From: "Ed Price"
Subject: Re: Resistance of a Sphere Revisited
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Date: Sat, 30 Nov 2002 23:10:07 GMT
NNTP-Posting-Date: Sat, 30 Nov 2002 18:10:07 EST
Organization: Cox Communications
"John Larkin" wrote in
> On 30 Nov 2002 12:43:30 -0800, email@example.com (bill) wrote:
> >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
> >for this.
> >Thanks, Bill
> As I recall, the resistance is hugely dependent on the contact area,
> so the 'reasonable assumption' you suggest is impossible. The contact
> area is not a 'nuance'; for a solid copper sphere, you can get
> micro-ohms to megohms depending on the contact area.
> The constriction resistance of a small round contact on a flat surface
> is Rc = r/d
> where r = resistivity of material
> d = diameter of contact.
> which goes infinite as d approaches zero.
> This will get weird on a sphere as the contact gets big and approaches
> covering a hemisphere.
> And it's not nice to argue with your brother on Thanksgiving.
Unless your large contact is appreciably crushing the sphere, you would have
to assume a very small point-contact to any sphere. Imagine the contact to
be an infinitely large, perfectly flat plane. An undistorted sphere can
touch the plane at only one, infinitely small point.
For the purpose of the original poster's question, lets assume the first
sphere has a diameter of one meter, and is compared to a sphere of two
meters diameter. Let's assume the probes each have a tiny contact area (if
you must, make it 0.0005 square cm) in relation to the surface of either a
one- or two-meter diameter sphere.
Now, let's get to the geometry of the problem.