Re: Wire length LC derivation,

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: jdwarshui@xxxxxxxxx 


Part 1

By Ampere, the inductance for a long air cored solenoid is:
L = (u Nsqrd A) / h                       (where h is the length of
the solenoid)
                                          (where A is Area )
                                          (where N is the number of
turns)
Since:
 2 pi r N = wire length*

  *actual wire length = sqrt ( (2pirN)sqrd + (h)sqrd )
    but we disregard (h)sqrd as an insignificant term.

Then:
N = wire length / 2 pi r N

Substituting we get:
L = u ( wire length)sqrd / 4 pi h

By Maxwell:
u = 1 / Csqrd  epsilon        (Where C = 3x10 to 8 m/s)

Then:
L = (wire length)sqrd / (epsilon Csqrd  4 pi  h )

Part 2

For sake of simplicity in the following description of LC we will use
the derived capacitance of an isolated sphere. This would be directly
applicable to a Tesla coil secondary.

 From Gauss:
V = q/4 pi epsilon integral a to b for 1/ (r sqrd) dr

Since:
 q = cap Volt

 Then:
cap. = 4 pi epsilon (ab / b-a)

Since we have a single isolated plate, b approaches infinity
we get:
cap. = 4 pi epsilon R           ( R is the radius of the sphere )

Now omega = 1/ sqrt (LC)
Substituting and canceling we get:

1/ (omega sqrd) = (wire length / 3x10 to 8 m/s)sqrd  x  ( R/h )

So:

 [omega = (C/ wire length) x sqrt(h/R)

*R must of course be decreased to accommodate the self capacitance of
the coil.

Notice that the distance units cancelled for sqrt (h/R).  Wire length
and the speed of light provide the units for 1/s

END.

Jared Dwarshuis, Lawrence Morris