Roy Lewallen wrote:
3. How do you resolve this with the graphs in Terman and your
explanation of the voltage and current being in quadrature everywhere
along the line?

The standing wave voltage is *ALWAYS* in quadrature with
the standing wave current at all points and at all times.
If the total voltage is not in quadrature with the total
current, there is a traveling wave in the mixture.
--
73, Cecil http://www.w5dxp.com

Denny wrote:
1. A standing wave is not 'standing' in time... It phase rotates at
the same rate as the excitation frequency...

Yes, but that rotation is occurring at a fixed point on
the wire. The standing wave is not moving - it is just
rotating in a fixed place on the wire. Standing waves
do not flow. They are basically an illusion created by
the superposition of two traveling waves.

2. It is real because I can measure it with a volt meter and I can
extract power from it with a lamp, simultaneously..

You cannot extract steady-state power from standing waves.
You can only extract steady-state power from traveling waves.

However, you can extract transient-state power from standing
waves. That's what happens at key-up - the energy in the
standing wave is radiated after the source is disconnected.

Try this thought... A Tesla coil (automobile spark coil) with no load
on the output is all standing wave voltage and no current - so
according to some has no power... Touch your finger to it...

When you touch your finger to it, it is no longer steady-state.
Standing waves can and do give up their energy as joules/sec
power during transient states. But they change during that
process.
--
73, Cecil http://www.w5dxp.com

On Dec 24, 12:01*am, Cecil Moore wrote:
Richard Harrison wrote:
Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

We can make Keith's assertion true by the addition of one
word.

"Therefore, the forward and reverse waves cannot be transferring
*net* energy across these points. As Ramo and Whinnery say about
the forward and reflected Poynting vectors:

With an open circuited line, I agree that there is no
net energy transfer at any point on the line. At most
points on the line, there is energy sloshing back and
forth, but netting to zero.

My statement about those 90 degree points where the
voltage or current is always 0 is much stronger: NO
energy transfer.

This follows inexorably from P(x,t) = V(x,t) * I(x,t).

If you disagree with the general applicability of this
equation, please indicate when it can and can not be
applied.

...Keith

Keith Dysart wrote:
And could someone who likes to write "standing
wave power" (Yuri perhaps?) please provide an
unambiguous definition? It does not have to be
the "right" definition, or agreed by all, just
any definition which is unambiguous.

Confusion reigns because of steady-state short cuts.

The power density (Poynting vector) of any EM wave
is ExH. EM waves cannot exist without a power density.

For pure standing waves, ExH = 0. Therefore, a pure
standing wave is technically NOT an EM wave. It
doesn't move and contains no power. In many ways,
it is an illusion.

A standing wave is a math model created in the human
mind as a useful shortcut. Shortcuts do NOT dictate
reality. Reality is supposed to do the dictating.

Standing waves are the results of the superposition
of two traveling waves. Any power extracted comes
from the component traveling waves, not from the
standing waves. For pure standing waves:

ExH = V*I*cos(A) = 0 watts (per unit area)
--
73, Cecil http://www.w5dxp.com

Dave wrote:
"Denny" wrote:
1. A standing wave is not 'standing' in time... It phase rotates at
the same rate as the excitation frequency...
If the phase is rotating then V and I are changing - else Feynman is
rotating in his grave...

true, it is 'standing' in space. it does not move along the line.

How can such a signal be used to measure the delay through
a mobile loading coil?

why do i even bother... time to start plonking more of the ones who refuse
to learn and reduce the noise level on here even more.

What you have to do is overcome their programming -
that math models dictate reality, not vice versa.
Just keep dripping on that stone - someday it will
wear away.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
Where do you get so many goofy ideas? Do you have any references at
all that support your contention that standing wave energy does not
meet the definition of EM energy? I have been in the wave business
professionally for about 40 years, and I have read many technical
papers, reference books, and text books. I have yet to encounter
anything that indicated the inferior nature of standing waves in the
energy community.

I guess the authors of the textbooks never thought anyone
would be so ignorant as to believe that EM waves can stand
still. :-)

EM waves are photonic in nature must travel at the speed of
light in the medium. A standing wave stands still and oscillates
in place. Therefore, A standing wave is not an EM wave - It is
something else, by definition.

Cecil,

Thanks, you completely confirmed my suspicion. This photonic limitation
is something that exists only in your head.

8-)

73,
Gene
W4SZ

On Dec 24, 9:26*am, "Dave" wrote:
"Keith Dysart" wrote in message

...
On Dec 23, 10:12 am, "Dave" wrote:

Are you really prepared to throw away P = VI?

yes, when the V and I are the superimposed voltage and current that you
insist are the real current and voltage on the line.

They are the real current and voltage. I can measure them
with voltmeters and ammeters.

Directional wattmeters measure the real current and
voltage and perform some arithmetic on these measured
values to display "forward power" and "reflected power".

"Forward and reflected power" are derived from the
real voltage and current.

In my analysis, P(x,t) = V(x,t) * I(x,t) is the equation that means
the power at any point and time can by obtained by measuring the
actual
voltage and current on the line at the point and time of interest.

try to look at it this way. *when you look at the forward and reflected
waves separately it is intuitively obvious how the power calculation shows
the flow along the line with each wave. *however, when you look at standing
waves you get spots every 1/4 wave where either V(x,t) or I(x,t) is ALWAYS
zero... by V*I this means the power at that point is ALWAYS zero. *since
power is just the measure of the flow of energy, and energy can neither be
created nor destroyed then in the traveling wave there can be no energy flow
past those points. *where it is obvious from the individual Vf(x,t) and
Vr(x,t) or If(x,t) and Ir(x,t) that are ALWAYS related by Z0 at every point
on the line that power does flow both directions. *an obvious contradiction

Absolutely. And many readers resolve the contradiction in the
wrong direction.

And then are willing to throw away P = VI.

and if you can't see it by this point i give up.

Are you sure you want to throw away this ability? Are you sure you
want to claim that instantaneous power can NOT be obtained by
multiplying the instaneous measured voltage by the instanteous
measured current?

i want to throw away this falicy and replace it with the real physically
correct calculation.

Throwing this away will invalidate much.

only in your mind.

I have said it enough times now, and hate repeating myself... so you can
live with your poor misguided assumptions and formula. *i have shown the
obvious errors

I'd suggest that remains to be seen.

...Keith

Keith Dysart wrote:
Are you really saying that if I measure the instantaneous
voltage and the instantaneous current then I can NOT multiply
them together to obtain the instantaneous power?

It certainly works some of the time.

If I can not do it all the time, when can I do it?

Actually, "multiply" is ambiguous. You need to take
the *dot product* of the voltage and current to
obtain power.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Keith Dysart wrote:
And could someone who likes to write "standing
wave power" (Yuri perhaps?) please provide an
unambiguous definition? It does not have to be
the "right" definition, or agreed by all, just
any definition which is unambiguous.

Confusion reigns because of steady-state short cuts.

The power density (Poynting vector) of any EM wave
is ExH. EM waves cannot exist without a power density.

For pure standing waves, ExH = 0. Therefore, a pure
standing wave is technically NOT an EM wave. It
doesn't move and contains no power. In many ways,
it is an illusion.

A standing wave is a math model created in the human
mind as a useful shortcut. Shortcuts do NOT dictate
reality. Reality is supposed to do the dictating.

Standing waves are the results of the superposition
of two traveling waves. Any power extracted comes
from the component traveling waves, not from the
standing waves. For pure standing waves:

ExH = V*I*cos(A) = 0 watts (per unit area)

Cecil,

Do you simply make this stuff up in some arbitrary fashion, or is there
a method to your madness?

With such profound statements as, "a pure standing wave is technically
NOT an EM wave", it you might either offer some sort of reference or
start planning your trip to Stockholm.

73,
Gene
W4SZ

Keith Dysart wrote:
And P = V * I seems rather fundamental, so V or
I is always 0, then P must always be 0.

Make that *NET* power and you will be correct. There is
zero *NET* power transfer in pure standing waves.

Since Pfor = Pref, then Pfor - Pref = 0
--
73, Cecil http://www.w5dxp.com