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Subject: Re: Another resistor problem
References: <firstname.lastname@example.org> <3DEEC9AB.F1EC3186@ieee.org>
NNTP-Posting-Date: Thu, 05 Dec 2002 07:38:55 GMT
Organization: AT&T Broadband
Date: Thu, 05 Dec 2002 07:38:55 GMT
Phantom, analog, Phantom 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 me take a measurement to confirm (pulls out meter and sets dial
to measure in dBS). Ok, all set up now.
>> Let's see, that would be 1/2 ohm times the non isotropic square
>> lattice conduction factor of 4/Pi. Hmmm...
Here goes: [...../] Yep, good call, Phantom. Pegged the dBS meter.
>> 2/Pi ohms.
>>> 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