Re: spark energies

To: tesla-at-pupman-dot-com
Subject: Re: spark energies
From: Richard Hull <hullr-at-whitlock-dot-com>
Date: Mon, 04 Nov 1996 18:40:15 -0800
Tesla List wrote:
> 
> > Subject: spark energies
> 
> >From bert.hickman-at-aquila-dot-comSun Nov  3 21:42:58 1996
> Date: Sat, 02 Nov 1996 00:52:06 -0800
> From: Bert Hickman <bert.hickman-at-aquila-dot-com>
> To: tesla-at-pupman-dot-com
> Subject: Re: spark energies
> 
> Tesla List wrote:
> >
> > >From hullr-at-whitlock-dot-comThu Oct 31 22:48:03 1996
> > Date: Thu, 31 Oct 1996 17:57:25 -0800
> > From: Richard Hull <hullr-at-whitlock-dot-com>
> > To: tesla-at-pupman-dot-com
> > Subject: spark energies
> >
> > I just got to thinkin' deeper over a "samich" at lunch.  In my post on
> > spark energy, I would literaly have to account for all energies created
> > by the sparks themselves in my scenario.
> >
> > Help me out guys.
> >R. Hullbig snip


> Some other thoughts:
> It actually may be easier to measure/calculate the isotropic capacitance
> of the output, and make a capacitive voltage divider (similar to that
> shown in Duane Bylund's Modern Tesla Coil Theory Supplement, pp. 22-24).
> If we know the degree of re-tuning necessary to get peak performance due
> to ion cloud capacitance, we can recompute the apparent isotropic
> capacitance of the toroid and ion cloud. With the calibrated voltage
> divider, we can determine the actual output voltage under discharge
> conditions AND the average Q of the secondary under heavy discharge
> conditions.
> 
> Assuming we've got a toroid with a large curvature, we should also be
> able to measure the Q of the secondary just prior to breakout, and the
> maximum output voltage without breakout, which would give us a measure
> of the maxiumum secondary energy available per bang (using 1/2(Ct
> +Cion)Vs^2). Since the difference between the two Q's is due to energy
> lost/cycle to the corona, we should be able to fairly closely estimate
> the total energy lost per bang in the discharge alone. If we know the
> PPS rate, we can calculate the actual average power being dissipated by
> the secondary discharge. The advantage of this appproach is that we
> could do the measurements/calculations at a variety of power levels up
> to full multi-kilowatt levels.
> 
> Finally, we could make a hot-wire ammeter to measure the "average"
> current flow of the corona discharge (like the earlier light bulb
> calculations??). Coupled with the voltage divider measurements, we
> should be able to compute the average power dissipated by the corona
> discharge. I need to think about this a little bit more...
> 
> Speculative measuring to ya!
> 
> -- Bert --

Bert, 

Thanks for the input and thoughts,  They are well taken and will be 
incorporated in some fashion as I go.

Richard Hull, TCBOR