The hour, minute, and second hands of a clock with continuous sweeping can never be spaced at exactly 120° intervals. This was shown in Part 1
Find the exact time when these angles are as close as possible and the order of the hands are hour, minute, second in a clockwise fashion.
['As close as possible' means the sum of the individual deviations from 120° is minimized.]
(In reply to Spec-tacular
by ed bottemiller)
Ahhh! I see I neglected the proviso that the clockwise order of the hands be hour, minute, second. That makes only the 9:05:25 + 325/719 sec solution true.
I assume "continuous sweeping" means the hands move smoothly--not just every second, nor even every 1/50 or 1/60 of a second, but as a smooth motion that even Zeno would love.
Posted by Charlie
on 2009-04-13 18:49:34