Wednesday, December 13, 2006

I haven't lost it...

Ben (a regular commenter on this blog) observed when we went round to his for lunch at the weekend that the shortest day of the year (21 December?) is not the same as the day with the latest dawn or earliest sunset (which was a few days ago, apparently).

At work today, somebody linked to an explanation. I didn't get to read it because it started off with an analogy about clocks that made me realise I hadn't done any mathematics for ages.

The example invited the reader to consider a regular analogue clock - one with a big hand and a small hand. The question was: assuming it starts at 12noon, at what time will the hands next line up exactly?

It's clear from inspection that it'll be somewhere shortly after five past one... but when, precisely?

I decided to find out.

After a bit of fiddling about, here is the solution I came up with. What I'm interested in knowing is: is there an easier way?

  • Let t be the time since noon in seconds
  • Let b(t) be how far the big hand has rotated at time t
  • Let s(t) be how far the small hand has rotated at time t

We can see the following:

b(t)=2.pi.t/3600

s(t)= 2.pi.t/(3600*12)

So, when the hands are aligned, they must be at precisely the same angle.

Thus, at alignment, s(b)=b(t)+2.pi.n, where n is an integer.

To see why we need the 2.pi.n term, consider that the big hand may perform several full circles before it aligns with the small hand. Therefore, we have to consider the case where the alignment occurs without any "extra" loops, when the big hand has looped once, when it has looped twice, and so on.

Thus, cancelling where possible, we get

t/3600 = t/(3600*12) + n

Rearranging, we get

t = 12*3600*n/11

For n=0 we get the trivial result.

For n=1 we get t = 3927 3/11, which means the answer to our question is: five past 1, 27 and 3/11 seconds.

So there we have it.

(Actually, now I come to think about it, there are eleven alignments in any given hour 12-hour period, so an appeal to symmetry would yield the same result... i.e. an alignment every 12*3600/11 = 3927 3/11 seconds. Sigh...)



[UPDATE 2006-12-15 15:58. Thanks to Lee for spotting the error... there are not 11 alignments per hour; there are 11 alignments per 12-hour period.]


[UPDATE 2006-12-15 16:01. Talk about pedantic... apparently it is necessary to multiple the numbers by 12 as well as make it clear in the text. Good grief!]

2 comments:

Ben said...

Presumably this is the web page you mentioned. You might also like this multimedia explanation (recently featured on Good Math, Bad Math).

Richard Brown said...

Thanks Ben