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

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

Original poster: "Mark Broker by way of Terry Fritz <twftesla-at-uswest-dot-net>" <broker-at-uwplatt.edu>



Tesla list wrote:

>
> 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? 

<snip>

>
> 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. 

According to: 

<http://members.tripod-dot-com/~schematics/xform/xformer6.htm>http://members.tri
pod-dot-com/~schematics/xform/xformer6.htm
The formula for AWG is: D(inches) = .005 (39<root>92)^(36-N)  where N is AWG
(4/0 is -4).  (that's the 39th root of 92, btw)

I used this formula and the one supplied in an earlier email, and get the same
number (to within a few femptometers ;-).

Mark B.