# Re: Math Doodling

```to: Dave

An interesting hypothesis.

Playing devil's advocate --- the magnetic field depends on current not
voltage, so a larger cap should be capable of delivering higher current if
the impedance is low enough.  The potential produced depends on rate change
of current (again not voltage) across the secondary inductor.  If you
presume the impedance to be a fixed value then the current actually "drawn"
by the load would be equal no matter how large a current source is
available.  We do need to be sure, however, this minimum value is achieved.

Comments -- ideas --- large flames???

Regards,

Dr.Resonance

-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
To: tesla-at-pupman-dot-com <tesla-at-pupman-dot-com>
Date: Thursday, June 10, 1999 3:11 AM
Subject: Math Doodling

>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.
>
>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.
>
>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???
>
>Regards
>
>DAVE SHARPE, TCBOR
>Chesterfield, VA. USA.
>
>

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