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

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

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. 
> --