Cecil Moore wrote in news:AmWFh.2314$tv6.372
@newssvr19.news.prodigy.net:

Owen Duffy wrote:
- that V/I=Zo for a travelling wave in a transmission line, and that
(Vf+Vr)/(If-Ir) at the load end of the line must equal Zload;

It seems that we can draw some conclusions from these
assumptions.

Vf*If*cos(0) = forward joules/sec

What is this "we" business? That is your conclusion

What is the cos(0) term for Cecil? Are you implying that the phase angle of
Zo is 0? That isn't true in the general case, it is true ONLY for lossless
or distortionless lines.

Owen

chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.



Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.


A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the
Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)



And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

73,

Chuck


Static fields, by definition, do not have a time varying divergence. No time
variation, no Poynting Vector. Nes Pas?

On 2 Mar, 11:35, Dave wrote:
chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the
Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)

And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

73,

Chuck

Static fields, by definition, do not have a time varying divergence. No time
variation, no Poynting Vector. Nes Pas?- Hide quoted text -

- Show quoted text -

Then what you do is add a time varying field to a static field and add
something like " during a period of time " to Gausses law.
What on earth is your problem? Show a bit of inititive !
Art

And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

What the Poynting vector actually tells us is that the flux into any closed
volume is equal to the rate of storage of energy within the volume. Since a
static field has no flux into any closed volume the poynting vector tells us
that no storage or dissipation of energy is taking place, ie it tells us
nothing and is therefore of no use in this situation.

In non-static fields it does mean something, but even here it may not
completely describe the energy flow. The Poynting Vector is just one of an
infinite number of way of describing energy flow. It is a convenient and
mathematically simple one, but not a complete solution. For instance an
alternative one is the Slepian Vector which is mathematically more
cumbersome but does include the static case.

73
Jeff

On Fri, 2 Mar 2007 19:56:27 -0000, "Jeff" wrote:

Slepian Vector which is mathematically more
cumbersome but does include the static case.

Sorry Jeff, but I can just hear hope re-awakening beneath the frost of
disappointment, and the birth of a new theory emerging. I look
forward to a new thread
"Why haven't I received my Nobel Award from Ed McMahon yet?"

73's
Richard Clark, KB7QHC

On Thu, 01 Mar 2007 17:36:12 -0600, Cecil Moore
wrote:

Basically nothing.

Richard Clark wrote:
When it is generally accepted that our sources do not exhibit 50 Ohms
source resistance/impedance, what resistance/impedance do they
exhibit?

To which I got no answers in stereo.

Some may sense frustration to which I once asked how much relief I
would need. Having already long anticipated a non-response (score
100% for accuracy in modeling); others may want to fumble rhetorically
with equally uninformed responses. In anticipation of that I provide
a multiple choice. Richard needs to take:
1. An aspirin;
2. A syringe of morphine;
3. Both of the above;
4. 50 Ohms of dissipation.
5. All of the above, except 3.
6. ____________

Now take that number, and divide it by the Sarcasm co-efficient.

Plot this on a graph and draw an asymptote. You now have a diagram
for a Gaussian Bundle. Find the Pointing Vector for this (it points
to the same answer you picked in the multiple choice) to verify the
authenticity of your newsreader's software design. ;-)

73's
Richard Clark, KB7QHC

Dave wrote:
chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.



Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.


A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector
relationship]. How do the physics type adjust their definition to
include the Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)



And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static
E- and H- fields?

73,

Chuck


Static fields, by definition, do not have a time varying divergence. No
time variation, no Poynting Vector. Nes Pas?


I think the Poynting vector can be calculated even when the E and H
fields are static. Of course, doing so would violate Poynting's
assumptions and thus be meaningless. But if one didn't know in advance
that an arbitrary closed surface contained static sources, and he found
the Poynting vector, S, for some small area, he could well get a
non-zero answer. Of course, the integral of S over the entire surface
would always be zero in the case of static sources.

To be applicable, the Poynting theorem requires that the E and H fields
arise from a single source, satisfying Maxwell's first two equations.
But that information may not be known in advance.

73,

Chuck








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Jeff wrote:
And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

What the Poynting vector actually tells us is that the flux into any closed
volume is equal to the rate of storage of energy within the volume. Since a
static field has no flux into any closed volume the poynting vector tells us
that no storage or dissipation of energy is taking place, ie it tells us
nothing and is therefore of no use in this situation.

My understanding is somewhat different (and quite possibly in error).
You can calculate a non-zero Poynting vector and it will not always tell
you that no storage or dissipation of energy is taking place. Assuming
you do not already know the source of fields inside a closed surface,
you can be misled by the Poynting vector. You need an additional step in
that case: i.e., integrating S over the entire surface. That will
produce a value of zero in the case of static fields.



In non-static fields it does mean something, but even here it may not
completely describe the energy flow. The Poynting Vector is just one of an
infinite number of way of describing energy flow. It is a convenient and
mathematically simple one, but not a complete solution. For instance an
alternative one is the Slepian Vector which is mathematically more
cumbersome but does include the static case.


There seems to be a divergence of views on the Slepian vector, but I am
not familiar with it.

73,

Chuck

73
Jeff



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On 2 Mar, 13:04, chuck wrote:
Dave wrote:
chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector
relationship]. How do the physics type adjust their definition to
include the Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)

And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static
E- and H- fields?

73,

Chuck

Static fields, by definition, do not have a time varying divergence. No
time variation, no Poynting Vector. Nes Pas?

I think the Poynting vector can be calculated even when the E and H
fields are static. Of course, doing so would violate Poynting's
assumptions and thus be meaningless. But if one didn't know in advance
that an arbitrary closed surface contained static sources, and he found
the Poynting vector, S, for some small area, he could well get a
non-zero answer. Of course, the integral of S over the entire surface
would always be zero in the case of static sources.

To be applicable, the Poynting theorem requires that the E and H fields
arise from a single source, satisfying Maxwell's first two equations.
But that information may not be known in advance.

73,

Chuck

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- Show quoted text -

Chuck,
I believe you would be much better off choosing Gausses law for
statics and adding a time element to the written law itself thus
providing all three cartesian coodinates reguired for the law.
I suppose one could say it is not Gaussian law anymore but
mathematically it fits quite well. This is the basis for Gaussian
antenna arrays for which I have submitted for consideration from the
patent office. As yet I have not heard any comment that invalidates
this concept other than it can't be done from psuedo experts and
frankly I feel that the addition speaks for itself.
Regards
Art

Owen Duffy wrote:
What is this "we" business? That is your conclusion

Hams have a habit of saying "we", Owen. Are you a ham?
We have an IC-756PRO. We are a Texas Aggie. We drive a
GMC pickup. We are not married.

What is the cos(0) term for Cecil? Are you implying that the phase angle of
Zo is 0? That isn't true in the general case, it is true ONLY for lossless
or distortionless lines.

We sometimes assume lossless lines for the sake of discussion
which is often a close enough approximation to real-world
lines. We don't have our books with us but seems we remember
that the Z0 phase angle for common Z0 lines is about 1-2
degrees.
--
73, Cecil http://www.w5dxp.com