How does the American Wire Gauge work? I've tried to get my head around it a few times, but have never really succeeded. And what about SWG(Standard Wire Gauge), how did this come about? How do these systems translate to Square Millimetres?. Please help.

Let's face it, these days if you're not young, you're old - Red Green

Trumpy, The problem with attempting to "memorize" the AWG is this: Originally, AWG was all based on #10 wire - it was .10 thick, with a circular mil area of 10,000. #11 was .9 thick, 12 - .8, and so on.

BUT - it has since been adjusted many times over the years because of scientific conferences that identified areas of inaccuracy in measurement, calculation, etc. to the point it is NOT able to be memorized.

Such as, #10 thickness is now officially .1020 with a cma of 10,380, and it just gets worse from there.

Remember- when these gauges were first used, stranded wire just was not around.

I've used Steel Wire Gauge, but I am unfamiliar with Standard. I have heard of the British Standard Wire gauge, is this the one you're talking about?

Heck, Whilst I'm here, I just GOTTA get a Nationalistic dig in..........

All of this is available in

"THE AMERICAN ELECTRICIANS HANDBOOK"

[This message has been edited by George Corron (edited 06-21-2003).]

BTW - ya know ya been around this biz awhile when you can actually remember #10 being 10,000 cma

[This message has been edited by George Corron (edited 06-21-2003).]

To figure out the diameter in millimeters, with x standing for the AWG gauge: Take the 39th root of 92. Then raise this to the power of (36-x), keeping in mind the goofy negative numbers for those big AWG gauges. Multiply the result by 0.127 to get diameter in millimeters. If you want Fred Flintstone units, figure out the proper number to use in the last step yourself. You won't get rid of that 39th root, however.

Here's a quick conversion for the smaller AWG sizes:

#18 = 0.823 sq. mm. #16 = 1.31 sq. mm. #14 = 2.08 sq. mm. #12 = 3.31 sq. mm. #10 = 5.261 sq. mm. #8 = 8.367 sq. mm. #6 = 13.30 sq. mm. #4 = 21.15 sq. mm.

By the way, I've found a lot of Brits these days who don't know the proper meaning of the term mils. They often use the word as a colloquialism for millimeters and don't realize that it really means a thousandth of an inch (more commonly called a "thou" here -- Not sure if you use that in NZ?).

The old British SWG (Standard Wire Gauge) works in a similar way to American Wire Gauge in that the larger the number the smaller the wire, but there's no direct correlation between the two. SWG wire sizes were commonly specified here in radio and electronics work before metric came into general use.

House wiring cables, however, were generally specified by strands and diameter, e.g. 7/.029 meant 7 strands each 0.029" diameter.

[This message has been edited by pauluk (edited 06-22-2003).]

American wire gauge, also called Brown & Sharp gauge, usually shortened AWG was invented 1857 by J. R. Brown. When manufacturing copper wire, you start with a thicker wire, which is gradually stretched to be thinner and thinner. AWG originally corresponded to the number of successive stretching operations necessary to reach a certain wire dimension.

The measures are not arbitrarily chosen, but follow a simple mathematical relation. Mr Brown choose AWG 0000 (4/0) to correspond to a diameter of 0.46 inches and AWG 36 correspond to a diameter of 0.0050 inches. In between are 38 AWG-measures. An in between AWG-measure is calculated as the geometrical (not to be confused with the aritmetical) mean value of the "surrounding" values. Between AWG 0000 and AWG 36 there are 39 steps, hence the relation between the diameter of a certain AWG-measure and the following measure is given by the expression :

(0.4600/0.0050) ^1/39 = 1.1229322

The exact mathematical relation between the diameter of the wire and the AWG-measure is :

D = (92 ^ ((36-AWG)/39))*0.005 inch

Due to reasons of rounding and that flexible AWG-wires are built of other AWG-wires the AWG-measures will in practice not exactly follow the mathematical formula.

How in the heck did anyone figure the 39th root of 92 in 1857? I sortaâ€™ remember in high school learning to do square roots of small numbers on paper, but the 39th root? The guy musta' been a mental case... if not before then certainly after that calculation.

Bjarney, One should not underestimate the capacity of mathematicians of the past. Before there were calculators, tables were used. Logarithms and Fourier transforms do magic to difficult calculations.

To calculate the diameter from the formula

D = (92 ^ ((36-AWG)/39))/200 inch

use logarithms

log(D)=(36-AWG)/39 * log(92) - log(200)

To calculate the diameter of 10 AWG you just look up lg 92 and lg 200 in a table, do some multiplication and subtraction and then convert the answer back to the diameter using the table.

Let's just hope I didn't screw up the math above. {Edit: I did screw up the math! Thanks to Peter for pointing out the error to me. He also pointed out that the formula can be rearranged to give the AWG from the diameter: AWG = 36 - (39 * log( 200 * D)) / log(92) }

[This message has been edited by C-H (edited 02-03-2005).]