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 Hands On A Clock Part 2 (Posted on 2009-04-10)
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.]

 No Solution Yet Submitted by Jer Rating: 3.0000 (1 votes)

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 It is discreet to be discrete -- solution | Comment 2 of 4 |

Since we are not told the mechanics of this clock, I shall suppose that time (or at least its measurement) is discrete, and that each hand moves on the second.  We may thus divide one circle of the clock into 43200 segments. In 12 hours, the hour hand will make one revolution, the minute hand 12 revolutions, and the second hand 720 revolutions.  Each second the minute and hour hands will make proportional movements.

From hh:mm:ss starting at 00:00:00 we can locate each hand as follows:

Hour hand: (hh * 3600) + (mm * 60) + ss

Minute hand: (mm * 720) + (ss * 12)

Second hand: (ss * 720)

We will generate each combination of hh (00 to 11), mm (00 to 59) and ss (00 to 59) -- except for the restrictions (since we are looking for a minmax where the seconds hand is highest, the minutes hand is next, and the hours hand lowest (counting from 00:00:00), so mm > (hh * 5), and ss > mm.

Since there are 43200 evenly spaced ticks, if the hands were evenly spaced (120 degrees) these numbers would be 14400 apart.  Failing that, we are to get "as close as possible" defined as lowest sum of the deviations.

For this latter, for each calculate-and-compare I shall take the minute hand as the base, and take the respective absolute values of the divergence of each of the other hands from 14400 added to (for seconds hand) or subtracted from (for hours hand) the position of the minutes hand -- then add those two divergences for the total divergence.

The lowest value is obtained at 03:37:58.  The hours hand is at 13078, the minutes hand 27336, and the seconds hand at 41760. The hours hand diverges from the minutes hand by 142, and the seconds hand diverges from the minutes hand by 24 -- for a total divergence of 166. (The next best is 02:32:53, with a total divergence of 187; then 00:21:42 with a total divergence of 294, etc.)

Clearly these are only meaningful if adopting the interpretation of discrete movements of the hands.

The computation procedure, as noted above, involved three loops (to generate discrete values for hh, mm, and ss), and for each calculating the position of each of the three hands, and the spacing of the other two from the minutes hand.

(I haven't yet looked at any other comments, so may look now to see how the "continuous sweeping" is interpreted.)

 Posted by ed bottemiller on 2009-04-13 14:11:56

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