"John KD5YI" wrote in
:

....
I seem to recall that a small loop has less inductance than a large
loop. I also seem to recall that the transformation between them is a
function of their inductances. So, a small loop would transform the
larger loop's impedance to an even lower impedance. Is this not so?

You can think of it in terms of inductance, but visit what inductance
really means (ie the relationship between inductance, current and flux).

I gave you an explanation in terms of area of the two loops. If your feed
loop os 20% of the area of the main loop, it intercepts 20% of the flux,
the voltage induced in the feed loop is 20% (1/5) of the main loop
voltage, the impedance is transformed by 1/(1/5^2)=25, so a 2 ohm main
loop is transformed to a 50 ohm load at the feed loop feed point.

Now you can expand that and write it in terms of inductance... but Mike
can measure the loop dimensions with a ruler and calculate area.
Measuring inductance of the components is fraught with problems.

So, if Mike works out what the radiation+loss resistance is of his main
loop, divides it by 50, takes the square root, it gives him the portion
of the main loop area to be occupied by the feed loop. Initially, one
might make the area of the feed loop 20%-50% larger and squash it to fine
tune the transformation ratio.

Owen