Walter Maxwell wrote:
Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your address
appeared as . What's the ONEDOT? I now see it in your return
address on the rraa.

What's going on?

It's a spam preventative, Walt. Change '.ONEDOT.' to '.'
to get

Cecil Moore wrote:

Jim Kelley wrote:
Therein lies part of the problem with thinking that the unit
(Joules/sec) moves along a transmission line. Energy in Joules moves.
(Joules/sec) of power does not.

What about the Poynting Vector and Power Flow Vectors?

What about them?

What about the 60 Hz "power generation" and "power
distribution" system?

What are you trying to imply about power generation?

"Power distribution system" is really a misnomer. In the present day
venacular it would be called exactly what it is - an "energy
distribution system".

Are you saying that the trailing edge of an ExH wave
is not moving?

I don't recall ever expressing the opinion that traveling waves don't
travel.
However mathematical formulas do not propagate along transmission
lines. Fields do, but there is no such thing as an ExB "field".

73, Jim AC6XG

On Thu, 03 Jun 2004 16:18:36 -0500, Cecil Moore
wrote:

Walter Maxwell wrote:
Cecil, this is exactly what I've been trying to persuade you of, but always said
no, there is no short developed. But you must also agree that under this
condition the current doubles.

Nope, for complete destructive interference, both the E-field and
H-field associated with the two interfering waves collapse to zero.
For the resulting complete constructive interference, the ratio of
the E-field to the H-field equals the characteristic impedance of
the medium.

The two corresponding rearward-traveling superposing currents might be:
(2a at 180 degrees) superposed with (2a at zero degrees)
The superposed sum of the two rearward-traveling currents is zero.
This acts like an open where currents go to zero.

Of course, but the voltage doubles.

Nope, again here are the two sets of reflected waves.

#1 100v at zero degrees and 2a at 180 degrees = 200W

#2 100v at 180 degrees and 2a at zero degrees = 200W

Superposing those two reflected waves yields zero volts and
zero amps.

Well, Cecil, here's where we part company to a degree. Unlike voltage and
current that can go to zero simultaneously only in the rearward direction, E and
H fields can never go to zero simultaneously.

For "complete destructive interference" as explained in _Optics_, by
Hecht, the E-field and B-field (H-field) indeed do go to zero
simultaneously. That is what causes a completely dark ring in a
set of interference rings. Of course, a resulting corresponding
complete constructive interference causes the brightest of rings.

If Steve understands the action of the fields in the EM wave it's hard to
understand why he finds it so erroneous to associate voltage and current with
the their respective fields in impedance matching. Apparently he can't conceive
that the voltages and currents in reflected waves can be considered to have been
delivered by separate generators connected with opposing polarities.

He pretty much ignored current. His power equations are exact copies of
the light irradiance interference equations from optics, but he apparently
didn't realize it until I pointed it out to him.

Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

This sequence must also prevail in optics--are you sure you are interpreting
your double zero at the correct point in the circles?

Walt

Jim Kelley wrote:
... there is no such thing as an ExB "field".

Good grief, Jim, ExB is proportional to the irradiance of a light beam.
I'm sorry if my ASCII-limited character set hairlips you.
--
73, Cecil http://www.qsl.net/w5dxp



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Walter Maxwell wrote:
Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave
encounters a short circuit the E field goes to zero, but in its change from
normal level to zero, that changing field develops a corresponding H field that
adds to its original value, causing it to double--the H field does NOT go to
zero. But as it returns to its normal value that change in turn develops a new E
field propagating in the opposite direction. From there on the EM field
reconstitutes itself with normal E and H fields of equal value, each supporting
one half of the propagating energy.

I absolutely agree, Walt, "when an EM wave encounters a short circuit ...".
But when "complete destructive interference" occurs, something else happens.
The E-field goes to zero AND the H-field (B-field) goes to zero at the same
time. If you concentrate on the voltage, it looks like a short. If you
concentrate on the current, it looks like an open. It is both or neither.

Complete destructive interference requires that both fields go to zero
simultaneously and emerge as constructive interference in the opposite
direction obeying the rule that E/H=V/I=Z0. It is an energy reflection
that is also an impedance transformation at an impedance discontinuity.

Let's assume that 100v at zero degrees with a current of 2a at 180 degrees
encounters another wave traveling in the same rearward direction of 100v
at 180 degrees with a current of 2a at zero degrees. These two waves cancel.
The voltage goes to zero AND the current goes to zero. Each wave was associated
with 200 watts. So a total of 400 watts reverses directions. Assuming the
destructive interference occurred in a 50 ohm environment and the resulting
constructive interference occurred in a 300 ohm environment, the reflected
wave would be 346 volts at 1.16 amps. It's pretty simple math.

346*1.16 = 400 watts 346/1.16 = 300 ohms

The above quantities represent the destructive/constructive interference.
These quantities must be added to the other voltages and currents that are
present to obtain the net voltage and net current.
--
73, Cecil http://www.qsl.net/w5dxp



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It didn't bounce.
hmmmmm
I use spam filters.
Oh well, ........... takes all kinds.
73

BTW Walt, I've had one of your baluns for almost 20 years.
Damn thing just won't die.
H.

"Walter Maxwell" wrote in message
...
Cecil, H. Adam Stevens sent you an email earlier today, with a cc to me. I
replied to all, but the copy to you came back host unknown. But your
address
appeared as . What's the ONEDOT? I now see it in your
return
address on the rraa.

What's going on?

Walt

Cecil Moore wrote:

Jim Kelley wrote:
... there is no such thing as an ExB "field".

Good grief, Jim, ExB is proportional to the irradiance of a light beam.

Good grief, Cecil, irradiance isn't a field and doesn't propagates
either!

73, Jim AC6XG

On Fri, 04 Jun 2004 09:55:56 -0700, Jim Kelley
wrote:
Cecil Moore wrote:
Good grief, Jim, ExB is proportional to the irradiance of a light beam.

Good grief, Cecil, irradiance isn't a field and doesn't propagates
either!

Hi Jim,

Nearly every posting that Cecil pens with his rustic understanding of
Optics suffers from the obvious lack of experience.

There is a certain amount of pretense in this irradiance (an archaic
radiometric term used incorrectly for photometrics in an argument that
calls for luminous flux), much like quoting I²R and not knowing what
Ohms or Amperes are.

He's lucky there are so very few that appreciate the gaffs of his
unintended humor.

73's
Richard Clark, KB7QHC

Jim Kelley wrote:
Good grief, Cecil, irradiance isn't a field and doesn't propagates
either!

As I said earlier, irradiance is proportional to the cross product
of the E-field and B-field. Why do you think we call them fields
if they are not fields?

Exactly how does the irradiance from Alpha Centauri get to
us without propagating? Methinks you are playing word games.
--
73, Cecil http://www.qsl.net/w5dxp

Richard Clark wrote:
Nearly every posting that Cecil pens with his rustic understanding of
Optics suffers from the obvious lack of experience.

Everything I write about light/optics comes straight from
_Optics_, by Hecht. I assume he is more of an expert than
you are and he uses the term "irradiance" for the power
contained in a light beam. Hecht's interference equations
involving irradiance are identical in concept to Dr. Best's
power equations in his QEX article.
--
73, Cecil http://www.qsl.net/w5dxp