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

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

Original poster: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> 

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

 >  >  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!

As in the calculation of inductances of coils decomposed in
closed loops. For mutual inductances between distant loops
the calculation is simple. When the relation is between an
element and itself, something must be assumed about the shape
of the element.

 > 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]

If you consider the element as a disk with diameter Hj, the
relation v=Pjj*q leads to Pjj=0.25/(e0*Hj), very close to this
value, 0.28/(e0*Hj).

 > 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!

Good observation. I will verify.

 > 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]

I tried to consider the elements as infinitely thin rings. It's
(almost) easy to calculate the potential due to a ring as:
V=(q/(2*Pi^2*e0*r1))*K
where K is the complete elliptic integral of the first kind, with
modulus k=2*sqrt(a*r)/r1, r1=sqrt((r+a)^2+z^2).
a is the radius of the ring, r the radial distance from the center
of the ring, and z the axial distance from the center. q is the
total charge in the ring, assumed uniformly charged.

A shape with axial symmetry can then be decomposed in rings,
and the Pij coefficients calculated using the formula above.
For the Pii, I considered the potential at a distance R above
the ring, where R is half of the distance to the next ring,
or the inverse of the capacitance of a thin toroid.

I made an experiment considering a partial toroid, with major
and minor radii specified, and also considering initial and
final angles around the tube that forms the toroid. I compute
the [P] matrix, invert it, and add all the elements. This is
the free-space capacitance of the object.
The method appears to work reasonably for toroids, spheres, and
half-spheres.
An unknown factor is exactly where to place the rings, if exactly
at the surface (causes slightly larger capacitances), or with
the toroids of radius R just touching the surfaces (causes
slightly smaller capacitances). An average value between these
results in less than 1% of error with 20 rings.		

Ex: That 90 cm x 30 cm toroid:
C exact: 40.583973 pF
With 20 rings placed at the surface: 41.660502 pF
With 20 rings just touching the surface: 39.061591 pF
With half of the radius out of the surface: 40.3662370342 pF

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

I am using an algorithm that inverts a matrix in place. I am not
sure if it would be stable with so many equations. Takes about a
second with 200 rings.

 > 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.

With rings this is a bit more complicated. I didn't try yet to
deduce the field from the equation above, but it's not difficult.
This method doesn't seem very adequate to compute the surface field
precisely. Maybe something using a series of "image rings", producing
uniform potential where the surface of the object would be would
be better.

 > [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.

The thin ring approximation is a first step. The integral that
would give the inverse of the capacitance of a ring with uniform
charge seems solvable, but not trivial.

The ring algorithm is implemented in the Inca program:
http://www.coe.ufrj.br/~acmq/programs/inca.zip
See "other calculations".

Antonio Carlos M. de Queiroz