From: "Tesla list" <tesla@xxxxxxxxxx>
To: tesla@xxxxxxxxxx
Subject: Re: Safety gap issues
Date: Sun, 27 Nov 2005 15:03:19 -0700
Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>
On 27 Nov 2005, at 9:53, Tesla list wrote:
> Original poster: "JT Bowles" <jasotb@xxxxxxxxxxx>
>
> WOW, NICE. According to what you just said, my coil produces an
> absolute maximum of 533KV (if my toroid were perfectly smooth) Half a
> million volts seems too much for a mere 14" discharge average. But, if
> you say half a million, i'll go with that! :P
The maximum theoretical output voltage of any coil can be calculated
by:
Vo = Vgap x SQRT(Cp/Cs) where Cp is the primary capacitor and Cs is
the total capacitance of the secondary system. Cs is itself an
approximation but close enough. This value of Vo *assumes* the system
to be lossless which it isn't. A reasonable estimate would be the Vo
calculated above multiplied by a factor of 0.7 to 0.8 for the average
disruptive coil.
One of my coils gives a 12" (maybe a tad more) connected discharge in
single shot mode with a theoretical lossless output voltage of 340kV
(in reality probably about 300kV).
Malcolm
> >From: "Tesla list" <tesla@xxxxxxxxxx>
> >To: tesla@xxxxxxxxxx
> >Subject: Re: Safety gap issues
> >Date: Sat, 26 Nov 2005 12:51:45 -0700
> >
> >Original poster: Terry Fritz <vardin@xxxxxxxxxxxxxxxxxxxxxxx>
> >
> >Hi Jim,
> >
> >At 08:16 AM 11/26/2005, you wrote:
> >...........
> >>If you want to "estimate" the voltage on your tesla coil from it's
> >>physical design, your best bet is to measure the radius of
> >>curvature. The voltage won't be much higher than the radius of
> >>curvature in cm times 30 kV/cm, and will likely be lower (since
> >>that's the max voltage for smooth sphere with nothing around it).
> >
> >I have noticed that the "breakout voltage" does tend to correspond to
> >the radius of curvature and all. However, the top voltage can then
> >go substantially beyond that. If there is a lot of power behind the
> >arcs. Then the "breakout loading" is just not enough to hold the
> >voltage down. So it does not act like say a hard Zener diode, but
> >rather a Zener with a big resistor in series with it.
> >
> >Cheers,
> >
> > Terry
> >
>
>
>