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RE: New Inductance Formula
Original poster: "David Thomson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <dave-at-volantis-dot-org>
I am honored that you tried to improve the formula. I do believe there must
be a way to input meters and get meters. But your solution does not work,
at least not when the math is performed.
As for the intent of your effort, Wheeler's formula is already in meters.
When the formula
(N*R)^2 * 10^-6 * kg * m
henry = ------------------------
[(9*R)+(10*H)] * coul^2
is used, the input values are in inches but the inductance is output in
millihenries as based on the meter. Notice the "kg*m" in the numerator?
The m is meters. From this it would appear that Wheeler's formula has
always produced inductance in meters but nobody previously noticed. So my
formula did do something to clarify Wheeler's formula.
On this list and throughout the Internet, people have mistakenly concluded
the units output by Wheeler's formula was in was in thousand inches, but
it's in thousand meters.
It would be interesting for someone who has access to Wheeler's publications
to see if Wheeler had it right:
Harold A. Wheeler, "Formulas for the Skin Effect," Proceedings of the
I.R.E., September 1942, pp. 412-424
Harold A. Wheeler, "Simple Inductance Formulas for Radio Coils," Proceedings
of the I.R.E., October 1928, pp. 1398-1400.
>The unit of R^2/R is not metres, it is, as you said, inches.
The input unit is inches, the output unit is meters. You are right about N
being 1 and not inches. That was an oversight on my part. Thanks.
>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.
Even though you didn't get it on the first try, I would like to see a
version of Wheeler's formula that works for meters. Maybe with this new
knowledge you can do it?
> > 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.
If you follow the steps I outlined in a previous post you will see where the
inch to meter conversion took place.
>Anyway, you made a good effort with the units, and your work certainly adds
some clarity to the Wheeler formula.
Thank you Jan. All the trouble of posting this formula to this list has now
been worth it. It really does mean a lot to me that you took the time to
check into the formula and see what could be done to improve it.
>> I won't bother with the rest of your post since it was based on a
>*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.
I apologize if you weren't taking a cheap shot at me. But your post was
still founded on a mistake. There was another meter in the R^2/R term and
the units truly are in mH and not mH/in or mH/m. I'll read it anyway,
though, just to see what you were trying to convey.