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

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "John H. Couture by way of Terry Fritz <twftesla-at-uswest-dot-net>" <couturejh-at-worldnet.att-dot-net>


Bart -

I obtained the equation many years ago from the Simplex Wire & Cable Co. You
did a lot better than many other people in past years trying to come up with
a different equation.

Note that for 4/0 AWG wire that  x = 0  and  dia = .46 inches.
All equations have to take 1/0, 2/0, 3/0, and 4/0 (x=3,2,1,0) into
consideration.

 The AWG table then changes to MCM for larger wires. The MCM stands for
millions of circular mils. A circular mil is the area of a circle one mil in
diameter. A solid wire one inch in diameter has an area of one million
circular mils. One square inch equals  4/pi x 1,000,000 CM = 1,273,200 CM .
It can be confusing. However, it does avoid decimals in specifying wire
sizes, a lot of which I did in the past.

John Couture

----------------------------------

-----Original Message-----
From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Tuesday, February 06, 2001 6:49 AM
To: tesla-at-pupman-dot-com
Subject: Re: Awg formula, was "New formula for secondary resonant
frequency"


Original poster: "Barton B. Anderson by way of Terry Fritz
<twftesla-at-uswest-dot-net>" <tesla123-at-pacbell-dot-net>

John C.,
Thanks for posting. I remember trying to find this info a long time ago and
could never get data. I eventually got together with someone at work to
derive
a formula (which was successful). Now it seems there are three or four
equations floating around. Isn't this list great!

BTW, when you state "the standard", I assume you grabbed the equations from
wire manufacturers?

Take care,
Bart

Tesla list wrote:
>
> Original poster: "John H. Couture by way of Terry Fritz
> <twftesla-at-uswest-dot-net>" <couturejh-at-worldnet.att-dot-net>
>
> I understand the standard equation for the AWG is
>
>     Dia inches = .46/(1.122932)^x
>
> where x = wire gauge + 3
>
> Example  # 24 AWG    x = 27
>
>     dia ins = .46/(1.122932)^27 = .02010 ins
>     dia mm  = .02010 * 25.4     = .51054 mm
>
>           # 18 AWG   x = 21
>
>     dia ins = .46/(1.122932)^21 =  .04030 ins
>     dia mm  = .04030 * 25.4     = 1.02362 mm
>
> John Couture
>
> -------------------------------
>
> -----Original Message-----
> From: Tesla list [<mailto:tesla-at-pupman-dot-com>mailto:tesla-at-pupman-dot-com]
> Sent: Monday, February 05, 2001 6:31 AM
> To: tesla-at-pupman-dot-com
> Subject: Re: Awg formula, was "New formula for secondary resonant
> frequency"
>
> Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>"
> <paul-at-abelian.demon.co.uk>
>
> Bart wrote:
>
> > I'm still coming up with 17.5 using your formula (I assume your
> > using something other than 1.0236mm for 18 awg?).
>
> Check your intermediate steps:
>
>  awg = 1 + log(7.348e-3/wd)/0.115943     (use natural log)
>
>  wd = 1.0236e-3
>
>  7.348e-3/wd = 7.17859
>  log(7.17859) = 1.9711
>  1.9711/0.115943 = 17.0006
>  1 + 17.0006 = 18.0006
>
> Maybe you used 0.119543 instead of 0.115943 or something?
>
> > I kept the long decimal places for accuracy -  I saw no reason to
> > shorten them up since I used it simply as a formula in programs.
>
> Yes, I know what you mean, same here with the longish coefficients
> in the new formula. I try to stop before I reach the size of an atom,
> or in your case the atomic nucleus :)), eg your first factor begs to
> be rounded a smidgen (er, thats a UK smidgen BTW).
>
> > Nominal wire sizes taken from the Brown & Sharpe American Wire
> > Table. Possibly, this is where the discrepancy exist?
>
> 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.
>
> --
> Paul Nicholson,
> Manchester, UK.
> --