Angles In Time

Have you ever wondered at what time the three hands of a clock-the hour hand, the minute hand and the second hand trisect the circular face of the clock? In other words, at what time would the angles between each hand be exactly equal to 120 degrees?

One possible way of finding this is laying the three hands fixed at 120 degrees apart and rotating all three on a clock face. You must remember though that most orientations obtained like this do not occur in an actual clock because all three hands move relative to each other. For example at 10 minutes and 30 seconds past 10-o-clock (10:10:30/22:10:30), the movement of the minute hand through the 10 minutes will cause the hour hand to shift past 10, thus changing the angle slightly.

This problem can be solved by using the simple equations of motion for the clock hands. The speeds of each hand can be calculated in degrees per second as shown below.
speed of second hand = 360/60 = 6 degrees per second
speed of minute hand = 360/60*60 = 1/10 degrees per second
speed of the hour hand = 360/12*60*60 = 1/120 degrees per second

At a given time t, the angles through which each hand has revolved can be calculated as,
6t,  t/10,  and  t/120 respectively
thus the ratio of the angles revolved are 720 : 12 : 1

Assuming the there is a time where the three clock hands trisect the face of the clock, let the angle swept by the hour hand at this time be a. Then the angle swept by the minute and second hands are 12a and 720 a respectively. Then for trisection of the face,
12a - a = 120 degrees   or   12a - a = 240 degrees.
              12a - a = 120     =>     11a = 120     =>     a = 10.909090.. degrees
              12a = 130.9090... degrees
and        720a  = 7854.5454.. degrees
removing complete revolutions, 7854.5454 - 7560 = 294.5454      
but 294.5454 - 130.9090 = 163.6364 this is NOT equal to 120 degrees.

This is a contradiction and therefore the assumption, "a time where the hands trisect the face of the clock exists" should be false. It can be similarly shown that the case 12a - a = 240 also leads to a contradiction like this. Therefore the three hands of a clock will never trisect the circular face of the clock. For a more detailed explanation of this problem check the following link : http://mathforum.org/library/drmath/view/56788.html

The above answer was arrived at by assuming that all three hands of a clock move seamlessly with a certain velocity. But there are some clocks where the minute hand moves over one minute division after one full rotation of the second hand. In this case the problem  becomes easier and a solution can be found. In order to avoid the shifting of the hands by small amounts we need to make sure that the minute hand points to the 12 position because at that moment the hour hand will point exactly to that particular hour and since the second hand has not completed its revolution around the clock face, the minute hand would also remain there. The answer then is, 20 seconds past 8-o-clock. (08:00:20/20:00:20)

Clock dial obtained from : http://www.thewoodshop.20m.com/clockfaces.htm