Richard Harrison wrote:
"If the transmitting antenna were to radiate energy isotropically-that
... -the power flow through unit area at a distance R, ...

What virtually every engineer means when he says, "power flow",
is, "the power (energy per unit-time) associated with the energy
flow". Knowing Jim, he would also probably argue that the sun
doesn't rise - that instead, the earth rotates. However, most
weather forcasts on TV gives the times for sunrise and sunset.
--
73, Cecil http://www.qsl.net/w5dxp

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Cecil, W5DXP wrote:
"What is it about, "...all "lost" reflected intensity will appear as
enhanced intensity in the transmitted beam." that you guys don`t
understand?"

A quarter-wave intermediate impedance line section can perfectly match
different resistances at its ends.

When a match exists at the end of a line, there is no discontinuity and
the line appears as if infinite. There is no reflection from an
impedance match.

It`s a multiplicity of waves which seem gratuitous.

Best regards, Richard Harrison, KB5WZI

On Mon, 1 Aug 2005 11:13:34 -0500, (Richard
Harrison) wrote:

It`s a multiplicity of waves which seem gratuitous.
Amen brother!

Hi Richard,

This statement you respond to is one of those attempts to draw the
topic away from its focus to prove another point of match which
humorously was going to be proven with the original topic.

Now, if that sounds confusing, it is simply par for the course in how
we got here. And to extend the metaphor for par, we started out
Bowling for Dollars and have now ended up putting on the back nine.

Anyway, the optical analogue has proven to be a bust when I
demonstrated that reflection products do persist. This is not the
place to hash over that again. If you wish, you can consult my
original posting and the follow-ons for details.

73's
Richard Clark, KB7QHC

Richard Harrison wrote:

Jim Kelley wrote:
"That`s because power doesn`t propagate, I hasten to add - neither do
Poynting vectors.:

Some world-class experts disagree with Jim. Here is a sample quotation
using the words "power flow".

From E.M. Purcell writing about "Antenna Gain and Receiving Cross
Section" on page 19 of "Radar System Engineering" edited by Louis M.
Ridenour:
"If the transmitting antenna were to radiate energy isotropically-that
is, uniformly in all directions-the power flow through unit area at a
distance R, from the antenna could be found by dividing P, the total
radiated power, by 4piRsquared."

Best regards, Richard Harrison, KB5WZI

There's nothing wrong with calculating the power per unit area at any
point in a transmission line. I'm sorry if I gave some other
impression. The problem I have is with believing that the calculated
value propagates as if it were an electromagnetic field. The Poynting
vector is useful for making calculations, but it not a useful tool for
explaining the behavior of natural phenomenon.

73, ac6xg

Cecil Moore wrote:

Richard Harrison wrote:

"If the transmitting antenna were to radiate energy isotropically-that
... -the power flow through unit area at a distance R, ...


What virtually every engineer means when he says, "power flow",
is, "the power (energy per unit-time) associated with the energy
flow". Knowing Jim, he would also probably argue that the sun
doesn't rise - that instead, the earth rotates. However, most
weather forcasts on TV gives the times for sunrise and sunset.

Yea, somebody around here once said that I would probably step out of
the shower to take a pee. To that I say why would anyone step into it
to take one? ;-)

73, ac6xg

Walt, W2DU wrote:
"A semantic problem with the term "Power Flow" also fuels the erroneous
belief that reflected power is fictitious."

Another world-class expert has agreed that power flows.

Energy may or may not flow. It is the accumulation of flows that you pay
for on your monthly electric bill. Power is the rate of delivering
energy (doing wotk), so it can`t stand still. No power flow, no work.

Take it from W2DU. Reflections are important. Reflections occur at
discontinuities.

To the extent that voltage and current are in-phase, they mean real
power. Zo is resistive in a practical r-f transmission line. Incident
and reflected waves both propagate on the line with their individual
voltages and currents locked in-phase. The sum of incident and reflected
waves can`t be relied upon to indicate power, but infividually each wave
can indicate true power.

The reflected wave interferes with the incident wave to produce standing
waves. These represent the impedance distribution along the line. They
do not represent variation of power in the wave in eather direction.

Best regards, Richard Harrison, KB5WZI

Cecil Moore wrote:


If the power associated with
an EM pulse is not in the pulse, where is it?

According to definition, the 'effect' of power is something which can be
realized at a place where energy either is being transferred from, or
transferred to. The amount of effect is proportional to the rate at
which energy is transfered. But power is not the thing which is being
transferred from one place to another. Often times though we want to
know the rate at which energy is being transferred to or from one place
to another, and there are mechanisms we can use to measure things that
manifest themselves in proportion to that quantity. But some folks
forget that it's just a tool for making accurate predictions about
things, and start to believe some odd things about the behavior of
nature. One must be careful not to mistake a physical quantity for a
physical entity.

73, ac6xg

Walter Maxwell wrote:


The same problem exists with the term "power flow." Engineering
textbooks define power as the "quantity of energy passing a point per
unit time." Thus, power does not flow--energy flows. However, except
when reciting the definition of power, textbooks and journals on wave
propagation use the term "power flow" almost exclusively, with only an
occasional use of "energy flow." As with "current flow," we know what
is meant because of the common usage which generally overshadows the
strict definition."
Perhaps this explanation will satisfy Jim, but perhaps not. We'll see.

Walt, W2DU


Hi Walt,

I am familiar with the common usage of the expression 'power flow' and
of course the Poynting vector. I've been explaining that "energy flows -
not power" on this newsgroup for 4 years and have gotten nothing but
grief for it. I think it will be more interesting to see whether your
explanation will satisfy the others.

73, ac6xg

Jim Kelley wrote:
"That`s because power doesn`t propagate, and hasten to add - neither do
Poynting vectots."

One could make a long list of serious authors who freely treat power as
moving energy. I namrd Ktaus, Ridenour, Purcell, and Walter Maxwell.

Terman was another world-class expert who was unafraid of the word
"power". On page 76 of his 1955 edition he wrote:
"Alternatively, a load impedance may be matched to a source of power in
such a way as to make the power delivered to the load a maximum (The
power delivered to the load under these conditions is termed the
available power of the power source). This is accomplished by making the
load impedance the conjugate of the generator impedance as defined by
Thevenin`s theorem. That is, the load impedance must have the same
magnitude as the generator impedance, but the phase angle of the load is
the negative of the phase angle of the generator impedance."

I have not yet seen a copy of "Reflections", but would bet that W2DU
says the same thing, perhaps in fewer and shorter words.

Best regards, Richard Harrison, KB5WZI

Richard Clark wrote:
Anyway, the optical analogue has proven to be a bust when I
demonstrated that reflection products do persist. This is not the
place to hash over that again. If you wish, you can consult my
original posting and the follow-ons for details.

But in your "proof", you superposed powers which is a no-no.
When 111.1mW interfers with 87.78mW, the result is not
(111.1-87.78). Since the associated E-fields are 180
degrees out of phase, the power equation must take the
interference into account.

Pref1 = 111.1mW + 87.78mW - 2* sqrt(111.1*87.78)

Pref1 = 111.1mW + 87.78mW - 197.5mW = 1.38mW

You subtract 87.78 from 111.1 and get 23.32. That value is
almost 17 times too high. All your math after that is invalid.
All except 1.38mW of reflections are canceled by that first
internal reflection. Your value of 23.32 is simply wrong.

RF engineers usually convert to voltage, perform the superposition,
and then calculate the total power. One doesn't have that luxury
when dealing with light so the power (irradiance) equations must
be used to obtain the correct results.
--
73, Cecil http://www.qsl.net/w5dxp

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