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 $\theta(0) = 0$)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?
