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

["Tesla list" <tesla-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
            
.