Re: Calculating streamer breakout of top-loads

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Original poster: "Rikard Titus" <rikard_titus-at-hotmail-dot-com> 




Hi Gerry,

>Hi Rikard,
>
>Do you mean:  1/2 * (0.05uf x 13KV ^^2)  = 5.6 joules
>5.6J x 345 pps = 1940 W

Correct.Forgot about the square.
One way or another,some 1000 W is missing,(see dr Resonance's post).


>John's empirical spark length formula of 1.7 sqrt (power) is based on
>xformer input power not the power delivered thru the spark gap. Is this
>right???

I am not familiar with this formula.Provide me with link/more info .



If so, seems like1940 watts at the spark gap could mean
>3000W
>xformer input power


Hey wait a minute!
If PT losses 34 % of a real power,than it means it is mature for garbage  !

-Rik