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This made the popular news a few years ago. Summary: it's the first homogeneous, convex solid to be found with only one stable and one unstable mechanical equilibrium when resting on a flat surface. I found the notion pretty interesting and visited the discoverers' website, where they describe attempts to discover a smooth version which technically succeeded, but they were unable to build a physical model due to extremely high sensitivity to imperfections. The piecewise smooth version they ended up with is still very sensitive to such imperfections, but less so. So then I wondered: what if we no longer care about the number of unstable equilibria?

Let A be the set of smooth, bounded, convex solids (assumed to be homogeneous). An element of A can be thought of as an immersion X : S2 → R3; we can define oX to be the centre of mass of the solid interior. Now let B ⊂ A be those solids having only one stable equilibrium point (local minimum of |X-oX|). Then we can define:

d(x,Y) = infy∈S^2|X(x)-Y(y)|

r(X,Y) = supx∈S^2(d(x,Y)))

and finally rmin(X) = infY∈A\B(r(X,Y)), the size of the "safety margin" around X ∈ B. Then the question is: which unit-volume X ∈ B maximizes rmin(X), i.e. is the least sensitive to imperfections?

With regard to the tagging: I suppose this is really an exercise in calculus of variations, but this was the closest I could get using the arXiv tags.

Robin Saunders
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  • How about a nearly flat cone with a flat base (rounded appropriately near edges to make it smooth)? Then the wide base will be a safe stable equilibrium point, and that seems to be the only stable position. Perhaps I am misinterpreting your question... – Joseph O'Rourke Jul 15 '10 at 22:30
  • Technically you'll have a circle of "stable equilibria" near the apex of the cone. I know they're not truly stable, but informally they are "more like stable equilibria than anything else". More formally, they are still local minima of |X|. If placed on one of these points, the cone will roll around in circles, eventually coming to rest (due to friction) on a point other than the base. You'd also have to be careful rounding off edges in general, to avoid introducing extra local minima there. – Robin Saunders Jul 15 '10 at 22:47
  • @Robin: I see! Thanks for explaining. – Joseph O'Rourke Jul 15 '10 at 22:56
  • Can you perhaps rewrite the question to make is complete and precise? I am note sure I understand "What about...?" type of question. – Igor Pak Jul 15 '10 at 23:05
  • The precise question itself is stated further down; the "what if...?" was just used to introduce it. Sorry if that was a bit confusing. Ignoring the introductory background paragraph though, the rest of the question should be self-contained. Does it make sense, or is there something I could explain more clearly? – Robin Saunders Jul 15 '10 at 23:32
  • Perhaps restricting your question to surfaces of revolution would be a place to start. Do you know of the optimum 2D curve? Or good candidates? – Joseph O'Rourke Jul 16 '10 at 11:43
  • I expect that surfaces of revolution cannot be Gömböc objects. I don't know if there's a proof of this (if there isn't, it would be worth looking for one). – Peter Shor Jul 16 '10 at 18:19
  • @Peter: My understanding is that Robin was loosening the criteria for Gömböc objects, so that surfaces of revolution do make sense: he just seeks a unique local min of $|X - o_X|$ with a big "saftey margin." – Joseph O'Rourke Jul 16 '10 at 20:54
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    Well, it's pointed out on the creators' website (www.gomboc.eu) and also in the Wikipedia article that any plane curve must have at least two stable equilibria (as well as at least two unstable equilibria), as a consequence of the four-vertex theorem. So a surface of revolution wouldn't work even with my loosened criteria. – Robin Saunders Jul 16 '10 at 21:14
  • @Robin: Ah! That makes sense. Nice application of the 4-vertex theorem! – Joseph O'Rourke Jul 16 '10 at 21:20
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    @Robin: a three-dimensional surface of revolution isn't the same as a constant-mass plane curve, so it's not clear to me that this theorem directly applies. – Peter Shor Jul 17 '10 at 18:25
  • You're right, it doesn't follow directly. But we can put some constraints on what such an object could be like. Say we have a solid of revolution about the x axis, whose centre of mass is the origin. Take the cross section through the xy plane. Now the surface is a smooth, convex, plane curve, symmetric about the x axis; minima of the curve correspond to minima of the surface, so for there to be only one it must be one of the intersections with the axis. – Robin Saunders Jul 18 '10 at 22:17
  • So the problem reduces to finding a function f from [-1,M] to the non-negative reals with the following properties: smooth on (-1,M) with non-positive second derivative; passes through 0 vertically at -1 and M; (x^2+f(x)^2) has positive derivative; and the integral of xf(x)^2 dx is 0. – Robin Saunders Jul 18 '10 at 22:17
  • Since time has passed and some people may assume the ball's in my court, I should point out that I'm abysmal at calculus and am not going to be able to find an example (or proof of nonexistence thereof) myself. – Robin Saunders Jul 23 '10 at 01:09

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