"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
news:UTAbj.1170$OH6.803@trndny03...

"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
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"Yuri Blanarovich" wrote in message
...
So you are trying to say that there is standing wave voltage but no
standing wave current and therefore no power associated with
current???

i am saying that there is standing wave voltage, and standing wave
current, but there is no physical sense in multiplying them to
calculate a power. hence the concept of standing power waves is
meaningless.


K3BU found out that when he put 800W into the antenna, the bottom of
the coil started to fry the heatshrink tubing, demonstrating more
power to be dissipated at the bottom of the coil, proportional to the
higher current there, creating more heat and "frying power" (RxI2).
This is in perfect

right R*I^2 makes perfect sense. you are talking about ONLY the
current standing wave which makes perfectly good sense. and the R*I^2
losses associated with it make perfect sense. BUT, resistive losses
ARE NOT a result of power in the standing wave, they are resistive
lossed resulting from the current. remember the initial assumptions of
my analysis, a LOSSLESS line, hence there are not resistive losses. If
this requirement is changed then you can start to talk about resistive
heating at current maximums and all the havoc that can cause.

They seem to be real current and voltage, current heats up resistance,
voltage lights up the neons and power is consumed, portion is
radiated. The larger the current containing portion, the better
antenna efficiency.
Where am I wrong?

again on the voltage, it is exactly analogous to the current above.
voltage waves capacitively coupled to a neon bulb are losses and not
covered in my statements. But again note, that type of measurement is
measuring the peak voltage of the voltage standing wave, and any power
dissipated is sampled from that wave and is not a measure of power in
the standing wave.

I have a hard time to swallow statement that there is no power in
standing wave, when I SAW standing wave's current fry my precious coil
and tip burned off with spectacular corona Elmo's fire due to standing
wave voltage at the tip.

YOU HAVE IT RIGHT! the standing wave current causes heating. the
standing wave voltage causes corona. but neither one represents POWER
in the standing waves.


So we have better than perpetual motion case - we can cause heating
without consuming power. I better get patent for this before Artsie gets
it! There must be some equilibrium somewhere.

...like, there is standing wave current, there standing wave voltage,
but no standing wave and no power in it?

Richard, wake up your english majorettes and snort it out :-)


remember the relationship that must apply within the coax. and can be
extended to antennas, though it gets real messy taking into account the
radiated part and the change in capacitance and inductance along the
length of the radiating elements.

P=V^2/Z0=I^2*Z0

now remember that you only have to look at one or the other and at all
times the relationship in each wave must obey V=I*Z0.

Now consider the infamous shorted coax. at the shorted end the voltage
standing wave is always ZERO, by your logic the power would always be
zero at that point, but how can that be?? conversly, at a point where
the current standing wave is always zero there can be no power in the
standing wave, but at that point the voltage is a maximum so would say
the power was a maximum... an obvious contradiction.


I know one thing, when standing wave current is high, I get loses in the
resistance, wires heating up - power being used.
When standing wave voltage is high, I get dielectric loses, insulation
heats up and melts - power being used, ergo there is a power in standing
wave and is demonstrated by certain magnitude of current and voltage at
particular distance and P = U x I. I don't know what you are feeding
your standing wave antennas, but I am pumping power into them and some
IS radiated, some lost in the resistive or dielectric loses.

73 Yuri



you are SO CLOSE... open your eyes, turn off your preconceived notions
and read what i wrote again slowly and carefully.

FIRST remember the assumption was a LOSSLESS line so there are no
dielectric or resistive losses. But this is only useful because it makes
it easier to see that the power given by V*I in the standing waves
doesn't make sense.

YOU ARE CORRECT when you say that the R*I^2 loss is REAL... and YOU ARE
CORRECT when you say the V^2/R loss in dielectric is REAL.

Where you lose it is that the V*I for the standing waves is not
correct... this is because V and I are related to each other and you
can't apply superposition to a non-linear relationship. If you think you
have a way to do it then please take the given conditions of a lossless
transmission line, shorted at the end, in sinusoidal steady state, and
write the equation for power in the standing wave at 1/4 and 1/2 a
wavelength from the shorted end of the line.


So let me get this straight:
I describe real antenna situation, which is a standing wave circuit, with
real heating of the coil, which is consuming power demonstrably
proportional to the amount of standing wave current.
You are not answering rest of my argument.
You bring in lossless transmission line to argue that there can not be
power and no standing waves.
I still don't get it.

73 Yuri


ok, last try.. YOU ARE RIGHT!!! "consuming power demonstrably proportional
to the amount of standing wave current". LISTEN, YOU ARE RIGHT!!! the loss
in your coil is proportional to the square of the standing wave CURRENT. as
long as you keep saying that you are RIGHT. now repeat after me... the loss
in your coil is proportional to the square of the standing wave CURRENT.
emphasize CURRENT every time. DO NOT start talking about POWER in the
standing wave, then you will be wrong. Use the CURRENT young Yuri, Use the
CURRENT... forget the POWER of the standing wave.

On Dec 24, 6:49*am, "Dave" wrote:
"Yuri Blanarovich" wrote in message

...







"Dave" wrote in message
news:UTAbj.1170$OH6.803@trndny03...

"Yuri Blanarovich" wrote in message
...

"Dave" wrote in message
news:mozbj.844$si6.691@trndny08...

"Yuri Blanarovich" wrote in message
...
So you are trying to say that there is standing wave voltage but no
standing wave current and therefore no power associated with
current???

i am saying that there is standing wave voltage, and standing wave
current, but there is no physical sense in multiplying them to
calculate a power. hence the concept of standing power waves is
meaningless.

K3BU found out that when he put 800W into the antenna, the bottom of
the coil started to fry the heatshrink tubing, demonstrating more
power to be dissipated at the bottom of the coil, proportional to the
higher current there, creating more heat and "frying power" (RxI2).
This is in perfect

right R*I^2 makes perfect sense. *you are talking about ONLY the
current standing wave which makes perfectly good sense. *and the R*I^2
losses associated with it make perfect sense. *BUT, resistive losses
ARE NOT a result of power in the standing wave, they are resistive
lossed resulting from the current. *remember the initial assumptions of
my analysis, a LOSSLESS line, hence there are not resistive losses. *If
this requirement is changed then you can start to talk about resistive
heating at current maximums and all the havoc that can cause.

They seem to be real current and voltage, current heats up resistance,
voltage lights up the neons and power is consumed, portion is
radiated. The larger the current containing portion, the better
antenna efficiency.
Where am I wrong?

again on the voltage, it is exactly analogous to the current above.
voltage waves capacitively coupled to a neon bulb are losses and not
covered in my statements. * But again note, that type of measurement is
measuring the peak voltage of the voltage standing wave, and any power
dissipated is sampled from that wave and is not a measure of power in
the standing wave.

I have a hard time to swallow statement that there is no power in
standing wave, when I SAW standing wave's current fry my precious coil
and tip burned off with spectacular corona Elmo's fire due to standing
wave voltage at the tip.

YOU HAVE IT RIGHT! *the standing wave current causes heating. *the
standing wave voltage causes corona. *but neither one represents POWER
in the standing waves.

So we have better than perpetual motion case - we can cause heating
without consuming power. I better get patent for this before Artsie gets
it! There must be some equilibrium somewhere.

...like, there is standing wave current, there standing wave voltage,
but no standing wave and no power in it?

Richard, wake up your english majorettes and snort it out :-)

remember the relationship that must apply within the coax. and can be
extended to antennas, though it gets real messy taking into account the
radiated part and the change in capacitance and inductance along the
length of the radiating elements.

P=V^2/Z0=I^2*Z0

now remember that you only have to look at one or the other and at all
times the relationship in each wave must obey V=I*Z0.

Now consider the infamous shorted coax. *at the shorted end the voltage
standing wave is always ZERO, by your logic the power would always be
zero at that point, but how can that be?? *conversly, at a point where
the current standing wave is always zero there can be no power in the
standing wave, but at that point the voltage is a maximum so would say
the power was a maximum... an obvious contradiction.

I know one thing, when standing wave current is high, I get loses in the
resistance, wires heating up - power being used.
When standing wave voltage is high, I get dielectric loses, insulation
heats up and melts - power being used, ergo there is a power in standing
wave and is demonstrated by certain magnitude of current and voltage at
particular distance and P = U x I. I don't know what you are feeding
your standing wave antennas, but I am pumping power into them and some
IS radiated, some lost in the resistive or dielectric loses.

73 *Yuri

you are SO CLOSE... open your eyes, turn off your preconceived notions
and read what i wrote again slowly and carefully.

FIRST remember the assumption was a LOSSLESS line so there are no
dielectric or resistive losses. *But this is only useful because it makes
it easier to see that the power given by V*I in the standing waves
doesn't make sense.

YOU ARE CORRECT when you say that the R*I^2 loss is REAL... and YOU ARE
CORRECT when you say the V^2/R loss in dielectric is REAL.

Where you lose it is that the V*I for the standing waves is not
correct... this is because V and I are related to each other and you
can't apply superposition to a non-linear relationship. *If you think you
have a way to do it then please take the given conditions of a lossless
transmission line, shorted at the end, in sinusoidal steady state, and
write the equation for power in the standing wave at 1/4 and 1/2 a
wavelength from the shorted end of the line.

So let me get this straight:
I describe real antenna situation, which is a standing wave circuit, with
real heating of the coil, which is consuming power demonstrably
proportional to the amount of standing wave current.
You are not answering rest of my argument.
You bring in lossless transmission line to argue that there can not be
power and no standing waves.
I still don't get it.

73 Yuri

ok, last try.. *YOU ARE RIGHT!!! *"consuming power demonstrably proportional
to the amount of standing wave current". *LISTEN, YOU ARE RIGHT!!! *the loss
in your coil is proportional to the square of the standing wave CURRENT. *as
long as you keep saying that you are RIGHT. *now repeat after me... the loss
in your coil is proportional to the square of the standing wave CURRENT.
emphasize CURRENT every time. *DO NOT start talking about POWER in the
standing wave, then you will be wrong. *Use the CURRENT young Yuri, Use the
CURRENT... forget the POWER of the standing wave.- Hide quoted text -

It seems to me that a bit more precision in the use of
language might help. So, strictly:

It is consuming power proportional to the square of
the RMS current at that point on the line, and the
constant of proportionality is the resistance of
the line over the length of interest.

When an author writes "standing wave current", do
they always mean the RMS current at a point on the
line? Or do they mean the envelope of the RMS
current at each point along the line? Or the
envelope of the peak current at each point? Or the
RMS value of the spatially distributed peak currents
along the line? Or? What is THE standing wave
current? No wonder there is so much dispute.

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.

...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

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

Gene Fuller wrote:
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.

Gene, here is a true/false quiz for you. If you have a
reference that disagrees with the obvious answers, please
quote it.

1. Is EM wave energy photonic in nature? ________

2. Do photons move at the speed of light in a medium? ______

3. Do standing waves move at the speed of light? _______

If the answers are yes, yes, and no, then standing waves
have been eliminated from the set of EM waves.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Gene Fuller wrote:
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.

Gene, here is a true/false quiz for you. If you have a
reference that disagrees with the obvious answers, please
quote it.

1. Is EM wave energy photonic in nature? ________

2. Do photons move at the speed of light in a medium? ______

3. Do standing waves move at the speed of light? _______

If the answers are yes, yes, and no, then standing waves
have been eliminated from the set of EM waves.

Cecil,

Too easy.

1. Yes, but who cares? If you want to go beyond ordinary classical
models, then most of your standard formalism about transmission lines
would need a bit of work. Classical works pretty well for HF.

2. Yes.

3. Yes.


Do I win the prize?

73,
Gene
W4SZ

Gene Fuller wrote:
Cecil Moore wrote:
3. Do standing waves move at the speed of light? _______

3. Yes.

Do I win the prize?

Of course not. Please explain how standing waves can
move at the speed of light yet only change phase by 1/10
of one degree in 90 degrees of transmission line or antenna?
Please be technically specific and include the math.

We cannot wait to hear your answer to that one.
--
73, Cecil http://www.w5dxp.com

Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

compare and comment on the relation between pf, pr, and p.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.

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, 10:30 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.