Re: I've lost my k. Can someone help me find it?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>

John Couture wrote:

> The results of K factor tests of one of my coils. I calc a
> K of .1720 compared to .1708 using your calcc method. About
> 0.7 percent difference.
>  F1 = 425
>  F2 = 505
>  Fr = 465

>  K = (F2-F1)/Fr = (505-425)/465 = .1720

The above is correct when F1 and F2 are the two secondary
resonant frequencies obtained with the secondary top is
open-circuit and short-circuit, in the way John describes
below:

> Look for a resonant frequency with the secondary coil ends
> open with no connection and also the resonant frequency with
> the secondary coil ends connected together.

Please don't confuse these with the two resonant mode frequencies
which comprise the beat envelope.  The mode frequencies are the
two resonances that you will detect if you sweep the coupled coils
with a signal generator (primary gap closed and the secondary with
its normal topload and grounding).  This latter method is much
better than the one you suggest, because it reports the actual mode
frequencies that appear during operation.  When F1 and F2 
refer to the mode frequencies (as opposed to the short/open test
frequencies) then k is calculated with

   x = F2/F1; 
   k = (x^2-1)/(x^2+1);

This represents the real k factor that is in effect during operation.

I hope everybody has got this clear now!  There are two different
methods here, one which the original question referred to, and the
other which John is advocating.  Two different formulas.
--
Paul Nicholson
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