684. We shall explain by the model shown in Fig. 101, how the hour-hand
and the minute-hand are made to revolve with different velocities about
the same dial.
[Illustration: FIG. 101.]
G is a handle by which I can turn round the shaft which carries the
wheel F, and the hand B. There are 20 teeth in F, and it gears into
another wheel, E, containing 80 teeth; the shaft which is turned by
E carries a third wheel D, containing 25 teeth, and D works with a
fourth C, containing 75 teeth, C is capable of turning freely round
the shaft, so that the motion of the shaft does not affect it, except
through the intervention of the wheels E, F, and D. To C another hand
A is attached, which therefore turns round simultaneously with C. Let
us compare the motion of the two hands A and B. We suppose that the
handle G is turned twelve times; then, of course, the hand B, since
it is on the shaft, will turn twelve times. The wheel F also turns
twelve times, but E has four times the number of teeth that A has, and
therefore, when F has gone round four times, E will only have gone
round once: hence, when F has revolved twelve times, E will have gone
round three times. D turns with E, and therefore the twelve revolutions
of the handle will have turned D round three times, but since C has 75
teeth and D 25 teeth, C will have only made one revolution, while D
has made three revolutions; hence the hand A will have made only one
revolution, while the hand B has made twelve revolutions.
We have already seen (Art. 681) how, by a train of wheels, one wheel
can be made to revolve once in an hour. If that wheel be upon the shaft
instead of the handle G, the hand B will be the minute-hand of the
clock, and the hand A the hour-hand.
685. The adjustment of the numbers of teeth is important, and the
choice of wheels which would answer is limited. For since the shafts
are parallel, the distance between the centres of F and E must equal
that between the centres of C and of D. But it is evident that the
distance from the centre of F to the centre of E is equal to the sum
of the radii of the wheels F and E. Hence the sum of the radii of the
wheels F and E must be equal to the sum of the radii of C and D. But
the circumferences of circles are proportional to their radii, and
hence the sum of the circumferences of F and E must equal that of C and
D; it follows that the sum of the teeth in E and F must be equal to the
sum of the teeth in C and D. In the present case each of these sums is
one hundred.
686. Other arrangements of wheels might have been devised, which would
give the required motion; for example, if F were 20, as before, and
E 240, and if C and D were each equal to 130, the sum of the teeth
in each pair would be 260. E would only turn once for every twelve
revolutions of F, and C and D would turn with the same velocity as E;
hence the motion of the hand A would be one-twelfth that of B. This
plan requires larger wheels than the train already proposed.