On Apr 23, 11:16*am, Art Unwin wrote:
On Apr 23, 10:07*am, Cecil Moore wrote:



steveeh131047 wrote:
I'm inclined to try to understand it better, because it's this derived
Characteristic Impedance, along with the axial Velocity Factor, that
generates the reactance values which seem such a good match to
experimental and modeled results.

Steve, you will find some old-fashioned concepts here
based on the lumped-circuit model rather than the
distributed network EM wave reflection model. One can
easily disprove the assertion that a single wire
in free space doesn't have a characteristic impedance
by asking the question: Does a single electromagnetic
wave traveling through free space (without a wire)
encounter a characteristic impedance? If so, why doesn't
a single wave traveling through a wire in free space
encounter a characteristic impedance? Of course, the
ratio of the electric field to the magnetic field,
whatever that turns out to be, is the characteristic
impedance of a single wire in free space. It, like
the characteristic impedance of free space, seems
to be a few hundred ohms.

There are lots of old wives tales asserted by the gurus
on this newsgroup. One must be careful what one accepts
as technical fact.

"A single conductor doesn't have a characteristic impedance."
is a preposterous assertion. If free space itself has a
characteristic impedance, what are the chances that a
single wire in free space would not have a characteristic
impedance??? Zero, at best??? :-)

Some will say: "Where is the return path for the current?"
I will respond: Where is the return path for the "current"
arriving from the Sun that can be captured by a solar
panel? Good Grief!
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

Cecil, reference you comment that a straight wire does NOT have a
characteristic impedance, this is one place where you misunderstanding
things. A charge rests on the surface and when it is radiating it
instantly is removed from the surface by the displacement current in
coordination with the applied current. If the radiator is not a full
wave length there is no surface for a displacement current to exist
thus the direction of charge is not elevated away from the surface but
continuing the parallel to the surface direction which is the observed
as "end effect"
If the concept of a bounce back of charge was maintained then the
amount of charge
must also change as time revolves around a full period where
eventually the charge
totally reaches the scource when the bouncing around coincided with a
period.
Thus if the charge is in "standing wave" form the impedance changes
during every circuit of the charge back to the source and that can
never be. Characteristic impedance is that seen only with a closed
anti resonant point or in other words at the point of equilibrium
which is represented by a period.
Looking at things from a different angle, when the time varying field
becomes a constant which is then the application of DC then you have a
tesla coil where the spark or energy and thus radiation is parallel to
the conductor and where the period covered by over shoot, a one time
event, where radio radiation is shown by the area of the curve during
the time of that event.
Best regards
Art

Cecil,
You based your proof of a magnetic wave in a vacuum but it is an
accelerating charge
which obviously must have mass, that is radiation ala the particle.
If you have a Tesla set up in a vacuum the speed of the particle/spark/
light is the approximation of the speed of light.( I say approximation
since I am using the metric
of Earth's vacuum and not that of the Universe) The velocity factor is
the true ratio of the mismatch with the travel of a electric current
on Earth with all its relavent factors and comparing it to the speed
of light in the average metric of vacuum of the Universe.
Bottom line is particles are part of radiation as is light, "waves"
are not involved other than a bevy of particles separated by a
fraction of a period.
Art