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>

Hi Paul, 

Thanks for taking a moment to spell out the calc. I made 2 errors. First was
using log to base (of which I didn't specify) instead of the NL, and the other
was a decimal placement with wd (due to how I use the wd cell in an Excel
sheet). Anyway, all is well. I now get 18.000447. 

Thanks for clarifying the "Natural Log". If a base isn't specifically called
out, Excel and calculators assume log to base 1. If you specified that in the
text then I simply missed it. It's funny how the error came so close so I
assumed the formula was entered correctly. 

Thanks again, 
Bart 

Tesla list wrote: 
>
> 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. 
> --