Re: infinite square lattice of resistors

[brian whatcott <inet@INTELLISYS.NET>]

At 16:11 10/11/99 -0700, you wrote:
> > Okay, somebody let me in on the trick for the resistance across a single
> > diagonal of the infinite square lattice of say one-ohm resistors. I'd like
> > both the elegant solution from symmetry which Leigh implied exists, and I'd
> > like to know the formal Fourier method to go between any pair of
junctions --
> > or at least a sketch of the basic idea behind the latter if it's very
> > complicated.
>
> Sorry, Carl. I don't have a simple, elegant solution from symmetry.
> As I hinted (and David Bowman knew), the solution is transcendental.
> For just that reason I'm quite sure that no solution as simple as
> the finite superposition trick for adjacent vertices exists. I just
> gave Herb the problem because he was so pleased with himself to have
> figured out the cube.
>
> Leigh

I began in this way:
to find the resistance across opposite ends of a diagonal across a
square cell of an infinite array, i draw a line through these two
vertices and ignore all of one side as mirrored by the other.
I next draw the line perpendicular to the first line, joining the
other two diagonally opposite vertices of the same square.
All points intersecting this line are at the same potential by
inspection, so the diagonal can be pictured as a conductor with no
loss of fidelity.
 I then attempt to enumerate paths from the first given vertex to
this conductive line in just one quadrant.
 I appear to see a progression like this:

1 ohm in parallel with
2 x 3 ohm in parallel with
3 x 5 ohm in parallel with....

This is an expansion for.....

And that's where I left it.

(But as this is a superposition method, perhaps it's as well I gave
up quickly, if as Leigh suggests there is no such approach   :-)



brian whatcott <inet@intellisys.net>
Altus OK