Re: How do you measure couplin

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

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

Tesla list wrote:

 > Original poster: Peter Lawrence <Peter.Lawrence-at-Sun.COM>

 > a coil operating in 9:10 mode has a K of ~0.10497 from one of your
 > formulas: k=(b^2-a^2)/(b^2+a^2), and takes b/2 cycles => 5 to the
 > "first notch".

This is correct. 5 full cycles in the primary voltage until the
first notch, or 10 full cycles in a complete beat. The interpretation
of cycles in this case, where there ate two frequencies in the waveform,
is "interval between two peaks of the same polarity". But this is
also 10 cycles of the higher frequency.
The time must be this, because the two oscillations start with the
same polarity and the same amplitude. The first notch occurs when they
add destructively, at 5 cycles of the higher frequency and 4.5 cycles
of the lower frequency. Note that this formula only makes sense
when there is complete energy transfer (b and a integers with odd
difference).

 > someone else recently posted a formula that says the number of oscillations
 > it takes for the energy to transfer from the primary to the secondary is
 > sqrt(1-k^2)/k, which for 0.10497 computes to ~9.42, which is about double
 > the b/2 that you state.

The correct formula would be:
Number of full cycles until first notch=(k+1+sqrt(1-k^2))/(4*k).
I just substituted a=b-1 and found b/2. This relation is valid for
a=b-1, but this is the usual case. It's possible to have a=b-3,
for example, but this results in a "false" first notch at 2.5 cycles,
without complete energy transfer, before the true complete energy
transfer at 5 cycles.

 > I can see how it is real easy to lose track of whether you're counting
 > full cycles or half cycles, whether you're measuring cycles relative to Fa,
 > Fb, or Fres, and I'm sure there are other simple things to accidentally drop
 > out of a description of how to use a formula.

Another formula, more basic: If the two resonance frequencies are
a*f0 and b*f0, the energy transfer takes 1/(2*f0) seconds.

Antonio Carlos M. de Queiroz