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Clock Part II

In the solution for "clock": "if you just think about it, intuitively you'll see the minute hand passes the hour hand 11 times every 12 hours"

I think that "intuitively" is the wrong word here, since I would argue that, intuitively, the minute hand should pass the hour hand 12 times in a 12-hour period. I believe that you're starting erroneously at 12:00. The minute hand has passed _half_ of the hour hand already (and will pass the other half 12 hours later).

To prove this, set the starting position as 11:58. When the minute hand passes the hour hand at 12:00, it has passed once. It will have passed twice at 1:00 and will have passed twelve times at 11:00. The minute hand will not, of course, pass the hour hand again before 11:58.

Kevin
Tuesday, October 08, 2002

Uh, no. The hour hand goes around once in twelve hours. The minute hand goes around twelve times in twelve hours. So once the minute hand pulls away from the hour hand after 12:00, the difference in speed is 11 times. So the minute hand will catch up every time one-eleventh of 12 hours has elapsed. So the hands synch at 12:00, 1:05:27, 2:10:54, 3:16:22, 4:21:49, 5:27:16, 6:32:44, 7:38:11, 8:43:38, 9:49:06, and 10:54:33.

Hugh Brown
Wednesday, October 23, 2002

What do you mean the "difference in speed"? This should be a ratio, not a difference. The minute goes around the clock 12 times for every 1 time the hour hand goes around the clock. Therefore, the minute hand should pass the hour hand one twelfth of the time that it rotates around the clock face.

Kevin
Monday, October 28, 2002

Okay, nevermind ... I see your point. My mistake was in assuming a ration since that would imply that the minute hand passed the hour hand at each hour.

Kevin
Monday, October 28, 2002

Moreover, you solution indicates that there is no overlap during the 12th hour, which cannot be true as there is atleast one overlap at each hour.

Sanjay
Sunday, November 03, 2002

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