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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.

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Duplicate?… – Gjergji Zaimi Jul 20 '11 at 9:19

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

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Yes indeed -- by as brush width half the width of the rope. This seems to give an optimal painting for countably many widths. Very nice! – Jean-Marc Schlenker Jul 20 '11 at 10:38

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 $2\sqrt{2}\pi$.

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Of course $d=\pi/4$ is very special, since for this painting each point of the sphere is painted exactly once (except for a curve). It's not clear whether there is another value of $d$ for which this is possible. – Jean-Marc Schlenker Jul 20 '11 at 10:02
indeed your comment there was very good too. – Pietro Majer Jul 20 '11 at 10:04
Thanks! But the answer by Joseph O'Rourke above is much more complete. – Jean-Marc Schlenker Jul 20 '11 at 10:46

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