On Dec 24, 9:39*am, "Dave" wrote:
"Keith Dysart" wrote in message

...
On Dec 23, 11:33 am, Roger wrote:

You are basing this conclusion on the observation that Power = V*I, and
because we can not detect V or I at some points in the standing wave,
then V*I is zero at these points. Correct math, but wrong conclusion.
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?

you can do it when it makes physical sense. *

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

At a minimum, the rules should cover DC circuits,
AC circuits, and transmission lines with various
excitations and terminations. Be sure that the
rules cover 60 Hz circuits and transmission lines
since this is a common application of P = V * I.

it does not make sense in
standing waves for all the obvious reasons that i have pointed out. *it does
make sense in the individual traveling waves. *just accept what your little
swr meter tells you, it shows the forward power and reflected power, that is
all you need and the only powers that make sense.

And for a challenging use case, please consider two
circuits connected together. The circuits are in black
boxes so you do not know their details, but the voltage
on the connection between the circuits is measured as
10 V RMS at 4 MHz. The current is measured as 0.

How much energy is being transferred between the
circuits?

1) P = VI, so 0.
2) P = VI some of the time, so there is insufficient
detail to answer the question.

Do you choose 1), 2) or perhaps some third answer?

...Keith

PS. The connection between the circuits is very small
so that it is possible that it is part of a
transmission line that continues into each box,
but you can not be sure.

On Dec 24, 10:49*am, Cecil Moore wrote:
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.

Since the discussion concerns real numbers and not
vectors, "multiply" is exactly correct.

But you haven't answered the hard part of the question.
When does P(x,t) not equal V(x,t) * I(x,t)?

...Keith

On Dec 24, 10:30*am, 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)

I am having great difficulty matching the words you
wrote with the request for an unambiguous definition
of "standing wave power".

Are you saying the concept is meaningless?
Or do you think you provided a definition?

I really wanted a definition from someone
who thought the "standing wave power" had
meaning.

...Keith

On Dec 24, 10:52*am, Cecil Moore wrote:
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.

Oooppps. Make that "so *if* V or I is always 0".

And if I had meant *NET* I would have written "net".

Since Pfor = Pref, then Pfor - Pref = 0

Isn't that superposition of power? Something that
most agree is not a legal operation.

In any case, for sure it is 'average' power.

All my examples are using instantaneous power.
If the instantaneous power is always 0, the
average power can not be different.

...Keith

"Keith Dysart" wrote in message
...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. use traveling waves then V*I works everywhere all
the time. use standing waves and it fails. period, end of story.

On Dec 24, 10:30 am, Cecil Moore wrote:

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)

OH NO! even cecil has it right! carry on cecil, you are more persistent
than i am. i have enough other stuff to do right now... i had some fun
yesterday while the wx was bad here, today there are better things to do,
like scrape ice and sand the driveway, and fix a yagi that got tipped in the
ice/wind last weekend. so you all carry on, i'm sure cecil will set you all
straight now so i'll leave it in his capable hands.

On Dec 24, 11:18*am, "Dave" wrote:
"Keith Dysart" wrote in message

...

Can you kindly articulate the rules you use to know
when it is appropriate to use P = V * I?

it is extremely simple. *use traveling waves then V*I works everywhere all
the time. *use standing waves and it fails. *period, end of story.

What happens on a line that is terminated in a real
impedance that is not equal to Z0?

There are aspects of both travelling waves and
standing waves present on the line.

Is it appropriate to use P = V * I?

...Keith

"Dave" wrote in message
news:cwPbj.1073$ML6.117@trndny04...


you can do it when it makes physical sense. it does not make sense in
standing waves for all the obvious reasons that i have pointed out. it
does make sense in the individual traveling waves. just accept what your
little swr meter tells you, it shows the forward power and reflected
power, that is all you need and the only powers that make sense.


Little SWR meter shows forward AND reflected power in one direction, and
reflected power only in reverse direction. Why is the Bird wattmeter
calibrated in Watts, measuring power (forward and reverse) and has chart to
calculate SWR, when there are no standing waves and no power in them?
Laying waves or sitting waves???

Seems to me that the PROBLEM is that some consider standing wave to be some
imaginary, stopped, frozen wave, no good, while some of us consider standing
wave to be the result of superposition of forward and reverse waves, that
can be (their components) measured, current heats when flowing through
resistance, voltage "burns" when poor dielectric.
Like there is standing wave current, but no standing wave, huh????
Or are we forgetting that we are dealing with electromagnetic waves?
Can someone sort out the terminology and definitions?

Yuri, K3BU

"Keith Dysart" wrote in message
...

What happens on a line that is terminated in a real
impedance that is not equal to Z0?

There are aspects of both travelling waves and
standing waves present on the line.

Is it appropriate to use P = V * I?

it is always appropriate to use P=V*I on the forward and reflected traveling
waves. it is never appropriate to use it on the standing wave voltage and
current. period... plonk.

Roger wrote:
Roger wrote:
Keith Dysart wrote:
clip ....

In the setup above used for "standing waves"
it can be seen that there is zero power in
the line every 90 degrees back from the open
end. At a zero power point, no energy is
being transferred. Therefore, the forward
and reverse waves can not be transferring
energy across these points. Conclusion:
forward and reverse waves do not always
transport energy.

....Keith

Hi Keith,

You are basing this conclusion on the observation that Power = V*I,
and because we can not detect V or I at some points in the standing
wave, then V*I is zero at these points. Correct math, but wrong
conclusion.

What you are forgetting is that power is also found from Power = V^2/Zo
and Power = I^2*Zo. More accurately, on the standing wave line,
Power = (V^2 + I^2)/Zo. This is why a SWR power meter detects both
current and voltage from the standing wave.

This will also be true on the quarter wave stub, which is really 1/2
wave length long electrically, when you consider the time required for
the wave to go from initiation to end and back to beginning point.
Power is stored on the stub during the 1/2 cycle energized, and then
that stored power acts to present either a high or low impedance to
the next 1/2 cycle, depending upon whether the stub is shorted or open.

I think you did a very good job in building your theory. It was only
at the end (where I think we need to consider additional ways of
measuring power) that we disagree.

73, Roger, W7WKB

Haste makes waste, and errors as well. The standing wave power
equation is incorrect. It should read "Power = V^2 / Zo + I^2 * Zo"

Sorry for any inconvenience, and for the several postings it will
probably stimulate.

73, Roger, W7WKB

Contributing to this news group always carries the risk of making errors
and this was a DUZZY! "Power = V^2 / Zo + I^2 * Zo" IS VERY WRONG AND
SHOULD NOT BE USED AS DESCRIBED. Thanks to several folks for pointing
this out.

I was writing to describe a concept, a concept that was obviously wrong
upon closer examination.

73, Roger, W7KB