Re: Charge distribution on a Toroid (was spheres vs toroids)

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

Original poster: Paul Nicholson <paul-at-abelian.demon.co.uk> 

Antonio wrote:

 >  How do you compute the "self-potentials" of the tiles?

This is a tricky bit.  What we're asking for is the
potential at some point (usually the centroid) of the tile,
as a function of the charge distributed over its area.

Its almost is if we've reproduced the original problem again,
only in miniature!

The simple trick I use to is make the tiles approximately
square, and make them small enough so that they can be
considered flat, and assume the charge density over the tile
is uniform.   Then the coefficient of self potential Pjj for
tile j is given by a simple formula:

     Pjj = ln(1+sqrt(2)) / (Hj * pi * epsilon)

where Hj is the length of a side of the tile j. [1]

A little cheating is worthwhile:  I allow slightly rectangular
tiles - by choosing a tiling that gives reasonably square tiles
but not trying too hard!   To get away with this I use for Hj
the mean length of the two sides of the rectangular tile, which
seems to work quite nicely.

You can tell when the self-potential is causing a problem. One of
its effects when you get it wrong is that the C distribution
as you approach an edge begins to oscillate up and down, rather
than rising monotonically to the edge.  So compute for say a
disc.  Then plot the C distribution across the diameter and
make sure you have a bathtub with no ripple on the end slopes.
If you have a nice bathtub, and the total C agrees with the
formula for a disc, then your self potentials are good and
you're up and running!

 > Your description is valid for a tridimensional problem. What
 > happens in an axisymmetrical problem, as in a typical Tesla
 > coil?

The potential Pjk and self potential Pjj calcs all involve
the Coulomb potential, which is of course a 3d formula, and
the coefficients are just

   Pjk = 1 / ( 4 * pi * epsilon * Djk)

where Djk is the 3d pythagorean distance between the centroids
of tiles j and k.

So far, everything said applies to a full 3d solution.  But we
can exploit the cylindrical symmetry of the TC to reduce the
work needed.  With the restriction to this symmetry we can
consider the rectangular (-ish) tiles to belong to circular
(tape-like) rings. Then we can say

a) The potential on all tiles in a given ring is the same.
b) All tiles in a given ring have the same charge density.

Then instead of constructing the potential matrix relating
tiles to tiles, we can apply it to the rings instead.

So Pjk becomes the coefficient giving the potential along the
centre line of the tape ring j as a function of the charge
distributed around ring k.

So, these new Pjk are just elliptic coefficients.   In fact
I compute these by numeric integration (straight sum) of the
contributions from each tile of the 'source' ring. [2]

In the end, say a toroid is divided into 50 rings, each
ring into 50 tiles, so there are 50*50 tiles involved for this
object.  If I were to set up the potential matrix for the tiles
themselves, it would be 2500*2500 elements - a big matrix!
Exploiting the cylindrical coefficients and setting up a matrix
which applies to the rings instead gives a 50*50 potential
matrix - much more tractable. [3]

Tssp when modelling the entire system of coil, topload,
surroundings, etc, uses between 500 and 2000 rings for the
whole system.

The matrix equations are solved quickly by using the generalised
minimum residuals algorithm.

The running time is much quicker than say, relaxation of
Poisson's equation.  This is because we're not having to compute
the field throughout a volume.  Instead, we're only computing the
field at the surfaces of our objects.   Hence the name 'boundary
element method', since we're only working with the boundary
surfaces of the field, not the volume.

If you want the surrounding field too, then you can produce that
as a 2nd phase of calculation after the tile charges have been
computed.  Simply sum the Coulomb contribution from each source
tile, at each point in space.

Hope that helps!

[1] This self potential formula is derived on p39 of
     http://www.ttc-cmc-dot-net/~fme/plates.12-24-00.ps.gz

[2] Please someone give me a formula for the potential due to a
     tape ring!!! It would make capacitance extraction much
     quicker.  I'm sure a closed formula must exist.
     Oh, you'll also need to come up with a self-potential formula
     for the ring too.

[3] Without a magic formula in [2] I therefore still have to
     compute 2500*2500 tile-to-tile potential coefficients, but
     these are then summed into the elements of a 50*50 matrix.

--
Paul Nicholson
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