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Subject: Re: Resistance of a Sphere Revisited
NNTP-Posting-Date: Mon, 02 Dec 2002 07:39:57 GMT
Organization: AT&T Broadband
Date: Mon, 02 Dec 2002 07:39:57 GMT
Phantom, John Woodgate wrote:
>> 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.
Both problems seem non trivial.
> 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.
Who knows what resistance lurks in the spheres of men? Surely the
Phantom knows. Or at least his finite element solver might.
Since I don't, I'll have to take a reasoned guess. Let's start with
a long cylinder (a piece of wire ) such that what happens at the end
connections is not an issue. Since resistance is proportional to length
divided by cross sectional area, doubling all dimensions would be like
putting four 2x resistors in parallel. So, at least for wire anyway,
double all dimensions and get half the resistance.
My best guess for the sphere would be that doubling all aspects of the
sphere (including the size of the end contacts) would result in half
the resistance just like with the wire. However, if the end contacts
were held to the original size then perhaps resistance might possibly
remain the same. Who knows? Oh. Right. The Phantom knows. Or at
least he hopefully knows someone who knows, 'cause now I sure want to
know. -- analog :)
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