222 
Mr. Atwood’s Disquisition on 
be represented by the letter t, it is inferred, from the construe- 
tion in p, 220, that SA = and SB == . The 
v h-\-V ac Vh-SfVac 
value of the line ET having been thus determined, if ER ~dx 
sin. ASH or d s be subtracted from it, the result will be GZ, 
the measure of the vessel’s stability. 
Suppose the sides BY, AH, (fig. 4.) to project outward, at an 
angle of 15 0 inclination to the parallel sides B C, AW, so as to 
make the angle SAH = 105°. Let the vessel’s inclination from 
the upright be the angle ASH — 15°; and therefore AHS = 6o°, 
and SCB = 75 0 . Let the breadth BA or t = 100 equal parts, of 
which d or GE = 13. Then, by calculating from the analy- 
tical values just determined, it is found that KL = SL + SK = 
68,017: the area ASH — 347.44, and the entire volume 
immersed V, being, as in the former case, = 3600, E T — 
'' ° X = 6.57. And, since ER or d x sin. ASH is = 
3.36, if the latter value be subtracted from the former, the re- 
sult will be GZ = 3-2i, or the measure of the vessel’s stability. 
The force of stability, to restore the vessel to the upright po- 
sition, will be precisely the vessel’s weight, or fluid’s pressure, 
acting in the direction of a vertical line, which passes at a dis- 
tance of 3.20 from the axis, estimated in a horizontal direction. 
And this force is equivalent to, and will counterbalance, -~* 
parts of the vessel’s weight, applied to act in a contrary direc- 
tion, at the distance of 50 from the said axis. So that, if the 
vessel’s weight should be 1000 tons, the force of stability would 
balance a weight or force of = 64.2 tons, applied to 
act at the distance 50 from the axis. 
