Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

A wave is defined as a progressive vibrational disturbance propagated
through a medium, such as air, without progress or advance of the parts
or particles themselves, as in the transmission of sound, light, and an
electromagnetic field. Light, for example, is also calld luminous or
radiant energy. Sound and radio waves are also examples of energy in
motion.

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

Best regards, Richard Harrison, KB5WZI

"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.

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.

"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

Evidently, you haven't done enough reading. Yuri is right this time.
73,
Tom Donaly, KA6RUH

Whoaaaa!
When wasn't Yuri right? :-)))))

Yuri, da BUm

"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.

On Sun, 23 Dec 2007 13:34:49 -0600, (Richard Harrison) wrote:

Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

A wave is defined as a progressive vibrational disturbance propagated
through a medium, such as air, without progress or advance of the parts
or particles themselves, as in the transmission of sound, light, and an
electromagnetic field. Light, for example, is also calld luminous or
radiant energy. Sound and radio waves are also examples of energy in
motion.

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

Best regards, Richard Harrison, KB5WZI

It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

In addition to another comment above that implies that reflected power is reactive power, this is not
true--reflected power is as real as forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in phase in the forward direction, they are
traveling 180° out of phase in the rearward direction. Multiplying voltage and current while 180° different in
phase results in the same power as when they are in phase.

Walt, W2DU

There's been a lot of confusion between average and instantaneous power.
Let me try to clarify a little.

Keith Dysart wrote:
. . .
Now power is really interesting. Recall that

P(x,t) = V(x,t) * I(x,t)
. . .

This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important information
is lost. (As was the case of the statistician who drowned crossing a
creek whose average depth was only two feet.)

Roger wrote:

I would suggest that you add a caveat here. The power equation is true
if the measurements are across a resistance. If we are also measuring
reactive power (or reflected power), then we need to account for that.

And here's where one of the common errors occurs. The fundamental
equation given by Keith does "account for" reactive power. If I(t) and
V(t) are sinusoidal in quadrature, for example, then V(t) * I(t) is a
sinusoidal function (at twice the frequency of V or I), with zero
offset. This tells us that for half the time, energy is moving in one
direction, and for the other half the time, energy is moving in the
opposite direction. The average power is zero, so there is no net
movement of energy over an integral number of cycles. This is what is
called "reactive power".

On the other hand, if V(t) and I(t) are in phase, the product is again a
sinusoid with twice the frequency of V or I, but this time with an
offset equal to half the peak value of V times the peak value of I. What
this tells us is that energy is always moving in the same direction,
although its rate (the power) increases and decreases -- clear to zero,
in fact, for an instant -- over a cycle. The average power equals this
offset. So here the "reactive power" is zero. I've described this as
energy "sloshing back and forth", which Cecil has taken great delight in
disparaging. But that's exactly what it does, as the fundamental
equations clearly show.

When V(t) and I(t) are at other relative angles, the result will still
be the sinusoid at twice the frequency of V and I, but with an amount of
"DC" offset corresponding to the net or average power and therefore the
average rate of energy flow. The periodic up and down cycles represent
energy moving back and forth around that net value.

No additional equation or correction is needed to fully describe the
power or the energy flow, or to calculate "real" (average) power or
"reactive power".

A careful look at what the power and energy are doing on an
instantaneous basis is essential to understanding what the energy flow
actually is in a transmission line. Attempting to ignore this cyclic
movement and looking only at average power can lead to some incorrect
conclusions and the necessity to invent non-existent phenomena (such as
waves bouncing off each other) in order to hold the flawed theory together.

Roy Lewallen, W7EL

Roy Lewallen wrote:

...
This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important information
is lost. (As was the case of the statistician who drowned crossing a
creek whose average depth was only two feet.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Regards,
JS

"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

Roger wrote:

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

Sorry, neither is correct.

Z0 is the ratio of V to I of a traveling wave. It's not the ratio of V
to I of the total of a forward and reverse wave (which has been
carelessly called "the standing wave" in this thread). If you want to
calculate power as V^2 / R or I^2 * R, you have to use Re(V/I) as the
value of R -- then you will, as you must, get the same result using V *
I, I^2 * R, or V^2 / R. As any text can tell you, the value of Z (ratio
of V to I) varies along a line which has reflected waves (i.e., has a
standing wave). If you use Z0 for the calculation in those cases, you
get results which have no meaning or physical significance.

V^2 / R + I^2 * R doesn't give power, using either Z0 or V/I as R. I'm
curious as to where that equation came from or how it was derived.

Roy Lewallen, W7EL