which might be considered as a perpetual motion, though it was not so. But this requires some explanation. The ingenious author of this clock employed the variations in the state of the atmosphere, for winding up his moving weight: various artifices might be devised for this purpose; but this is no more a perpetual motion, than if the flux and reflux of the sea were employed to keep the machine continually going; for this principle of motion is exterior to the machine, and forms no part of it.

But enough has been said on this chimera of me chanics. We sincerely hope that none of our readers will ever lose themselves in the idiculous and unfortunate labyrinth of such a research.

To conclude, it is false that any reward has been promised by the European powers to the person who shall discover the perpetual motion; and the case is the same in regaid to the quadrature of the circle. It is this idea, no doubt, that excites so many to attempt the solution of these problems; and it is proper they should be undeceived.

PROBLEM LIII.

To determine the height of the arched ceiling of a church, by the vibrations of the lamps suspended from it.

FOR this invention we are indebted, it is 'said, to Gallileo, who first ascertained the ratio of the duration of the oscillations made by pendulums of different lengths*. But in order that this method

* Indeed, it seems it was by that author accidentally observing the uniformity in the intervals of the swing of the

may have a certain degree of exactness, the weight of the lamp ought to be several times greater than that of the cord by which it is supported.

This being supposed, put the lamp in motion by removing it a very little from its perpendicular direction, or carefully observe that communicated to it by the air, which is very common; and with a stop-watch find how many seconds one vibration continues, or, if a stop-watch is not at hand, count the number of vibrations performed in a certain number of minutes: the greater the number of minutes, the more exact will the duration of each vibration be determined; for nothing will then be necessary, but to divide those minutes by the number of vibrations, and the quotient will be the duration of each in minutes or seconds.

We shall here suppose that it has been found, by either of these methods, that the time of each vibration is 5 seconds; square 5, which is 30, and multiply by it 39 inches, the length of a pendulum that swings seconds in the latitude of London, the product will be 98 ft. 7 inc. 6 lin., which will be nearly the height from the point of suspension to the bottom or rather centre of the lamp.

If the distance from the bottom of the lamp to the pavement be then measured, which may be done by means of a stick, and added to the former result, the sum will give the height of the arch above the pavement.

suspended lamps, that he first took the hint of employing the oscillations of pendulous bodies, or pendulums, for the purpose of measuring time. And hence the invention of pendulum clocks.

This solution is founded on a property of pendulums, demonstrated in mechanics; which is, that the squares of the times of the vibrations are as the lengths; so that a pendulum four times the length of another, performs vibrations which last twice as long.

But on account of the irregular form of the lamp, and the weight of the rope, which sustains it, we must confess that this method is rather cu rious than exact. We shall however present the reader with another problem of the same kind.

PROBLEM LIV.

To measure the depth of a well, by the time elapsed between the commencement of the fall of a heavy body, and that when the sound of its fall is conveyed to the ear.

HAVE in readiness a small pendulum that swings half seconds, that is to say, 9 inches in length, between the centre of the ball, and the point of suspension. You must also employ a weight of some substance as heavy as possible, such for example as lead; as a common stone or pebble experiences a considerable retardation in falling, and therefore would not answer the purpose so well.

Let go the weight and the ball of the pendulum at the same moment of time, and count the number of the vibrations the latter makes, till the moment when you hear the sound. We shall here suppose that there were ten vibrations, which make 5 seconds.

As a heavy body near the earth's surface falls about 16 feet in one second of time, or for this

T

purpose 16 feet will be exact enough; and as sound moves at the rate of 1142 feet per second; multiply together 1142, 16 and 5, which will give 91360, and to 4 times this product, or 365440 add the square of 1142, which is 1304164, and the sum will be 1669604; and if from the square root of the last number 1292 the number 1142 be subtracted, the remainder 150 divided by 32 will give 469 for the number of seconds which elapsed during the fall of the body: if this remainder be subracted from 5, the number of seconds during which the body was falling and the sound returning, we shall have 0 31 for the time which the sound alone employed before it reached the ear; and this number multiplied by 1142, will give for product 354 feet the depth of the well.

This rule, which we must allow to be rather complex, is founded on the property of falling bodies, which are accelerated in the ratio of the times, so that the spaces passed over increase as the squares of the times *. of the times *. But as the resistance of

a

*For the sake of our algebraical readers we shall here shew how to find the formula from which the above rule is deduced: Let a = 5, b = 1671, c = 1142, and let > be the time which the body employs in falling, consequently x will be the time of the sound returning. Then as 12: b : : x2 : b x2 = depth of the well; and, I c:: a -x: ca - cx = depth of the well also; therefore b2 cx, and by transposition and division, x2 +

ca

Completing the square, x2 +-+ = + =

462

462

Hence, x + √4bea+2 and x = √4bca +2

26

462

462

x=

ca

==

4bca +c

462

26

the time of descent. Con

the air, which in considerable heights, such as those of several hundred feet, does not fail to retard the fall in a sensible manner, has been neglected, the case of this problem is nearly the same as with the preceding; that is to say, the solution is rather curious than useful.

is obtained for the time of the descent, which is as follows: Multiply 1142 by 5, which gives for product 5710; then multiply also 16 by 5, which gives 80, to which add 1142, this gives 1222, by which sum divide the first product 5710, and the quotient 4'68 will be the time of descent, nearly the same as before. This taken from 5 leaves 0.32 for the time of the ascent; which multiplied by 1142, gives 536 for the depth, differing but little from the former more exact number.