The equation of the watch
I have a round analogue watch. The only problem with it is that the
hour hand is almost as long as the minute hand. Many a time when looking
at it I have some doubts as to which of them is which. It is not always
so. Sometimes the situation is quite clear. For example, at five o'clock
sharp no error is possible: the only sensible hour indicated by the hands
with interchanged meaning (i.e. with the hour hand indicating minutes and
viceversa) would be twentyfive past twelve. But then, assuming that the
lower hand is pointing exactly at five, the upper would be expected to
point somewhere near the midpoint between twelve and one. If, in turn,
the real time were very close to half past eleven, one might wonder whether
we are not at twoandahalf minutes to six.
How many of such misleading handle positions can there be? Let's formulate
the problem as follows: determine, for an analogue watch with indistinguishable
hands, how many times in twelve hours can there be uncertainty about the
time indicated by the hands.
We shall intend to tackle the problem with a very interesting tool which
is de Moivre's formula
It comes in very handy for describing the position of hands. Let's place
the watch within the complex plane in such a way that the origin of the
system coincides with the watch axis and the ends of both hands lie on
the circumference of the unit circle.
We may also assume, to remain in agreement with the "point zero"
(i.e. with the position of twelve o'clock) and the movement of the hands
in traditional watches, that the real axis is directed upwards and the
imaginary axis rightwards. In our system twelve o'clock will be hour one.
Now, let h be the complex number in the unit circle which corresponds
to the position of the hour hand, whereas m will stand for the position
of the minute hand. The minute hand turns around the center of the watch
twelve times quicker than the hour hand, so the angle it covers is twelve
times bigger than that covered by the hour hand. This leads us to the equation
of the watch:
since according to de Moivre's formula multiplication of an angle corresponds
to raising to the appropriate power.
Let's get back to our problem. The case which may arouse doubts as to
the time indicated by the hands corresponds to the simultaneous fulfillment
of two watch equations:
Substituting in the second we get h=h^{144} and after
dividing by h (not equal to zero, since it lies on the unit circumference)
we eventually have h^{143} =1. Thus the solutions of this
equation are all the 143 roots of the unity. i.e. the vertices of a regular
143gon with one vertex at 1 (at twelve o'clock). This means that every
12/143 hours we have a situation of uncertainty.
In principle, we should discard all the cases when the two hands coincide
(it is rather hard to make a big mistake then!). How many of them can there
be? We can deduce that with our magic tool, too. The hands coincide when
the hour hand assumes the same position as the minute hand, which means
h=h^{12}. After dividing by h we get h^{11}
=1; now the solution is provided by all the eleven roots of the unity of
order 11.
Eventually, the really misleading situations only occur 143  11 = 132
times in twelve hours.
Incidentally, I have another watch, too, which has perfectly distinguishable
hands. However, when I wake up at night and look at it I can't tell whether
the watch is lying upside down or not. Is there any chance of telling one
position from the other just by watching the face of the watch?
A rotation of 180 degrees corresponds on the complex plane to the multiplication
of a number by 1. If any doubts as to the orientation of the watch arise,
the reason must be the simultaneous satisfaction of the following two equations:
since in both cases  normal and upside down  we can read a correct
time. This pair of equations, however, does not have any solutions on the
unit circumference, so only one of them can be true. This means that with
a sight good enough it is impossible to commit this kind of error. I recommend
to all of you who have a watch with hands to look at it upside down. I
can assure you that you will not see anything that can be reasonably interpreted
as some correct time.
Surprisingly, an Indian acquaintance of mine, Raghu, when we discussed
this observation, told me that there is a method well known in India to
establish quickly the actual time in the capital of the British Empire.
You have to look at the watch upside down and move in your mind the hour
watch to the nearest sensible position. For instance, if it's 10:25 in
Bombay, you rotate the watch by 180 degrees and see that the minute hand
points at "5 to something", whereas the hour hand lies more or
less midway between 4 and 5. So, in your mind you push the latter forwards
and now you can guess that the time in London is five to five.  Why forwards
and not backwards?  asked I.  If you can move it either forwards or backwards
to get a reasonable time, we prefer to push it forwards  answered Raghu
and we set to work on the verification of this phenomenon.
The time difference between Bombay and London is 5.5 hours. (In India,
which covers an area ranging from longitude 67 to 97 degrees east, the
official time during the whole year is that of the line of longitude 82
degrees; roughly speaking, the line divides the country in two equal parts).
Let h_{L} and m_{L} represent the positions
of the hands in London, h_{B} and m_{B} 
the positions of the hands in Bombay. We shall prove that whenever in Bombay
the watch is turned upside down and the hour hand is moved forwards by
exactly half an hour (which corresponds to a multiplication by exp{\pi
i/12}), the resulting time is that of Greenwich. To get that effect, the
following two equations must be satisfied:
and
and simultaneously the watch equation m_{B}=h_{B}^{12
}must hold true for Bombay. Let's see whether in such a case this
last equation will be satisfied for London too. Applying Euler's formula
we get
Terrific! We got it! This just means that proceeding in Bombay as it
has been described above we always get a correct time. Moreover, taking
into account that in our algorithm the hour hand has been moved backwards
by exactly 5.5 hours (6 hours backwards when the watch was rotated and
then half an hour forwards) we always get the time at Greenwich. It should
be clear now what should be done in London to get the time in India!
Let's quickly solve a few more little problems. First, looking at the
watch in a mirror we will always see a sensible hour (assuming, of course,
that the face carries no numbers). The reason for this nice phenomenon
is that the mirror reflection is just a symmetry with respect to the real
axis, which corresponds to transforming a complex number z into its conjugate
z*. Since h^{12}=m if and only if (h*)^{12}=m*,
the equation of the watch will be satisfied by the "conjugate"
face, too. Just take a look in the mirror to check it.
Next, assuming that the second hand coincides with the other two at
twelve o'clock, we can deduce that a similar situation will only occur
after twelve hours. To see this just observe that if s represents the position
of the second hand, then the equation of the watch can be extended by a
new condition: s = m^{60}. But then it is easily checked
that the only solution of the system of three equations: h = m = s,
m = h^{ 12}, s = m^{60} is h=m=s=1, i.e.
twelve o'clock.
In the epoch of digital watches it seems worthwhile to make a sentimental
journey back to those good old times when time was being measured in a
continuous way. It may be true that this continuity was not as continuous
as it should be. In fact, the hands were pushed forwards in tiny "quanta"
of motion, be it following the rythm of falling sand grains or that of
the pendulum or finally that of the mainspring. Nevertheless, to our imperfect
eyes it was an ideal approximation of continuity. Besides, there used to
be some mechanisms which indeed had very much in common with continuity,
like burning candles or flowing water. Flashing digits split the time needlessly.
Piotr Chrzastowski
