Jim Kelley wrote:

Cecil Moore wrote:
Measuring the current at the mid-point of that
1/2WL of feedline will prove the feedline is filled with
EM wave energy which must travel at the speed of light.

So it follows that measuring voltage at the wall outlet proves there's
energy filling the wall. As always, it's important to remember that any
such energy would of course be traveling at the speed of light - and no
faster. :-)

There are thousands of unterminated wall outlets and hundreds of
terminated wall outlets spaced only a fraction of a wavelength from
each other. By all means, if you cannot understand the simplest of
examples, create an example that is so complicated that nobody can
understand. This is an example of someone trying to obfuscate things
in order to reduce everyone down to his/her low level of understanding.

Noting of course that EM energy can't normally put itself back into the
source after it's done bouncing around. So, since there's no load and
the system is lossless, no energy is produced or transferred, which
means zero power.

Zero *NET* power has nothing to do with the component powers. All it means
is that the forward power and reflected power are equal. The losses in
a real world transmission line depend upon the magnitude of the forward
and reflected powers which pretty much shoots your illogical argument in
the foot.

The higher the voltage applied to a real-world 1/2WL transmission line,
the greater the losses absorbed by the feedline. How the heck do you
explain that one by neglecting forward and reflected energy?
--
73, Cecil http://www.qsl.net/w5dxp


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