Re: Inductance of a conical coil

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

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <teslalist-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
 >
 > Original poster: "Godfrey Loudner by way of Terry Fritz 
<teslalist-at-qwest-dot-net>" <ggreen-at-gwtc-dot-net>
 >
 > Hello Antonio
 >
 > Well the Mathematica software failed to numerically integrate the
 > Neumann integral last night. I thought I was doing Mathematica
 > a favor by shrinking the limits of integration to (0, Pi/2) by a change
 > of variables. This caused the cosine and sine functions to oscillate
 > wildly on the short interval. The resulting convergence was so slow
 > that Mathematica gave up the task. I'll have to make a change of
 > variables to get the trigonometric functions completely out of the
 > integral. I used parametric representations for the spirals of
 > both the primary and the secondary. I suppose the next problem
 > will be the long intervals for the integration. This must be why you
 > broke the conductors up into 20 segments. But I'm really
 > surprised that Mathematica balked at the form of the integral I
 > presented. After all, Mathematica will compute Pi to 30,000
 > places on a PC in a few moments!

I don't have experience with Mathematica, but I think that it is
a symbolic calculus program. Maybe it doesn't know a solution for
that integral. I it can run a numerical integration, I don't expect
problems. The integral that I am using can be seen at:

http://www.coe.ufrj.br/~acmq/tesla/neumann.pdf

That form is good enough for numerical integration. I didn't even
try to simplify it, but just put all calculations of the x, y, z,
dx, dy, and dz in the integration loops. A simplification may
reduce the number of operations.

If mutual inductances can be accurately calculated, there is a
possible method to test experessions for self-inductances.
Split a coil with inductance L in two sections, with inductances
L1 and L2, and mutual inductance M. L, L1, and L2 are calculated
by the formula in test, and M by Neumann's formula. If the formulas
are ok, L = L1+L2+2*M.
This works quite well for Wheeler's formulas, for solenoidal or
flat coils. A problem is that the calculation of M must be made
with great number of segments, because of the small distance
between the coils when they are joined.
Ex:
A flat coil with rmin=0.1 m, rmax=0.4 m, 40 turns.
Splitting the coil in two:
rmin1=0.1 m, rmax2=0.2 m, rmin2=0.2 m, rmax2=0.3 m, n1=n2=20 turns.
I get there results:
               Inca     Fantc
L1           0.1541    0.1517
L2           0.3175    0.3148
M            0.0961    0.0961
L            0.6631    0.6595
L1+L2+2*M    0.6638    0.6587

Not bad. I will make a flat coil to see what Nature says.

Antonio Carlos M. de Queiroz