228 
Mr. Atwood’s Disquisition on 
Let BWA (fig. 10.) represent a vertical section of the ves- 
sel. Through the extremity B of the line BA, draw BO in- 
clined to BA, at the angle ABO, equal to the vessel’s inclina- 
tion from the upright. In this line, take any point R, and in 
B R take B I to B R, as the sine of the angle WBR to radius. 
Also take BF to BR as the sine of BRW to radius; and let 
BG be a geometrical mean proportional between the lines BF 
and B I ; from the point G, set off GZ equal to BF ; join Z A, 
and, through G, draw GS parallel to ZA; and, through S, 
draw CH parallel to BZ. The area ASH will be equal to the 
area SBC. 
By the construction, the angle ARB =AHS, and the angle 
WCH = WBR ; 
also BR : BI : : rad. : sine SCB, 
and BF : BR : : sine AHS : rad. 
Joining these ratios, BF : BI : : sine AHS : sine SCB. 
By the construction, and the similarity of the triangles 
BGS, BZA. 
BF : BI : : BF 2 : BG 2 : : GZ 2 : BG 2 : : SA 2 : SB 2 . 
Wherefore SA 2 : SB 2 : : sine AHS : sine SCB 
By trigonometry, SH : SA : : sine SAH : sine AHS 
and SB : SC : : sine SCB : sine SAH = sine SBC 
Joining these ratios, SA x SH : SB x SC : : 1 : 1. 
Therefore SA x SH — SB x SC. 
But the angle ASH being equal to the angle BSC, the area 
of the triangle ASH will be to the area of the triangle BSC, as 
SA x SH is to SB x SC ; and, since SA x SH is equal to SB 
x SC, the area of the triangle ASH is equal to the area of the 
triangle SBC. 
