Re: Query on formula existence

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

Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz> 

Hi Jared,

On 24 Oct 2004, at 18:48, Tesla list wrote:

 > 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.

It's not even as complex as that. You can make any coil at all turn
into a "1/4 wave" coil by simply popping sufficient extra capacitance
on top. I tried this years ago and found there was no benefit at all.

Malcolm