Page:EB1911 - Volume 14.djvu/131

HYDROSTATICS] and the lowering of the surface is

(20)

as before in (17).

16. Centre of Pressure.—A plane area exposed to fluid pressure on one side experiences a single resultant thrust, the integrated pressure over the area, acting through a definite point called the centre of pressure (C.P.) of the area.

Thus if the plane is normal to Oz, the resultant thrust

R = ∬ pdxdy, (1)

and the coordinates $\overline{x}$, $\overline{y}$ of the C.P. are given by

$\overline{x}$R = ∬ xpdxdy, &emsp; $\overline{y}$R = ∬ ypdxdy. (2)

The C.P. is thus the C.G. of a plane lamina bounded by the area, in which the surface density is p.

If p is uniform, the C.P. and C.G. of the area coincide.

For a homogeneous liquid at rest under gravity, p is proportional to the depth below the surface, i.e. to the perpendicular distance from the line of intersection of the plane of the area with the free surface of the liquid.

If the equation of this line, referred to new coordinate axes in the plane area, is written

x cos + y sin  − h = 0, (3)

R = ∬ (h − x cos  − y sin ) dxdy, (4)

$\overline{x}$R = ∬ x (h − x cos − y sin ) dxdy, (5)

$\overline{y}$R = ∬ y (h − x cos − y sin ) dxdy.

Placing the new origin at the C.G. of the area A,

∬ xdxdy = 0, &emsp; ∬ ydxdy = 0, (6)

R = hA, (7)

$\overline{x}$hA = −cos ∬ x2 dA − sin  ∬ xydA,  (8)

$\overline{y}$hA = −cos ∬ xydA − sin  ∬ y2dA.  (9)

Turning the axes to make them coincide with the principal axes of the area A, thus making ∫∫ xy dA = 0,

$\overline{x}$h = −a2 cos, $\overline{y}$h = −b2 sin , (10)

where

∬ x2dA = Aa2, &emsp; ∬ y2dA = Ab2, (11)

a and b denoting the semi-axes of the momental ellipse of the area.

This shows that the C.P. is the antipole of the line of intersection of its plane with the free surface with respect to the momental ellipse at the C.G. of the area.

Thus the C.P. of a rectangle or parallelogram with a side in the surface is at of the depth of the lower side; of a triangle with a vertex in the surface and base horizontal is  of the depth of the base; but if the base is in the surface, the C.P. is at half the depth of the vertex; as on the faces of a tetrahedron, with one edge in the surface.

The core of an area is the name given to the limited area round its C.G. within which the C.P. must lie when the area is immersed completely; the boundary of the core is therefore the locus of the antipodes with respect to the momental ellipse of water lines which touch the boundary of the area. Thus the core of a circle or an ellipse is a concentric circle or ellipse of one quarter the size.

The C.P. of water lines passing through a fixed point lies on a straight line, the antipolar of the point; and thus the core of a triangle is a similar triangle of one quarter the size, and the core of a parallelogram is another parallelogram, the diagonals of which are the middle third of the median lines.

In the design of a structure such as a tall reservoir dam it is important that the line of thrust in the material should pass inside the core of a section, so that the material should not be in a state of tension anywhere and so liable to open and admit the water.

17. Equilibrium and Stability of a Ship or Floating Body. The Metacentre.—The principle of Archimedes in § 12 leads immediately to the conditions of equilibrium of a body supported freely in fluid, like a fish in water or a balloon in the air, or like a ship (fig. 3) floating partly immersed in water and the rest in air. The body is in equilibrium under two forces:—(i.) its weight W acting vertically downward through G, the C.G. of the body, and (ii.) the buoyancy of the fluid, equal to the weight of the displaced fluid, and acting vertically upward through B, the C.G. of the displaced fluid; for equilibrium these two forces must be equal and opposite in the same line.

The conditions of equilibrium of a body, floating like a ship on the surface of a liquid, are therefore:—

(i.) the weight of the body must be less than the weight of the total volume of liquid it can displace; or else the body will sink to the bottom of the liquid; the difference of the weights is called the “reserve of buoyancy.”

(ii.) the weight of liquid which the body displaces in the position of equilibrium is equal to the weight W of the body; and

(iii.) the C.G., B, of the liquid displaced and G of the body, must lie in the same vertical line GB.

18. In addition to satisfying these conditions of equilibrium, a ship must fulfil the further condition of stability, so as to keep upright; if displaced slightly from this position, the forces called into play must be such as to restore the ship to the upright again. The stability of a ship is investigated practically by inclining it; a weight is moved across the deck and the angle is observed of the heel produced.

Suppose P tons is moved c ft. across the deck of a ship of W tons displacement; the C.G. will move from G to G1 the reduced distance G1G2 = c(P/W); and if B, called the centre of buoyancy, moves to B1, along the curve of buoyancy BB1, the normal of this curve at B1 will be the new vertical B1G1, meeting the old vertical in a point M, the centre of curvature of BB1, called the metacentre.

If the ship heels through an angle or a slope of 1 in m,

GM = GG1 cot = mc (P/W), (1)

and GM is called the metacentric height; and the ship must be ballasted, so that G lies below M. If G was above M, the tangent drawn from G to the evolute of B, and normal to the curve of buoyancy, would give the vertical in a new position of equilibrium. Thus in H.M.S. “Achilles” of 9000 tons displacement it was found that moving 20 tons across the deck, a distance of 42 ft., caused the bob of a pendulum 20 ft. long to move through 10 in., so that

(2)

also

cot = 24,  = 2°24′. (3)

In a diagram it is conducive to clearness to draw the ship in one position, and to incline the water-line; and the page can be turned if it is desired to bring the new water-line horizontal.

Suppose the ship turns about an axis through F in the water-line area, perpendicular to the plane of the paper; denoting by y the distance of an element dA if the water-line area from the axis of rotation, the change of displacement is ydA tan, so that there is no change of displacement if ydA = 0, that is, if the axis passes through the C.G. of the water-line area, which we denote by F and call the centre of flotation.

The righting couple of the wedges of immersion and emersion will be

wydA tan ·y = w tan  y2dA = w tan ·Ak2 ft. tons, (4)

w denoting the density of water in tons/ft.3, and W = wV, for a displacement of V ft.3

This couple, combined with the original buoyancy W through B, is equivalent to the new buoyancy through B, so that

W.BB1 = wAk2 tan , (5)

BM = BB1 cot = Ak2/V, (6)

giving the radius of curvature BM of the curve of buoyancy B, in terms of the displacement V, and Ak2 the moment of inertia of the water-line area about an axis through F, perpendicular to the plane of displacement.

An inclining couple due to moving a weight about in a ship will heel the ship about an axis perpendicular to the plane of the couple, only when this axis is a principal axis at F of the momental ellipse of the water-line area A. For if the ship turns through a small angle about the line FF′, then b1, b2, the C.G. of the wedge of immersion and emersion, will be the C.P. with respect to FF′ of the two parts of the water-line area, so that b1b2 will be conjugate to FF′ with respect to the momental ellipse at F.

The naval architect distinguishes between the stability of form, represented by the righting couple W.BM, and the stability of ballasting, represented by W.BG. Ballasted with G at B, the righting couple when the ship is heeled through is given by W.BM. tan; but if weights inside the ship are raised to bring G above B, the righting couple is diminished by W·BG.tan, so that the resultant righting couple is W·GM·tan. Provided the ship is designed to float upright at the smallest draft with no load on board, the stability at any other draft of water can be arranged by the stowage of the weight, high or low.

19. Proceeding as in § 16 for the determination of the C.P. of an area, the same argument will show that an inclining couple due to