The Rest of the Story
 

On Mar 26, 1:37*pm, Cecil Moore wrote:
Keith Dysart wrote:
Now if the 200 W from the wave from source 1 and the
50 W from the wave from source 2 represent actual
energy flows, then the "interference energy" must
also be an actual energy flow to satisfy conservation
of energy.

One other observation: Although the interference model
will work for a lumped circuit example, there is no
reason to use it as it complicates the computations
and adds nothing to the solution. The wave reflection
model also works for circuits but there is simply no
good reason to use it for lumped circuit analysis.

Except when it can be called into service to disprove
some aspects of *your* "wave reflection model".

Where interference becomes a useful tool is when it
happens away from any compensating source. An analysis
of the interference of two EM light waves in free space
far removed from any source leaves us with two constant
sources of energy, the total energy of which has to go
somewhere. The following two web pages tell us exactly
where the energy goes.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

http://micro.magnet.fsu.edu/primer/j...interference/w...

It is unfortunate that they did not expect someone to use
their website to try to understand the behavior of transmission
lines, or they might have taken more care to explain what is
happening at the sub-wavelength level.

Your theory seems to require that the EM waves must
know beforehand whether to carry energy or not from
a star light years away. You apparently have invented
a rather curious "smart wave theory".

That would be nice, but no, not required. Though it is
easy to see how one might be misguided into thinking so.

The question is, can you set aside the mis-guiding
influences long enough to learn about the alternative
explanations?

The question is: How did those two interfering waves
from Alpha Centauri know whether to arrive at the
planet Earth ten years later carrying ExH energy
or not carrying ExH energy?

It is a question, but only seems relevent if *your* model
is incorrect.

...Keith

The Rest of the Story
 

On Mar 26, 2:53*pm, Cecil Moore wrote:
Keith Dysart wrote:
Let us build a slightly better example that complies with
your "NOT =" expression above.

Your constant voltage sources are NOT a better example
and not even a good example. To confront the subject
being discussed, you should use constant power sources.

Well, if that is all it takes to stop you, here is a
variation that should satisfy. It aligns well with your
example of Fig 1-1......

Let us build a slightly better example that complies with
your "NOT =" expression above.

50 ohms 50 ohms
+------/\/\/\----------------+---/\/\/\---+
| | |
| \ |
Vs1(t) = 282.8cos(wt) Rload / Vs2(t) = 141.4cos(wt)
| = 200 Vrms 50 ohms \ | = 100 Vrms
| / |
| | |
+----------------------------+------------+

Using superposition the contribution from source 1 is
Vload.s1 = 100 Vrms
Iload.s1 = 2 Arms
and from source 2 is
Vload.s2 = 50 Vrms
Iload.s2 = 1 Arms
combining
Vload = 150 Vrms
Iload = 3 Arms

From
Pload = Vload * Iload
= 150 * 3
= 450 Waverage

As can be seen, this example satisfies your requirement
for interference:
[V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2

Computing the imputed powers for the waves from each source
we have
Pload.s1 = 100 * 2
= 200 Waverage
Pload.s2 = 50 * 1
= 50 Waverage
To obtain the power in the load from these imputed powers
we need to use
Pload = Pload.s1 + Pload.s2 + Pload.correction
450 = 200 + 50 + Pload.correction
200 = Pload.correction

From previous analysis
Pcorrection = 2 * sqrt(P1 * P2)cos(theta)
(the cos(theta) term is appropriate here because these
are average powers being used)
Pcorrection = 2 * sqrt(10000) * 1
= 200
as required from above.

So according to your energy analsysis, the power in
the load comes from
the wave from source 1 = 200 W
the wave from source 2 = 50 W
"interference energy" = 200 W
for a total of
450 W as required.

Now if the 200 W from the wave from source 1 and the
50 W from the wave from source 2 represent actual
energy flows, then the "interference energy" must
also be an actual energy flow to satisfy conservation
of energy.

What element provides the energy for this "interference
energy" flow?

In other posts, you have suggested that this would be
a constructive interference energy and that there would
be an equal destructive interference energy to provide
it. If you still claim this, where is this destructive
interference happening?

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Or perhaps the element you have identified does not have
the appropriate energy flow function? (It doesn't.)

Please prove your assertion.

This requires that the sum of the flows out of the
elements providing energy equals the sum of the flows
into the elements receiving the energy.

True for energy. Not true for power.

And we are still waiting for the energy flow function
for the element that you claim is doing the storing
of the energy.

If you cannot understand the reference I gave you,
I don't know what to tell you.

Does it detect energy? Are you sure?
Or is it voltage that it detects? Or current?

Please provide proof that voltage or current can
exist without energy.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
The question is, can you set aside the mis-guiding
influences long enough to learn about the alternative
explanations?

Sorry, I'm not interested in those religious
concepts.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
In other posts, you have suggested that this would be
a constructive interference energy and that there would
be an equal destructive interference energy to provide
it. If you still claim this, where is this destructive
interference happening?

I have said a source can match any destructive interference
by supplying less power and match any constructive interference
by supplying more power. If you have to falsify what I have
said to try to win the argument, you have already lost.
Since you have ample sources available in your example, my
assertion about interference far removed from any source
doesn't apply - but you know that.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
In fact, the thing you need to do is forget the transmission
line and deal with light waves encountering boundaries with
different indexes of refraction. The problem is identical,
but dealing with light out in the open prohibits you from
pushing your mashed-potatoes energy religion.

No. Light, in a 3 dimensional space and at such high frequency
makes the math and measurements so complicated that it is
extremely difficult to follow the energy.

Now how did I know that you would refuse to expose your
strange concepts to the light of day?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sun, 09 Mar 2008 21:33:41 -0400, Chuck
wrote:

On Sun, 9 Mar 2008 15:07:26 -0700 (PDT), K7ITM wrote:

snip


Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers

snip


It is said he used sodium vapor gas light (~589 nm). Coherent enough.

Monochromatic is not the same as coherent or in phase such as a
laser.


Chuck

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
Roger Halstead (K8RI & ARRL life member)
(N833R, S# CD-2 Worlds oldest Debonair)
www.rogerhalstead.com

The Rest of the Story
 

Cecil Moore wrote:
Roger Sparks wrote:
You need to take a look at the spreadsheets.

Roger, in a nutshell, what is the bottom line?

The bottom line in a nutshell? I'll try.

First, I added a note to both spreadsheets indicating that zero degrees
is CURRENT zero degrees. This because the source turns out to be
reactive, with current peak 45 degrees from voltage peak.

http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf

The spreadsheet addresses the following issues:

Does the traveling wave carry power? Yes. The spreadsheet was built
assuming that power is carried by traveling waves. Because the
resulting wave form and powers seem correct, the underlaying assumption
seems correct.

Is power conserved on the transmission line, meaning, can the energy
contained in power be conserved and located over time on the
transmission line? Yes, the spreadsheet was built assuming that power
could be conserved and traced over time so the underlaying assumption
seems correct.

Does interference occur in this example? The spreadsheet was built
assuming that voltage and currents from superpose in a manner consistent
with constructive and destructive interference, so the underlaying
assumption seems correct.

Is power stored in the reactive component for release in later in the
cycle or during the next half cycle? Yes, power is stored on the
transmission line during the time it takes for power to enter the line,
travel to the end and return. The time of wave travel on the
transmission line is related to the value of the reactive component.

Does the direction of wave travel affect the measurement of voltage and
the application of power to a device? Yes. A wave loses energy (and
therefore voltage) as it travels through a resistance. As a result,
power from the prime source is ALWAYS applied across the sum of the
resistance from the resistor AND transmission line. The spreadsheet was
built using this assumption and seems correct. (At times during the
cycle, the forward and reflected waves oppose, resulting in very little
current through the resistor. During those times, the power applied to
the transmission line is much HIGHER because the reflected wave reflects
from the load and source, and merges/adds to the forward wave from the
source.)

Is the power interference equation Ptot = P1 + P2 +
2*SQRT(P1*P2)cos(theta) valid? The equation was not reviewed on this
spreadsheet.

The bottom line, but maybe not in a nutshell.

73, Roger, W7WKB

The Rest of the Story
 

Roger wrote:

Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers

It is said he used sodium vapor gas light (~589 nm). Coherent enough.

Monochromatic is not the same as coherent or in phase such as a
laser.

Just a slight addition here. Before lasers, the way to get a
coherent light source was to bottle-up a high-intensity,
monochromatic source, such as the aforementioned sodium-
vapor light, in a reflective cavity with a very small pinhole in
its side. As the photons dribble out through the pinhole, they
are forced into a somewhat phase-coherent wave train. This
source was used in optical processors for synthetic-aperture
radar imagery back in the 50's....

Jim, K7JEB

The Rest of the Story
 

Roger Sparks wrote:
The bottom line in a nutshell? I'll try.

Thanks Roger, good stuff and much appreciated.
My digesting of your spread sheets is about to
be interrupted by surgery.

During those times, the power applied to
the transmission line is much HIGHER because the reflected wave reflects
from the load and source, and merges/adds to the forward wave from the
source.)

May I suggest that you use the word "redistributed"
instead of "reflected" as does the FSU web page at:

http://micro.magnet.fsu.edu/primer/j...ons/index.html

For the purposes of this discussion, I would suggest
that the word "reflection" be reserved for something
that happens to a single wave. When two waves are
superposed, energy can be redistributed but technically
it is not an ordinary reflection. I once used the
word "reflection" to describe both phenomena and it
confused people. Now I am careful to call the
reversal of energy flow due to superposition a
"redistribution" instead of a "reflection".

For instance, the multi-colored patterns seen when
a thin film of oil is on top of a puddle of water
is not an ordinary reflection but a combination
of multiple reflections and interference.

In addition, the reflection coefficient seen by the
reflected wave in our examples is 0.0 since the source
impedance equals the characteristic impedance of the
transmission line. There are no ordinary reflections
when the reflection coefficient is zero.
--
73, Cecil http://www.w5dxp.com