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Re: Sparklength inquiry
In a message dated 98-09-23 03:03:30 EDT, you write:
<< >
> > I posted the square law suggestion some time ago, based on my
> > results with the sync gap coils in which I obtained 42" sparks using
> > 620 watts, and 64" sparks using 1570 watts. I was relating input
> > power to spark length. I then scaled up the spark lengths using the
> > square law, and your 25" spark TC fit well on the curve. This
> > posting was before I improved the efficiency of my coils, the old
> > figures were:
>
>>> Power input (revised) spark length toroid dia (inches)
> > 680 W 620W actual 42" 20
> > 2100 W 1570W actual 64" 30
> > 8400 W 6280W 128" 60
> > 33.6kW 25kW 21' 120
> > 67kW 31'
240
> > 134kW 100kW 42' 480
> > 538kW 400kW 84' 960
> > 1.6MW 168' 1920
> >***new*** 5.1MW 300'
> Thanks John. I agree that it's possible to do somewhat better
> than the curve that you have defined above, but your basic curve
> itself seems to give a realistic, conservative estimate of a
> coil's performance vs power level. I had calculated that ALF
> would need 5.1MW to develop 300ft arcs. This curve when inter-
> polated gives 6.3MW -at- 300ft.
Greg,
Agreed.
I noted in my posting above that this chart was created before I
improved the efficiency of my coils. I added a new column in the
chart above, and plugged in the values for my present efficiency
of my coils (see chart). Using these new values, I show a need
for only 5.1MW to develop the 300 foot spark, which by coincidence(?)
agrees exactly with your figure.
John Freau
>However, if you use your same
> square law and interpolate from a good performer such as Ed
> Wingate's coil (15ft at 10kW I believe), you get:
> (300/15)^2 * 10kW = 4MW.
> Likewise, if you increase Ed's power to 26kW, then:
> 15ft * SQRT(26/10) = 24.1ft, which is very close to my present
> coil specs (26kW, 25ft).
> Given these results, it seems like 5.1MW is in the ballpark
> for a 300ft discharge.
> -GL
> www.lod-dot-org >>