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Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-uswest-dot-net>" <evp-at-pacbell-dot-net>
Tesla list wrote:
> Original poster: "BunnyKiller by way of Terry Fritz
> Tesla list wrote:
> > Original poster: "Jason Johnson by way of Terry Fritz
> <twftesla-at-uswest-dot-net>" <hvjjohnson13-at-hotmail-dot-com>
> > Does anyone have a formula for calculating the AWG size of a wire from the
> > diameter (preferably in inches), and/or a formula for the number of
> > inch based upon the AWG? I need these formulas for an excel spreadsheet
> so the
> > simpler they are the easier it is for me to put them in. They need to have
> > fairly good accuracy (+/- 1% or less). Also I haven't really looked
> around for
> > them yet, but if anyone has any formulas for inverse conical (saucer
> > primaries I could use those too.
> > Jason Johnson
> Hi Jason....
> what you are asking for is nearly impossible... the AWG size of wires
> produced by the professionals are
> at an accuracy of 5% for common usage. In certian cases, there are
> situations in which the manufacturer
> will design and release wire at 2% accuracy but the cost is for groups of
> buyers like NASA, the armed
> forces, and anyone else who can spend a fortune.
> Also it seems that each major manufacturer has its own standards for the
> AWG compared to wire dia.
> Scot D
My experience is that the wire I have used is made much more
accurately, with the resistance/foot seemingly corresponding within a
couple of percent at worst to that in various wire tables. Here is some
stuff on wire sizes which I found in a very old book:
SOLENOIDS, ELECTROMAGNETS, and ELECTROMAGNETIC WINDINGS
SECOND EDITION, 1921
Charles N. Underhill
D. VAN NOSTRAND COMPANY
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)
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)
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)