comes to A, a communication is opened between AM and the boiler, and between MB and the condenser, and is closed again when M has described one third of AB; and when M comes to B, a communication is opened between MB and the boiler, and AM and the condenser, and closed, as in the former case, when M has described one third of BA. Suppose M to ascend from B to A, the space below M being filled with steam from the boiler, as soon as M arrives at A, the communication is opened between MB and D, and the steam in MB flows into D, and is there condensed leaving a vacuum in MB; at the same time, a communication being open between AM and the boiler, steam flows into AM, and M is forced downwards by the full pressure of the steam, during one third of its descent, and after the communication between AM and the boiler is cut off, by the diminished pressure of the steam in the cylinder. In the same manner, when M arrives at B, a vacuum is produced in AM by the condensation of the steam, and M is pressed upwards by the steam admitted below. The condensation of the steam in D is promoted by a jet of cold water, which is removed as fast as it collects by a pump P. A description of the contrivances for regulating the supply of steam and water, and for making the extremity of the piston rod describe a curve approaching to a straight line; as well the enumeration of the advantages of this construction over the atmospheric engine, would be improper in this place, on account of its length. K APPENDIX. ART. 116. THE proposition which forms the subject of (Art. 5.) may be proved by experiment, in the following manner. Let a vessel ABCD (fig. 60.) closed on all sides, and exactly filled with fluid, be placed with the side AD horizontal, and therefore free from any pressure arising from the weight of the fluid. At E and F make two equal orifices, and apply any pressure to the surface of the fluid at E by means of a piston; then, in order to prevent the fluid from escaping at F, another piston must be applied and pressed with exactly the same force as that at E. Thus the pressure communicated perpendicularly downwards at E, has, by the intervention of the fluid, been made to act with the same force perpendicularly upwards at F. If the orifice F, instead of being made in the upper horizontal surface of the vessel, be made at any point G in the inclined side, and a force applied sufficient to counteract the effort made by the fluid to escape; then, if any pressure be applied at E, it will be found, as before, that an additional pressure equal to that at E, must be applied at G perpendicular to the side CD, to preserve the equilibrium. We may conclude therefore that a force impressed on a given surface in any part of a fluid, produces an equal pressure on an equal surface in any other part of the fluid. 117. To find the pressure of a fluid on any surface. Suppose the surface (S) divided into an indefinite number of portions A, B, C, &c. so small that every point of any one of them may be considered as at the same perpendicular depth below the surface of the fluid; and let their respective perpen dicular depths be a, b, c, &c.; then (Art. 12.) the pressure of the fluid on any one of them A = gp Aa; p being the density of the fluid, similarly, the pressure on B = gp Bb, &c.; therefore the sum of the pressures = gp (Aa + Bb + Cc+ &c.) But if the depth of the centre of gravity of S below the surface of the fluid = X; then, since A, B, C, &c. may be considered as bodies whose perpendicular distances from the surface of the fluid are a, b, c, &c., (Wood's Mechanics, 172.) Aa + Bb+Cc+ &c. = (A + B + C + &c.) X = SX; ... the pressure of the fluid on S = gpSX. A vessel of the form of a cone with its base downwards, is filled with fluid; to compare the pressure on the base of the vessel with the weight of the fluid contained in it. Let the cone be generated by the revolution of the rightangled triangle ABC (fig. 61.) round AC, p the density of the fluid; then, the area of the base of the cone T. CB2, and the depth of its centre of gravity below the surface of the fluid AC; therefore the pressure on the base of the conegрT.AC.BC; and the weight of the fluid in the cone = gp = π 3 AC.BC2; therefore pressure on the base of the cone = 3.(weight of the fluid). Since the pressure on the base of a vessel filled with a given fluid, depends only on its area, and the depth of its centre of gravity below the surface of the fluid, the pressure on the base of the cone BAB′ (fig. 61) is equal to the pressure on the base of the cylinder BAA'B' (fig. 62), or the pressure on the base of the truncated cone BAA'B' (fig. 63); the area of the base and the depth of its centre of gravity below the surface of the fluid being the same in each case. 118. A hollow sphere is just filled with fluid; to compare the pressure on the internal surface of the sphere with the weight of the fluid. Let a be the radius of the sphere; then, the area of the surface of the sphere = 4πa, and the depth of its centre of gravity below the surface of the fluid = a, therefore the pressure on the surface of the sphere = gp4πa3; and the weight of the fluid contained in the sphere = gpa3; therefore the pressure on the surface of the sphere = 3 (weight of the fluid). 119. To find the centre of pressure of the triangle AOB (fig. 64.) having the side OB perpendicular to the surface of the fluid, and the side OA in the surface. Draw HR parallel to AO; and let X, Y be the distances of the centre of pressure from OA, OB respectively; OH = x. Then, (Art. 17.) X √ √yx = √ √yx2, Y fx fy x = fx Syx y; the integrals being taken between the limits corresponding to the boundary of the figure. .. se sy∞, between the proper limits, = 40.OB2. ffy, between the proper limits, = .40.OB3. ... fywy, between the proper limits, = 40. OB2. .. X = 1 OB, Y = 10A. 12 120. To find the centre of pressure of a semicircle ORA (fig. 65.) having its diameter OA perpendicular to the surface of the fluid, and the extremity O of the diameter in the surface of the fluid. Let the plane of the semicircle meet the surface of the fluid in Oy. Draw HR parallel to Oy; and let OA 2а, ОН = х. = (Sy = HR — Sy=0) x = x.HR = x √(2 ax − x2); (Sr=2a - Sx=0)√(2 ax − x2) = 2 fr Jy, between the proper limits, Jyv2 = a2y + C |