Vibration and Oscillation. Synonymous terms for every physicist, although I believe I see some difference; by vibration in particular I understand any alternating or reciprocal motion caused by elasticity alone. Such are the motions of vibrating strings and any resonant body in general, as well as the balance wheels of watches, which vibrate through the elasticity of the spiral springs fitted to them. See Regulator (elastic) .

On the other hand, I understand by oscillation any alternating or reciprocal motion caused exclusively by weight or gravity. Such are the motions of waves and all suspended bodies, whence derives the theory of pendulums. See Center of oscillation and Regulator.

The term center of oscillation is used, but not center of vibration; one measures time and the other sound. Bells, for example, make both vibrations and oscillations; the former derive from the body that strikes the bell and compresses it by virtue of its elasticity, making it oval in alternation and producing sounds; the latter are determined by the motion of the bell, which is subject to gravity.

It remains to be seen whether the sound of a bell carries farther as the rhythm of its oscillation is closer to coinciding with the rhythm of its vibration ; in other words, is the ratio of these rhythms harmonic or aliquot? This is an idea that I am not qualified to examine further. Since this article deals with vibrations in horology, I shall spend less time stating what they are in and of themselves than showing the use horologists make of them in watches and clocks.

We see in the horology article "Friction" how vibrations are to be considered in the planning of wheels and teeth, to achieve a given number of vibrations with the smallest number of revolutions possible. So I shall not repeat here the basic theorem I employed; I shall merely give an example for calculating them, followed by a table with several numbers for various wheel trains that can be used, and the number of vibrations and oscillations resulting from them.

Many treatises on horology have tables of length for simple pendulums, but there are none for the numbers of wheels and teeth corresponding to them, even though they are indispensable; for what use is it to the horologist to know that a given length makes a given number of oscillations, if the number is not a multiple of a certain number of aliquots suitable for use in wheel trains?

Mine is therefore a table of pendulum lengths that includes one for various corresponding wheel trains, a useful reference for the practice of horology; but since time does not allow me to draw it up as I imagine it, I shall simply give a few examples of wheel trains in watches and clocks for the most necessary and frequent cases.

I shall take as a reference the period of one hour, since that is the most familiar and most often used for calculating vibrations ; and to show that the number of vibrations requires more wheels and teeth as the number is larger in a given period of time, I shall give two examples where a single wheel may suffice, but is impractical because of the pendulum length required.

1) A pendulum making just one oscillation an hour would be 39,690,000 pieds long; a single wheel with 12 teeth would make 24 oscillations in 24 hours, for we know that each tooth acts on the pendulum twice. A simple pulley on the arbor of this wheel, hung with a weight as required by the bob, would maintain the motion as long as the weight could drop.

2) A pendulum making only 60 oscillations in an hour would be 11,025 pieds long; a single wheel with 30 teeth would oscillate 60 times per hour, and as in the previous example, a pulley and a weight related to the bob would maintain its motion as long as the weight could drop.

I have given these two examples to show that shortening the pendulum requires increasing the vibrations and consequently the wheels, to maintain them for 24 hours.

We know that a pendulum that beats seconds makes 3,600 oscillations an hour, and its length is 3 pieds 8 51/100 lignes ; now to keep it in motion for 24 hours several wheels are required, for 3,600 oscillations an hour multiplied by 24 gives 86,400 oscillations in 24 hours. This number shows us therefore that several wheels are needed, and if the usual method is followed, the divisors are sought in this way.

We see that almost 100 divisors result, but then the horologist cannot choose among them; nothing guides him as to the number of wheels or the respective numbers of teeth, and it seems almost arbitrary to him. He sees that he can answer the question with an indeterminate number of wheels, so long as the number is taken from the divisors that have been found; but by following my method I find not only the smallest number of wheels that can produce a given number of vibrations , but in addition the number of teeth that meet the goal while not needlessly multiplying the intermediate revolutions, as is likely when following the usual method.

Therefore I consider 86,400 as a power, from which I extract the various roots; first the square, for two wheels; the cube, for three; then the fourth, which would be for four; until I reach a root small enough to be multiplied by the number of pinion leaves with which they are to mesh. From this it follows that these numbers must be changed only when special circumstances require it, for when teeth are removed from one wheel and added to the preceding or following one with the same number of teeth, the number of vibrations must necessarily decrease by the square of the number of teeth removed, even though they are added to the other wheel. I have seen horologists make this error, as well as the one of adding a few teeth to the first and last wheels in order to have a greater or lesser effect on the number of vibrations . This is absolutely immaterial, because the wheels are multiplied by each other, and so the number of vibrations does not change, whatever the order of the multiplied factors. Therefore the only essential point in slightly increasing or decreasing the number of vibrations , without eliminating or adding wheels, is to make the number of teeth unequal in order to decrease the vibrations , and equal to increase them. For example, if we have two wheels and the sum of their teeth is 120, and they mesh with six-leaf pinions to produce as many revolutions as possible in a third wheel (like the fan of a strike train), we divide the sum of their teeth into two equal numbers, 60 teeth for each wheel. Multiplied together, they make 3,600, which divided by the product of the two pinions, 36, gives a quotient of 100 revolutions for the third wheel or fan. Now if four teeth are removed from one wheel and added to the other, we have 56 x 64, or 3,584; divided by the product of the two pinions, 36, this gives a quotient of 99 revolutions for the third wheel to one of the first. This number of revolutions differs from the first product by 4/9, the square of 2/3, because the four teeth I removed from one and added to the other must each be considered as sixths of a revolution, since they mesh with six-leaf pinions. Therefore four teeth are 2/3 of a revolution, which squared is 4/9.

If 17 teeth are removed from one wheel and added to the other, we have 77 x 43, or 3,311; divided by 36, the product of the two pinions, this gives a quotient of 91 35/36 revolutions for the third wheel for one of the first; and this latter number of revolutions differs from the first one, 100, by 8 1/30 of a revolution, the square of the 17 teeth, considered as 17/6 because of the 6-leaf pinions.

Finally, if we remove 59 teeth from one wheel and add them to the other, we have 119 x 1, whose product divided by that of the two 6-leaf pinions gives a quotient of 3 11/36 revolutions of the third wheel for one of the first; this quotient differs from the first, 100, by 96 25/36 revolutions, the square root of which is 59/6.

Plainly, the revolutions decrease as teeth are removed from one wheel even though they are added to the other. Therefore this question may be stated: if teeth are removed from one wheel, how many must be added to the other to keep the same number of revolutions? The question would soon be answered if there were fractions of teeth as there are fractions of revolutions in the examples above. If we do the operation we find

[lines to left of table:]
in the first case
in the second case
in the third case

The advantage of this method in knowing the result produced by the inequality given to the factor seems to me so useful to horology, where almost all motions act through multiplication and division in levers acting on each other, that I have resolved to give a further example with two small numbers.

There is one further comment to be made on wheel trains. Whenever possible, numbers must be used that are multiples of the pinion leaves they mesh with; this has the advantage that the same teeth always act on the same leaves, and when testing a train, a single turn of the wheel suffices. When the pinions do not divide their wheels exactly, the same teeth meet the same leaves only after a certain number of revolutions, which raises a question that is not at all difficult in and of itself but may be unknown to many, and since it is often necessary to mesh wheels of different numbers to produce an action with a certain part or a certain number of revolutions, the question is simply to show when the same teeth meet the same leaves.

If two wheels with the same number of teeth mesh together, however many revolutions they make, the same teeth will meet at every revolution, as is easily understood. But if one wheel has one additional tooth, the revolutions of one wheel will not equal those of the other; there will be one tooth fewer after the first revolution, two after the second, and so on until the revolutions of the first wheel equal the teeth of the second. For example, if we have two wheels, of 31 and 17 respectively, if 31 is driving 17 the same teeth will meet on the seventeenth revolution of the first wheel; if on the contrary the 17 wheel is driving the 31, they will meet on the thirty-first revolution of the first wheel. In short, the same teeth meet taking in alternation the number of teeth on one wheel and the number of revolutions of the other.

Finally, to discharge my obligations, it remains for me to give a series of trains all of which are composed to produce a given number of vibrations and oscillations.

Seconds

seconds wheel driven by the escape wheel pinion

Watch with two balance wheels, M. de la Roche's escapement

Seconds watch in a ring

Thirty-six hour watch beating seconds

Half-seconds thirty-two-hour watch

Center half-seconds eight-day watch

Center seconds eight-day watch

Center seconds one-month watch

Six-month watch beating seconds

Eccentric seconds one-year watch beating seconds

Train for a seconds clock to be wound monthly

Another thirty-day clock

Seconds clock to be wound weekly

Another, eight days and more

Eight-day clock beating half seconds, with a fusee as in a watch, a very accurate clock though the pendulum is short

Spring-driven one-month clock

Spring-driven two-week clock

Eight-day clock

In spring-driven clocks, where the pendulum is made as long as the case will allow, the train numbers vary little; only with the escape wheel is the number of teeth decreased when the pendulum increases in length, and vice versa. Thus without appreciable error one may take a ratchet or escape

This article is by M. Romilly, Horologist.