Richard Clark wrote:
Cecil Moore wrote:
You furnished less than none.

Short memory in long supply. The complete treatment in math was
offered successfully rebutting your proposition and you have shown
nothing new. The negation stands.

There is zero net refraction, given by definition. So all your
refraction math was irrelevant and negated nothing. All that
exists in the example is forward energy and reflected energy
which you chose not to deal with at all.

The reason that optical engineers know so much more about power
and energy in EM waves is because that's about all they could
measure for 100 years. They don't have the luxury of measuring
the voltage in an EM light wave. And using voltage to analyze
photonic EM energy waves doesn't reveal the whole story.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sat, 16 Jul 2005 17:15:58 -0500, Cecil Moore
wrote:

There is zero net refraction, given by definition. So all your
refraction math was irrelevant and negated nothing.

The math remains inviolate, exhibits the laws of conservation, and
negate your premise. So far you have added nothing to offset this.

The reason that optical engineers know so much more about power
and energy in EM waves is because

Uh-huh. Against myself, a practicing and successful optical engineer
with optical patents; you, a xerox jockey, are comparing yourself?
:-)

Bubba, you can't even answer simple power questions like:
You have a one square cm target that is irradiated with
64 microWatts of 660nM radiation at a distance of 1M
from a light bulb. (which is less than the power reflected from
one of your example interfaces and still visibly quite bright.)

How much power is found in the 555nM spectrum expressed
in Lux?

Both are power spectrums within a 30nM BW.

How much total power is the light bulb radiating?

Optical engineers can answer this, and I will by midnight. ;-)

No one expects a binary engineer can (example of a simple 1 or 0).

Richard Clark wrote:

Cecil Moore wrote:
There is zero net refraction, given by definition. So all your
refraction math was irrelevant and negated nothing.

The math remains inviolate, exhibits the laws of conservation, and
negate your premise. So far you have added nothing to offset this.

Most of the reflection examples in _Optics_, by Hecht, assume
zero net refraction and I will continue to follow Hecht's lead.
(Complicating examples beyond what is needed for understanding
the principles involved is a form of obfuscation.)
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Clark, KB7GHQ wrote:
"No one expects a binary engineer can (answer that)."

I have a different take on reflection of the reflected wave at the
generator. I think current results from potential difference. When there
is no difference, there is no current.

Suppose you have an ideal transformer with a perfectly centertapped
secondary. Volts are the same on both sides of the centettap but
180-degrees out of phase with respect to the centertap. Connect the
extremes together and a high current results from the short circuit.

Suppose you modify the secondary by untwisting the sentertap and this
results in two identical secondary coils. Connect these in the same
phase in parallel. No current flows between the coils because no
potential difference ever exists between the coils.

An ideal transmission line with a complete reflection at its far end is
much like the transformer with two identical secondaries. The incident
and reflected voltages are equal and at some points can completely add
or subtract.

Back at the generator connected to a line with a complete reflection at
the load end, there are line lengths which produces a reflection
arriving back at the generator 180-degrees out of phase with the
generator voltage. This produces the equivalent of a short circuit on
the generator.

If line length results in reflected volts exactly in-phase with
generated volts, no current flows into the line once steady-state
conditions are established. This is what happens in the short-circuited
1/4-wave stub (metal insulator). It is the standoff between identical
voltages which produces the extremely high impedance at the input to the
stub.

I have not tried my Bird wattmeter on this. Maybe Cecil has and will say
I am wrong.

Best regards, Richard Harrison, KB5WZI

On Sat, 16 Jul 2005 22:00:47 -0500, Cecil Moore
wrote:

Most of the reflection examples in _Optics_, by Hecht

Dear Readers, it isn't midnight and there is no chance that our
scribbler has actually had the experience to express any solution
bearing on power but his own cooked theory. After-all if we can see
the reflections from these anti-reflective layers, and that is
sufficient proof to invalidate this folderol; well, experience must be
a lie compared to tarted up references (which is so much sacred
hamburger).

So, for the sake of those following this (and sufficiently wise enough
not to jump into this sewer without a snorkel), I posed a simple
question as to the amount of power in the 555nM band (within a 30nM
BW) given a known power of 64microWatts illuminating a square cM
target. I further asked that this be expressed in Lux.

Well, this problem is no more difficult than being able to simply take
the one power already known, the characteristic of a tungsten lamp and
transforming it into the other wavelength. This is a commonplace of
optical engineering unknown to the binary engineer who finds the sum
total of his entire instruction in two pages xeroxed from a library
book.

I will skip the expression in Lux simply because that would be showing
off, and cut to the chase of power expressed in conventional terms
(the binary engineer will be wholly lost in the arcana of practical
measures of light and couldn't be trusted to answer if any power were
actually visible - commonplace experience is a mystery) = 220 nW.

This answer reveals there is more to the characteristic of the
tungsten lamp, than meets the eye. 220 nW is actually quite bright,
and yet we are being handed sloppy work that is acknowledged to
dismiss nearly a thousand times as much power as a trivial difference
that doesn't invalidate a claim of "total" cancellation. I wouldn't
trust such a personality as a bank teller, nor a goldsmith, nor a
surgeon, bridge builder, ... in short, no one serious about the
subject. This kind of slop is what CFA and EH antennas are built
from.

So, this practiced optical engineer has delivered what the binary
engineer could not. Nothing amazing about that - experience clearly
differentiates knowledge from wishing. I could continue with turning
this into Lux/Lumens/Foot-Candles/Candelas (terms of confusion to our
scribbler), shifting the wavelength again, expressing the total power
radiated, expressing the total light seen (or power in the BW selected
light), and so on that are complete mysteries that utterly wipe out
these facades breathlessly offered as compelling proofs.

The single most embarrassing question I've offered to this correlation
of "glare-proof" optics was to ask the obvious:
What wavelength is Glare?

Even here there is a practical answer that stumps the binary mentality
limited by the lack of experience, and the dearth of exposition from
two pages of thumb worn xeroxes.

So, we have these "Can you solve this?" howlers where the author is so
utterly unversed on the topic that he cannot describe power; fails to
acknowledge if that glare's reflection could be seen at the typical
values found in perfect math solutions; what wavelength we are talking
about; why the problem is posed out to 5 places of precision and with
only 1 place of accuracy results; why a light level of 1/1000th of the
typical perfect math solution is still visible but is dismissed as a
correct result.

It has been amusing nonetheless.

73's
Richard Clark, KB7QHC

Richard Harrison wrote:
I have not tried my Bird wattmeter on this. Maybe Cecil has and will say
I am wrong.

No, I agree with your analysis, Richard, as far as it goes
but an expanded look will reveal some interesting facts.

Concerning open-circuited or short-circuited transmission
lines, if the voltages add up to zero, that looks like a short,
and current is at a maximum. If the currents add up to zero,
that looks like an open and the voltage is at a maximum. If
these conditions occur at the source, 100% re-reflection of
the incident reflected wave energy is guaranteed. The Bird
reads equal forward and reflected power all up and down the
feedline.

If an open-circuited 1/2WL of 50 ohm lossless feedline is
connected to a simple class-A source having a series source
impedance of 50 ohms and a voltage source of 141.4v (as in
W7EL's "Food For Thought #1) the steady-state impedance seen
by the source is an open-circuit so the source voltage is
141.4v and the source current is 0 amps. The implication is
that it is like having nothing connected at all but that's
not correct. Measuring the current at the mid-point of that
1/2WL of feedline will prove the feedline is filled with
EM wave energy which must travel at the speed of light.

The Bird will read 100w forward and 100 reflected on the
feedline. An RF current meter at the center of that 1/2WL
of feedline will read 2.828 amps. That's the sum of the
forward current and reflected current in phase, 1.414 amps
in the forward direction and 1.414 amps in the rearward
direction. 1.414^2*50 = 100w for both forward and reflected
powers. The Bird is right. Those powers are really there
supporting the forward and reflected waves and cannot be
used for any other purpose. Since the feedline is lossless,
there's no lost energy to replace so the source power output
is zero. Anything else would violate the conservation of
energy principle. Note that at the above current maximum
point, the net flow of energy is zero since the Poynting
vectors for forward and reflected power add up to zero.
--
73, Cecil http://www.qsl.net/w5dxp

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Richard Clark wrote:

Cecil Moore wrote:
Most of the reflection examples in _Optics_, by Hecht, assume
zero net refraction and I will continue to follow Hecht's lead.

After-all if we can see
the reflections from these anti-reflective layers, and that is
sufficient proof to invalidate this folderol;

Following exactly the same reasoning, if we can measure loss in
all transmission lines, the concept of a lossless transmission
line is also folderol. The example I gave was designed for
clarity of understanding. If you choose to obfuscate clarity,
don't expect me to join you.
--
73, Cecil http://www.qsl.net/w5dxp

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On Sun, 17 Jul 2005 07:55:32 -0500, Cecil Moore
wrote:

The example I gave was designed for clarity of understanding.
It is a poor example of understanding when you purposely inject error.
There is nothing clear about intentional mistakes. Any rejection of a
complete solution is a suspect agenda from the beginning.

Understanding does not come of clouded data and murky results. What
has been offered is evidence of poor knowledge, and no experience.
That poor foundation built on sand has translated into outrageous
conclusions that are nothing more than castles in the sky.

I can demonstrate this in that when posed with a REAL power
application, silence typically falls for the simplest of computations.
Like how much power does a light bulb radiate to illuminate in the
660nM region a target of 1 cm square, at 1 M with 64 microWatts/30 nM
of bandwidth?

The inability to do such trivial power models reveals every thing else
offered has the makings of superstition that builds CFAs and their
ilk.

Dear Readers,

Knowing the binary result beforehand (0) to this simple appeal, I will
render that answer before the day is out. It will exhibit how little
optical knowledge is contained in this "Can you solve this" banality.

You may all note that my embarrassing question:
What is the wavelength of Glare?
remains without comment or response, even though there is a practical
answer and a perfectly reasonable explanation. The gulf of silence
that attends this remarks how complete the void of experience is. Can
you imagine basing an entire exposition around such a commonplace
problem and not knowing the basics? In short, this "Can you solve
this?" is more an appeal for knowledge than a demonstration of skill.
For one, you need to know WHO needs this glare cancellation capacity,
and then you would ask WHY; and it then follows that the wavelength
falls within these particular aspects. Fairly simple stuff for the
optical engineer, but wholly outside of the binary engineer's
experience and education.

The greater embarrassment is that apparently it is outside of the
skill of performing a simple Google search to fill that gap of
knowledge. As I offered, once that knowledge came to mind, the WHY
and WHO would explain the WHEREFORE instead of this rummaging through
text to xerox formulas to force-fit a presumed theory of "total"
cancellation.

It is still entertaining tho', as a burlesque of engineering. ;-)

73's
Richard Clark, KB7QHC

On Sun, 17 Jul 2005 11:07:12 -0700, Richard Clark
wrote:
I can demonstrate this in that when posed with a REAL power
application, silence typically falls for the simplest of computations.
Like how much power does a light bulb radiate to illuminate in the
660nM region a target of 1 cm square, at 1 M with 64 microWatts/30 nM
of bandwidth?

Dear Readers,

It comes as no great surprise that this simple example of optical
power is so powerfully baffling to a neophyte. Optical studies are
not nearly as simple as some might believe, and on the other hand, the
field yields answers as easily as any if you simply observe first
principles.

Those principles are as simple as knowing the area of the total
surface illuminated by the standard light bulb (we will ignore the
shadow cast by the base). This is quite easily determined as I
already offered the distance of 1M and the equivalent power within a
1cM area of that total surface.

The math is hardly as arcane as "interference math," but as I have
demonstrated how poorly that was performed, perhaps a little walk
through here is in order.

The area of sphere, at 1M radius quickly resolves to 4·pi M².
Conversion to cM² "should" be a knock off, but when you use 5 place
math to perform 1 place accurate results.... Well, let's just cut to
the chase and skip all the so obviously difficult parts and just
answer the question above.

Our source is a 10W bulb. Common, mud-ordinary situation offering
smaller values of power than supposedly encountered in the
"reflections" of this "Can you solve this?" drek. And as I offered,
such reflections as you would observe (much like the mathematical
impossibility of a bee's flight) - quite bright.

However, this was on scale with 1cM² and when the original question
posed a 1W laser - well you can appreciate that I do not hesitate to
point out that the power of those "totally" cancelled reflections are
easily 10,000 time brighter than my insipid Christmas tree bulb - and
AT LEAST 10 TIMES BRIGHTER THAN THE SUN!

One must truly imagine quite hard to dismiss such brilliance as being
"totally" canceled out to the point of invisibility. :-)

It has been enjoyable, and yet there are questions remaining that
relate to the point of this origin in "Glare." I notice that this too
draws a vacuum of response and yet it was so central to "illustrating"
a thesis. No wonder such papers find their way to the round file at
the editor's desk. The only chance for publication, however, is
strictly meeting the academic strictures of a vanity press - its
vaudeville quality. In this regard, the opus of "total cancellation"
is destined for serialization and possibly a Hollywood movie for next
summer. Myself, I enjoyed the new "Hitchhiker's Guide to the Galaxy,"
but I found the new Ford Prefect rather lame. Oh well, I went to the
matinee showing to save a buck or two - so let's call it break-even.

73's
Richard Clark, KB7QHC

Cecil, W5DXP wrote:
"Concerning open-circuited or short-circuited transmission lines, if the
voltages add up to zero, that looks like a short, and current is at a
maximum. If these conditions occur at the source, 100% re-reflection of
the incident reflected wave energy is guaranteed. The Bird reads equal
forward and reflected power all up and down the feedline."

Yes. A bird is indicatihg power in one direction only and this is the
same along a low-loss line everywhere along the line regardless of SWR.

A short does not kill line energy. It merely transfers electric field
energy into the magnetic field for an instant producing an SWR in which
voltage and current patterns are 90-degrees apart.

When the load end of a transmission line is shorted, 100% of the
incident wave is reflected but the reflected wave has a complete
reversal in phase of its voltage with no change in the phase of its
associated current.

When the load end of a transmission line is open-circuited, 100% of the
incident wave is reflected but the reflected wave has a complete
reversal in the phase of its current with no change in the phase of its
associated voltage.

When the source end of a transmission line is effectively a short or
open circuit, ir re-reflects the reflected wave.

Best regards, Richard Harrison, KB5WZI