From: John Woodgate
Subject: Re: Resistance of a Sphere Revisited
Date: Sat, 30 Nov 2002 22:16:13 +0000
Organization: JMWA Electronics Consultancy
Reply-To: John Woodgate
NNTP-Posting-Date: Sun, 1 Dec 2002 05:20:00 +0000 (UTC)
X-Newsreader: Turnpike (32) Version 4.01 <5Z8C9wtxbnpWyFnyfFzqmVF739>
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.
But it's one H**L of a problem if the points are not very small and/or
not diametrically opposite.
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
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