# Re: half wave coil

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• Subject: Re: half wave coil
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Tue, 22 Mar 2005 08:10:17 -0700
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Original poster: jdwarshui@xxxxxxxxx

`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`