Re: Safety gap issues

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Malcolm Watts" <m.j.watts@xxxxxxxxxxxx>

On 27 Nov 2005, at 18:30, Tesla list wrote:

> Original poster: "JT Bowles" <jasotb@xxxxxxxxxxx>
>
> Whoa, whoa, whoa. People, which is it?
>
> *Vo = Vgap x SQRT(Cp/Cs)
>              OR
> *Vo = Vgap x SQRT(Cp/2Cs)

The first is for the lossless (ideal) case. The second takes typical
losses into account. If you read the messages carefully you will find
no disagreement. I do believe 1/SQRT(2) = 0.7 approximately.

Malcolm

>
>
>
> >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
msnip...