On Tue, 03 Oct 2006 10:47:58 -0700, Jim Kelley
wrote:

Pressure of a reflected wave = 2*S(t)/c

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? :-)

Hi Jim,

Who needs reasoning when a Xeroxed formula, like a "4th mechanism of
reflection," is simpler to cut and paste than actually offer a
practical answer to? This is akin to his sophomoric allusion to
anti-glare coating yet again (yawn), he couldn't work the math on that
one anywhere closer than pi = 22/7.

Soon we will be down the path of 66% allowable error to prove a
concept.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Who needs reasoning when a Xeroxed formula, like a "4th mechanism of
reflection," is simpler to cut and paste than actually offer a
practical answer to? This is akin to his sophomoric allusion to
anti-glare coating yet again (yawn), he couldn't work the math on that
one anywhere closer than pi = 22/7.

Richard, I wasn't the one who, through superposition of powers,
came up with an irradiance brighter than the surface of the sun
at the non-reflective surface interface. Did you ever figure
out that superposing powers is a no-no?
--
73, Cecil http://www.w5dxp.com

OK, so we've established you don't know the G forces for changed
momentum, only how to sniff toner at the Xerox.

On Tue, 03 Oct 2006 18:34:17 GMT, Cecil Moore
wrote:

Richard, I wasn't the one who, through superposition of powers,
Yes, you did have a problem translating power to energy and back. I
could offer any number of common scenarios that would have you gasping
for air:
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?
or:
How much power is being supplied to the bulb?

came up with an irradiance brighter than the surface of the sun
at the non-reflective surface interface.

No, true to your form, you rounded errors and fudged numbers to prove
light was black.

In that regard I will offer you a third choice question from above:
Can you see this amount of light on the target?
(choose this one, you might guess it right - it doesn't demand any
math skill.)

:-0

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

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?

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

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