2 , 5 ° 
Mr. Atwood's Disquisition on 
under the water-line, should be equal to the angle at which the 
sides are inclined inward above the water-line, according to 
Case in. all the other conditions being the same, the stabilities 
of the two vessels will be equal, at all equal inclinations from 
the upright. The solution of Case ix. is therefore to be derived 
from that of Case nr. 
CASE X. 
The sides of a vessel coincide with the surface of a cylinder, 
the vertical sections being equal circles. 
Let QBOAH (fig. 19.) represent a vertical section of the 
vessel. The surface of the water coincides with the line BA, 
when the vessel floats upright. Suppose the vessel to be in- 
clined from the quiescent position, through an angle ASH, so 
that the water’s surface shall intersect the vessel’s, when in- 
clined, in the line CH. Bisect the line BA in D, and the line 
CH in Y ; and, through the points D and Y, draw OD, FY, 
perpendicular to the lines BA, CH, respectively, and meet- 
ing, when produced, in the point M, which is the centre of 
the circle. The angle ASH is the inclination of the vessel from 
the perpendicular ; and, being the inclination of the lines BA, 
CH, which are perpendicular to the lines OM, FM, respec- 
tively, the inclination of the lines OM, FM, or the angle 
DMF, will be equal to the angle ASH. Let E be the centre of 
gravity of the area BOA, representing the volume displaced, 
when the vessel floats upright, and quiescent. In the line MF, 
take MQ equal to ME ; Q will be the centre of gravity of the 
area CFH, representing the volume displaced when the vessel 
is inclined. Let G be the centre of gravity of the vessel ; and, 
through E, draw ET perpendicular to MF ; and, through G, 
draw GZ perpendicular to MF, intersecting that line in the 
