A mango made me wonder about this. (See also this question, which is in a similar spirit.)

Fix $L >0$ and a smooth body (possibly nonconvex—pears or bananas are fair game!) $B \subset \mathbb{R}^3$ (and assume w/l/o/g below that $L$ is sufficiently large since we can dilate $B$). For $\gamma:[0,L] \rightarrow \mathbb{R}^3$ smooth and parametrized by arclength and $\theta:[0,L] \rightarrow S^1 $ smooth, let $k(\gamma, \theta,s)$ denote a copy of the unit interval centered at $\gamma(s)$ and in the plane orthogonal to $\dot \gamma(s)$, and at the angle $\theta(s)$ in that plane (we require $k(\gamma,\theta,0)$ to be tangent to $B$, say, and w/l/o/g that this sets $\theta(0) = 0$; angles in planes away from $s=0$ can be sensibly defined via parallel translation). Let $K(\gamma,\theta):= \{ k(\gamma, \theta,s) \cap B : s \in [0,L] \} $. If $K$ contains the boundary of a body $C_K \subset B$ then say that $(\gamma, \theta)$ is a peeling of $B$.

For $L$ fixed, is there an effective way to determine a peeling that minimizes $\mbox{vol}(B \backslash C_K)$?

Followup: can the best peeling of the unit ball for a given value of $L$ be explicitly constructed?

  • 6
    $\begingroup$ +1 for the charismatic title and way of explaining the idea. $\endgroup$ – Joel David Hamkins May 28 '10 at 2:35
  • 13
    $\begingroup$ Unfortunately, as you might expect when peeling fruit, it's a messy problem. $\endgroup$ – Greg Kuperberg May 28 '10 at 3:09
  • 3
    $\begingroup$ Reminds me of a paper I spotted a couple of days ago: arxiv.org/abs/1005.4609 $\endgroup$ – Dan Piponi May 28 '10 at 3:50
  • 7
    $\begingroup$ The title is sexy but I got lost in line 3 of the mathematical description. Is there way to restate it in plane language without excessive notation? $\endgroup$ – Victor Protsak May 28 '10 at 6:45
  • 4
    $\begingroup$ @Victor: here is my interpretation of what the notation means: $k(\gamma,\theta,s)$ is describing the knife, where $\gamma$ is the movement of the center of the knife while $\theta$ is the rotation of it (I didn't understand why it is only allowed to rotate orthogonal to the motion of the center, is that sufficient?). You are allowed to just make a limited movement (L) and want to minimize the pulp you are cutting away. $\endgroup$ – Michael Bächtold May 28 '10 at 9:10

Sorry, I'm still not allowed to comment. So I use the "Answer" window...

I'm not completely sure to understand your formulation, but for the case of a 2-dimensional sphere and some fixed width of the pealing, you may find your answer in the sphere-filling ropes of Gelrach and von der Mosel. These are ropes with a certain fixed width that are going on a sphere and trying to cover the greatest area. For some width, it is possible to cover everything.

  • Heiko von der Mosel et Henryk Gerlach On sphere-filling ropes. Amer. Math. Monthly 118 (2011), no. 10,
  • Heiko von der Mosel et Henryk Gerlach : What are the longest ropes on the unit sphere ? Arch. Ration. Mech. Anal. 201 (2011), no. 1, 303–342.
| cite | improve this answer | |

If the path is allowed to be piecewise smooth (see the comments above), and the fruit is convex, then you can cover the surface with a large number of small patches, and use very short circular trajectories to peel each patch, roughly spinning the blade in place to remove the piece of peel. As the size of patches decreases, this will approach a perfect peel, even if the total length $L$ is chosen to be arbitrarily small. This is like peeling a fruit by bouncing it off a belt sander.

If the fruit is non-convex, we still approach a perfect peeling, as long as we allow $C_K$ to have more than one connected component.

This suggests that the problem is only interesting if you put a bound on the number of jumps.

| cite | improve this answer | |

After a bit of thought I have a sketch of a very simple case. Suppose that $B$ is a 3-polytope; let $B^*$ denote its dual and consider the graph $G$ associated to the 1-skeleton $B^*_1$ of $B^*$. Now vertices of $G$ correspond to faces of $B$, and edges of $G$ correspond to adjacent faces of $B$. So if $G$ admits a Hamiltonian path then we can use it to get a (nearly?) optimal peeling for $L$ appropriate.

Google results for Hamiltonian circuits on 3-polytopes are here.

  • $\begingroup$ Thinking a bit further, it's not clear that the arclength cost incurred by a Hamiltonian path will be small; it might be more efficient to revisit already-peeled faces. So this casts the near-optimality into doubt. $\endgroup$ – Steve Huntsman May 28 '10 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.