On Wed, 20 Jul 2005 17:28:03 -0400, "Fred W4JLE"
wrote:

It takes at least two stooges interacting to be funny.

Now taking applications.

Hi Fred,

It takes a condition called irony-deficit disorder to wade into these
things and say that. ;-)

Let's just give an example of the straight-man's lead into the joke:

"I flipped the switch to a 100W light Bulb.
What direction vector is the optical power?"

The Scientist would ask for this in standard notation, but we all know
that he isn't going to get that - hence, the subject from the
beginning is a joke. Now, only to wait for the punchline:
(drumroll) Ta-ta-dum....

73's
Richard Clark, KB7QHC

Richard Clark jested:
"I flipped the switch to a light bulb. What direction is the optical
power?"

Seriously, away from and toward are directions. We expect a light bulb
to be an energy source. If it becomes a sink it has a negative effect.

From John E. (Here`s Johnny!) Cinningham`s "The Complete Broadcast
Antenna Handbook", page 243:
"Again, if the base impedance is a negative number, this merely means
that energy is flowing out of a tower (toward the transmitter) instead
of into it (from the transmitter)."

Sign is certainly used to indicate the direction of energy movement or
the same thing, power flow.

Best regards, Richard Harrison, KB5WZI

On Wed, 20 Jul 2005 22:06:35 -0500, (Richard
Harrison) wrote:

"I flipped the switch to a light bulb. What direction is the optical
power?"

Seriously, away from and toward are directions. We expect a light bulb
to be an energy source. If it becomes a sink it has a negative effect.

Hi Richard,

We are speaking of a load, often times an observer for optics, not a
source.

Then there is a serious answer to the observer standing in the field
of illumination and noting that there is power towards him, and power
away from him. This is a very ordinary occurence with a light bulb
that is within everyones common experience.

The light bulb in a room illuminates all the walls, the ceiling, and
the floor. The observer occupies some portion of that 3-space and can
confirm the "away from and toward directions." Still and all, the two
powers do not vectorally add to zero. If this is confounded by
stating that there are reflections, remove all such artifices and do
this in the void of space. The net result is that there is still no
vectoral addition that blacks out the light bulb simply because you
can exhibit "away from and toward directions."

There is still the mathematical representation of the direction using
standard notation if you want to take a swing at that. Again, what is
the vector of direction for the light bulb? Absurdities abound in
this discovery.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
"Again, what is the vector of direction for the light bulb?"

Electromagnetic waves include light and heat whicjh have extremely short
wavelengths. The light bulb may not be a perfect point source but the
waves travel away from the source with the velocity of light and consist
of electric and magnetic fields that are at right angles to each other
and also at right angles to the direction of travel. Wave energy is
divided 50-50 between the electric and magnetic fields.

Many frequencies (colors) make up the radiation from a light bulb. Much
more heat is radiated than visible light.

In a radio wave the essential properties are frequency, intensity,
direction of travel, and plane of polarization, For the constituents of
light bulb radiation, it is the same.

300 million m/sec is the velocity and this equals the product of
frequency X wavelength. Emissions of a light bulb are of extremely high
frequency but of extremely short wavelenggth too.

All points on a wavefront are equidistant from the source and emerged
simultaneouslly so they share the same phase.. From a point source light
bulb we would be in the far field.

The field is transverse. The power flow (J.D. Kraus` words), or Poynting
vector, is entirely radial.

Best regards, Richard Harrison, KB5WZI

On Thu, 21 Jul 2005 01:24:36 -0500, (Richard
Harrison) wrote:

Many frequencies (colors) make up the radiation from a light bulb. Much
more heat is radiated than visible light.

Hi Richard,

Actually that is quite wrong. IR is not heat, it is radiation. Heat,
actually phonons, constitutes something less that 10% of the
conversion of electrical power in a light bulb. Lest we take off on
the tangent of IR bulbs being used for heating, it is the load of that
IR radiation (directed upon a dissipative surface) that renders
phonons, otherwise IR is radiated in exactly the same manner as any
radiation. There are any number of simple, practical tests to confirm
this. For one, IR passes through most glass without heating it. You
have to go out of your way to obtain IR blocking glass (which doesn't
even absorb that much either). There are some IR wavelengths that go
right through water, and others that are entirely absorbed.

However, this is not about heat, nor IR, nor even the loss of a
principle vector property, its angle notation, or even the whole
absence of the vector property from the solution to wave interference
powers. Rather, it is about the facade of complete cancellation

Entirely ignoring all these other trivial details, that cancellation
is incomplete in and of necessity for real or imagined initial
conditions. This is revealed in any mathematical solution, and
certainly by examination.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
Entirely ignoring all these other trivial details, that cancellation
is incomplete in and of necessity for real or imagined initial
conditions.

That's not true, Richard. If zero reflected energy reaches the
source in a system with reflections, a Z0-match has been
achieved. For a Z0-match to be achieved, 100% wave cancellation
is necessary. For all the nearly perfectly Z0-matched systems
out there, near perfect wave cancellation of reflected waves
has been achieved.
--
73, Cecil http://www.qsl.net/w5dxp


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Richard Clark wrote:
"Actually that is quite wrong. IR is not heat."

He got me. According to Lincoln`s Industrial Reference, from a 100-watt
MAZDA lamp the amount of energy emanating as light is 10%, and as
infrared is 72%. The rest is lost to gas end loss, etc. The loss would
be only 18% You can`t see infrared. The eye is most sensitive to a
yellow-green color around 5550 Angstrom units. Lamps are made to
emphasize white or "daylight" which is rated at about 2400 to 3100
degrees Kelvin.

Best regards, Richard Harrison, KB5WZI

On Thu, 21 Jul 2005 09:47:12 -0500, (Richard
Harrison) wrote:

Richard Clark wrote:
"Actually that is quite wrong. IR is not heat."

He got me. According to Lincoln`s Industrial Reference, from a 100-watt
MAZDA lamp the amount of energy emanating as light is 10%, and as
infrared is 72%. The rest is lost to gas end loss, etc. The loss would
be only 18% You can`t see infrared. The eye is most sensitive to a
yellow-green color around 5550 Angstrom units. Lamps are made to
emphasize white or "daylight" which is rated at about 2400 to 3100
degrees Kelvin.

Hi Richard,

Well, your ability to research the topic continues well in advance of
other's effort. Some may note the congruence of the specified
emission peak and my statements earlier choosing exactly this same
wavelength. This is called the eye's photopic response, but at night
it shifts slightly to become more sensitive in its scotopic response.
This is rod vision and occurs around the 510nM (5100Å) wavelength or a
pale blue.

The unintended consequence of this is that it suppress the eye's
ability to perceive red light at night (why you see them used in dark
rooms and WWII movies) which is something of a bummer for traffic
lights and taillights (they have to be brighter than they would be
normally).

Also, this discussion bears upon the answer to
"What is the wavelength of Glare?"
that has remained undiscovered by binary engineers. I bet our Readers
can catch this clue. :-)

73's
Richard Clark, KB7QHC

Richard Harrison wrote:
All points on a wavefront are equidistant from the source and emerged
simultaneouslly so they share the same phase.. From a point source light
bulb we would be in the far field.

The big difference is that a light bulb doesn't emit
coherent light. Coherence is a requirement for
superposition and wave cancellation. The light bulb
example is simply irrelevant to what happens with
a single frequency coherent RF transmitter or a
single frequency coherent laser.
--
73, Cecil http://www.qsl.net/w5dxp


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Richard Clark wrote:
The net result is that there is still no
vectoral addition that blacks out the light bulb simply because you
can exhibit "away from and toward directions."

A light bulb does not emit coherent light so your
statement is 100% irrelevant to coherent RF sources
and/or coherent laser sources.
--
73, Cecil http://www.qsl.net/w5dxp


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