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*To*: tesla-at-pupman-dot-com*Subject*: Re: Awg formula, was "New formula for secondary resonant frequency"*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Mon, 05 Feb 2001 07:30:46 -0700*Resent-Date*: Mon, 5 Feb 2001 07:42:08 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <2fCr2.A.47.9urf6-at-poodle>*Resent-Sender*: tesla-request-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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