Re: New Inductance Formula

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

Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Mddeming-at-aol-dot-com>

(SEE BOTTOM)

In a message dated 4/28/02 5:01:58 PM Eastern Daylight Time, tesla-at-pupman-dot-com
writes:



>
> Original poster: "Terry Fritz" <twftesla-at-qwest-dot-net>
>
> Hi Dave,
>
> I have my small coil here so let's give it a try.  I just fixed things for
> the pi^2 error you just mentioned...
>
> >
> >             (N * R)^2
> >mH = ---------------------------
> >     L * c * Cd * 16 *pi^2 *10^-4
> >
> >N is the number of turns
> >R is the average radius
> >L is the length of the wire (not length of the windings)
> >c = 2.99 x 10^8
> >Cd = 2.112 x 10^-4
> >pi = 3.14159
> >
>
> For my coil...
>
> N = 1175.5 turns
> R = 2.125 inches (0.053975m)
> L = 1316 feet (401.117m)
> The winding length is 26.125 inches.
>
> I am assuming all the dimensions should be in meters since that is how the
> numbers on your site are.
>
>
>              (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
>
> The actual inductance is 22.1mH
>
> So it is not working for me.  Maybe I messed up in dimensions or number
> systems some where.
>
> We can try wheeler's...
>
> L (mH) = (N x R)^2 / (9R + 10H)
>
>        = (1175.5 x 2.125)^2 / (9 x 2.125 + 10 x 26.125)
>
>        = 22.254mH
>
> So that works.
>
> 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
>
> Maybe you can find where I messed up here.
>
> Cheers,
>
>     Terry
>



Hi Dave, Terry, Paul, All,

I just tried my coil with the following specs:
N=820 t #23 single-build wire
R=2.14"=0.0544 m
h=21.16" =0.5375 m
L=2piRN=6.28318*2.14*820=11025.7"=918.81 ft=319.75 m

by Wheeler:
mH= (NR)^2/ (9R+10H)  = 13338.5 = 13.34mH
Measured value = 13.97 mH
error= 4.7%
"new" method:
mH=(NR)^2/(L*c*Cd*16*pi^2*10^-4)
         = 0.007105 mH
I rechecked Terry's calcs and got the same value he did  0.0100640.

Looks like "new method" is 0 for 2 in our attempts to corroborate the findings.

Also, please note that if mH(wheeler)=mH(new) as alleged, then the numerators
differ only by a dimensionless units conversion factor(1 in=0.0254 m), and
therefore the denominators must agree dimensionally and differ only by a
conversion of units factor (k). Now,
         (NR)^2 /(9R+10H)wheeler= k^2 * (NR)^2/(k*L*2.99*2.112*16*pi^2) New
                         1/(9R+10H) = k/(*L*6.315*16*pi^2)
                         (9R+10H) = L*997.2/k   if we let k'=k/997.2 then
                   L=k' * (9R+10H) for EVERY solenoidal coil where the
measurements are in meters, which is obviously false.

Matt D.
G5-#12
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