77 
of floating Bodies, and the Stability of Ships. 
It is required to ascertain through what angle WGS, the 
solid will be inclined round its axis, when the centres of gra- 
vity of the solid and of the part immersed are again in the 
same vertical line. As in the former cases, this problem will 
be solved, by referring to the general expression for the dis- 
tance between the two vertical lines which pass through the 
centres of gravity of the solid and of the part immersed. 
Suppose then the solid to be inclined from its former position 
of equilibrium in an angle WGS, so as to become transferred 
from the position ECDF into the position YWHV ; the part 
immersed will now be ZHVR ; the line AB will also be trans- 
ferred to PQ, and the space QXR, which was before above the 
fluid’s surface, will now be immersed under it ; and the space 
PXZ, which was before under the surface, will now be above 
it. Bisect the lines PZ, QR, in m and n, and join mX, nX; 
and take Xa = y of Xm and Xd = of Xn ; so shall a and d 
be the centres of gravity of the triangles PXZ, QXR, respec- 
tively ; draw the lines ah, cd , perpendicular to the horizontal 
line AB. Referring to the quantity expressing the distance 
between the vertical lines which pass through the centres of 
gravity of the solid and of the part immersed, namely, — 
ds, there will be applicable to the present case, the space 
QXR = A ; the space ZHVR or ACDB = V ; bc = b ; OG 
= d; the sine of the angle of inclination or WGO=5 : let t 
be the tangent of the same angle to radius = 1 ; then, since the 
triangles ZXP, QXR, are similar, and the areas are equal by 
the supposition, the sides of the two triangles will be respec- 
tively equal; that is, QX will be equal to XP; ZP equal to 
OR ; and ZX to XR. Let the height of the solid SL = c , 
