Jim Kelley wrote:
Cecil Moore wrote:

What I said is the voltage and
current in a standing wave are *always* 90 degrees out of
phase and it is impossible to generate heat when the voltage
and current are 90 degrees out of phase.

So then shouldn't one expect coax to be heated uniformly along its
length at a high SWR?

No, any and all heat is work done by traveling waves, not
standing waves. As long as standing waves exist as standing
waves, they are incapable of doing work or heating anything.

You, of all people, should appreciate that since V*I*cos(90)
equals zero for standing waves, absolutely no work can be
performed by a standing wave. If the energy in a standing
wave is used to provide work, the standing wave ceases to
exist as it does at key-up.

In trying to get any work out of V*I*cos(90)=0, blood out
of a turnip comes to mind.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

...
In trying to get any work out of V*I*cos(90)=0, blood out
of a turnip comes to mind.

I am not at the point where I would dismiss/cease-to-listen your
argument(s), far from it.

However, I see no reason why standing waves on a string, standing waves
in water, acoustic standing waves in the medium of a tuning fork, etc.
should be expected to behave in any manner inconsistent to em waves in a
medium--which can contain them and whereas standing waves will result ...

While most techs are "rote leaned", I am simply a hobbiest in these
areas, I have no "indoctrinated beliefs."

It is pretty apparent where I would attempt connection with any of these
mechanical waves in an attempt to extract work (power/energy/joules),
and from any "container" capable of containing them.

I have no problem in forming questions about the validity of the math
which drives your argument(s)--in one of my hands lays the
formulas/equations, in the other what my eyes suggest--NO, "what my eyes
tell me" ... close examination of either suggests the other is lie.

You will have patience if this takes a bit--I am sure ...

Regards,
JS

Cecil Moore wrote:
Jim Kelley wrote:

Cecil Moore wrote:

What I said is the voltage and
current in a standing wave are *always* 90 degrees out of
phase and it is impossible to generate heat when the voltage
and current are 90 degrees out of phase.


So then shouldn't one expect coax to be heated uniformly along its
length at a high SWR?


No, any and all heat is work done by traveling waves, not
standing waves.

Perhaps you misunderstood the purpose of the question. Traveling
waves are basically uniform in amplitude along the length of a
transmission line. Why would any heating associated with a traveling
wave anything other than uniformly distributed?

As long as standing waves exist as standing
waves, they are incapable of doing work or heating anything.

You, of all people, should appreciate that since V*I*cos(90)
equals zero for standing waves, absolutely no work can be
performed by a standing wave. If the energy in a standing
wave is used to provide work, the standing wave ceases to
exist as it does at key-up.

In trying to get any work out of V*I*cos(90)=0, blood out
of a turnip comes to mind.

I have a piece of coax around here somewhere that I once burned up. I
recall telling you about it. The insulation is bubbled and melted at
half wavelength intervals. Please explain what particular aspect of a
traveling wave might have caused that to happen.

Thanks and happy new year de ac6xg

On Wed, 02 Jan 2008 11:19:47 -0800, Jim Kelley
wrote:

I have a piece of coax around here somewhere that I once burned up. I
recall telling you about it. The insulation is bubbled and melted at
half wavelength intervals. Please explain what particular aspect of a
traveling wave might have caused that to happen.

Head on collision?

73's
Richard Clark, KB7QHC

Jim Kelley wrote:

I have a piece of coax around here somewhere that I once burned up. I
recall telling you about it. The insulation is bubbled and melted at
half wavelength intervals. Please explain what particular aspect of a
traveling wave might have caused that to happen.

It's the consequence of having *two* traveling waves, which occurs any
time the line isn't terminated with its characteristic impedance. Betcha
yours wasn't.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Jim Kelley wrote:
I have a piece of coax around here somewhere that I once burned up. I
recall telling you about it. The insulation is bubbled and melted at
half wavelength intervals. Please explain what particular aspect of a
traveling wave might have caused that to happen.

It's the consequence of having *two* traveling waves, ...

Funny how Roy changes his tune from posting to posting, huh?

Consider his exactly opposite response to measuring the
delay through a 75m loading coil using those same two
traveling waves. :-)
--
73, Cecil http://www.w5dxp.com

Jim Kelley wrote:
I have a piece of coax around here somewhere that I once burned up. I
recall telling you about it. The insulation is bubbled and melted at
half wavelength intervals. Please explain what particular aspect of a
traveling wave might have caused that to happen.

It was the simple scalar addition of two traveling waves
that caused it to happen. Since standing waves contain no
real power, they cannot directly supply any real power.

Only traveling waves, with their voltages and currents
in phase, can supply real power. Each traveling wave
delivers some of its energy which is converted into heat.

If you believe that steady-state standing waves can supply
real power, please explain how real power can be obtained
when the voltage and current are 90 degrees out of phase.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Jim Kelley wrote:

I have a piece of coax around here somewhere that I once burned up. I
recall telling you about it. The insulation is bubbled and melted at
half wavelength intervals. Please explain what particular aspect of a
traveling wave might have caused that to happen.


It was the simple scalar addition of two traveling waves
that caused it to happen.

'Addition' is not a cause. What is the actual cause - i.e. what
exactly causes coax to melt at half wavelength intervals?

Since standing waves contain no
real power, they cannot directly supply any real power.

To be consistent with the definitions EM waves don't actually
'contain' power, but it is certainly true that interference patterns
don't propagate and transfer energy.

73, ac6xg

Jim Kelley wrote:
'Addition' is not a cause.

Superposition is not a cause????
Superposition *IS* addition of phasors.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:

Jim Kelley wrote:

'Addition' is not a cause.


Superposition is not a cause????


Superposition *IS* addition of phasors.

Like addition, superposition is a mathematical operation. There is
without question a mathematical result to most mathematical
operations. But what does this operation itself actually cause in our
case? A physical result has been obtained, but what is its exact
cause? I don't think 'addition' is the best answer.

73, ac6xg