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Three Hands on a Clock


Date: 07/03/2001 at 10:01:57
From: Lucas
Subject: A question from trig principle

At what time after 12:00 are the three hands on the clock on top of 
each other? (At 12:00, the hour hand, the minute hand, and the second 
hand are all pointing to 12). 

I need to know all the way to a fraction of second. Help!


Date: 07/03/2001 at 14:39:34
From: Doctor Rob
Subject: Re: A question from trig principle

Thanks for writing to Ask Dr. Math, Lucas.

The minute hand moves 12 times as fast as the hour hand, and the
second hand moves 60 times as fast as the minute hand. That leads to 
the formulas for the angle traversed by the three hands after t hours 
time:

   Hour hand:  30*t degrees
   Minute hand:  360*t degrees
   Second hand:  21600*t degrees

For the hour hand and the minute hand to coincide, it must be that, 
for some integer n, 

   30*t = 360*t - 360*n.  

n is the number of whole revolutions the minute hand makes in t hours.  
It is the greatest integer less than or equal to t. Solving for t, one 
gets

   t = 12*n/11,

for some integer n. 

For the minute and second hand to coincide, it must be that, for some 
integer m,

   360*t - 360*n = 21600*t - 360*m.  

m is the number of whole revolutions the second hand makes in t hours.  
It is the greatest integer less than or equal to 60*t. Solving for t, 
one gets

   t = (m-n)/59,

for some integer m. That means that when all three hands coincide,
there must be integers m and n such that

   12*n/11 = (m-n)/59,
   12*59*n = 11*(m-n).

Now since 11 is a prime number and isn't a divisor of either 12 or 59, 
it must be that 11 is a divisor of n, so n = 11*k for some integer k.

I leave the rest to you.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/20/2001 at 11:43:35
From: David Helinek
Subject: 3-Handed Clock Crossings

Given an analog clock with three very thin moving hands (hour, minute
and second), how many times during the day will they line up exactly
(to within observable capabilities) other than the trivial 12 o'clock 
case?

It's been a long time since high school algebra, so I worked out the
answer using brute force, but I know there must a more formal yet
elegant proof of the answer.  

The answer I came up with was that there are no other times during the
day when this happens. One can think of the obvious time,
approximately 6 minutes after 1 am, 12 minutes after 2 am, etc., but
in each case, the second hand is nowhere near the minute and hour
hands during this overlap.


Date: 08/20/2001 at 13:32:19
From: Doctor Rob
Subject: Re: 3-Handed Clock Crossings

Thanks for writing to Ask Dr. Math, David.

Let the angle which the three hands make with vertical at time t in
hours be h, m, and s, all in degrees.  Then

   h = 30*t
   m = 360*t
   s = 21600*t

(Why?) Then if, after a time t hours, they all point in the same
direction, you have that

   m - h = 360*n
   s - h = 360*m

for some integers n and m. (Why?)

Putting these together, you get

   360*t - 30*t = 330*t = 360*n
   21600*t - 30*t = 21570*t = 360*m
   11*t = 12*n
   719*t = 12*m
   12*n/11 = t = 12*m/719
   719*n = 11*m

Now 11 and 719 have no common factor, so that n must be a multiple
of 11, say n = 11*x, for some integer x. Then m = 719*x, and t = 12*x. 
That shows that the only time when this happens is after an integer
multiple of 12 hours, that is, at 12 o'clock.

(The equation 11*t = 12*n gives you the times when the hour and
minute hand coincide, t = (12/11)*n hours after 12 o'clock.)

Is that what you had in mind?

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/20/2001 at 14:02:31
From: David Helinek
Subject: Re: 3-Handed Clock Crossings

Yes, although I though you would go about it locking at the periodicity 
and ratios of sine wave, as each hand swipes out integer numbers of 
cycles throughout the day.

Yours is a more numerical approach, which works well. Do we know
immediately that 11 and 719 have no common factor because 11 is prime,
and 719 is not a multiple of 11? I'm assuming so. It would be
interesting to compute the possible crossings for a 24-hour clock.

Thanks for the refresher.

David Helinek


Date: 08/20/2001 at 15:05:59
From: Doctor Rob
Subject: Re: 3-Handed Clock Crossings

Thanks for writing back, David.

11 and 719 have no common factor because both 11 and 719 are prime
(and so, in particular, 719 = 11*65 + 4 is not a multiple of 11).

To modify the above argument for a 24-hour clock, replace the formula
for h by h = 15*t, and proceed as before. This has the effect of
replacing 11 by 23, 12 by 24, and 719 by 1439. Both of these are 
primes, so the answer remains the same: only at midnight do the
hands coincide.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/20/2001 at 15:54:38
From: David Helinek
Subject: Re: 3-Handed Clock Crossings

What's funny to me is that everyone I've ever asked this question of 
has always answered that there are many times during the day that all 
three hands line up, without even thinking too hard about it. People
intuitively think that somehow clocks are magically 'resonant'.

Again, thanks for the answer to a somewhat interesting question.

David H.


Date: 08/21/2001 at 09:11:58
From: Doctor Rob
Subject: Re: 3-Handed Clock Crossings

One could ask for the closest bunching not at 12 o'clock. I find that 
this occurs at about 5:27:27.3, when all the hands are within a
1.0014 degree sector. I believe that this is distinguishable to the
naked eye, being 1/3 of the distance between two consecutive second
marks around the clock (and so not "within observable capabilities"),
but quite close, nonetheless.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
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