Re: Calculating streamer breakout of top-loads

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

Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net> 

Hi Rikard,

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

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???  If so, seems like1940 watts at the spark gap could mean 3000W
xformer input power and potentially result in 94 inch sparks using the
John's equation.

Gerry R
Ft Collins, CO

 > Put it in another way:
 > 0.05 uF x 13 kV /2 = 4.225 J
 > 4.225 J x 345 pps = 1457 W
 >
 > I think you are running this coil about 3000 W to get 80 " sparks.Right?
 > Logical conclusion:there are shots exceeding 13 kV primary voltage,and
 > consequently higher potentials V2 than just 148 kV you measured.
 >
 > But still I do think the information of V2/V1 cca 11 in single shot test
of
 > your coil is of interest.Most coils loseless conversion ratios exceed 25.
 >
 > kind regards
 >
 > -Rik
 >
 > _
 >
 >