RE: New Inductance Formula

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

Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-qwest-dot-net>" <m.j.watts-at-massey.ac.nz>

On 29 Apr 2002, at 0:15, Tesla list wrote:

> Original poster: "David Thomson by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <dave-at-volantis-dot-org>
> 
> Hi Terry,
> 
>              (N * R)^2
> mH = ---------------------------
>      L * c * Cd * 16 *pi^2 *10^-4
> 
> 
> = (1175.5 * 0.053975)^2
>   -----------------------------------------------
>   401.117 x (2.99 x 10^8) x (2.112 x 10^-4) x 16 x (3.14159^2) x (10^-4)
> 
> = 4025.6 / 127322.65 == 0.0100640 mH
> 
> When I did the above math I came up with 10.036 mH.  Did you multiply the
> numbers I outlined in parenthesis above before multiplying with the rest of
> the equation?

Oh my! Multiplication actually has commutative properties. That is 
basic primary school stuff. Your calculator didn't run out of digits 
did it?
  
> >My coil is:
> 
> 26.125 inches long
> 1175.5 turns of #24 enamel wire
> 4.25 inches in diameter
> 1316 feet of wire
> inductance is 22.1mH
> 
> I have a question on your coil.  Are your turns counted?  Is the wire gage
> right?  According to the dimensions you gave, your coil is space wound or
> has about 10% space between the windings.

What possible difference could that make? Your formula appears to 
take no account of wire gauge; at least I don't see a variable 
specifying wire gauge in there.  

> >The actual inductance is 22.1mH
> 
> The value for the Wheeler formula that I got was 22.156 mH.
> 
> I went back to see why the values were so far off and I discovered I had not
> finished cleaning up my fudge factors (I used them as markers while trying
> to get the right unit proportions.)  I concede, the formula I presented is
> wrong.  I still have the right idea, though.

Well if you were using the wrong formula with claimed success, what 
is the correct formula going to do? Give wrong answers I suppose ;)
  
> The Wheeler formula gives inductance as length.  His values are correct but
> his units are incomplete.  If you look at inductance in relation to
> Coulomb's constant it is:
> 
>                     m
> henry = --------------------------
>         c * Cd * 16 * pi^2 * 10^-7
> 
> What this equation says is that 1 meter is equal to 1 henry.  Also .003
> meter is equal to 3 mH and so on.  So looking at inductance from Coulomb's
> constant it makes sense to express inductance in length (just as long as we
> realize there is a denominator that converts the length to inductance.)
> 
> Wheeler's formula gives inductance in henry as inch times 1,000,000.
> Wheeler did not see that inductance was related to Coulomb's constant.  If
> he did he would have written the full equation for inductance as...
> 
>                          (N*R)^2
> henry = ------------=-----------------------------
>         ((X*R)+(Y*H)) * c * Cd * 16 * pi^2 * 10^-7
> 
> where
> 
>            (N*R)^2
>       m = ---------
> 	   (X*R)+(Y*H)
> 
> and X and Y are factors for meters instead of inches. N, R, and H are as in
> Wheeler's formula.
> 
> It's getting late, but I came up with some close factors for X and Y.  They
> are not correct.  I'll give this more time tomorrow.  Maybe others on this
> list would like to take up the challenge and come up with the correct
> conversion factors?
> 
>                                 (N*R)^2
> henry = --------------------------------------------------------
>         ((354331*R)+(3937001.79*H)) * c * Cd * 16 * pi^2 * 10^-7
> 
> Dave

I am fully satisfied that the correct answers were worked out by a 
number of geniuses and diligent workers many years ago. I for one 
have not found them to be wanting. So why this major push to change 
that which has worked well for so many years and has yet to be shown 
to be in error?

Malcolm