| |
BGonline.org Forums
OT Analog clock math problems
Posted By: Nack Ballard In Response To: OT Analog clock math problems (Jason Lee)
Date: Tuesday, 4 May 2010, at 11:17 a.m.
I thought this thread was about backgammon clock settings, so I had ignored it.
Forgive me if this has already been answered, but Problem 1 seems simple if looked at the right way. The minute hand travels 12 times as fast as the hour hand. During a 12-hour cycle, the former passes the latter 12 - 1 = 11 times. Since the angle formed by the two hands is 0 degrees 11 times, then it must be at every other angle 11 times. So, it’s at +/-90 degrees twice as often, or 22 times in 12 hours, which is 44 times in a 24-hour period (day).
By similar logic, the second hand passes the minute hand 60 – 1 = 59 times per hour, so it will be at +/-120 degrees 59*2 = 118 times per hour, or 118*24 = 2832 times per day.
At midnight, all three hands line up (are at 0 degrees). Over the next 24 hours, 44 times hour/minute hands are at +/-120, and 2832 times minute/second hands are at +/-120. The 44 and 2832 share prime factoring of 2*2, which suggests that at 4 times during the 24 hour period the situations exist simultaneously (I'm not quite certain of this step, because there's an implication of regular interval). However, 2 of these 4 times the +120 and -120 offset (instead of an even +120 +120 spread) so it seems to me that the answer to Problem 2 is twice a day.
Nack
| |
BGonline.org Forums is maintained by Stick with WebBBS 5.12.