Subject: Re: Resistance of a Sphere Revisited
Date: 1 Dec 2002 19:11:11 -0600
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On Sat, 30 Nov 2002 22:16:13 +0000, John Woodgate
>I read in sci.electronics.design that bill wrote (in
>) about 'Resistance of a
>Sphere Revisited', on Sat, 30 Nov 2002:
>>If a solid sphere of diameter D has a resistance of 1 ohm (as measured
>>across the diameter), what will be the resistance of a sphere with
>>diameter 2D? We use the same meter and probes for both measurements,
>>so the contact area should be the same, yes?
>>I've seen other posts on this subject where all sorts of nuances
>>affecting the measurement are postulated, but how 'bout skipping all
>>that, make reasonable assumption about the contact area, the meter and
>>its leads, and give an answer within 10%.
>>I'm in an arguement with my brother about this and we do not agree.
>>Seems like there should be a simple way to compute an approximation
>With point contacts at the ends of diameters, the resistance of the
>larger sphere is twice that of the first. You just divide the spheres up
>into thin discs of equal thickness.
Hellmut Sennewald and yourself have fallen into a big gotcha that I
also learned about the hard way. Dividing the sphere into discs and
integrating won't work. There is a hidden assumption when you do
this. You are assuming that you can compute the resistance of each
disc and add the individual resistances together. To do this is to
assume that the flat surfaces of each disc could be a: plated with
silver (or preferably, superconductor) and the discs then stacked, or
b: connected with a zero ohm wire without stacking the discs. This
only gives the right answer if the flat surfaces of the discs are
already equipotential surfaces before plating them; plating them with
superconductor would MAKE this true, but in the real problem with a
solid sphere of finite conductivity, the flat surfaces of the discs
without plating would almost certainly not be equipotentials.
This is, of course, is in addition to the complication of computing
the resistance of the contacts.
To solve problems of resistance between contacts applied to continuous
resistive media, you have to be sure that the little dx's, dy's and
dh's you use have equipotential surfaces where they stack up. In
situations with extremely high symmetry, you can do this. For
example, the resistance between the circular faces of a right circular
cylinder of resistive material with highly conductive material plated
on the faces. Otherwise, you have to bite the bullet and solve
Poisson's equation throughout the volume. Nowadays, this can be done
fairly easily with a finite element solver.
>But it's one H**L of a problem if the points are not very small and/or
>not diametrically opposite.