Re: Awg formula, was "New formula for secondary resonant frequency"

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

Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>" <paul-at-abelian.demon.co.uk>

Bart wrote:

> I'm still coming up with 17.5 using your formula (I assume your
> using something other than 1.0236mm for 18 awg?).

Check your intermediate steps:

 awg = 1 + log(7.348e-3/wd)/0.115943     (use natural log)

 wd = 1.0236e-3
 
 7.348e-3/wd = 7.17859
 log(7.17859) = 1.9711
 1.9711/0.115943 = 17.0006
 1 + 17.0006 = 18.0006

Maybe you used 0.119543 instead of 0.115943 or something?

> I kept the long decimal places for accuracy -  I saw no reason to
> shorten them up since I used it simply as a formula in programs.

Yes, I know what you mean, same here with the longish coefficients
in the new formula. I try to stop before I reach the size of an atom,
or in your case the atomic nucleus :)), eg your first factor begs to
be rounded a smidgen (er, thats a UK smidgen BTW).

> Nominal wire sizes taken from the Brown & Sharpe American Wire
> Table. Possibly, this is where the discrepancy exist?

Nope, the AWG sizes are fairly well defined, decreasing by a factor 
1.122932 with each step. This factor is the sixth root of two,
which means therefore that six AWG increments will exactly halve
the wire size.

--
Paul Nicholson,
Manchester, UK.
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