Re: nobody knows whos maggie at fla teslathon?

["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: "D&M's High Voltage" <DMsHV.DavMcKin-at-verizon-dot-net>
 >
 > Antonio,
 >
 > Thanks for running the numbers.  Could you run the numbers using 400 nF.  As
 > for the coupling, we were running it really loosely coupled as your numbers
 > show.  We plan on making a much closer coupled system for next year.
 >
 > Could you also run some numbers to determine what would be the optimum
 > inductance on the tertiary coil (using either 133nF or 400 nF), and then
 > also what would be the "worst case" (physically realizable) inductance.
 > We'll then see if we can wind a couple of coils and try them out at next
 > years thon (Labor day weekend 2004).

My programs are adjusted to calculate the elements from a given
resonator,
so what I can do now is to try some designs keeping the present third
coil.
The Magsim program (http://www.coe.ufrj.br/~acmq/programs)
can be set to generate a table with many designs. Fixing C3=105 pF,
L3=65.97 mH, and setting C1 to 133 nF and 400 nF, from the table it's
easy to find realizable values for L2, C2, and k12. The problem then
is how to build them. It's better to select a low C2, that
can be a distributed capacitor, L2 similar to what you have, and a
coupling coefficient not too high.

Looking at some possibilities (not necessarily the best):

With C1=133 nF:
Mode=  3, 4,13: L1=0.07042 mH L2= 23.226 mH L3= 65.970 mH C1=133000.0 pF
C2=  50.2 pF C3= 105.0 pF k12=0.510 time:   33.07 us
Mode=  3, 4,15: L1=0.07090 mH L2= 23.831 mH L3= 65.970 mH C1=133000.0 pF
C2=  36.7 pF C3= 105.0 pF k12=0.515 time:   33.07 us
Mode=  5, 6,23: L1=0.06276 mH L2= 13.526 mH L3= 65.970 mH C1=133000.0 pF
C2=  50.2 pF C3= 105.0 pF k12=0.412 time:   49.61 us
Mode=  5, 6,25: L1=0.06288 mH L2= 13.677 mH L3= 65.970 mH C1=133000.0 pF
C2=  42.0 pF C3= 105.0 pF k12=0.414 time:   49.61 us

With C1=400 nF:
Mode=  3, 4,17: L1=0.02368 mH L2= 24.235 mH L3= 65.970 mH C1=400000.0 pF
C2=  28.1 pF C3= 105.0 pF k12=0.518 time:   33.07 us
Mode=  3, 4,19: L1=0.02375 mH L2= 24.518 mH L3= 65.970 mH C1=400000.0 pF
C2=  22.3 pF C3= 105.0 pF k12=0.521 time:   33.07 us
Mode=  5, 6,29: L1=0.02096 mH L2= 13.892 mH L3= 65.970 mH C1=400000.0 pF
C2=  30.7 pF C3= 105.0 pF k12=0.417 time:   49.61 us
Mode=  5, 6,31: L1=0.02098 mH L2= 13.970 mH L3= 65.970 mH C1=400000.0 pF
C2=  26.7 pF C3= 105.0 pF k12=0.418 time:   49.61 us
Mode=  8, 9,36: L1=0.01947 mH L2=  8.214 mH L3= 65.970 mH C1=400000.0 pF
C2=  66.7 pF C3= 105.0 pF k12=0.333 time:   74.41 us
Mode=  8, 9,38: L1=0.01949 mH L2=  8.270 mH L3= 65.970 mH C1=400000.0 pF
C2=  59.5 pF C3= 105.0 pF k12=0.334 time:   74.41 us

The complete table can be generated by the program. It's not easy to
find modes that keep C2 and k12 low,
and are not too high. Note that I selected the modes k:k+1:k+n. These
produce waveforms similar to
what is seen in a Tesla coil, with perfect energy transfer at the first
notch. When the third factor
goes to infinity, C2 goes to zero, and the result is an equivalent of a
Tesla coil, imperfect because
C2 is not zero in practice.

Antonio Carlos M. de Queiroz