Re: ballast formula

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

Original poster: "James T by way of Terry Fritz <twftesla-at-uswest-dot-net>" <jamest2000-at-att-dot-net>

Hi Deano,
 Thanks for the response. That is just what I was looking for. I think i
need those drawings to
figure out the mean length.
Whenever you get a chance to scan those pages would be great. There should
be a clickable email
next:
mailto:jamest2000-at-att-dot-net

 I think the MU(permaebility) should be around 60,000. Does that sound right?

Thanks James



Tesla list wrote:

> Original poster: "David Dean by way of Terry Fritz <twftesla-at-uswest-dot-net>"
<deano-at-corridor-dot-net>
> Hi James,
> I have an old text book with some of that stuff in it.  It says:
> L = N^2*mu*A/10^8*l
> (L equals N squared mu A over ten to the eighth l)
> where L is in henries,
> and   N = number of turns
>       mu = permeability of core (in English units)
>       A = area of core in square inches
>       l = mean length of core in inches
> The mean length would be {twice the length of the center leg (with the I
> part on there) plus twice half the width of the core} plus [the perimeter of
> one window} / 2.
> Trying to draw pictures with words is not my strong point. If you want, I
> can scan the pages and email them to you.
>
> later
> deano