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*From*: Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>*Date*: Fri, 9 Feb 2001 12:31:17 -0500

Thanks JohnM for answering my question about JohnD's

first example. It makes sense. It is clear that I must change

my 2-dim code to cylindrical coordinates in order to solve

the disk problem (when I have time). But it would be

useful to have the averaging formula for "a" surrounded

by six cells (b,c,d,e,f,g). My intuitive guess is:

a=[(b+c+d+e+f+g) + ((c-e)/2*r(a)]/6

which is the same as in 2-dim, except that f and g cells

were added to account for the third dimension. I would

justify this by noticing that for any given r(a) all cells

have the same size. Is this acceptable? We do not need

this for solving the disk problem but it would be nice to

write down a formula for a possiblefuture reference.

Ludwik Kowalski

John Mallinckrodt wrote:

On Fri, 9 Feb 2001, Ludwik Kowalski wrote:

And what is the corresponding formula for "a" when it has

six neighbors (3-dim)? Suppose f and g are added in the phi

direction. I do not trust myself in trying to answer this

question on the basis of the general theoretical formula.

Ludwik Kowalski

You don't want to (and really can't, I think) go there with a 2-d

spreadsheet. Fortunately, the isolated disk capacitor has

azimuthal symmetry so all derivatives in the phi direction vanish.

The spreadsheet universe for this problem is a plane that includes

the symmetry axis and the full 3-d picture emerges by rotating it

around that axis.

Similarly BTW, In the usual 2-d cartesian relaxation scheme, we

implicitly assume no out-of-the-plane variations. John D's "dog"

is not, therefore, "one block thick." It is a cross sectional

slice of a structure that is extended infinitely in both

directions in and out of the plane.

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