired
output frequency, since at this frequency such a section has an
impedance ideally infinite, actually about 400,000 ohms."

How much current flows into 400.000 ohms?

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

Cecil, W5DXP wrote:
"Net current doesn`t flow."

I`m inclined to agree, but at first current flows with a volts to amps
ratio rqual to the Zo of the line until a reflection returns to the
connection point of the stub. Then the total phase rotation within the
stub has reached 360-degrees. The complete reflection supplies a
reflected voltage wqual to the incident voltage. There is no difference
of potential or phase to evoke current. It`s the equivalent of a very
high impedance. Almost no more current is motivated to flow, once the
steady-state condition is reached.

Richard, some people use that exact same argument to try to prove
that no current is flowing in the middle of a transmission line
where forward power equals reflected power. The strange thing is
that the current 1/4WL away from that zero net current point is
sky high.

If the current 1/4WL away from your above source output terminals
is indeed sky high, it is because the forward and reflected currents
are in phase at that point which means they are 180 degrees out of
phase at the source output which means they are both still there.
If the net current is zero, all it means is that |Ifor|-|Iref| = 0.
It tells us nothing about the magnitudes of Ifor and Iref.

Trying to treat a distributed network as a lumped circuit can lead
to mistakes.

Alexander Wing wrote on page 29 of "Transmission Lines, Antennas, and
Wave Guides":
"Suppression of Even Harmonics.- An application of a short-circuited
quarter-wavelength line is to suppress any unwanted even harmonics in
the output of a radio transmitter. A short-circuited one-quarter
wavelength long at the desired output frequency may be connected across
the output terminals or across the antenna feeder at any point without
placing much load on the transmitter at the fundamental or desired
output frequency, since at this frequency such a section has an
impedance ideally infinite, actually about 400,000 ohms."

How much current flows into 400.000 ohms?

How much current flows in the shorted end of the stub? Let's say we
measure it at 2 amps. Where does that current come from?

The 400,000 ohms is (Vfor+Vref)/(Ifor+Iref) at the mouth of the stub
where V's and I's are phasors. Knowing that (Ifor+Iref) is a small
value doesn't tell us anything about the magnitudes of Ifor and Iref
except that they are nearly equal as is always the case for a
low-loss stub. Knowing (Inet=|Ifor|-|Iref|) is small doesn't tell us
a thing about the magnitude of (|Ifor|+|Iref|) that exists at the
shorted end of the stub. Does that make sense?
--
73, Cecil http://www.qsl.net/w5dxp

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