On Mon, 18 Jul 2005 16:25:56 -0500, Cecil Moore
wrote:
it makes no sense to complicate things with additional details
They are your details and you can't do the math. You instead beg it
doesn't make sense to? Certainly not from the perspective of your
theories. ;-)

There are 10 types of people, those that understand binary and those that
don't...

"Richard Clark" wrote in message
...
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

Jim Kelley wrote:
"In other words a system in which all of the power from the source
reaches the load and none is reflected back to the source without first
reflecting then re-reflecting would violate conservation of energy."

Conservation of energy means that energy is neither created nor
destroyed, but that heat and other forms of energy are quantitifiable
and convertable in their equivalence. The total amount of mechanical,
thermal, chemical, electrical, and other forms of energy in any isolated
system remains constant. A century ago, Einstein broadened the law to
include equivalence of mass and energy.

Regardless of reflections and re-reflections, all the energy sourced
into a transmission line ends up in the load if it isn`t lost in
transmission by radiation or conversion into heat. There`s no place else
for it to go.

Best regards, Richard Harrison, KB5WZI

Richard Clark wrote:
Cecil Moore wrote:
The problem is that if you cannot understand
the simplest of examples involving lossless, refractionless, laser
systems, you cannot possibly understand anything more complicated.

Um, yes. Can tell us why your example exhibits a reflection product
TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled
completely? :-)

It doesn't. All reflections are eliminated by wave cancellation.
That is a given boundary condition for the simple example.

You can argue that there's no such thing in reality as a
dimensionless point or a line of only one dimension or a
plane of only two dimensions. That doesn't keep such from
being taught as concepts in every plane geometry class.
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Clark wrote:

Cecil Moore wrote:
it makes no sense to complicate things with additional details

They are your details and you can't do the math.

No, they are your details and I choose not to waste my
time with your logical diversion of the basic issue.
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Harrison wrote:
Regardless of reflections and re-reflections, all the energy sourced
into a transmission line ends up in the load if it isn`t lost in
transmission by radiation or conversion into heat. There`s no place else
for it to go.

Hi Richard, does that statement assume that all energy
dissipated as heat in the source was never sourced?
--
73, Cecil http://www.qsl.net/w5dxp

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On Tue, 19 Jul 2005 01:45:26 -0500, Cecil Moore
wrote:

Um, yes. Can tell us why your example exhibits a reflection product
TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled
completely? :-)

It doesn't. All reflections are eliminated by wave cancellation.
That is a given boundary condition for the simple example.

What a larf. :-)

If you can't explain it, you simply re-write the math to suit the
outcome? This is neither demonstrable nor provable, unless, of course
you simply fudge the numbers until they agree with your result. Given
that every reader here has had the opposite experience as your
prediction, this is classic thumb on the scale work - quite shabby.

Richard Clark wrote:

Cecil Moore wrote:
it makes no sense to complicate things with additional details

They are your details and you can't do the math. You instead beg it
doesn't make sense to? Certainly not from the perspective of your
theories. ;-)

It seems you have forgotten the purpose of the laser example.
It was supposed to be as much like a transmission line example
as is possible. Refraction is irrelevant in a transmission line
so there's no logical reason to introduce refraction into the
simple laser example. A laser was chosen to make the example
single frequency and coherent like an RF source. Occam's Razor
is used to trim off all irrelevant stuff.
--
73, Cecil http://www.qsl.net/w5dxp

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On Tue, 19 Jul 2005 01:48:00 -0500, Cecil Moore
wrote:

No, they are your details
All the math was per your explicit design. If that embarrasses you,
fine, I will accept that as "my" details.

and I choose not to waste my
time with your logical diversion of the basic issue.
No one is forcing you to explain what you cannot. What you choose to
respond to is strictly your choice and wholly your responsibility. If
you don't care to answer for why the reflections of your model are
TEN TIMES BRIGHTER THAN THE SUN, and call this power cancellation
instead, no one really expects you to anyway.

Why don't you give us another hymn instead about understanding and
such? Is your paper in the hands of a devotional publication? Do
they cater to a herd of sacred cows? Would this make you a holy
cowboy? :-)

Richard Clark wrote:

Cecil Moore wrote:
Um, yes. Can tell us why your example exhibits a reflection product
TEN TIMES BRIGHTER THAN THE SUN; when in your words it has canceled
completely? :-)

It doesn't. All reflections are eliminated by wave cancellation.
That is a given boundary condition for the simple example.

If you can't explain it, you simply re-write the math to suit the
outcome?

The purpose of the laser example is to make it as much like a
transmission line example as possible. A Bird wattmeter indicates
that all reflections are eliminated in the T-line example so
the laser example assumes that as a boundary condition. Your
insistance on rewriting the math to make it as different from a
T-line example as possible is simply a diversion away from the
original purpose.
--
73, Cecil http://www.qsl.net/w5dxp

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