Crossing Hands
February 25, 2014
Since the minute hand spins twelve times while the hour hand spins once, the hands cross eleven times. There are 60 × 60 × 12 = 43200 seconds in twelve hours, so the hands cross every 43200 / 11 = 3927 3/11 seconds, or 1 hour, 5 minutes, and 27 3/11 seconds. We write a function that converts a number of seconds to hours, minutes and seconds:
(define (sec->hms sec)
(let* ((h (quotient (floor sec) 3600))
(m (quotient (- (floor sec) (* h 3600)) 60))
(s (- sec (* h 3600) (* m 60))))
(list h m (round s))))
Then we call the function eleven times:
> (for-each
(lambda (sec) (display (sec->hms sec)) (newline))
(map (lambda (s) (* s 43200 (/ 11))) (range 11)))
(0 0 0)
(1 5 27)
(2 10 55)
(3 16 22)
(4 21 49)
(5 27 16)
(6 32 44)
(7 38 11)
(8 43 38)
(9 49 5)
(10 54 33)
Notice that zero seconds is at zero hours, not twelve hours; if sec is zero, no time has elapsed. We used range from the Standard Prelude. You can run the program at http://programmingpraxis.codepad.org/qd2PCeys.
Programming Praxis 
Here’s a piece of code that does this.
#!/usr/bin/perl -w use strict; use warnings; my $hour = 0; my $minute = 0; my $second = 0; while($hour <= 11 && $minute <= 59 && $second <= 59) { my $h_pointer = ($hour*3600+$minute*60+$second); my $m_pointer = int(43200*($minute*60+$second)/3600); if($m_pointer >= $h_pointer) { print "$hour hours $minute minutes $second seconds\n"; $hour += (1+($minute=0)+($second=0)); next; } $second++; $minute += (1+($second=0)) if $second == 60; $hour += (1+($minute=0)) if $minute == 60; }Here’s the output:
The idea here was to divide the clock into 43200 positions (the number of seconds in 12 hours), then calculate at each time the position where the hour and minute hands are and finally being able to detect when the minute hand comes ahead of the hour hand.
Note that the two hands are hardly ever at the same position, so what I tried to do was to detect when the minute hand passes from being behind the hour hand to being ahead of it.
As an example, in the output shown above, at 01:05:27 the minute hand was slightly behind the hour hand while at 01:05:28 it was slightly ahead.
Cheers!
Problem with your problem and solution. Many analogue clocks jump the hands by the minute not by the second, even if they have a second hand. So the solutions include valid answers that only have movements by the minute. This gives and interesting result at 9:48 and 9:49, depending on the mechanism used for moving the clock hands. It is possible for the hands to cross at both times.
Hi Hill, would you elaborate on the 9:48 case? I don’t understand what you mean by the mechanisms used for moving the clock hands.
On the basis of mechanically moving by the minute, during the ninth hour, the hands are at the following positions
Time: 9:00 9:12 and later 9:24 and later 9:36 and later
Hour hand: Major mark 9 Minor mark 46 Minor mark 47 Minor Mark 48
Minute hand: Major mark 12 and .. minor mark 12 and … Minor mark 24 and … Minor mark 36 and …
Now when the minute hand reaches minor mark 48, the hour hand is on 48 for a fraction of second before travelling to minor mark 49, on which it will rest until hour hand moves to major mark 10 and minute hand is on Major mark twelve. Hence there is a crossover of the hands at this point in time, however briefly.
Depending on the way the mechanism works, there is a small overlap due to (generally) the minute hand operation preceding the hour hand operation. The internals of most mechanical “analogue” clocks are relatively cheap so the easy way is taken out and movement of hour/minute hands is based on minutes not seconds as this would require greater accuracy and more complexity in the internal workings of the clock. This would increase the expense of the construction of the clock mechanism.
The best way to see this is to actually watch the movements of the hands as the second hand moves around the clock face.
The problem describes above is fine when modelled ideally, that is everything is purely analog and there are no steps within the mechanism. But when looking at the real world, models do have their limits. An example that has just come to mind is the mobius strip. Any real world example, where the thickness is very much less than the width actually has two faces, the main face and the edge face. To get an real world example that only has one face requires the width of the strip to be equal to the thickness of the strip and requires an extra quarter turn to get the single face. Try showing smart children an example and someone will ask about the edge, I know, I did.
If you see a problem in my description above, please feel free to point out what it is. After all, I am only human.
In the previous reply, the additional spaces were removed so add additional spaces after the words “and later” and after “and …” to make sense of the three lines
Here’s some Racket:
(define f (curry ~a #:min-width 2 #:align ‘right #:pad-string “0”))
(for* ([h (in-range 12)]
[m (in-range 60)]
#:when (= m (+ (* h 5) (quotient m 12))))
(printf “~a:~a\n” (f h) (f m)))
Technically, it returns 11:59 as well, but for 30 seconds of code (most of which was looking up the particulars of
~a), that’s not half bad.Ok Hill, I understand what you’re saying and it is pretty interesting, actually, from a mechanical point of view.
I do not think however that it was the purpose of the question to go mechanical about it, but I guess that’s for the poster to tell.
By the way, not sure how my name shows up for everybody but, just to clarify, “sabapcblog” and “Juan Ignacio Saba” are the same person (me). It’s just that I changed my display name at some point but the older posts are not dynamic with regards to that (too bad).