Subject: Re: Another resistor problem
Date: 5 Dec 2002 00:32:10 -0600
References: <email@example.com> <3DEEC9AB.F1EC3186@ieee.org>
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On Thu, 05 Dec 2002 03:36:12 GMT, analog wrote:
>> Many of us have seen the old problem of the infinite planar lattice of
>> 1 ohm resistors, the nodes being arranged in a square array. Assume
>> that the nodes are the points whose coordinates are the integers on
>> the plane. Then it's easy to find the resistance between nodes (1,1)
>> and (2,1), or any other adjacent nodes where either the ordinate or
>> abscissa is constant.
>Er, um 1/2 ohm?
>> But what about a pair of diagonally adjacent nodes? That is, what is
>> the resistance between nodes (1,1) and (2,2)?
>Ok, take a uniform sheet and distort its conductance preferentially
>along two orthogonal axes.
Distort it preferentially how much? Just enough so that the non
isotropic square lattice conduction factor is 4/Pi?
Let's see, that would be 1/2 ohm times the
>non isotropic square lattice conduction factor of 4/Pi. Hmmm...
>> Show your work.
>Sorry. I never show my work on a first post.
>> Extra credit: If we have a sheet of isotropic resistive material,
>> infinite in extent, and using circular contacts (use silver paint on
>> teledeltos paper), measure the resistance between the contacts with
>> the distance between contacts 1000 times the diameter of the contacts.
>> Now move the contacts sqrt(2) times as far apart as they were. What
>> is the ratio of the resistance between the contacts now to what it was
>As I recall, it should be one over one, but I'm only going by fading
>memories from double E classes long past, so forgive me if I may have
>inadvertently inverted my answer. -- analog