Roy Lewallen wrote:

A single conductor doesn't have a characteristic impedance -- it's the
impedance between the two conductors of a transmission line. You can
measure a characteristic impedance between, say, a coil and ground, but
its value depends on the spacing between the two. If the coil is tilted
with respect to the ground, the impedance of this two-conductor system
will change with the position along the coil.

Roy: I understand what you are saying. But the derivation of
Characteristic Impedance in the Corum Bros. paper depends only on the
coil dimensions and number of turns; it is independent of any
relationship to other conductors or groundplanes. I also note that
ON4AA's inductance calculator predicts the "Characteristic impedance
of n=0 sheath helix waveguide mode at design frequency" based purely
on the coil geometry. The maths is a bit beyond me (trying to solve
Maxwell's equations for a solenoidal helix), but seems to bear analogy
to the derivation of the characteristic impedance of a waveguide.

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 modelled results.

Regards,
Steve G3TXQ

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

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

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

On Apr 23, 12:22*pm, Art Unwin wrote:
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

When students perform an experiment to proove the laws of Nature it
really does belittle seeing is believing. To change the statistics of
what we are seeing which is the situation on Earth, this alludes
the"relative" term of Einstein, then to bring what we deduced by
seeing by the conversion of weight to mass. This correction thus
brings in to focus what Einstein meant by relativity because it
depends on the gravitational pull relative to what part of the
Universe the experiment was performed. What we term as Classical
physics is the behavior of the Universe and the laws that govern it.
Thus mass is the carrier of potential energy where decay is synonamous
with the break off of a particle which contains a portion of the
potential energy where the brake off is the decelleration of the
partical when it enters a different gravitational field
and thus turns to kinetic energy and where this change is seen as
light i.e Kinetic energy that is transformed to heat which also
governs light. Thus when considering
a perfect conductor ie zero resistance which is also a measure of the
datum level of zero movement of electrons within mass there is zero
movement within mass to affect the passage of current and thus the
current travels at the speed of light. When temperature in not at the
datum level it is the movement of electron within mass that provides
the resistance to current flow and thus we have what is known as the
"velocity factor", and it is the circular movement of displacement
current which is also a movement of current flow that applies what we
know as displacement current.
Thus there is a Universal law of nature because all things revolve
about the relative movement of particles compared to that of a static
particle which if the change is instantaneous we have what Hawkings
calls the BIG BANG.
All of the above emphasises where all the participants of this thread
are argueing about the same problem but from different relative
positions within the Universe

Lesson.
All scientific debate is correlated to the whole of the Universe and
not the metric datum of vacuum as represented by the size of a
arbitrary fieldwithin the Universe
This is what is meant by CLASSICAL PHYSICS.
Enuff said.
Art Unwin

Art Unwin wrote:
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.

The accelerating charges are slow-moving electrons.
The RF current moves at the speed of light in the
(conductive) medium. Therefore, the RF current is
associated with photons emitted by the electrons.
Photons have zero rest mass and zero electric charge.
Photons are the particles associated with RF waves.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Apr 23, 3:22*pm, Cecil Moore wrote:
Art Unwin wrote:
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.

The accelerating charges are slow-moving electrons.
The RF current moves at the speed of light in the
(conductive) medium. Therefore, the RF current is
associated with photons emitted by the electrons.
Photons have zero rest mass and zero electric charge.
Photons are the particles associated with RF waves.
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

Cecil there you go again with the idea that zero mass and zero energy
is something that can exist which some have termed as "photons"
Point to the laws of nature that support that notion. At one time it
was the same
as a particle without mass. Presence science now state that particular
particle does have mass which is why the World spent millions for the
CERN project. Your technical expertise is built on the state of
science 50 years ago and you are now building a castle on sand or
excuses to justify your unwillingness to embrace change. It took 7
days to build the Universe by the initial expenditure of kinetic
energy which embraces the laws of nature and the concept of a cycle or
equilibrium. Remember the words" let there be light" which aligns with
energy expenditure upon mass ie everything starts with the expenditure
of energy upon or from mass it is not a chicken or egg analogy.
And it is expenditure of energy upon mass that makes it a particle
unconnected to all
mass around it such that the particle is unbound and cannot be
absorbed by another state. Lets face it, Adam and Eve knew nothing
regarding equilibrium and the notion of frequency or period. All the
work had been completed way belong that came upon the scene,, which is
why religeon exists as the sole explanation of who and what was and is
in charge with respect to the laws of nature. Again it is impossible
for something to exist without mass.
Art

Art Unwin wrote:
Cecil there you go again with the idea that zero mass and zero energy
is something that can exist which some have termed as "photons"
Point to the laws of nature that support that notion.

It's part of the standard model, Art, with which I am
not about to disagree.

http://en.wikipedia.org/wiki/Standard_Model
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Apr 23, 3:22*pm, Cecil Moore wrote:
Art Unwin wrote:
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.

The accelerating charges are slow-moving electrons.
The RF current moves at the speed of light in the
(conductive) medium. Therefore, the RF current is
associated with photons emitted by the electrons.
Photons have zero rest mass and zero electric charge.
Photons are the particles associated with RF waves.
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

You are quoting the aproach of the bookswhere it is acknowledged that
radiationn cannot fully be explained
For me I am quoting an alternative that does provide the explanation.
If current moves at the speed of light within the Universe ( the
speed of light relative to Earth is slower which creates delay
compared to the former. Insertion of Plank's constant I believe is
a metric that represents the ratio of that delay)
it imparts the same speed to a static particle when impacted, where
the acceleration is determined by Newton's law u.t +f.t sq/2. Since
the particle is static the "u.t" portion equals zero and f.t.sq/2 is
the acceleration from zero to that of the speed of current of the
particle, which is a measure of the expended kinetic energy that
creates the initial format of radiation. I state again without mass
there can be no acceleration.Period
Regards
Art

"Art Unwin" wrote in message
...
I state again

and again, and again, adding bafflegab and gobbledygook with every
iteration... I love it art, how much deeper can you go with this?