Richard Harrison wrote:
When the source end of a transmission line is effectively a short or
open circuit, ir re-reflects the reflected wave.

Are we then supposed to infer that it [the source] doesn't re-reflect
the wave if anything other than a short or open circuit appears there?

73, ac6xg

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:
"Are we then supposed to infer that it (the source) doesn`t re-reflect
the wave if anything other than a short or open circuit appears there?"

The reflection may be incomplete unless either a hard short or complete
open-circuit appears at the source where the generator meets the
transmission line. If the source appears as a complete short or open the
reflection is total.

Best regards, Richard Harrison, KB5WZI.

Cecil Moore wrote:

Jim Kelley wrote:

Richard Clark wrote:

Jim Kelley wrote:

Born and Wolf has an interesting comment in the section on total
reflection. "...the electromagnetic field in the second medium does
not disappear, only there is no longer a flow of energy across the
boundary."


your source, and yet unable or unwilling to confront this single
observation.


Apparently that would mean the waves aren't traveling at the speed of
light and it would violate his "waves cannot exist without energy" law
of physics, so therefore the book is wrong.


Wrong. All it means is that reflected energy doesn't make it across
the match point.

It says, and means more than just that, obviously.


"... when two waves of equal amplitude and wavelength that are 180-degrees
out of phase with each other meet, they are (canceled but) not actually
annihilated. All of the photon energy present in these waves must somehow
be recovered or redistributed in a new direction, according to the law of
energy conservation. (There are only two directions available in a
transmission
line.) Instead, upon meeting, the photons are redistributed to regions that
permit constructive interference, so the effect should be considered as a
redistribution of light waves and photon energy (back toward the load)
rather
than the spontaneous construction or destruction of light." (Words in
parentheses
are mine added for clarity.)

Interesting to note that the interference phenomenon is often described
as a redistribution, but is never described in any reference as a
reflection, or re-reflection as you have done. The reason is that when
light destructively interferes in some direction, it simply does not go
in that direction. It doesn't, contrary to your assertion, first go in
the reflected direction, say oooops, then turn around in mid-air and go
in the other direction. The reflection is prevented. Comprende senor?
The Bird wattmeter can be misleading in this regard. It measures the
effect of a field (sometimes like the one in Born and Wolf that doesn't
have transfer of energy associated with it), and in every case assumes
energy and power. But it's simple minded so it has an excuse.

73, ac6xg

On Sun, 17 Jul 2005 21:37:50 -0700, Richard Clark
wrote:

there are questions remaining that
relate to the point of this origin in "Glare." I notice that this too
draws a vacuum of response and yet it was so central to "illustrating"
a thesis.

Dear Readers,

Taking a deep breath and wading into the hits found at Google comes
one very typical observation:

" The methods discussed above will work with varying degrees of
success, depending on the details of the manufacturer's process.
The glare reduction factors claimed by many manufacturers should
not be taken too seriously. Some claim glare reduction factors of
up to 250 to 1. Such numbers are either dreamed up in the
advertising department or are the result of unrealistic test
conditions. A good quality anti-glare filter with a multi-layer
optical coating will reduce glare and reflections by about a
factor of 20. Visually compare glare reductions before you buy
anything."

I leave the WHO undisclosed (being it is such a tantalizing overture
to yet another, unanswerable question). We may all note that what is
described here is "multi-layer" which suggests more than one layer for
more than one BW of glare components (yet another tantalizing,
unanswerable question: What wavelength is Glare?). I will pose that
the necessity for multiple layers is derived from the same need for
one (which is actually quite useless in ordinary life).

As I offered, this is all a very common arena for the optical
engineer, and thin-films are tools for SPECIFIC needs (the WHEREFOR,
or yet another tantalizing, unanswerable question whose origin is one
of practicality) rather than as sacred cows for the theory of
moonbeams.

There are correlatives in RF and antennas such that this is not an
alien discussion. However, it takes little more effort to be correct
instead of simply clowning through the math, because these side-show
results we have been subjected with can be stretched to cover any kind
of bizarre theory. As the final line of the quote above suggests:
"Visually compare glare reductions before you buy anything."
rather ordinary advice that is so aggressively shunned by, instead,
shoveling formulas at us instead.

73's
Richard Clark, KB7QHC

Richard Harrison wrote:

Jim Kelley wrote:
"Are we then supposed to infer that it (the source) doesn`t re-reflect
the wave if anything other than a short or open circuit appears there?"

The reflection may be incomplete unless either a hard short or complete
open-circuit appears at the source where the generator meets the
transmission line. If the source appears as a complete short or open the
reflection is total.

Best regards, Richard Harrison, KB5WZI.

Hmmm. I wonder if the phase change on re-reflection the same as it does
on reflection. If it does, and the transmission line is a half wave
long, the Bird wattmeter readings would be real hard to explain.

And isn't it true that if there were actually a real hard short or a
complete open circuit at the source, there wouldn't even be a signal on
the transmission line?

73, ac6xg

11001000 nW

"Richard Clark" wrote in message
...
On Sat, 16 Jul 2005 17:15:58 -0500, Cecil Moore
wrote:

Bubba, you can't even answer simple power questions like:
You have a one square cm target that is irradiated with
64 microWatts of 660nM radiation at a distance of 1M
from a light bulb. (which is less than the power reflected from
one of your example interfaces and still visibly quite bright.)

How much power is found in the 555nM spectrum expressed
in Lux?

Both are power spectrums within a 30nM BW.

How much total power is the light bulb radiating?

Optical engineers can answer this, and I will by midnight. ;-)

No one expects a binary engineer can (example of a simple 1 or 0).

On Mon, 18 Jul 2005 16:53:37 -0400, "Fred W4JLE"
wrote:

11001000 nW


Hi Fred,

Well, actually 11011100 nW, but as you are within 1010% you have
certainly come closer than any other binary engineer. ;-)

73's
Richard Clark, KB7QHC

On Mon, 18 Jul 2005 13:46:02 -0700, Jim Kelley
wrote:

Hmmm. I wonder if the phase change on re-reflection the same as it does
on reflection. If it does, and the transmission line is a half wave
long, the Bird wattmeter readings would be real hard to explain.

Hi Jim,

In fact it is explainable (done it several times here in fact) and is
called mismatch uncertainty which is quantifiable error derived from
attempting to measure power between two mismatches. When the
generator mismatches the line by as little as 2:1 and so does the
load, you are already pushing 20% error.

However, the quantification is resolved through interference math.
No, not like the stuff presented in this "Can you solve this," but
close enough (sans the howling errors of commission).

And isn't it true that if there were actually a real hard short or a
complete open circuit at the source, there wouldn't even be a signal on
the transmission line?

Ah, Reciprocity! The first law ditched over the side when a new
"theory" hits the boards.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
It is a poor example of understanding when you purposely inject error.
There is nothing clear about intentional mistakes. Any rejection of a
complete solution is a suspect agenda from the beginning.

Richard, since you seem to be incapable of understanding the
simplest lossless, refractionless, laser example, it makes
no sense to complicate things with additional details.
--
73, Cecil http://www.qsl.net/w5dxp


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