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. :-)

The Bird will read 100w forward and 100 reflected on the
feedline. An RF current meter at the center of that 1/2WL
of feedline will read 2.828 amps. That's the sum of the
forward current and reflected current in phase, 1.414 amps
in the forward direction and 1.414 amps in the rearward
direction. 1.414^2*50 = 100w for both forward and reflected
powers. The Bird is right. Those powers are really there
supporting the forward and reflected waves and cannot be
used for any other purpose.

A veritable black hole of logic: it's inescapable! :-)

Since the feedline is lossless,
there's no lost energy to replace so the source power output
is zero. Anything else would violate the conservation of
energy principle. Note that at the above current maximum
point, the net flow of energy is zero since the Poynting
vectors for forward and reflected power add up to zero.

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. In this instance, the readings on the Bird wattmeter
are not at all helpful toward understanding the flow of EM energy -
which as Cecil is always kind enough to remind us, travels not just at
any speed, but at the speed of light.

73, ac6xg

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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