Calculating numbers in mechanics and horology means calculating the numbers [of teeth] on the wheels and pinions of a machine, so that they may make a given number of revolutions in a given time. That can only be achieved by moderating the velocity of the wheels with a pendulum or balance wheel with isochronous vibrations. See Pendulum, and Clockmaking Plates, Plate I, Figures 2 and 3 , showing a clock movement. The contrate wheel is d , the crown wheel c, and the great wheel is b ; it must make one revolution in an hour. It is driven by wheel A , backed by a pulley that weight G turns by pulling downward; this wheel engages a pinion set on the same arbor as wheel b , which must make one revolution in an hour. This wheel meshes similarly with the pinion on the arbor of crown wheel c ; the latter meshes with the pinion of contrate wheel d , whose velocity is moderated by the vibrations of the pendulum, which lets only one tooth of the contrate wheel go by on each vibration of the pendulum. But since in a whole revolution each tooth of the contrate wheel encounters the pendulum pallets twice, it follows that the number of vibrations during one revolution of the contrate wheel is twice that of the teeth on the wheel. So if each vibration of the pendulum lasts one second, and the contrate wheel has 15 teeth, the period of its revolution will be 30", or one-half minute. If we assume that pinion x of contrate wheel d has six leaves or teeth, and that the crown wheel driving it has 24, since the pinion leaves go through the wheel teeth only one at a time, plainly while crown wheel c makes one revolution, pinion x must make four, since its 6 leaves are contained 4 times in the 24 teeth of the wheel. Now we have seen that the contrate wheel, and consequently pinion x , set on the same arbor, takes 30" to make a revolution; consequently, crown wheel c must take four times longer to make a complete revolution: 3"x 4 = 120" = 2'; thus the period of its revolution is two minutes.

Now if we assume that pinion y on the crown wheel arbor has six leaves, and that long-arbor wheel b has 60 teeth, the pinion must make ten revolutions before wheel b has made one; but the pinion on the arbor of crown wheel c takes as long as the wheel to make one revolution, and the time is 2'; wheel b will therefore take 10 times longer, that is 20' or 1,200" or vibrations of the pendulum. So we see that the time it takes for one revolution is only a third of 3,600" or an hour, which it was supposed to take. The assumed numbers are thus lower than the real ones since they do not satisfy the problem posed; thus we feel the necessity of a reliable method for finding suitable numbers.

First we must know the number of the pendulum vibrations to be had during the time any of the wheels is to make one revolution. See Pendulum for the method of determining the number of vibrations, by this rule: the square of the number, in a given time, is inversely proportional to the length of the pendulum. Divide the number by two, and you will have the product of all the exponents; we call exponents the numbers showing how many times the teeth of a wheel contain the teeth of the pinion it meshes with. So if a wheel has sixty teeth and meshes with a pinion of six, the exponent will be 10, showing that the pinion must make ten revolutions for one of the wheel. The pinions are written above the wheels and the exponents between them, thus:

When there are several pinions and wheels, they are written one after the other, separating the exponents by x ( times ); one side represents the arbor with a pinion and a wheel as a single part and revolving together. Example:

1, 2, 15, 6, 5, 7 1/2, are the exponents or quotients of the wheels divided by their pinions. 7, 7, 8 are the pinions. 15, 42, 35, 60 are the wheels meshing with the pinions written above. The Xs, as we have said, show that pinion 7 and wheel 15 are on the same arbor, like the second pinion 7 and wheel 42; similarly pinion 8 is on the arbor of wheel 35.

Theorem: The product of the exponents doubled equals the number of vibrations of the pendulum during one revolution of wheel b .

Demonstration: The contrate wheel of 15, as explained above, lets only one tooth past at each vibration of the pendulum, but since each tooth passes under the pallets of the pendulum twice, the number of vibrations during one revolution of the contrate wheel is double the number of teeth on the wheel, so 30 vibrations, or 2 x 15, must be counted; but pinion 7 on the arbor of the contrate wheel makes its revolution in the same time as the wheel makes its own, and it must make six revolutions for the wheel of 42 to make one; the number of vibrations of the latter wheel of 42 will thus be six times the number for pinion 7, which takes 2 x 15 to make its revolution; thus wheel 42 will take 2 x 15 x 6 vibrations to make a complete revolution. The second pinion of 7, on the arbor of this wheel, will take the same time as the wheel to make a revolution, but five revolutions of this pinion are required for one of the wheel of 35; so the number of vibrations during a revolution of the latter wheel will be (2 x 15 x 6) x 5 vibrations; pinion 8 will take the same time, and wheel 60 7 1/2 times longer, since pinion 8 must make 7 1/2 revolutions for wheel 60 to make one; thus the number of vibrations during one revolution of the latter wheel will be (2 x 15 x 6 x 5) x 7 1/2, the product of all the exponents multiplied by 2. Q.E.D.

In a wheel train the smallest pinions are usually located towards the escapement, and the heaviest towards the power; similarly for the wheels with the largest numbers of teeth, which puts the largest exponents towards the escapement. So in the preceding example, wheels 35 and 42 should trade places so that the exponents may decrease from A towards B thus:

which makes a train that can be used to good advantage for all its parts. The number of vibrations or product of the exponents is placed at the end with the sign =, thus:

which expresses the number of vibrations during a complete revolution of the last wheel, of 63.

So when one proposes to make a wheel train, the number of vibrations of the pendulum must be known for the time a wheel is to take for its revolution. Let us assume the time is an hour and that the pendulum beats seconds, that is, each vibration takes one second; one hour contains 3,600, so during one revolution of the wheel that is to make a revolution in an hour, the pendulum will make 3,600 vibrations, and this number, 3,600, is double the product of all the exponents, 2 x r x s x t , of the wheels and pinions that must be determined. Divide 3,600 by 2, the result is 1,800, the product of three unknown quantities, r, s, t ; we know they must decrease from r to t , and that exponent r representing the ratchet or contrate wheel may be double or triple exponent s , which must only exceed the third, t , by one unit at the most.

To find these three unknowns, assume a value for the first, r , a convenient number for a ratchet, and always an odd number for a contrate wheel. Assuming r = 30, it is easily derived from the equation 1,800 = r s t , and we have for the value of s t = 1,800 / 30 = 60. Now, since s and t are equal, or almost equal, assuming t = s , we have the equation s s = 60; thus we must extract the square root of 60. Since it is not exact, we take as exponent the nearest square root above or below, and we divide the product t = 60 by this root, and the quotient is the other exponent; the larger is placed first. Thus in the example 64 is the nearest square to 60, its root is 8; divide 60 by 8, and the other exponent is 7 4/8.

They are all arranged in this manner:

Observe: 1) When the exponent is a fraction, the pinion must always be the denominator of the fraction or a multiple of the denominator, if it is too small to be a pinion. 2) If there were three exponents, s t u , not counting the ratchet or contrate wheel, one must extract the cube root of their product; this cube root or that of the nearest cube will be one of the exponents.