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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 11:34:29 -0700*Resent-Date*: Tue, 6 Feb 2001 11:44:43 -0700*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <Vqb2xB.A.ui.XYEg6-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net> Tesla list wrote: > > 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: "by way of Terry Fritz <twftesla-at-uswest-dot-net>" > <paul-at-abelian.demon.co.uk> > > > 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. > > This was discussed in the list some time ago. I obtained the formula: > Wire diamater in mils for AWG "n" = exp(a*n+b) > Or: n=(Ln(diameter in mils)-bb)/aa; > Where: > a=-Ln(460/5)/39 > b=(36*Ln(460)+3*Ln(5))/39 > This is exact, by the definition of the scale. Use n below 0 as > negative. > 1 mil= 0.0254 mm > > Antonio Carlos M. de Queiroz Here is a quote from an neat old book of mine which covers the same subject: From: SOLENOIDS, ELECTROMAGNETS, and ELECTROMAGNETIC WINDINGS 1914 SECOND EDITION, 1921 Charles N. Underhill D. VAN NOSTRAND COMPANY NEW YORK page 215: (The rather quaint language is exact quote.) "109. American Wire Gauge (B.& S.) This is the standard wire gauge in use in the United States. It is based on the geometrical series in which No. 0000 is 0.46 inch diameter, and No. 36 is 0.005 inch diameter. 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 n may represent half, quarter, or decimal sizes. If d represent the diameter of the wire in millimeters, then log d = 0.9165312 - 0.0503535 n, (189) 0.9165312 - log d and n = ----------------- (190) 0.0503535 The ratio of diameters is 2.0050 for every six sizes, while the cross-sections, and consequently the conduc- tances, vary in the ratio of nearly 2 for every three sizes." In a reference on the following page these expressions are attributed to the "Supplement to Transactions of the American Institute of Electrical Engineers", October, 1893. As for wire resistance, the resistivity at a constant temperature can vary by several percent, depending on the purity of the copper and its mechanical treatment, so the values for resistance given in the wire tables are approximations. The resistivity of "pure annealed copper" is given as 1.584 x 10^-6 ohm-cm, while that of "hard-drawn copper" is given as 1.619 x 10^-6 ohm-cm. I have no idea of the tolerance on manufactured wire diameter, but can't imagine it being much better than a percent for large sizes and worse than that for very small sizes, so the above formulae have more precision than circumstances warrant. I, personally, find the standard wire tables quite adequate. By the way, this book is now available from Lindsay Publications, and I recommend it to anyone interested in the design of solenoids or other electromagnets. The equations above are exact quotes from the original and may get screwed up in transmission. These should not: n = (-0.4883027 - log d)/0.0503535 (d in inches) (188) n = (0.9165312 - log d)/ 0.0503535 (d in mm) (190)" As you can see, this standard goes back a long way. In another reference, now forgotten, is the statement that this wire guage definition is derived from a very much older standard for sheet metal gauges. Ed .

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