Gauge (Thickness) to inches

I need the formula to convert (wire) gague to inches! I knew it once, but well, it must be old age!

I need the formula to convert (wire) gague to inches! I knew it once, but well, it must be old age!

A good reference is at http://www.sizes.com/materls/wire.htm

That's useless to me and impossible to understand. Sigh.

For AWG (AMerican Wire Gauge) only, see
http://www.powerstream.com/Wire_Size.htm

The formula is

d = 0.005" * 92^((36-AWG)/39)

This is based on #36 wire being 0.005" and #0000 (treat as -3) being 0.46". There are 39 steps in a 92:1 ratio.

Note that 20 steps 92^(20/29) = 10.164 (~10), and
6 steps, 92^(6/39) = 2.005 (~2).
These are useful to do in your head, but are not an exact fit, and WRONG formulas also appear on the web based on powers of 2 or 10.

For efficient computation, this can be reduced to
d = b*e^(-an)
where n = AWG number, 00 = -1, 000 = -2, etc
b = 0.46 inch * 92^(-1/13) (about 0.32486")
a = (ln92)/39 (about 0.11594)

in mm, multiply b by 25.4 mm/inch

According to this article, http://www.ncbi.nlm.nih.gov/entrez/quer ... t=Abstract , the system being used is 'Birmingham'. This article appears incorrect in their approximation of equivalence, since the steps in gauge correspond more closely to 0.01 inches, not 0.001. This is mostly an intuitive response, but I can visually discriminate between 18 and 22 gauge needles, which might correspond to ~.04 inches, but certainly not ~0.004 inches (this is in the thousanths!!)

Here is a list of BWG equivalents: http://www.sigmaaldrich.com/Area_of_Int ... Chart.html

The numbers are indeed non-linear, and thus the above (AWG) exponential equation is probably a good model for the Birmingham numbers.

i am looking to convert 0.001 to gauge? any help

i am looking to convert 0.001 to gauge? any help

There are different gauges, which one?

i am looking to convert 0.001 to gauge? any help

There are different gauges, which one?

if it is Birmingham Gauge, you may try this one; I tried getting this approximate expression using linear regression of a log function

Gauge to mm
Y (mm) = 10.105 * E^(-0.116 X); where X is value in Gauge

mm to Gauge
X = [ln (10.105 / Y)] / 0.116; Round off X to the nearest whole number