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Re: AWG WIRE TABLE for Coilers (fwd)
---------- Forwarded message ----------
Date: Fri, 1 May 1998 02:24:50 +0500
From: "Alfred A. Skrocki" <alfred.skrocki-at-cybernetworking-dot-com>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: AWG WIRE TABLE for Coilers (fwd)
On Thu, 30 Apr 1998 20:32:50 -0700 Antonio C. M. de Queiroz
<acmq-at-compuland-dot-com.br> wrote;
> Do someone know what is the official formula? and the reason for it?
> All these expressions generate some error when compared with the values in
> my table in the Ref. Data for Radio Eng. My expression I obtained from the
> values for #10 and #30. It is precise for these gauges, but disagrees
> slightly (<0.1%) for other gauges.
I had searched far and wide for an "official" formula for the wire gauges
and whenever I found a book claiming to have the "official" relationship it
disagreed with the very table supplied in the same book! So I started
playing with the tables until I got insight to the relationship expressed
in the equations;
AWG gauge = -8.624487202999999 * natural LOG(d / .3245574964)
diameter = .3245574964 * 2.718281829 ^ (gauge / -8.624487202999999)
I have found this equation agrees almost perfectly with the tables given
in; The Machinist's Handbook, The CRC Handbook of Chemistry and Physics,
and The Radio Amateur's Handbook by the ARRL, soo long as you use the
integer value only when calculating the AWG gauge from a measured
diameter. BTW I was well aware that my equation;
diameter = .3245574964 * 2.718281829 ^ (gauge / -8.624487202999999)
Translates into;
diameter = exp(-0.1159489226967782 * gague - 1.125292573941649)
where diameter is in inches.
BUT I chose to use the division over the subtraction due to a quirk in
Microsofts compiler BASIC's that make division more precise and faster
to calculate then subtraction!
> To include a related question closer to Tesla coils: What is the thinnest
> wire that you members of the list have used in a successful Tesla coil?
I used #46 gauge on a 1/2 inch diameter X 2 inch long tesla coil that gave
1 1/2 to 2 inch sparks.
Sincerely
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Alfred A. Skrocki
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alfred.skrocki-at-cybernetworking-dot-com
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