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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 20:43:10 -0700*Resent-Date*: Mon, 5 Feb 2001 20:44:02 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <Azve1B.A.0xD.eL3f6-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "Mark Broker by way of Terry Fritz <twftesla-at-uswest-dot-net>" <broker-at-uwplatt.edu> Tesla list wrote: > > 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? <snip> > > 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. According to: <http://members.tripod-dot-com/~schematics/xform/xformer6.htm>http://members.tri pod-dot-com/~schematics/xform/xformer6.htm The formula for AWG is: D(inches) = .005 (39<root>92)^(36-N) where N is AWG (4/0 is -4). (that's the 39th root of 92, btw) I used this formula and the one supplied in an earlier email, and get the same number (to within a few femptometers ;-). Mark B.

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