Jim Kelley wrote:
Cecil seems to believe that a standing wave is
more than a superposition of voltages.

Exactly, it is also a superposition of currents in phase with
the voltages. V*I*cos(0)=power

Everyone, including you, gets into trouble by completely ignoring
the superposition of currents. When are you (and your boss) going
to stop violating the laws of physics?
--
73, Cecil http://www.qsl.net/w5dxp



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H. Adam Stevens wrote:
That's why I prefer antennas with no reflections.

Dipoles are reflected-wave antennas. You don't like them?
--
73, Cecil http://www.qsl.net/w5dxp



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Tam/WB2TT wrote:

"Cecil Moore" wrote:
The 'why' is conservation of energy. If there are only two directions
available and energy was traveling in one direction and now it isn't,
it's a no-brainer to realize that it must have changed directions.

Cecil, I think a more convincing argument is that I can take a slotted line
and directly measure a standing wave on it. A wave traveling in one
direction can not do that.

Yes, that is the basis of my earlier challenge. Nobody has been able to
provide a standing wave not composed of a forward-traveling wave superposed
with a rearward-traveling wave. I have offered anyone and everyone $100 who
can accomplish that in a single-source/single transmission line/single-load
system. So far, no takers. One wonders why, considering all the gurus on
this newsgroup. What I get instead are obfuscations of how waves can exist
without energy.
--
73, Cecil http://www.qsl.net/w5dxp



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H. Adam Stevens wrote:
when they're resonant and match the transmission line
they ain't no reflections back down the line
it's all gone to somewhere else

H., there are standing waves all over a dipole antenna.
Its traveling-wave feedpoint impedance is about 600 ohms.
Only the reflections from each end of a dipole lower the
feedpoint impedance to 50 ohms or so. The reflected current
arrives back in phase at the feedpoint and the reflected
voltage arrives back out of phase. You can thank destructive
interference for the low feedpoint impedance of a dipole.

If you don't like reflections, you don't like a dipole and
should probably try a terminated rhombic. :-)
--
73, Cecil http://www.qsl.net/w5dxp



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H., there are standing waves all over a dipole antenna.
Its traveling-wave feedpoint impedance is about 600 ohms.
Only the reflections from each end of a dipole lower the
feedpoint impedance to 50 ohms or so. The reflected current
arrives back in phase at the feedpoint and the reflected
voltage arrives back out of phase. You can thank destructive
interference for the low feedpoint impedance of a dipole.

If you don't like reflections, you don't like a dipole and
should probably try a terminated rhombic. :-)
--
73, Cecil http://www.qsl.net/w5dxp



I think this should be on the front of each antenna book. It would open eyes of
many "gurus" especially when looking at loading the shortened antennas.

Thanks Cecil!

Yuri

On Thu, 27 May 2004 17:24:40 -0500, Cecil Moore
wrote:



Hank, when reflected current flows backwards through a pi-net loading
coil, some of the reflected power is dissipated as I^2*R losses in
the coil. Other than that, a properly tuned pi-net causes a match
point that reflects all the reflected energy back toward the load.


This is a great thread, I have a question about your post. If the all
the reflected energy is reflected back toward the load, is it in phase
with the original or subsequent energy? or does it matter?

If a match point exists in a ham radio antenna system, no reflected
energy will reach the source. This is the great majority of amateur
radio systems and no-reflections-at-the-source is the goal of every
ham. The thing that you are worried about is the unusual case where
reflections are allowed to reach the source.

Ken wrote:
This is a great thread, I have a question about your post. If the all
the reflected energy is reflected back toward the load, is it in phase
with the original or subsequent energy? or does it matter?

I am from the old school, Ken. I believe that power is a scalar and
doesn't possess phase. IMO, any phase calculation that you see being
used on power originates from the voltage phase and/or current phase
associated with that power. It is done all the time in optics.

The answer to your question is, assuming PA stands for Phase Angle:
Since forward voltages and forward currents are in phase, V*I*cos(PA)
yields watts with no vars. Since reflected voltage and reflected current
are 180 degrees out of phase, V*I*cos(PA) yields watts with no vars.
So extremely loosely speaking, the "phase" of the power can be considered
to be the same as the phase of the voltage or current since they are
the same phase. However, such a consideration cannot be considered to
be good physics.
--
73, Cecil http://www.qsl.net/w5dxp




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Cecil Moore wrote:

I am from the old school, Ken. I believe that power is a scalar and
doesn't possess phase. IMO, any phase calculation that you see being
used on power originates from the voltage phase and/or current phase
associated with that power. It is done all the time in optics.

Interesting comment, however the "power" companies would have a great
difference of opinion with you.

And no, I'm not about to enter this very silly argument you wish to
continue, apparently until the protons all decay.

tom
K0TAR

Tom Ring wrote:
Cecil Moore wrote:

I am from the old school, Ken. I believe that power is a scalar and
doesn't possess phase. IMO, any phase calculation that you see being
used on power originates from the voltage phase and/or current phase
associated with that power. It is done all the time in optics.

Interesting comment, however the "power" companies would have a great
difference of opinion with you.

My degree is in Power Engineering and I learned the above at
Texas A&M. Of course, Volt-Amps have a phase, but power always
lies along the real axis. What would be the physical meaning of
25 watts at 45 degrees?
--
73, Cecil http://www.qsl.net/w5dxp



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