Shortest path to paint a sphere with a large square or circular brush Survey telescopes in space need to "paint" the celestial sphere with their apertures in order to cover it without wasting too much time. Screen shots below from two videos give some idea. These are certainly not optimal in a mathematical sense though they are probably optimal in terms of spacecraft management and scientific return.
My question is about the mathematics behind painting a sphere by moving a finite size "brush" over the surface with the shortest stroke length.
I think that if $\theta(t)$ and $\phi(t)$ describe the stroke for $0 \le t \le 1$ then the path length to be minimized can be expressed as:
$$\int_0^1 dt \sqrt{ \left(\frac{d \theta}{dt} \right)^2 +  \left(\sin \theta \frac{d \phi}{dt} \right)^2}$$
but that should be checked.
For example, if the brush is $\pi/10$ radians wide, you can perform 10 great circles around lines of latitude, stepping by $\pi/10$ in longitude each time. The total path length is then $20 \pi$ for a circular or square brush of width $\pi/10$. 
But there is certainly a more complex spiral-like pattern starting at one pole and ending at the opposite pole that would involve substantially less overpainting.


*

*Has this problem been solved, or even addressed? 

*Does this problem have a better name than "Shortest path to paint a sphere with a large square or circular brush"?



example videos are pretty interesting, especially TESS' orbital maneuvering after 05:20
 TESS, from video 
 Spektr-RG/eROSITA  from video
 A: No idea about what is the shortest path but among piecewise $C^1$ curves, the spiral can be chosen to be reasonably close to optimal.
We will limit ourselves to circular brush as rigorous bound of that is known.
Motivated by testing of a nonlinear parameter in a regression model, Hotelling (1939) has computed the volume of a tube of given radius around a curve in $S^{n-1}$$\color{blue}{{}^{[1]}}$. In 1986, Naiman studied another statistical problem and arrived at same geometric problem. Naiman has shown Hotelling's result is an upper bound for the volume of a tube of arbitrary radius$\color{blue}{{}^{[2]}}$.
Translate this to geometry and limit ourselves to unit sphere $S^2$, we have

Let $\gamma : [0,L] \to S^2$ be any piecewise $C^1$ curve on unit sphere, parametrized by arc length. Let $L$ be its length and $M$ be its image $\gamma([0,L])$. For any $\alpha \in (0,\frac{\pi}{2})$, let 
  $$M_\alpha = \left\{ y \in S^2 : \exists z \in M, y\cdot z \ge \cos\alpha \right\}$$ 
  When $\gamma$ is open (i.e. $\gamma(0) \ne \gamma(L)$), one has
  $$\verb/Vol/(M_\alpha) \le 2L\sin\alpha + 2\pi(1-\cos\alpha)
$$
  When $\gamma$ is smooth and doesn't cross itself, above inequality becomes an equality for sufficiently small $\alpha$.

For our problem, let's say we paint along a piecewise $C^1$ curve of length $L$ using a circular brush of width $2\alpha$. In order for the paint to completely cover the sphere, we need
$$2L \sin\alpha + 2\pi(1-\cos(\alpha)) \ge 4\pi\quad\implies\quad
L \ge \pi \cot\frac{\alpha}{2}$$
For example, if we are using a stroke with width $2\alpha = \frac{\pi}{10}$,
the length of the path is at least $\pi\cot\frac{\pi}{40} \approx 39.9177194541743$.
For comparison,let us parametrize the unit sphere by spherical polar coordinate
$$[0,\pi]\times[0,2\pi) \ni (\theta,\phi) \quad\mapsto\quad
(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \in S^2$$
and consider following spiral
$$[0,\pi] \ni t\quad\mapsto\quad  (\theta,\phi) = \left(t,\frac{\pi t}{\alpha}\right)$$
If we paint along it with a circular brush of width $2\alpha$, it is not hard to see the paint will completely cover $S^2$. By brute force, its length equals to
$$L_{spiral} = \int_0^\pi \sqrt{\dot{\theta}^2 + \sin^2\theta\dot{\phi}^2} dt
= 2\sqrt{1 + \frac{\pi^2}{\alpha^2}}E\left( \frac{\pi}{\sqrt{\pi^2+\alpha^2}}\right)
$$
where $\quad\displaystyle E(k) = \int_0^{\frac{\pi}{2}} \sqrt{1 - k^2\sin^2\theta}d\theta\quad$
is the complete elliptic integral of 2nd kind.
For $2\alpha = \frac{\pi}{10}$, $L_{spiral}$ evaluates to $40.24404463079659$. This is only about $0.8\%$ larger than above bound. In certain sense, such a spiral is reasonably close to optimal.
Since I didn't study statistics, I don't have a proper reference for Hotelling-Naiman's inequality. My understanding of this mostly comes from a paper by Johnstone$\color{blue}{{}^{[3]}}$. Look at refs there for details on the statistics side.
Refs


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*$\color{blue}{[1]}$ Hotelling, Harold (1939). Tubes and spheres in n-spaces and a class of statistical problems, Amer. J. Math. 61, 440-460.

*$\color{blue}{[2]}$ Naiman, D.Q. (1986). Conservative confidence bands in curvilinear regression, Ann. Statist. 14,896-906.

*$\color{blue}{[3]}$  Johnstone, Iain, and David Siegmund. On Hotelling's Formula for the Volume of Tubes and Naiman's Inequality. The Annals of Statistics, vol. 17, no. 1, 1989, pp. 184–194.
