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?