# Surfaces that can be rolled by a ball

Let $S$ be a smooth solid body in $\mathbb{R}^3$, and $B$ a ball of radius $r$. Say that $B$ is in contact with $S$ if (1) they share a point $x$ that is on the surface of each, $x \in \partial S$ and $x \in \partial B$, and (2) neither penetrates the other, $\operatorname{int} S \cap \operatorname{int} B = \varnothing$. Say that $S$ can be rolled by $B$ if, for every pair of points $x,y$ on the surface of $S$, there is a path $\rho$ connecting $x$ to $y$ on $\partial S$ such that $B$ can be placed in contact at every point $z \in \rho$. In other words, the ball $B$ can roll between any pair of points of the surface without ever penetrating $S$.

My question is:

Characterize the bodies $S$ that can be rolled by a ball of radius $r$.

A necessary condition is that the Gaussian curvature at any point on the surface of $S$ must be $\ge -\frac{1}{r^2}$. Could that also be sufficient, or are there global obstructions such that $B$ could be in local contact but penetrate away from the contact point? Is the characterization different for surfaces of genus zero than for surfaces with handles? Any insights, speculations, or literature pointers would be appreciated. Thanks!

Update. Here is my interpretation of Anton's example:

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I would guess your conditions would have to be nonlocal in some way. A cheesy r->0 limit of this is to write down conditions for a surface to be embedded. –  j.c. Sep 22 '12 at 14:43
It seems to me that the path $\rho$ plays no important role, other than to rule out disconnected surfaces. The set of $T$ "touchable" points either covers $\partial S$ or it doesn't. There is no way to get a $T$ that isn't path-connected without having $T$ fail to cover $\partial S$. –  Ben Crowell Sep 23 '12 at 5:04
@Ben: You are right; see also Alexandre's remarks. –  Joseph O'Rourke Sep 23 '12 at 13:33

You can call such bodies as "bodies of reach $\ge r$".

Clearly all the principle curvatures has to be $\ge -\tfrac1r$.

Plus you have a global condition which might be stated the following way: There is no segment between points on the surface which (1) meets the surface at normal direction at both ends (2) has length $<2\cdot r$ and (3) its interior lies in the complement of the body.

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@Deane, take a smooth simple curve with curvature $<1$ which comes $(2\cdot r)$-close to itself. The body is its $\varepsilon$-neighborhood (smoothed as needed). –  Anton Petrunin Sep 22 '12 at 15:12
I think your condition is equivalent to saying that the tubular neighborhood of radius $r$ is embedded, when parameterized by Fermi coordinates. –  Ian Agol Sep 22 '12 at 15:53
Note that there are disconnected surfaces which are rollable. You might consider incorporating them. Gerhard "Ask Me About System Design" Paseman, 2012.09.22 –  Gerhard Paseman Sep 22 '12 at 16:16
@Gerhard, no, by definition the rollable surfaces are connected. –  Anton Petrunin Sep 22 '12 at 18:37
Actually, by definition it is unclear. One assumes "image of the interval (0,1) under a continuous map" when one sees the words "path" and "connecting", but this is not the same as "path-connected". Joseph may intend the surface to be connected, but unless he clarifies, I see enough ambiguity in the wording to admit disconnected surfaces. A certain form of connecting may involve the surface of the ball as well. Even so, I don't think you need to change your characterization much, if at all, for this situation. Gerhard "Ask Me About System Design" Paseman, 2012.09.22 –  Gerhard Paseman Sep 22 '12 at 21:48

$S$ is the complement of the Minkowski sum${}^*$ of a connected body and the sphere of radius $r$.

${}^*$ Provided the Minkowski sum doesn't self-intersect.

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1. Is not your condition equivalent to this: for every boundary point if the body, there exists a closed ball containing this point and whose interior does not intersect the interior of the body? (Of course I assume that your surface is connected). Or perhaps I did not understand what you mean by "rolling"? Why do you need two points, and a path?

2. If I understand your definition correctly, here is a reference where this notion is studied in dimension 2: http://arxiv.org/pdf/1204.4283.pdf They call such things r-convex.

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@Alexandre: You (and Ben Crowell) are correct, in that rolling plays no role, even though that was my original motivation: to "paint" the surface with a rolling "paint ball." And $r$-convexity is a nice viewpoint, because of course if $r \rightarrow \infty$, $S$ must be convex. –  Joseph O'Rourke Sep 23 '12 at 13:31

The property you are asking for is called global radius of curvature. The notion combines the condition of lower bound on the principal curvatures as well as the global condition of no interpenetration of a neightborhood. This notion is very well adapted to the calculus of variations.

The heuristic definition from the paper of P. Strzelecki and H. von der Mosel: Global curvature for surfaces and area minimization under a thickness constraint

The main idea can be sketched as follows: Take a continuous parametric surface $X : \mathbb{R}^2\supset \mathbb{B}^2 \to \mathbb{R}^3$ (with possibly infinite area) which possesses a tangent plane on a dense subset $G \subset \mathbb{B}^2$ which may even have zero measure. Consider the radii of all spheres touching the surface $X(\mathbb{B}^2)$ in one of these points $X(\omega)$, $\omega \in G$, and containing at least one other point of the surface. We define the infimum of these radii as the global radius of curvature $\Delta[X]$ of the surface $X$ . It turns out that a positive lower bound on $\Delta[X]$ serves as an excluded volume constraint for the surface $X$.

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Great reference---Thank you, André! –  Joseph O'Rourke Sep 27 '12 at 17:21