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Re: visualizing fields near charged objects

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.