Re: Theoretically, Cltr/Cres = pi/2

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

Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>" <biomed-at-miseri.winnipeg.mb.ca>


Hello Terry,

Very interesting derivation.  Looking at it, seems to make a lot of short
cuts, probably good ones.  I like using the NST impedance  Vnst/Inst
simplification.  Your using it like XL which is a close approaximation.  I
worked it out for a 12Kv -at- 60 mA.  Z = 200Kohms, now R measured was
4.6Kohms across both HV leads, thats only a 2.3 % factor of the impedance
which is small enought to ignore.  This also works out perfectly for my
coil because it is almost bang on LTR.  I guess other things to consider
would be the primary inductance and transformed impedance from the
secondary, but those too are probably negligable at 60 or 120 Hz compared
to that of the NST.  I did find a boo boo in the math, probably when You
typed it out because it fixes itself later, see below.


Thanks,
Shaun Epp

------snip ------
>However, also we have a direct relationship for capacitor size, energy,
and
>voltage.

>Ebang = 1/2 Cltr x Vfire^2  but...

>Vfire = Vnst x SQRT(2) so...

**No -- >Ebang = Cltr / Vnst^2

this was - Ebang = Cltr x Vnst^2


>Equating this into Eq2 gives:

** No -- >Cltr / Vnst^2 = Vnst x Inst / (2 x BPS)

this way -  Clts x Vnst^2 = Vnst x Inst / (2 x BPS)

>Solving for Cltr...

correct >Cltr = Inst / (2 x Vnst x BPS)  ...Eq3

in the end, the equations work out ok.