_A._--The weight of a malleable iron rim of one square inch sectional area and 7 feet diameter is 21.991 feet x 3.4 lbs. = 74.76, one half of which is 37.4 lbs. Then by the same process as before, 8,000/37.4 = 213.9, the centrifugal force in terms of the weight: 213.9 x 7, the diameter of the wheel = 1497.3, the square root of which, 38.3 x 4.01 = 155.187 feet per second, the highest velocity of the rims of railway carriage wheels that is consistent with safety. 155.187 feet per second is equivalent to 105.8 miles an hour. As 4,000 lbs. per square inch of sectional area is the utmost strain to which iron should be exposed in machinery, railway wheels can scarcely be considered safe at speed even considerably under 100 miles an hour, unless so constructed that the centrifugal force of the rim will be counteracted, to a material extent, by the centripetal action of the arms. Hooped wheels are very unsafe, unless the hoops are, by some process or other, firmly attached to the arms. It is of no use to increase the dimensions of the rim of a wheel with the view of giving increased strength to counteract the centrifugal force, as every increase in the weight of the rim will increase the centrifugal force in the same proportion.

CENTRES OF GRAVITY, GYRATION, AND OSCILLATION.

31. _Q._--What do you understand by the centre of gravity of a body?

_A._--That point within it, in which the whole of the weight may be supposed to be concentrated, and which continually endeavors to gain the lowest possible position. A body hung in the centre of gravity will remain at rest in any position.

32. _Q._--What is meant by the centre of gyration?

_A._--The centre of gyration is that point in a revolving body in which the whole momentum may be conceived to be concentrated, or in which the whole effect of the momentum resides. If the ball of a governor were to be moved in a straight line, the momentum might be said to be concentrated at the centre of gravity of the ball; but inasmuch as, by its revolution round an axis, the part of the ball furthest removed from the axis moves more quickly than the part nearest to it, the momentum cannot be supposed to be concentrated at the centre of gravity, but at a point further removed from the central shaft, and that point is what is called the centre of gyration.

33. _Q._--What is the centre of oscillation?

_A._--The centre of oscillation is a point in a pendulum or any swinging body, such, that if all the matter of the body were to be collected into that point, the velocity of its vibration would remain unaffected. It is in fact the mean distance from the centre of suspension of every atom, in a ratio which happens not to be an arithmetical one. The centre of oscillation is always in a line pa.s.sing through the centre of suspension and the centre of gravity.

THE PENDULUM AND GOVERNOR.

34. _Q._--By what circ.u.mstance is the velocity of vibration of a pendulous body determined?

_A._--By the length of the suspending rod only, or, more correctly, by the distance between the centre of suspension and the centre of oscillation.

The length of the arc described does not signify, as the times of vibration will be the same, whether the arc be the fourth or the four hundredth of a circle, or at least they will be nearly so, and would be so exactly, if the curve described were a portion of a cycloid. In the pendulum of clocks, therefore, a small arc is preferred, as there is, in that case, no sensible deviation from the cycloidal curve, but in other respects the size of the arc does not signify.

35. _Q._--If then the length of a pendulum be given, can the number of vibrations in a given time be determined?

_A._--Yes; the time of vibration bears the same relation to the time in which a body would fall through a s.p.a.ce equal to half the length of the pendulum, that the circ.u.mference of a circle bears to its diameter. The number of vibrations made in a given time by pendulums of different lengths, is inversely as the square roots of their lengths.

36. _Q._--Then when the length of the second's pendulum is known the proper length of a pendulum to make any given number of vibrations in the minute can readily be computed?

_A._--Yes; the length of the second's pendulum being known, the length of another pendulum, required to perform any given number of vibrations in the minute, may be obtained by the following rule: multiply the square root of the given length by 60, and divide the product by the given number of vibrations per minute; the square of the quotient is the length of pendulum required. Thus if the length of a pendulum were required that would make 70 vibrations per minute in the lat.i.tude of London, then SQRT(39.1393) x 60/70 = (5.363)^2 = 28.75 in. which is the length required.

37. _Q._--Can you explain how it comes that the length of a pendulum determines the number of vibrations it makes in a given time?

_A._--Because the length of the pendulum determines the steepness of the circle in which the body moves, and it is obvious, that a body will descend more rapidly over a steep inclined plane, or a steep arc of a circle, than over one in which there is but a slight inclination. The impelling force is gravity, which urges the body with a force proportionate to the distance descended, and if the velocity due to the descent of a body through a given height be spread over a great horizontal distance, the speed of the body must be slow in proportion to the greatness of that distance. It is clear, therefore, that as the length of the pendulum determines the steepness of the arc, it must also determine the velocity of vibration.

38. _Q._--If the motions of a pendulum be dependent on the speed with which a body falls, then a certain ratio must subsist between the distance through which a body falls in a second, and the length of the second's pendulum?

_A._--And so there is; the length of the second's pendulum at the level of the sea in London, is 39.1393 inches, and it is from the length of the second's pendulum that the s.p.a.ce through which a body falls in a second has been determined. As the time in which a pendulum vibrates is to the time in which a heavy body falls through half the length of the pendulum, as the circ.u.mference of a circle is to its diameter, and as the height through which a body falls is as the square of the time of falling, it is clear that the height through which a body will fall, during the vibration of a pendulum, is to half the length of the pendulum as the square of the circ.u.mference of a circle is to the square of its diameter; namely, as 9.8696 is to 1, or it is to the whole length of the pendulum as the half of this, namely, 4.9348 is to 1; and 4.9348 times 39.1393 in. is 16-1/12 ft.

very nearly, which is the s.p.a.ce through which a body falls by gravity in a second.

39. _Q._--Are the motions of the conical pendulum or governor reducible to the same laws which apply to the common pendulum?

_A._--Yes; the motion of the conical pendulum may be supposed to be compounded of the motions of two common pendulums, vibrating at right angles to one another, and one revolution of a conical pendulum will be performed in the same time as two vibrations of a common pendulum, of which the length is equal to the vertical height of the point of suspension above the plane of revolution of the b.a.l.l.s.

40. _Q._--Is not the conical pendulum or governor of a steam engine driven by the engine?

_A._--Yes.

41. _Q._--Then will it not be driven round as any other mechanism would be at a speed proportional to that of the engine?

_A._--It will.

42. _Q._--Then how can the length of the arms affect the time of revolution?

[Ill.u.s.tration: Fig. 1.]

_A._--By flying out until they a.s.sume a vertical height answering to the velocity with which they rotate round the central axis. As the speed is increased the b.a.l.l.s expand, and the height of the cone described by the arms is diminished, until its vertical height is such that a pendulum of that length would perform two vibrations for every revolution of the governor. By the outward motion of the arms, they partially shut off the steam from the engine. If, therefore, a certain expansion of the b.a.l.l.s be desired, and a certain length be fixed upon for the arms, so that the vertical height of the cone is fixed, then the speed of the governor must be such, that it will make half the number of revolutions in a given time that a pendulum equal in length to the height of the cone would make of vibrations. The rule is, multiply the square root of the height of the cone in inches by 0.31986, and the product will be the right time of revolution in seconds. If the number of revolutions and the length of the arms be fixed, and it is wanted to know what is the diameter of the circle described by the b.a.l.l.s, you must divide the constant number 187.58 by the number of revolutions per minute, and the square of the quotient will be the vertical height in inches of the centre of suspension above the plane of the b.a.l.l.s' revolution. Deduct the square of the vertical height in inches from the square of the length of the arm in inches, and twice the square root of the remainder is the diameter of the circle in which the centres of the b.a.l.l.s revolve.

43. _Q._ Cannot the operation of a governor be deduced merely from the consideration of centrifugal and centripetal forces?

_A._--It can; and by a very simple process. The horizontal distance of the arm from the spindle divided by the vertical height, will give the amount of centripetal force, and the velocity of revolution requisite to produce an equivalent centrifugal force may be found by multiplying the centripetal force of the ball in terms of its own weight by 70,440, and dividing the product by the diameter of the circle made by the centre of the ball in inches; the square root of the quotient is the number of revolutions per minute. By this rule you fix the length of the arms, and the diameter of the base of the cone, or, what is the same thing, the angle at which it is desired the arms shall revolve, and you then make the speed or number of revolutions such, that the centrifugal force will keep the b.a.l.l.s in the desired position.

44. _Q._--Does not the weight of the b.a.l.l.s affect the question?

_A._--Not in the least; each ball may be supposed to be made up of a number of small b.a.l.l.s or particles, and each particle of matter will act for itself. Heavy b.a.l.l.s attached to a governor are only requisite to overcome the friction of the throttle valve which shuts off the steam, and of the connections leading thereto. Though the weight of a ball increases its centripetal force, it increases its centrifugal force in the same proportion.

THE MECHANICAL POWERS.

45. _Q._--What do you understand by the mechanical powers?

_A._--The mechanical powers are certain contrivances, such as the wedge, the screw, the inclined plane, and other elementary machines, which convert a small force acting through a great s.p.a.ce into a great force acting through a small s.p.a.ce. In the school treatises on mechanics, a certain number of these devices are set forth as the mechanical powers, and each separate device is treated as if it involved a separate principle; but not a t.i.the of the contrivances which accomplish the stipulated end are represented in these learned works, and there is no very obvious necessity for considering the principle of each contrivance separately when the principles of all are one and the same. Every pressure acting with a certain velocity, or through a certain s.p.a.ce, is convertible into a greater pressure acting with a less velocity, or through a smaller s.p.a.ce; but the quant.i.ty of mechanical force remains unchanged by its transformation, and all that the implements called mechanical powers accomplish is to effect this transformation.

46. _Q._--Is there no power gained by the lever?

_A._--Not any: the power is merely put into another shape, just as the contents of a hogshead of porter are the same, whether they be let off by an inch tap or by a hole a foot in diameter. There is a greater gush in the one case than the other, but it will last a shorter time; when a lever is used there is a greater force exerted, but it acts through a shorter distance. It requires just the same expenditure of mechanical power to lift 1 lb. through 100 ft., as to lift 100 lbs. through 1 foot. A cylinder of a given cubical capacity will exert the same power by each stroke, whether the cylinder be made tall and narrow, or short and wide; but in the one case it will raise a small weight through a great height, and in the other case, a great weight through a small height.

47. _Q._--Is there no loss of power by the use of the crank?

_A._--Not any. Many persons have supposed that there was a loss of power by the use of the crank, because at the top and bottom centres it is capable of exerting little or no power; but at those times there is little or no steam consumed, so that no waste of power is occasioned by the peculiarity.

Those who imagine that there is a loss of power caused by the crank perplex themselves by confounding the vertical with the circ.u.mferential velocity.

If the circle of the crank be divided by any number of equidistant horizontal lines, it will be obvious that there must be the same steam consumed, and the same power expended, when the crank pin pa.s.ses from the level of one line to the level of the other, in whatever part of the circle it may be, those lines being indicative of equal ascents or descents of the piston. But it will be seen that the circ.u.mferential velocity is greater with the same expenditure of steam when the crank pin approaches the top and bottom centres; and this increased velocity exactly compensates for the diminished leverage, so that there is the same power given out by the crank in each of the divisions.

48. _Q._--Have no plans been projected for gaining power by means of a lever?

_A._--Yes, many plans,--some of them displaying much ingenuity, but all displaying a complete ignorance of the first principles of mechanics, which teach that power cannot be gained by any multiplication of levers and wheels. I have occasionally heard persons say: "You gain a great deal of power by the use of a capstan; why not apply the same resource in the case of a steam vessel, and increase the power of your engine by placing a capstan motion between the engine and paddle wheels?" Others I have heard say: "By the hydraulic press you can obtain unlimited power; why not then interpose a hydraulic press between the engines and the paddles?" To these questions the reply is sufficiently obvious. Whatever you gain in force you lose in velocity; and it would benefit you little to make the paddles revolve with ten times the force, if you at the same time caused them to make only a tenth of the number of revolutions. You cannot, by any combination of mechanism, get increased force and increased speed at the same time, or increased force without diminished speed; and it is from the ignorance of this inexorable condition, that such myriads of schemes for the realization of perpetual motion, by combinations of levers, weights, wheels, quicksilver, cranks, and other mere pieces of inert matter, have been propounded.

49. _Q._--Then a force once called into existence cannot be destroyed?

_A._--No; force is eternal, if by force you mean power, or in other words pressure acting though s.p.a.ce. But if by force you mean mere pressure, then it furnishes no measure of power. Power is not measurable by force but by force and velocity combined.

50. _Q._--Is not power lost when two moving bodies strike one other and come to a state of rest?

_A._--No, not even then. The bodies if elastic will rebound from one another with their original velocity; if not elastic they will sustain an alteration of form, and heat or electricity will be generated of equivalent value to the power which has disappeared.