Clock Quiz - Solution
Here is the solution for the math quiz presented in Hugi 28. I'll first quote the original question and then show the answer, so that you do not need to look up the original text. The result itself isn't that spectacular, but the various solutions and their implications may be quite surprising. Thanks to all those who tried out!
At 12 o'clock the minute hand and the hour hand of an analogue clock point to the same position. This happens at other times as well. Can you figure out a rule for calculating those times?
First, let's take a look at the clock and make two helpful observations:
a) The minute hand makes 12 rounds within 12 hours, the hour hand just a single one. Thus, the speed of the hour hand is 1/12 the speed of the minute hand.
b) When both hands overlap for the first time after 12 o'clock we can rotate the clock until both hands point to the top again, resulting in practically the same situation as for 12 o'clock. Thus, if we know the time between two adjacent overlaps, we can directly calculate all of them.
Approach using series:
Let's start at 12 o'clock = 0. The minute hand takes an hour until it reaches its old position again. However, in the meantime the hour hand has moved by 1/12 of the minute hand speed. The minute hand will thus need an additional 1/12 hour to catch up. Meanwhile, the hour hand moves again by 1/12*1/12, and so on. Some old Greeks may now conclude that there won't be any overlap again at all, but we better trust the guy called Al Gebra here:
1 1 1 1 1 1
1 + -- + -- * -- + -- * -- * -- + ...
12 12 12 12 12 12
= /__ (1/12)^n
= 1 = -- = 1,0909090909.... = approx. 65min 27sec
1 - -- 11
Approach using linear equations:
You might initially try to use something like 1 * x = 1/12 * x to obtain the right times. However, you'll only get the original overlapping position: 0 = 12 o'clock. This is because there is no difference between a minute and the same minute one hour later. So the correct equation would better be (1 * x) mod 1 = (1/12 * x) mod 1 Phew - modulo arithmetic is way harder to do! But since we only need the first overlapping position after 12 o'clock we can get around it (there are many possible variations of this approach):
(1 * x) mod 1 = (1/12 * x) mod 1
<=> (1 * x) mod 1 = (1/12 * x) mod 1 + 1 mod 1
<=> (1 * x) mod 1 = (1/12 * x + 1) mod 1
=> x = 1/12 * x + 1
=> 11/12 x = 1
=> x = 12/11 = 1,0909090909.... = approx. 65min 27sec
Using one of the results above we can now create a formula for the times the hands will overlap. This one was proposed by Stas Myasnikov and returns the times in seconds:
t = 60 * 60 * -- * n
where n = 0, 1, 2, ... , 11
Completely different approach:
It would have been a lot easier if we did not regard the absolute positions, but the relative positions instead! If we take the hour hand as the origin, we see that the hour hand won't move at all, and the relative speed of the minute hand comes out as the difference between both absolute speeds: 12-1 = 11 times faster than the absolute speed of the hour hand.
Relative to the hour hand, the minute hand has to make a complete turn, thus the resulting time is 1 - in relative positions, though!
Translating this back to an absolute position is easy, just multiply it with the relation of the absolute speed and its relative counterpart. This gives us:
x(absolute) = x(relative) * speed(absolute)/speed(relative) = 1 * 12/11 = 12/11
Pretty simple, isn't it?
Additionally, it may also be possible that all three hands (hours, minutes and seconds) overlap. When does this happen?
All three hands overlapping implies that the minute and the hour hand overlap. Since we already know a formula for those times we just have to test whether the seconds hand is at the same time or not. Unfortunately, this only happens at 12 o'clock (in theory). Mario Suvajac was the first to realize this, he also wrote a small Fortran program which you can find in the appendix (see below).
In real-world clocks, the hands do not move completely continuously, though. Thus, depending on the movement granularity of the clock, you might still see all hands overlapping each other at other times as well.
Note: This clock-riddle is pretty close to the relationship between day length, the length of a whole year and the duration of a single turn of the rotating earth.
You might wonder what an analogue clock and the movement of the earth have in common. Well, that's easy: Both are described by rotations with the same axis direction.
If you ask someone for the duration of a single rotation of the earth, most people will claim it is exactly 24 hours. This is true if you take the position of the sun as a reference. However, aren't the stars at night a way better indicator due to their larger distance (resulting in less relative movement)? If you have watched the stars for some time you'll know that not only they're moving like the sun but that they are also moving at a slightly different speed as well, resulting in different stars to be seen at night according to the various seasons.
The explanation is easy: Just like the movement of the hour hand in our quiz decreases the relative speed of the minute hand, the absolute movement of the earth around the sun affects the day length in our sun-relative perception. As we know now, the difference is a whole rotation period, resulting in a rotation time of roughly (365.24 rotations / 365.24+1 rotations)*24 hours = 23 hours 56 minutes. Doesn't sound much, but during the year the difference sums up to a whole day...
Solution submitted by Mario Sujavac (including a Fortran program)
Solution submitted by Stas Myasnikov