RE: New Inductance Formula

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

Original poster: "Jan Wagner by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jwagner-at-cc.hut.fi>

Hi Dave, and list,

I got a new, provocative version... hehe... ;)

     (N*R)^2 * 10^-6 * kg * m
 L = --------------------------
     [(350*R)+(390*H)] * coul^2

 (R and H in _metres_, N is unitless)

contra

> >>         (N*R)^2 * 10^-7 * kg * m
> >> henry = ------------------------
> >>         [(9*R)+(10*H)] * coul^2
> >> The units N, R, and H are still in inches.
> >Because a Henry is defined as m^2 kg / C^2 and C=s*A, your above
> >formula reduces to, after obviously fixing 10^-7 to 10^-6:
> 
> >  L = 10^-6 * (N*r)^2 / (9*r + 10*h) H / m
> >  L =         (N*r)^2 / (9*r + 10*h) uH / m
> 
> I don't want to embarrass you too much, but you did take a cheap shot at me.
> You neglected to account for the meter in R^2/R. 

I was just in the middle of starting to feel heavily embarrased ;)) but
then  suddenly noticed that my oversight and seems like of others on
this list, too, lies elsewhere: 

The unit of R^2/R is not metres, it is, as you said, inches. (By the
way, N has a unit of 1, and not inches. Obviously a small typo.)


You can fix the units of your formula by rewriting it as

              (N*R)^2 * 10^-6 * kg * m
 L = 0.0254 * ------------------------
              [(9*R)+(10*H)] * coul^2

 (derivation: 1 inch = 2.54cm = 0.0254m, and [R^2/(R+H)]=inches )

Of course, the formula doesn't look so esthetic that way. But at least
one can now give R and H in metres, so you can scrap those non-SI inches.


A cleaner version is, by multiplying the divisor by 1/0.0254 ~ 39.37, and
rounding to tens (oooh yeah, precision just gets worse and worse... ;o) :

     (N*R)^2 * 10^-6 * kg * m
 L = ------------------------
     [(350*R)+(390*H)] * coul^2

(someone else please check if this went right... I didn't do it on
paper and neither did plug in any numbers yet...)


It was too tempting for me to forget this inches stuff, because then the
units would have worked out very beautifully, without the inches or any
conversion factors... :o( oh dammit.

And while at it I do admit my oversight
> >  L =         (N*r)^2 / (9*r + 10*h) uH / m
which should be corrected to
     L = (N*r)^2 / (9*r + 10*h) uH / inch
and also want to add that in your formula the units are wrong by
actually (m/inch) and not (1/m) as I claimed previously.


Anyway, you made a good effort with the units, and your work certainly
adds some clarity to the Wheeler formula. Juggling around with units
SI<=>non-SI is just a very gnarly matter, and without checking 3+ times
there's just bound to be some error creeping in... :o(


> I won't bother with the rest of your post since it was based on a mistake.

*sigh* In my opion you should have read it, it was not only about 1/m (or,
now, m/inch), neither was it a cheap shot. At least IMHO.
 
But you were either too hasty condemning, or just otherwise
conventiently wanted to ignore the rest... oh well... Not much to be done
about that. But feel free to re-read that post, any time. ;o)


cheers,

  - Jan

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