"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