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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*: Tue, 06 Feb 2001 20:33:41 -0700*Resent-Date*: Tue, 6 Feb 2001 20:42:29 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <JOxIs.A.mIG.XQMg6-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-uswest-dot-net>" <acmq-at-compuland-dot-com.br> Tesla list wrote: > > Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net> > Here is a quote from an neat old book of mine which covers the same > subject: > No. 0000 is 0.46 inch diameter, and No. 36 is 0.005 inch diameter This is what I used. 0000 would be n=-3: Assuming d=10^(a*n+b), and taking decimal logarithms: log(d)=a*n+b n=-3 -> d=0.460 inches -> log(0.460)=a*(-3)+b n=36 -> d=0.005 inches -> log(0.005) =a*36+b Subtracting: log(0.460)-log(0.005)=-39*a -> a=-log(92)/39 = -0.05035353 >From the relation for n=36: b=log(0.005)-36*a -> b=(36*log(0.460)+3*log(0.005))/39 = -0.48830277 > Let n = number representing the size of wire. > > d = diameter of the wire in inch. > > Then log d = 1.5116973 - 0.0503535 n, (187) > > - 0.4883027 - log d > n = ------------------- (188) > 0.0503535 All the other formulas are just this same formula written in different ways. Antonio Carlos M. de Queiroz

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