232 Mr. Atwood’s Disquisition on 
found, the sum will be the line FM; from which, if FE, or 4 of 
FD, be subtracted, there will remain the line ME ; which is 
to ET as radius is to the sine of the inclination EMT, or 
ASH. ET will therefore be expressed in known terms ; from 
which, if ER be subtracted, the remaining line will be RT, 
or GZ, the measure of the vessel’s stability, analytically ex- 
pressed. 
By the construction, the area FBA is equal to the area 
FCH ; and, since the area BAF is to the area IKF in the 
same * * proportion which the area FCH bears to the area FOP, 
it follows, that the area FIK is equal to the area FOP. Also, 
because CN is equal to NH, and OP is parallel to CH, it fol- 
lows, that OQ is equal to QP. For brevity, let the angle KYP, 
or ASH, be denoted by the letter S; FPO = FHC by P; 
POF = HCF by O ; also let the angle PFO be made = F. 
Because the areas IFK, PFO, are equal, 
FO x FP x sine F 
FE*x tang.i F, radius being = 1 : wherefore, 
; FE-Xtang ^F = Fgxsec.^ because 
FOxsineF FO 
FO by substitution FP»= FE ' xsec ^ F / • — ; 
sine O ’ J sine P 
and therefore FP = FEx sec. AF x \/ s -^ : but sine FWP = 
■" sine ir 
cos. S. 
Wherefore, FW : FP : : sine P : cos. S ; or 
FW : FE x sec. {Fxv/ ; sine P : cos. S ; consequently, 
* 2 v sine P 
FW = 
FE X sec. |Fx^ sine O x sine P 
cos. S 
By investigation, ■f founded on the rules of trigonometry, it 
* Each of these proportions being as 9 to 4. 
f See Appendix. 
