Inductance (fwd) [correction]
From: Malcolm Watts [SMTP:MALCOLM-at-directorate.wnp.ac.nz]
Sent: Wednesday, April 01, 1998 6:05 PM
To: Tesla List
Subject: Re: Inductance (fwd)
Sorry to say this but there is a formulaic error in your
> Date: Tue, 31 Mar 1998 06:01:40 -0500
> From: Alan Sharp <AlanSharp-at-compuserve-dot-com>
> To: "INTERNET:tesla-at-pupman-dot-com" <tesla-at-pupman-dot-com>
> Subject: Inductance (fwd)
> It checks!
> Erik wrote:
> >L : inductance in uH, r : radius in inches, N : number of turns,
> >b : length of coil in inches, d : turns per inch, w : length of wire in
> >L = (r * r * N * N) / (9 * r + 10 * b)
> Nasty Maths omitted here :)
> >So the most inductance for a cylinder coil is when
> >r = sqrt((5 * w) / d) / (3 * sqrt( Pi ))
> >w = r * r * Pi * d * 9 / 5
> Daune Bylund goes through this in his "Modern Tesla Coil Theory"
> and you can go one step further:
> b = w / ( 2 * pi * r * d )
> the hieght of the coil is the length of wire divided by the circumference
> the number of turns per inch:
> b= (r * r * Pi * d * 9 / 5) / ( 2 * pi * r * d ) = 0.9 * r
> b = 0.9 * r
To use more familiar terms, this says h = 0.9 * r
Since r = d/2, we can substitute for r
So h = 0.9 * d/2 = 0.45 * d
Therefore, h/d = 0.45 which is pretty close to what I originally said
if the error above is of no account.
NB - this does not correspond to the coil shape that gives maximum Q
which is h/d = 1.