TITLE: Hands of the clock, Revisited.
QUESTION [8 upvotes]: It has already been answered (here) that it is impossible for the (continuously moving) hands of a clock to trisect the face of said clock. Even ideally the hour, minute, and second hand can never pairwise form 120 degree angles with each other. Can we find the time of day where they most nearly do?
To put a metric on this one could take the area of the smallest sector divided by the largest sector. (One being the unatainable optimum.) Or add up the amount by which the three angles are off.
$$\left|\frac{2\pi}{3} - \measuredangle H M\right| + \left|\frac{2\pi}{3} - \measuredangle H S\right| + \left|\frac{2\pi}{3} - \measuredangle M S\right|$$
where zero is now the unatainable optimum.

REPLY [2 votes]: Expanding on Han de Bruijn's work I think we can do better. (To be gentlemanly, I'll award the bounty.)
the two functions he sets out to minimize are:
$F_\pm(\theta) =  4 - 2\cos(\frac{2 \pi}{3} \mp 11 \theta)-2\cos(\frac{2 \pi}{3} \pm 719 \theta)$
Differentiating and simplifying, we find that the extrema of these functions are found at the roots of 
$G_\pm(\theta) = 719 \cos(\frac{\pi}{6} \pm 719\theta)-11\sin(\frac{\pi}{3}+11\theta)$
There are 1437 such roots for each of $G_\pm$ over the interval $[0\le \theta<2\pi]$. Using Mathematica I found each one to high accuracy and substituted back into $F_\pm$. I found a unique minimum for $F_\pm$. I also found that the minima were the same but got a lower value:
{$\min F_\pm$, $\theta$} = { $8.483 \times 10^{-6}$, $1.52346119$ (radians of hour hand)}
So this happens at about $1.52346119*720/(2\pi) = 174.575$ minutes
Or at 02:54:34.54