267. exerted upon the particle. Hence the squares of the times being as the spaces, or the times simply as the square roots of the spaces, BG is to GD as the time employed in describing BG to the time required to reach the horizontal plane DF. But in the time employed in describing BG, the particle would be carried uniformly and horizontally by the velocity thus acquired, through a space equal to 2 BG; therefore, to find the amplitude or horizontal range DE of the jet, we have the proportion Geom. 215. BG: GD: 2 BG: DE = 2 BG GD VBG =2BG GD = 2 GH. As the same reasoning may be used with respect to any other point in BD, if upon the height of the fluid BD as a diameter we describe a semicircle BKD, the horizontal distance to which the fluid will spout from any point will be twice the ordinate of the circle drawn through this point, the distance being measured on the plane of the bottom of the vessel. 486. It will hence be perceived, that if apertures be made at equal distances G, L, from the top and bottom of the vessel, the horizontal distances DE to which the fluid will spout from these apertures will be equal; and that the point I, bisecting the altitude, is that from which the fluid will spout to the greatest distance, this distance DF being equal to twice the radi us of the semicircle or to the altitude BD of the fluid. 487. If the fluid issue obliquely instead of horizontally, the curve described will still be parabolic, and the horizontal range, &c. of the jet may be calculated as in the case of other projectiles. Fig.237. Let the aperture C be inclined, for example, upward at different angles. CB will be equal to s, the space through which a body must fall to acquire the velocity of projection, and equal to the dis477. tance CF, CF', of the foci of the several parabolas, traced by particles issuing with different angles of elevation. Hence BE Trig.172 is the directrix to these parabolas, and the circle described from 'the centre C, and with the radius BC, will pass through the several foci F, F, &c. Let CE, for instance, be the direction of the jet, and draw CF making the angle ECF equal to BCE; let fall the perpendicular FH, and take HG equal to HC, the distance CG will be the horizontal range of the jet. But CG 2 CH 2 CF X cos FCH 2 CB × sin 2 ECF. 45°, the focus of the Therefore, when the angle of elevation is Of the Motion of Gases. 488. To determine with what velocity the air or any other gas will rush into a void space, when urged by its own weight, we proceed according to a method analogous to that by which the motion of liquids is determined. When the moving force and the mass or matter to be moved vary in the same proportion, the velocity will continue the same, since v = P. m Thus, if there be similar vessels of air and water, extending to the top of the atmosphere, on the supposition of a uniform density throughout, they will be discharged through equal and similar apertures with the same velocity; for in whatever proportion the quantity of matter moving through the aperture be varied by a change of density, the pressure which forces it out acting in circumstances perfectly similar will vary in the same proportion. Hence it follows that the air rushes into a void with the velocity which a heavy body would acquire by falling from the top of the atmosphere, this fluid being supposed to be of a uniform density throughout. The height of a uniformly dense or homogeneous atmosphere being 27807 feet, according to article 467, and g = 32,2, we shall have for the velocity in question v = √/2gh = √2 × 32,2 × 27807 = 1338. 489. But as the space into which the air rushes becomes more and more filled with air the velocity must be diminished continually. Indeed whatever be the density of this rarer air, its elasticity varying with its density, will balance a proportional part Mech. 48 28. 277. of the pressure of the atmosphere, and it is the excess of this pressure only which constitutes the moving force, the matter to be moved being the same as before. Let n be the natural density of the atmosphere, and ▲ the density of that which opposes itself to the motion in question. Let p be the pressure of the atmosphere, or the force which impels it into a void, and the force with which this rarer air would rush into a void; from the proportion we shall have for the moving force sought p PA D Again, let v be the velocity of air rushing into a void under the pressure p, and u the velocity of air under the same pressure rushing into rarefied air of the density A. Since the pressures are as the heights producing them, the fluid being supposed of a uniform density through, we shall have whence uv x A D :: 1: no allowance being made for the inertia of the rarer air, which being displaced must oppose a certain resistance. 490. Let it be proposed to determine the time t in seconds in which the air will flow into a given exhausted vessel, until the air shall have acquired in the vessel a certain density A. Suppose h the height due to the velocity v, b the bulk or capacity of the vessel, and 6 the area of the aperture, the measure in each case being in feet. Since the quantity of air necessary to fill the vessel will depend upon the size of the vessel, and also upon the density of the air, b▲ will represent this quantity, the differential of which is b d 4. The velocity of influx at the first instant is v2 gh; and when the air in the vessel has acquired the density A, that is, at the end of the time. t, the velocity is Hence the rate of influx, which may be measured by the infinitely small quantity of air passing the aperture during the instant dt with this velocity, will be denoted by D Δ 2gh X X Dodi odt/2gDh (D-4). D Putting these two values of the rate of influx equal to each other, we have To find the constant C, it will be observed, that when t = 0, ▲ = 0, and rected integral D-AD. We have, therefore, for the cor 491. When DA, the motion ceases, and the value of t, or the time of completely filling the vessel, becomes Suppose, for example, the capacity of the vessel to be 8 cubic feet, or nearly a wine hogshead, and that the aperture by which air of the ordinary density, or 1, enters, is an inch square, or of a foot. In this case 4/4/27807668, nearly; and hence 8 t= 1152 = =1", 72 nearly. 1668 668 479. If the aperture be only of a square inch, or the side, the time of completely filling the vessel will be 172′′ nearly, or a little less than 3'. If the experiment be made with an aperture, cut in a thin plate, we shall find the time greater nearly in the ratio of 62 or 63 to 100, as we have already remarked with respect to water flowing through small orifices. 492. We can find, in like manner, the time necessary for bringing the air in the vessel to any particular density, as of that of air in its ordinary state. For the only variable part of the integral, above found, is DA, which in this case becomes =, and gives D D4=; hence, if the aperture were a square, each side being of an inch, the time sought would be 172", or 86", nearly. A 493. If the air in the vessel be compressed by a weight acting on the moveable cover AD, the velocity of the expelled Fig. 238 air may be determined thus. Let the additional pressure be denoted by q, and the density thence resulting by D'; we shall then have Now, since the pressure which expels the air is the difference between the force which compresses the air in the vessel and that which compresses the internal air, the expelling force is q; whence, the forces being as the quantities of motion, m, n, being the masses expelled, v the velocity with which air rushes into a void, and u the velocity required. But the masses or number of particles which issue through the same orifice in |