"Yuri Blanarovich" wrote in message
...

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

"Yuri Blanarovich" wrote in message
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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.