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Old October 11th 06, 06:43 PM posted to rec.radio.amateur.antenna
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Default Where Does the Power Go?



Cecil Moore wrote:
If politics is a subset of physics and only Jim
Kelley's definitions are allowed, then President
Bush doesn't have any power.


Welcome back Cliffy!

:-)

ac6xg

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Old October 11th 06, 08:49 PM posted to rec.radio.amateur.antenna
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Default Where Does the Power Go?

On Wed, 11 Oct 2006 10:43:12 -0700, Jim Kelley
wrote:



Cecil Moore wrote:
If politics is a subset of physics and only Jim
Kelley's definitions are allowed, then President
Bush doesn't have any power.


Welcome back Cliffy!

:-)

ac6xg


Ah! the proof of reputation informing us all. :-0
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Old October 11th 06, 09:07 PM posted to rec.radio.amateur.antenna
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Default Where Does the Power Go?

Jim Kelley wrote:

Cecil Moore wrote:
If politics is a subset of physics and only Jim
Kelley's definitions are allowed, then President
Bush doesn't have any power.


Welcome back Cliffy!


Had a good time playing with my grand niece and grand nephew
in TN whose IQ's are impossible to measure in public schools.
It runs in the family. :-)

Back to the present thread. I infer that you believe
that Hecht's total irradiance equations are in error?
Before you reply, let me remind you that Dr. Best was
the first one, to the best of my knowledge, to publish
the irradiance equations in an amateur radio
publication, QEX.

In the following fixed font diagram, IR is the Index
of Refraction.

air | 1/4WL thin-film | Glass
1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---...
Ifor=1W | Ifor=1.0101W | Ifor=1W
Iref=0.01W | Iref=0.0101W | Iref=0

Note that I is "irradiance", not current.

Given: The irradiance reflection coefficient is 0.01
at both interfaces. The irradiance transmission coefficient
is 0.99 at both interfaces. Please describe your theory of
the wave cancellation process occurring at the air to thin-
film interface without using the superposition and interference
principles that I have been using to which you object.
--
73, Cecil http://www.w5dxp.com
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Old October 11th 06, 09:26 PM posted to rec.radio.amateur.antenna
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Default Where Does the Power Go?

Cecil Moore wrote:

The irradiance reflection coefficient is 0.01
at both interfaces.


"It's a little known fact that cows were domesticated in Mesopotamia
and were also used in China as guard animals for the forbidden city."

:-)
ac6xg

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Old October 11th 06, 11:27 PM posted to rec.radio.amateur.antenna
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Jim Kelley wrote:
Cecil Moore wrote:
The irradiance reflection coefficient is 0.01
at both interfaces.


"It's a little known fact that cows were domesticated in Mesopotamia and
were also used in China as guard animals for the forbidden city."


Perhaps, should I have used the term, "reflectance"?
Allow me to refer you to page 120, "Optics", by Hecht, 4th
edition, equation for reflectance, R, "ratio of the reflected
power to the incident power". (References are a bitch to
deal with, huh?) Note that reflectance, irradiance reflection
coefficient, and RF engineering's power reflection coefficient
are all the same thing.
--
73, Cecil http://www.w5dxp.com


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Old October 12th 06, 03:32 PM posted to rec.radio.amateur.antenna
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Jim Kelley wrote:
Cecil Moore wrote:
The irradiance reflection coefficient is 0.01
at both interfaces.


"It's a little known fact that cows were domesticated in Mesopotamia and
were also used in China as guard animals for the forbidden city."


air | 1/4WL thin-film | Glass
1W Laser---IR=1.0---|----IR=1.222-----|--IR=1.493---...
Ifor=1W | Ifor=1.0101W | Ifor=1W
Iref=0.01W | Iref=0.0101W | Iref=0


The reflectance (irradiance reflection coefficient) at
the air to thin-film interface is
[(1.222 - 1.0)/(1.222 + 1.0)]^2 = 0.01

The reflectance at the thin-film to glass interface is
[(1.493 - 1.222)/(1.493 + 1.222)]^2 = 0.01

1% of the irradiance is reflected at either surface and
99% of the irradiance is transmitted through the surfaces.

Would you like to engage in a technical discussion of
how non-reflective glass works or would you like to
keep avoiding that technical discussion?
--
73, Cecil, http://www.qsl.net/w5dxp
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Old October 12th 06, 03:50 PM posted to rec.radio.amateur.antenna
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Default Where Does the Power Go?



Cecil Moore wrote:
Would you like to engage in a technical discussion of
how non-reflective glass works or would you like to
keep avoiding that technical discussion?


Your memory must be failing you, Cecil, as it was I who originally
explained to you how anti-reflective coatings work and provided you
with the Melles-Griot web site, the irradiance equations, and several
references on the subject. I know how these coatings work. I have
the optics bible, "Principles of Optics" by Born and Wolf sitting on
my desk right here in front of me, and have referred to it frequently
during our many previous technical discussions on the subject. Your
pretense here is ridiculous.

Please cite a page number where your 1 watt laser problem can be
found. Thanks.

73, Jim AC6XG


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Old October 12th 06, 04:09 PM posted to rec.radio.amateur.antenna
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Jim Kelley wrote:
Please cite a page number where your 1 watt laser problem can be found.


It can be found in my posting. Why are you afraid to
discuss it?
--
73, Cecil, http://www.qsl.net/w5dxp
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Old October 12th 06, 05:40 PM posted to rec.radio.amateur.antenna
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On Thu, 12 Oct 2006 07:50:37 -0700, Jim Kelley
wrote:

Please cite a page number where your 1 watt laser problem can be
found.


Hi Jim,

Is that sad example being trotted out again?

73's
Richard Clark, KB7QHC
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Old October 12th 06, 05:56 PM posted to rec.radio.amateur.antenna
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Default Where Does the Power Go?



Richard Clark wrote:
On Thu, 12 Oct 2006 07:50:37 -0700, Jim Kelley
wrote:


Please cite a page number where your 1 watt laser problem can be
found.



Hi Jim,

Is that sad example being trotted out again?


I hadn't seen the one with a 'forward laser power' reading, as if he
was reading it from an SWR meter before. Seems suspicious.

tnx de ac6xg

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