254 Mr. Atwood's Disquisition on 
line BA is parallel to XV, and the line CH parallel to NW, 
by the construction, it follows, that the angle ASH is equal 
to the angle XVN ; wherefore the angle ASH is also equal to 
the angle MOI, or the given angle of inclination from the 
upright. VK being parallel to DL, and therefore a diameter of 
the curve to the point V, and CH being drawn parallel to 
NVW, which is a tangent to the curve in the point V, it 
follows, that VK bisects the line CH in the point K ; KH 
therefore will be an ordinate to the diameter VK : and, since 
VK is by construction equal to DL, and DL, VK, are ab- 
scissas of the segments BLA, CVH, respectively, it is known, 
from the properties of the figure, that the area of the segment 
BLA is equal to the area of the segment CVH ; and conse- 
quently the area of the figure ASH will be equal to the area of 
the figure BSC. And since, when the vessel floats upright, the 
line AB coincides with the water's surface, and the area of 
the segment ALB is equal to the area of the segment CVH, 
it follows, that when the vessel is inclined from the perpendi- 
cular, through an angle ASH, equal to the given angle MOI, 
the surface of the water will intersect the vessel in the line CH. 
Moreover, since LE is to LD as 3 to 5, by the construction, and 
VQ is to VK in the same proportion of 3 to 5, by the proper- 
ties of the figure, E is the centre of gravity of the area BLA, 
and Q is the centre of gravity of the area CVH, which repre- 
sents the total volume displaced, when the vessel is inclined 
through an angle ASH, or MOI ; and the line rQP being, by 
construction, drawn perpendicular to the water’s surface CH, 
will be a vertical line passing through the centre of gravity Q of 
the volume displaced CVH : and GZ, drawn through the centre 
of gravity G, perpendicular to this line, will be the measure of 
