Discoveries and Inventions of the Nineteenth Century

this conception of mass, remarking that it is a something quite distinct from that of weight. A given lump of metal, for instance, would have exactly the same mass in any part of the universe, whereas its weight would depend upon its position; as, for instance, at the distance from the earth of the moon’s orbit, it would weigh only as 13600th part of its weight at the earth’s surface, and if it could be carried to the very centre of the earth it would there have no weight at all. Though the lump of metal will have different weights at different parts of the earth’s surface, it has been found (by experiment) that the weights of bodies at any one place are proportional to their masses. Therefore the same numbers that express the weights of bodies might also express their masses; but for certain good reasons these quantities are referred to different units. In England a piece of metal weighing 32 lbs. under standard conditions is said to have mass = 1; and so on. As with the same pressure acting for the same time, the velocities imparted are inversely proportional to the 173masses, it follows that the number expressing the velocity multiplied by that representing the mass in each such case of action and reaction will give the same product, or in other words the amount of motion (momentum) will be the same. This is what Newton meant by saying the reaction is equal to the action. We may now by way of illustration calculate the velocity of recoil of a rifle under conditions similar to those that might occur in practice. Let us suppose that the rifle, including the stock and all attachments, weighs 10 lbs., and that from it is fired a bullet weighing one-sixteenth of a pound, with a velocity at the muzzle of 1,200 ft. per second. To obtain the amount of motion or the momentum, we should here multiply the number expressing the mass of the bullet by 1,200, but for our present purpose the weight numbers may be used for the sake of simplicity; therefore 116 x 1,200 = 75 will represent (proportionately) the forward momentum of the bullet, and according to Newton’s law the backward momentum of the rifle will be, on the same scale, 75 also. We must therefore find the number which multiplied by 10 will give 75, and this obviously is 7·5. That is as much as to say that at the instant the bullet is leaving the muzzle, the rifle itself, if free to move, would be moving backward at the speed of 7½ ft. per second. Observe that this result would be the same if the rifle were fired where weight is non-existent; nor is the recoil due, as sometimes is erroneously supposed, to the resistance of the air to the passage of the bullet along the barrel, for even if the air were abolished, the recoil, so far as due to the masses and velocities, would remain the same, as indeed may be seen from the fact of our calculation taking no account of the bore of the rifle or of the shape of the bullet, circumstances of the utmost importance where atmospheric resistance is concerned. The foregoing calculation however involves an assumption not in exact conformity with actual conditions, by taking for granted that the centre of gravity of the rifle is in the line of the axis of the barrel, while in fact this centre is almost always lower, and therefore the kick of the recoil acts in part as a turning-over push, tending to tilt up the muzzle of the gun, and for that reason the firer must hold the weapon very firmly or he will miss his aim. When such a rifle as we have supposed is fired, say from the shoulder, it would follow from the above calculation that the backward kick of the recoil is equivalent to a blow from a 10–lb. weight moving at the speed of 7½ ft. per second. This would certainly be a very uncomfortable experience, but the backward momentum must be met somehow. We have supposed that the gun is free to move, but we know the firer presses it firmly against the muscles of his shoulder, and the stock of the gun is spread out and provided with a smooth hollow heel plate, so that any pressure from it is felt as little as possible, especially as the muscle against which it is applied acts as an elastic pad. With the rifle thus firmly held we may regard the marksman and his rifle as forming only one mass, and the centre of gravity of this being now much below the axis of the barrel, the effect of the recoil tends to overthrow the man backwards; but he learns to resist this by standing firmly, so that the elasticity of his whole frame comes into play; and besides this, the mass factor of the momentum being now so large, the velocity factor becomes comparatively insignificant. Although the momenta of gun and projectile are, according to Newton’s law, equal and opposite, the case is very different with regard to their energies, or powers of doing work, for the measure of these is jointly mass 174and the square of the velocity. The energy (vis viva) of a body of weight in pounds = W, moving with the velocity of v feet per second is always Wv264·4, that is, it will do this number of foot-lbs. units of work before it comes to rest. It would require too much space to demonstrate and fully explain here what this means, but the reader may refer to our index under the entries “Energy” and “Work,” or to any modern elementary treatise on dynamics. If the calculation be made of the energies of the ball and of the rifle due to our calculated velocities of recoil, it will be found that that of the ball is 160 times greater than that of the other, and the ball possesses this energy in a much smaller compass. Fig. 80.—Trajectory of a Projectile. The course or track of a projectile through the air after it leaves the gun is called the trajectory, and this has been studied both experimentally and theoretically, with interesting results. Assuming that the shot passed through empty space, or that the air offered no resistance to its passage, it would be very easy to trace the path of a projectile. Let us suppose that Fig. 80 represents a gun elevated at a high angle. The moment the projectile leaves the muzzle, gravity begins to act upon it, causing it to move vertically downwards with ever-increasing velocity until it finally reaches the ground; the onward uniform movement parallel to the axis of the piece being continued all the time. We could find the position of the projectile at the end of successive equal periods of time by drawing a straight line AC, a prolongation of the axis of the piece, or a line of the same inclination; on this we mark off equal distances representing by scale the velocity of the projectile per second, the points B, C, D, E being the positions the projectile would be in at the end of each successive second if gravity did not act. In order to bring the diagram within moderate compass, we suppose the projectile to have only the small velocity of 115 ft. per second. At the end of the first second it would be at B, but now suppose that gravity is allowed to act for one second, it would at the end of that time have fallen 16 ft. vertically below B and have arrived at b. Similarly we may set off by scale on verticals through C, D, and E distances representing 64 ft., 144 ft., and 256 ft. respectively. Because, for instance, the ball, without gravity acting, would at the end of 3 seconds be at D, where we may suppose its course arrested and gravity then allowed to act for 3 seconds to pull the ball down from its position of rest at D; at the end of this period, gravity alone acting, its position would be 144 ft. vertically below D, because gravity pulls a body that distance in 3 seconds, and the actual position 3 seconds after the ball had left the muzzle would be at d, after it had described the curved 175path A, b, c, d. Supposing d to be the highest point of the trajectory, another 3 seconds would bring the ball along a downward curve, and at the end of 6 seconds from the discharge it would be at a point on the same level as A. Now the complete curve would be symmetrical on each side of a vertical line through its highest point, and it would be in fact a regular parabola with its vertex at d. The foregoing presupposes that the air offers no resistance to the passage of the projectile through it. The fact however is quite otherwise, for no sooner does the projectile begin its flight than its velocity is constantly diminished by the air’s resistance. Now this resistance is complex, depending upon a number of different conditions, the effect of which can be taken into account only by extremely complex calculations. Obviously it will vary according to the area of the section presented by the projectile to the line of its flight, and again by the shape of its front, for a pointed shot will cleave the air with less resistance than one with a flat front. Then the density of the air at the time will also enter into the calculation. The mass of the projectile and also its velocity, upon which depend its vis viva, energy, or power of overcoming resistance in doing work, will also have to be considered. Most complex of all is the law, or rather laws (i.e. relations), which connect the air resistance with the velocity; for this relation no definite expression has been found. It is a function of the velocity (known only by experiment under defined conditions), and varying with the velocity itself. Thus for velocities up to 790 ft. per second, it is a function (determined experimentally) of the second power or square of the velocity; between 790 ft. per second and 990 ft. per second the law of resistance is changed and becomes a function of the third power of the velocity; between 990 ft. and 1,120 ft. velocity the law again changes and is related to the sixth power of the velocity; between 1,120 ft. and 1,330 ft. the resistance is again related to the third power of the velocity; and with higher speeds than that last named it is again more nearly related to the square of the velocity. It will be seen that to calculate the path of a projectile is really a very difficult mathematical problem, and indeed one which can be solved only approximately when all the known data are supplied. The air resistance to the motion of a projectile is much greater than before trial would be supposed. Let us take an experiment that has actually been recorded, in which a bullet three-quarters of an inch in diameter, weighing one-twelfth of a pound, was found to have a velocity of 1,670 ft. per second at a distance of 25 ft. from the gun, and this 50 ft. farther was reduced to 1,550 ft. per second. Now if the reader will calculate, according to the formula we have given above, the energy due to the bullet’s velocity at these points, he will find it must have done 500 foot-lbs. units of work in traversing the 50 ft., and as this could have been expended only in overcoming the resistance of the air, we learn that this last must have been equivalent to a mean or average pressure of 10 lbs. thrusting the bullet backwards. It will be interesting to compare the difference in the trajectory of a projectile under defined conditions, worked out with the air resistance taken into account, compared with the trajectory when the air is supposed to be non-existent. We find an example of the former problem fully worked out by many elaborate mathematical formulæ