Let $S$ be a rigid body in $\mathbf{R}^3$ of finite diameter. Assume that there is a plane which divides the space in two regions, one containing $S$ in its entirety.

The question is: What is the smallest radius of the circular hole to be in the plane so that $S$ can move in the other region?

Examples: If $S$ is a sphere, it is obviously the radius of the sphere. If $S$ is an ellipsoid with two axes (= rugby ball) is clearly the semi-minor axis. If $S$ is an ellipsoid with 3 axes $a> b> c$ the wanted radius is $b/2$ ... For a right cylinder, it is the generating circle radius. For an oblique cylinder, I do not know.

Second question: how is called this problem in the literature (if any ...)?

Third question: How to calculate the radius for any $S$?

Correction: for the right cylinder of circular section, it is the radius of the disk generator, regardless of the length of the cylinder.

For a helical spring of circular cross section of radius r and major radius R without ties, it is obviously r (since R> r).

What does not work: it is not in general the radius of the circle circumscribing the minimum section (non-convex solid) of radius r or the radius of the circle circumscribing of the largest section.

I have a partial answer to my third question:

x is a point in S and m(x) is the sup of the radius of sphere of center x included in S.

Then the smallest radius of the circular hole is $\ge \sup_{x \in S} m(x)$.

(counter-example for = is the cylinder with radius > height)

Another inequalities is that the smallest radius of the circular hole is greater or equal to the biggest radius of cylinders included in S.

but there is a counter-example to equality : a cylinder with radius > height and small blisters.

  • For an oblique cylinder, I'd imagine that it would be the width of when you align on a vertical axis the endpoints on the longest line segment contained inside the cylinder that you can draw. – Lost Dec 6 '13 at 9:44
  • 2
    The problem can be very hard for non-convex bodies. Imagine a coiled spring; if the coil is very loose, the size of the hole is essentially given by diameter of the wire used to make the spring. If the coil is very tight, the size of the hole is the same as the convex hull of the object, which is now a cylinder. – Willie Wong Dec 6 '13 at 9:49
  • This type of problem is hard even in 2D. The moving sofa problem gives an idea – Ross Millikan Dec 7 '13 at 0:54
  • For a tetrahedron this problem has been solved here: brand.site.co.il/riddles/201008q.html . But try it on your own; its one of the cutest geometric problems I have ever met. – Christian Blatter Sep 14 '14 at 10:00

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.