Gene Fuller wrote:
Sorry, I guess my closing was not clear. Try this.

8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-)
8-) 8-) 8-) 8-) 8-) 8-) 8-) 8-)

Sorry, I thought your happy face was a localized
variable.
--
73, Cecil http://www.w5dxp.com

Hi All,

Even armed with college texts on optics, Cecil hasn't got a whisper of
a chance in actually answering a direct question such as has been
posed below. A Xerox is not, after all, a calculator (nor does it
impart knowledge). As such, I will offer direct answers and let it go
at that.

On Tue, 03 Oct 2006 12:03:34 -0700, Richard Clark
wrote:

There is a common bare light bulb 1 meter away;
it illuminates a cm² target with 3µW @ 55nM of POWER;

what is:
the number of candela per steradians,
at the target,
from total bandwidth radiation?

5

or:
How much power is being supplied to the bulb?

100W

Can you see this amount of light on the target?

Depending upon the infirmities claimed: none to some.

Aside: only 0.7% as bright as the sun, in the tropics, at noon, on an
equinox.

73's
Richard Clark, KB7QHC

Cecil Moore wrote:
Jim Kelley wrote:

I'll bet the reason no one can measure the radiation pressure
resulting from your "4th mechanism of reflection" is because it
cancels out with the radiation pressure from the cancelled reflection
in the other direction. Right, Cecil? :-)


There is certainly radiation pressure pushing outward
from a 1/4WL thin-film non-reflecting surface even
though the reflected waves cancel each other.

Actually I agree, but it's all from ordinary reflections, rather than
from backscattered interference or anything else from the 'square root
of negative one' axis.

73, ac6xg

On Tue, 03 Oct 2006 13:33:26 -0700, Richard Clark
wrote:

Aside: only 0.7% as bright as the sun, in the tropics, at noon, on an
equinox.

Also given that Cecil cannot compute the radiation pressure, the
answer with respect to the question offered in the thread above is:
3.2 pico Newtons
or
3.2 nano g·m/s²

Dare I ask the G forces again?

wrote:
Do you mean the mashed-potatoes energy equations where
the wave energy components are completely ignored?


Calling the steady state solution "mashed-potatoes" is ridiculous. All
other terms besides the steady-state ones have left the picture in the
steady state. This is true of the waves, their amplitudes, the energy
in the line, everything. That's what makes steady-state analysis
useful... you can do it and be right.

If you want to know what happens during the transient, you MUST start
from the original differential equation that describes how the fields
were set up in the line in the first place, but it doesn't matter in
the steady state. The answer is the same. If you start with the full,
time dependent equations for turning the source on into some line,
whatever it is that happens with the power and where it goes will
become clear from the transient solution, which I know you haven't
worked out. Neither have I.

I don't want to have to do it but one of these days I think it might be
necessary.

This is not a problem that will be solved in with Logic and English,
Cecil. There is no argument if you start with a full , time dependent
mathematical description of the waves. That is what will answer the
question "Where Does the Power Go" You'll end up with a time dependent
Poynting vector that will tell you. I hypothesize, for no particularly
good reason, that such a description will lead to the excess power
having been delivered to the load through the interaction of the
transient solutions on the line.

- - - - - -

Copied and pasted from the rest of my post at eHam:

In a matched line, none of us would disagree that the power flux out of
one end of the line is the power flux into the other end.

From what I understand, the electromagnetic energy contained in that
line is related to the Poynting vector. The integral of the Poynting
vector over the cross sectional area of the line gives you the time
averaged power flowing in the line. The power flows at the group
velocity of the waves in the line, for normal transmission lines, this
is somewhere between 0.6 - 1.0 times the speed of light.

The power flow divided by the group velocity gives you the energy
density per unit length in the line.

The power flow is the Poynting vector integrated over the cross
sectional area of the line.

The Poynting vector in a mismatched line has been worked out by me,
here, and I'd appreciate comments on the MATH.

http://en.wikipedia.org/wiki/Useran_Zimmerman/Sandbox

My claim is that that's it. That's all. No handwaving, no setting up
steady state startup and claiming that that energy remains in the line.
The real part of the Poynting vector represents real power flux and
the imaginary part represents reactive power flux. To get the energy
that exists in the line you just add it all up...

Take the real part, divide it by the propagation velocity on the line,
integrate it over the area of the line (I think for TEM, the fields
are uniform and so multiplying by the area of the line is sufficient)

This gives you the energy density per unit length for the power that's
flowing from source to load. Integrate that over the length of the
line. Set aside.

Take the imaginary part, do the same thing. This gives you the energy
contained in the line of the reactive, circulating power. Add in the
previously calculated energy of the flowing power, and you're done.

Am I wrong, Cecil et. al? If so, Cecil, could you please write down
the description mathematically or get someone to help you out, and
could you please look at and check my math?

I started with the assumption that there are two counterpropagating
waves with some arbitrary electric field amplitudes and an arbitrary
phase relationship. I didn't use anything about Thevenin. I didn't
use anything about virtual open circuits.

The forward and reverse waves have been included, from the beginning.
They're both there.

When the SWR is 1:1, there is no energy in the line associated with
reactive stored power. None at all. The reflection coefficient is
zero, and all energy in the line is associated with flowing power.

When the SWR is infinite, there is no energy in the line associated
with flowing power. None at all. The reflection coefficient is 1, and
all energy in the line is associated with circulating power (reactance
only, as you would expect from a stub)


Dan


Don't expect much math from Cecil. He always neglects the phase
difference of the reflected wave, and won't do the algebra, no matter
what.
73,
Tom Donaly, KA6RUH

Jim Kelley wrote:
Actually I agree, but it's all from ordinary reflections, rather than
from backscattered interference or anything else from the 'square root
of negative one' axis.

Jim, the reflected energy and momentum changes direction.
Walter Maxwell calls it a virtual short or open with a
virtual reflection coefficient of 1.0. I agree with "Optics",
by Hecht, that it is total destructive interference in the
source direction accompanied by total constructive interference
in the load direction. What do you call it? What reflection
coefficient does your reflected wave see? Please give it a
name.
--
73, Cecil http://www.w5dxp.com

Richard Clark wrote:
Even armed with college texts on optics, Cecil hasn't got a whisper of
a chance in actually answering a direct question such as has been
posed below.

Actually, I figure someone who superposes power to the
extent that non-reflective glass is brighter than the
surface of the sun doesn't know enough to ask a decent
question.
--
73, Cecil http://www.w5dxp.com

Tom Donaly wrote:
Don't expect much math from Cecil. He always neglects the phase
difference of the reflected wave, and won't do the algebra, no matter
what.

Simply not true, Tom. For a Z0-match, the phase is always
zero or 180 degrees. Only addition or subtraction is
ever required at a Z0-match. I've got the answer while
you guys are still trying to figure out the cosine of
zero degrees. :-)
--
73, Cecil http://www.w5dxp.com

On Wed, 04 Oct 2006 00:32:17 GMT, Cecil Moore
wrote:

Richard Clark wrote:
Even armed with college texts on optics, Cecil hasn't got a whisper of
a chance in actually answering a direct question such as has been
posed below.

Actually, I figure someone who superposes power to the
extent that non-reflective glass is brighter than the
surface of the sun doesn't know enough to ask a decent
question.


Sounds like a lame answer from you given you cannot perform ANY of the
math.

Richard Clark wrote:
Sounds like a lame answer from you given you cannot perform ANY of the
math.

Richard, you really need to disguise your snipe
hunts a little better.
--
73, Cecil http://www.w5dxp.com