226 Mr. Atwood’s Disquisition on 
Let BA, or t = 100, d = 13, ASH = 15 0 , SAH = 105°, V 
=3600, as in the former cases : then, 5 = sine 1 5 0 , a =sine 105 0 , 
h = sine 6 o°; by referring to the solution, GZ = 3.5q; and 
the stability will be the weight of the vessel, suppose 1 000 tons, 
acting at the distance 3.59 from the axis, to turn the vessel; 
which force is equivalent to a weight of 71.7 tons, applied at 
the distance of 50 from the axis. 
case v. 
The sides of a vessel are inclined inward, and at equal 
angles of inclination to the plane of the masts, both above and 
beneath the water-line. 
BA (fig. 8.) is the breadth of the vessel coinciding with 
the water’s surface, when floating upright. XE represents the 
plane of the masts, bisecting B A in the point S. U P, WQ, 
are lines drawn through the extremities of the line B A, parallel 
to XE. BY, AH, are the sides of the vessel above the water- 
line, inclined inward to the plane of the masts, at the angle 
QAH = YBP. BC, AD, are the sides under the water-line, 
inclined inward to the plane of the masts, at the angle DAW 
or CBU, which are equal to HAQ or YBP. The other condi- 
tions are as in the former cases. Through the point S, draw 
the line CH inclined to BA, at the angle ASH, equal to the 
vessel’s inclination from the upright. Since the triangles ASH, 
BSC, are similar and equal figures, it follows, that when the 
vessel is inclined to the angle ASH, it will be intersected by 
the water’s surface in the line CH. The remaining part of 
this construction is similar to that of the preceding cases. 
The same notation being adopted with that which was used 
