On Sep 4, 7:04*am, Cecil Moore wrote:
Dale Parfitt wrote:
Thank you Cecil,
That's all I was looking for.

You're welcome and I agree with 95% of what W8JI says.
(For instance, he is mistaken about the delay through
a 100T, 10TPI, 2" diameter 75m loading coil.)

Some may or may not understand why random folding
of antenna radiators tends to change the radiating
conductors into non-radiating conductors. (The same
effect is at work in loading coils.)

When two conductors are carrying differential coherent
currents with no common-mode current, there is negligible
radiation when the two conductors are parallel to
each other and the spacing is a very small fraction of
a wavelength. It's called a transmission line and most
of the losses at HF are I^2*R. Usually, one of the goals
of a transmission line is not to radiate. Transmission line
fields tend to cancel in the near field due to destructive
interference.

A single straight wire in free space is a very efficient
radiator because interference occurs mostly in the far
field. Fold it back upon itself and unless the
second conductor is positioned perfectly, there will exist
differential currents between the two conductors which
will tend to cancel the radiation - leaving mostly I^2*R
losses at HF.

Small folded/loaded antennas tend to cancel the radiating
fields. The only other avenue for a lot of the energy is
conversion to heat.
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

Cecil I must respectively disagree. Your arguement is based on the
presence of common mode current. When there is a state of equilibrium
there is no vector that represents common mode. Since the radiator is
a full WL that represents a period it is of closed circuit form. In
such a case any radiator bend is accompanied by a bend that is equal
and opposite per Newtons laws. The moment you introduce common mode
currents you have strayed from the concepts of equilibrium, where all
forces are accounted for. Maxwells laws are based on the position that
all forces involved are accounted for where the summation of such
equals zero.
Regards

Art Unwin wrote:
Since the radiator is
a full WL that represents a period it is of closed circuit form.

My comments were about a one-wavelength straight
wire dipole in free space. The reflections from
the ends are what result in that large resonant
resistance at the center.

--Vf
Open --If
----------------------fp----------------------
Circuit Vr--
Ir--

Zfp - feedpoint impedance, Vf - forward voltage,
Vr - reflected voltage, If - forward current,
Ir - reflected current

Zfp = (Vf+Vr)/(If-Ir) = thousands of ohms

However, if we fold the 1WL dipole into a circular
1WL loop it is still a standing-wave antenna but
the phase of the reflections is reversed.

Zfp = (Vf-Vr)/(If+Ir) = ~100 ohms.

Where are those reflections coming from in a
circular 1 WL loop? Why is the phase of the
reflections reversed?
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 4, 12:53*pm, Cecil Moore wrote:
Art Unwin wrote:
Since the radiator is
a full WL that represents a period it is of closed circuit form.

My comments were about a one-wavelength straight
wire dipole in free space. The reflections from
the ends are what result in that large resonant
resistance at the center.

* * * * * * * * * --Vf
Open * * * * * * --If
----------------------fp----------------------
Circuit * * * * *Vr--
* * * * * * * * * Ir--

Zfp - feedpoint impedance, Vf - forward voltage,
Vr - reflected voltage, If - forward current,
Ir - reflected current

Zfp = (Vf+Vr)/(If-Ir) = thousands of ohms

However, if we fold the 1WL dipole into a circular
1WL loop it is still a standing-wave antenna but
the phase of the reflections is reversed.

Zfp = (Vf-Vr)/(If+Ir) = ~100 ohms.

Where are those reflections coming from in a
circular 1 WL loop? Why is the phase of the
reflections reversed?
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

Equilibrium means equilibrium thus there are no reflections. Actions
have an equal and opposite reaction. What are you going to draw upon
for an equalizing vector?

Art Unwin wrote:
Equilibrium means equilibrium thus there are no reflections.

No reflections on a standing-wave antenna?
Where do the standing waves come from?
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 4, 3:00*pm, Cecil Moore wrote:
Art Unwin wrote:
Equilibrium means equilibrium thus there are no reflections.

No reflections on a standing-wave antenna?
Where do the standing waves come from?
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

There are no standing waves either

On Sep 4, 3:40*pm, Art Unwin wrote:
On Sep 4, 3:00*pm, Cecil Moore wrote:

Art Unwin wrote:
Equilibrium means equilibrium thus there are no reflections.

No reflections on a standing-wave antenna?
Where do the standing waves come from?
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

There are no standing waves either

Remember my WL is a closed circuit

Art Unwin wrote:
Remember my WL is a closed circuit

Believe it or not, there are reflections at the
feedpoint caused by the Z0 physical impedance
discontinuity. You would be correct only if your
"closed circuit" was a traveling wave antenna.
But if it was a traveling wave antenna, its
feedpoint impedance would be in the hundreds
of ohms which, I assume, it is not. The fact that
your antenna doesn't have a feedpoint impedance
equal to the Z0 of the antenna wire proves that
reflections are present.

W8JI made essentially the same mistake in his loading
coil delay measurements so don't feel bad about it.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

Art Unwin wrote:
There are no standing waves either

If there were no standing waves, a current pickup would
read a constant current when moved up and down the conductor,
but it doesn't. A current pickup proves there are standing
waves. You can see it with your own eyes using RF current
meters available from MFJ. If your theory rests on "no standing
waves" being present, it can easily be disproved.
--
73, Cecil, IEEE, OOTC, http://www.w5dxp.com

On Sep 4, 6:57*pm, Cecil Moore wrote:
Art Unwin wrote:
There are no standing waves either

If there were no standing waves, a current pickup would
read a constant current when moved up and down the conductor,
but it doesn't. A current pickup proves there are standing
waves. You can see it with your own eyes using RF current
meters available from MFJ. If your theory rests on "no standing
waves" being present, it can easily be disproved.
--
73, Cecil, IEEE, OOTC, *http://www.w5dxp.com

I will wait for that day.
Regards