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*To*: tesla@xxxxxxxxxx*Subject*: Re: half wave coil*From*: "Tesla list" <tesla@xxxxxxxxxx>*Date*: Tue, 22 Mar 2005 08:10:17 -0700*Delivered-to*: testla@pupman.com*Delivered-to*: tesla@pupman.com*Old-return-path*: <teslalist@twfpowerelectronics.com>*Resent-date*: Tue, 22 Mar 2005 08:10:35 -0700 (MST)*Resent-from*: tesla@xxxxxxxxxx*Resent-message-id*: <0bJZPB.A.KoD.pVDQCB@poodle>*Resent-sender*: tesla-request@xxxxxxxxxx

Hi: Lee:

You have about 821 meter, this represents 1/2 of a wavelength, both your secondary and primary should be at C/ ((2) (820.91)) or 183,000 Hz

Using:

n C / 2 wire length = 1 / 2 pi sqrt (((u x (N/2n)sqrd x Area) x (2n/l)) x capacitance )

( a half wave represents a complete voltage node from two half nodes so n =1 )

Then:

C / (2 (wire length)) = 1/ (2 pi sqrt ( L/2 x cap))

Where L/2 means calculate L from exactly 1/2 of the entire solenoid

L/2 = .0236 Henry (for your coil).

Solving the above equation for total capacitance we get 32.2 pf Subtracting the inter nodal self capacitance ( the self capacitance from the same length considered for inductance) we get approximately 10.5 pf

Then we have 20.85 pf remaining, this value will need to be split between both ends.

We have used these equations for half waves, full waves, one and a half waves, two waves and three wave coils. They always work the first time, and they always corectly predict node locations.

Good luck from Jared

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