Consider now a gas, or compressible fluid, satisfying the law of Boyle and Mariotte, P=a1).
Accordingly, (3) --= a log p+ mast., P P and (6) log p, _ a.
Thus as we ascend to heights in arithmetical progression the density decreases in geometrical progression, becoming zero at an infinite height. From equation (7) the barometric formula is obtained by which heights are found from barometer readings. The law of Boyle assumes constant temperature. It is, however, more likely that the temperature varies in accordance with what is called con vective equilibrium, so that if a portion of air' is hotter than the stratum in which it lies it will rise, and, cooling and expanding, will eventually find a layer of the same density and temperature as its own. The principles of thermodynamics give us the relation between pressure and temperature when the rarefaction is adiabatic, that is, when no heat is lost or gained by the air, p=bpk where K is a constant • for the gas, whose value is about 1.4. We then have (8) — — 1 Since K> 1, p diminishes as s increases, and is equal to zero when gs= c, so that on this hypothesis the atmospheric has an upper limit.
Let us now consider the equilibrium of a solid body floating in a liquid. If we consider the body removed and the space that it occu pied filled with water, since this water is in equilibrium, its weight is borne up by the pressure of the surrounding water, the effect of which is accordingly to apply to each por tion of the water in question an upward force just equal to its weight. Now just the same forces must be the resultant of the pressures on the solid when it is substituted for the dis placed water, so that it is borne up by a force equal to the weight of the displaced water. This is the Principle of Archimedes. Since the resultant of the weight of all the displaced water is a single force applied at its centre of mass, the resultant upward thrust on the floating body is applied at a point coinciding with the centre of mass of the displaced water. This point is called the centre of buoyancy of the body. If the body is to be. in equilibrium, according to the principles of statics of a rigid body, its weight must be equal to that of the displaced body, and its centre of mass and centre of buoyancy must be in the same ver tical line. If the first condition is satisfied, but not the second, the body will float, but will be subject to a turning couple.
Suppose the body floats without being wholly immersed. A plane which cuts off from the body a volume equal to the volume of water having an equal weight is called a plane of flotation, and if we draw all such planes they will envelope a surface called the surface of flotation. For every plane of flotation there will be a centre of buoyancy, and the locus of all these points is called the surface of buoy ancy. Suppose the floating body is displaced from its position of equilibrium by rotation through a small angle de about an axis OX through 0, Fig. 4, and let WL, W'L' be the original and final planes of flotation. By turn ing the figure until either is horizontal both positions of the body may be shown. Let B be the original centre of buoyancy, G the centre of mass of the floating body. Then if B were the centre of buoyancy in the second position, the body would be acted on by the couple of which either arm would be the weight W= mg, and the arm the horizontal projection of BG, MB, where b = BG. The moment of the couple tending to further displace the body will then be Wb68. But this is not the only couple, for the immersed part is not the same as before, the volume of the wedge EOE' having become immersed, giving rise to an upward thrust, and the wedge DOD' having emerged and lost its buoyancy, both these causes giving rise to a turning moment in the same direction, and opposite to that previously found. Since the volume under water is to be the same in both positions, the volume of the wedges of immersion and emersion must be equal. Since the wedges are infinitely thin, the thickness at any point x, y in the plane of flotation is z = yd0. The condition for equality of volumes is then (9) f f acixdy=f50 f f ydxdy=0, the integral being taken over the plane of flotation. This will be the case if the axis passes through the centre of mass of the area of flotation. The thrust on any element of vol ume d.r= zdxdy being gpdr, the moment about the X-axis will be (10) ffgpydr=--tpdeffyidxcly =gPdOSKA where is the square root of the mean of the squares of the distances of the elements from the X-axis, or the so-called radius of gyration of the area of flotation about the X-axis, and S is the area of flotation. In like manner the moment about the Y-axis is (11) M'.= —fffgpxdr--- — gpeOffxydxdy.