Richard Clark wrote in
:

On Sat, 30 May 2009 07:19:26 GMT, Owen Duffy wrote:

The main loop isn't anything like 2 ohms at resonance, it is much much
higher, and the voltage is quite high for a given power.

Hi Owen,

By your latest comment, it appears you have jumped off the rails
completely. Let's look at the spec:

On Wed, 27 May 2009 13:36:53 -0400, Michael Coslo
wrote:
It's Octagonal, .75 inch copper tubing, around 8 feet in diameter. It's
designed to just reach into the 80 meter voice segment with it's
capacitor setup.
which through any of several methods reveals radiation resistance
equals 0.27 Ohms. The need for large conductors, high voltage
components, and loop (which, for 80M would be half a meter) coupling
to obtain the proper match are all found in their proper
relationships.

My reckoning:

I ran up a model in NEC of a vertical circular loop of 2.4m diameter,
made of 19mm dia copper conductor, and located with its centre 3m above
average ground.

The feedpoint impedance is 0.13+j168 at 3.5MHz.

If that was resonated with a lossless capacitor, the parallel load
impedance would be about 220k+j0.

To transform that to 50+j0, the impedance ratio is 4400:1, the voltage
ratio is 4400^0.5:1 =66:1, the areas need to be in a ratio of 1:66, the
diameters (for circular feed loop) 1:66^0.5 =1:8.1. The feed loop would
be about 12% of the diameter of the main loop or about 0.3m.

....
Reggie covered all of this years ago. Unfortunately, he buried the
best of his legacy in a drift of bantering making it an extreme chore
to mine Google for the relevant postings.

Reg did leave us a calculator for this problem (MAGLOOP4) but without
(AFAIK) documentation of the method used. MAGLOOP4 comes up with slightly
different figures to my NEC model, slightly higher loss, slightly lower
main loop // Z, slightly larger feed loop.

Owen