Subject: Re: Resistance of a Sphere Revisited
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Date: Sun, 01 Dec 2002 01:26:03 GMT
NNTP-Posting-Date: Sat, 30 Nov 2002 20:26:03 EST
Organization: Cox Communications
"John Larkin" wrote in
> On 30 Nov 2002 12:43:30 -0800, firstname.lastname@example.org (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.
I hadn't seen this problem before (or if I had, I had pleasantly forgotten
about it), but sure enough, a point contact leads to an infinite resistance.
Imagine that the current equally distributes itself through the sphere (it
doesn't, but that won't matter, and it makes the math tractable). Then the
voltage drop across any cross section orthogonal to the diameter is just dV
= I*dR, where dR is (r/A)dx, and A is pi*(R(x))^2. A = area, R(x) = radius
as a function of x = sqrt(2*D*x - x^2). This can be integrated:
integral(0,2D)(I*r/(pi*(2*D*x - x^2))dx = infinity
If we fool around a little more, and assume that the contact is made by
slicing off the end of the sphere and using the flat portion as a contact,
then the limits of integration change from (0,2D) to (a,2D-a), and the
integral evaluates to I*r/(pi*D)*ln((2*D-a)/a), where a is the radius of the
contact. The problem is that as a -> 0 the total resistance goes to
Even though the original assumption, that the current is equally distributed
throughout the sphere, is invalid, it's only the contact points that make
the resistance go to infinity. At the contacts, especially as their area
goes to zero, the assumption that the current is equally distributed
throughout the contact area is valid. So, if the contact area is zero, the
resistance is infinite.
But, since the original post said that the existing sphere was 1 ohm, we can
assume that the contact points are non-zero radius, and use the existing
solution to predict what happens when we increase D by 2. The ratio of the
new resistance to the old is given by:
(I*r/(pi*2*D)*ln((4*D-a)/a)) / (I*r/(pi*D)*ln((2*D-a)/a))
= (1/2)*(ln((4*D-a)/a) / ln((2*D-a)/a))
As a -> 0, the ratio of logs approaches 1, and the overall result is 1/2.
This is still assuming uniform current density in each cross section of the
sphere, but I suspect I could make a good symmetry argument that even
without uniform current density, the overall resistance would still change
by a factor of 1/2 as a -> 0.
So, _given_ that the initial sphere had a measured resistance of 1 ohm,
_and_ that the contact size was small enough that the ratio of logs is
approximately 1, then the resistance of the 4D diameter sphere is 1/2 the
resistance of the 2D sphere.
Now, what if it wasn't a solid sphere, but an infinitely thin shell, so that
the current travels only over the surface...
-- Mike --