86 Mr. Atwood’s Propositions determining the Positions 
triangle XAD, and L will be the centre of gravity of the tri- 
angle BXE ; through the points M, L, draw the lines MP, QL, 
perpendicular to the horizontal line DE ; make PQ = b, the 
sine of BXE = s; the tangent of BEX = t to radius = 1, 
and let EC = a. 
Then CD = ta ; and CB = s /tef; CH = %/~^\ CO = 
— x CH = the area ABC= CH = — - ; put the area 
3 ^9 a r 
BXE = u ; then to find the distance OT, the following pro- 
portion is to be made ; as the area CDE or ABC is to the area 
BXE: : so is PQ* to OT;oras — : u: : b: OT = -^;and 
OG 
2 bll 
; and since 
CO = y/ it follows that CG = 
ter s ' *9 
2 — ; and therefore CV = -~~. bu -j- 
9 ta z s 1 
and the specific gravity being 
3 ta 1 s 
3 tla* s 
A /72 x bu -f- v' 4 1 3 a 6 s z 
3 tefs 
K /—_ CH _ 
C V 2 A/72 x bu -f- “S 4 1 3 a 6 s z 1 2 bu -f- 2 \/ 2 t\ a 3 s 
thus, if the angle at which the diagonal line IC is inclined to 
the vertical line TN or OGT = BXE should be 15 0 , the angle 
XEC = 30°; wherefore in the preceding expression, t = tan- 
gent 30°to radius 1 ; s = sine 15 0 ; if CE or a is assumed = 1, 
on making the proper trigonometrical computations, the area 
BXE — u — .039395, and PQ = b = 0.73089 ; from substi- 
tuting these quantities for their values in the equation s/ n = 
3 /I a 3 s 
it appears that s/ n = 
•34063 
1 zbu + 2 a/ zt\ a 3 s •34S5 2 + -3 2II 4 
0.51094, and n — 0.261 the specific gravity which causes the 
solid to float on the fluid in a position of equilibrium with a 
* Page 59. 
