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Re: Math Doodling

Hi Dave,
         Comments interspersed:

> Original Poster: David Sharpe <sccr4us-at-erols-dot-com> 
> 
> Terry, ALL
> 
> I've been doing some doodling, and off-line discussion with
> Terry, Barry Benson, and John Freau.  Here is an interesting
> math derivation to try over a cup of coffee...
> --------------------------------------
> 
> Here is a simple math analysis situation that blew Richard Hull
> and Alex Tajnsek away.  Based on equations in the Heise paper and
> assuming lossless transfer of power:
> 
> Vo = Vin * sqrt ( Ls/Lp )  Where        Vo = max Vout from resonator
>                                         Vin = Vin applied to tank circ.
>                                         Ls = Inductance of resonator
>                                         Lp = Inductance of tank pri.
> 
> If the following equation is assumed to be correct in the time domain:
> 
> Vin = Iin * sqrt ( Lp/Cp ) Where        Vin = Vin applied to tank circ.
>                                         Iin = peak tank current
>                                         Lp = Inductance of tank pri.
>                                         Cp = Capacitance of tank C
> 
> AUTHORS NOTE:  This is RMS tank current times Surge Impedance equals
>                applied voltage to tank circuit.

Halve Cp. Then Lp has to be doubled to maintain tune. That means Iin 
halves if the same input voltage is applied as the primary surge 
impedance has doubled. 

> Then substituting equation 2 into 1 and simplifying results in:
> 
> Vo = Iin * sqrt ( Ls/Cp )  Variables as listed above
> 
> This suggests that Cp should be made a small as possible, and
> to maximize Vo, as high a Vin as possible should be employed.  This
> makes sense because Iin will go up with higher Vin, and bang energy is
> .5*C*V^2.

If you plug the halved current and doubled capacitance into the 
combined equation it shows a SQRT(2) reduction in output voltage. Vin 
has to double to make a SQRT(2) gain in output voltage.

I reached the exact same conclusion from a very different point of 
view which considered losses in the primary. The reasoning:

If gap current is kept the same, gap losses are also kept the same.
Hence, halving Cp, doubling Lp and doubling Vin gives a greater 
energy throughput for the same gap loss.
 
> Also, if C is made smaller, dielectric losses maybe REDUCED, with a
> given capacitor (since dielectric area and volume are reduced).
> This is the first time that in doodling with the equations, a
> possible mathematical validation of what has been touted by the TCBOR
> all along is derived, make tank capacitors small, and leverage energy
> by the use of very high voltages.
> 
> FYI and discussion. Am I full of it or does this make sense???

>From every point of view (bar insulation difficulties) I think small C 
and high Vin is the way to go. I showed in another post how the 
linear decrement in the primary also makes this approach a desirable 
one.

Regards,
Malcolm