5

Let $S$ be the sphere in $\mathbb{R}^3$ and $C:[0,1]\to S$ a continuously differentiable curve on $S$. Let $T:[0,1]\to\mathbb{R}^3$ denote the tangent vector of $C$. Let $P(t)$ be the plane containing $C(t)$ and having normal vector $T(t)$.

Given a size $d$ of the "paint brush" we define the "brush" $b:[0,1]\to \mathcal{P}(S)$ by letting $b(t)$ be the points of $S$ that are at most a distance $d$ (metric on the sphere) from $C(t)$ that are contained in $P(t)$.

We can think of this as saying the "brush" $b(t)$ is an arc on the sphere that is "orthogonal" to the motion $C(t)$ of the "paint brush".

Given $d$ what is the arclength of the shortest curves such that $\cup_{t\in[0,1]} b(t) = S$. This says that the "paint brush" covered the sphere.

Martin Sleziak
  • 4,608
user16557
  • 1,513
  • 2
  • 14
  • 14

2 Answers2

17

The model is that used by Henryk Gerlach and Heiko von der Mosel in their 2010 paper "On sphere-filling ropes" arXiv:1005.4609v1 (math.GT) may be relevant. Their question is different: What is the longest rope of a given thickness on a sphere? But their explicit solutions are packings, and it seems they could be converted to painting paths. Here is a piece of their Fig. 6:
           Rope on sphere

Joseph O'Rourke
  • 149,182
  • 34
  • 342
  • 933
5

This question is somewhat related to this recent one. More precisely, the comment by Gjergji Zaimi in the earlier question gives a painting of length $2\sqrt{2}\pi$ for $d=\pi/4$, which, as explained in another comment there, is optimal for a path at constant distance from the sphere. So for $d=\pi/4$ the optimal length should be $2\sqrt{2}\pi$.

Community
  • 1
Jean-Marc Schlenker
  • 1,700