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.