Re: Query on formula existence

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: jdwarshui-at-emich.edu 

There is an exact solution for top end capacitance as a function of
inductor height for coils that have been wound to match the wire
length to the wave length. The exact solution takes into account the
length of wire above and beyond (2 pi r N) needed to reach the top of
a coil. This amount of wire is negligible for all but analytic work.
So if we substitute the close approximation (2 pi r N) = wire length
into the Ideal Resonant Inductor Formulae, solving for capacitance
using the fact that  (the speed of light C) = 1/ sqrt ( (u) (e) ) or
1/ Csqrd (u) = (e) we get.

8 (e) / n pi = cap/h
for a quarter wave (n) = 1/2,  for a half wave (n) = 1, for a full
wave (n) = 2

This boils down to 45.01 pf per meter height for a quarter wave.

If you want an exact solution you can substitute
Sqrt [(h)sqrd + (2 pi r N)sqrd] = (wire length)  into the Ideal
Resonant Inductor Formulae
(http://people.emich.edu/jdwarshui/correspondence.html).

The solution of 45 pf / meter (for a quarter wave) will be the total
sum of capacitance. So  you will need to subtract the Medhurst self
capacitance of the coil from the above figure before sizing your top
end capacitor.

If you are building a half wave where (n) = 1 You get 22.5 pf / m.
This is the total sum of capacitance. You will need to have the
Medhurst self capacitance found between a single voltage and current
node subtracted from it. ( calculate only half the coils worth of self
capacitance) Once this adjusted capacitance value has been found you
split the value between the two ends of the coil.

If you are building a full wave you get 11.25 pf /  m, you subtract
the Medhursts self capacitance found in a quarter of the coil before
splitting the value.

And of course, you do not split the capacitance for grounded coils
such as 1/4 wave and 3/4 wave.