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Where would one put $n$ telescopes on the surface of the earth to see the whole sky as well as possible ?
Use the cosine metric to define how well we can see in direction $x$:
$ \qquad \text{cansee}( x; x_1 \dots x_n ) = \text{max}_i \ x \cdot x_i $
The worst direction for all the telescopes is then
$ \qquad \text{worstsee}( x_1 \dots x_n ) = \text{min}_x \ \text{cansee}( x )$
and we want $n$ telescope positions that maximize that, i.e. that can see pretty well even in the worst direction.

That's in $R^3$. What I really want to do is generate approximate solutions on a sphere in $R^d$ for $3 <= d <= 10$ and $d+1 <= n < 2d$.
"Very approximate" would do; with < 20 points, an iterative method would do.

(Feel free to change the metric if min-max is intractable.)

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Where does this come from (the restriction on $n$ is a little peculiar...)? –  Igor Rivin Feb 24 '12 at 14:30
2  
IIRC, equally spacing points on spheres generally is hard (there are easy symmetric cases). Typically one treats the points as electric charges and lets them equilibrate to get a practical solution. –  Steve Huntsman Feb 24 '12 at 15:57
    
@Igor, from playing with noisy optimization: Nelder-Mead tracks d+1 points. (You're welcome to unrestrict ...) –  denis Feb 24 '12 at 17:06

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