OF THE POSITIONS OF EQUILIBRIUM. 339 



and moreover, the area of the immersed part HE CD, is expressed by 

 half the sum of the parallel sides drawn into the perpendicular dis- 

 tance between them ; consequently, we obtain 



abs= \bs\x + y), 

 and finally, by multiplication and division, it is 



2s s'O + y)- (265). 



The equation which we have just investigated, involves one of the 

 conditions of equilibrium, viz. that in which the area of the immersed 

 part, and that of the whole section, are to each other, as the specific 

 gravity of the solid is to that of the fluid ; but in order to discover the 

 equation which involves the other condition, we must have recourse to 

 a separate construction, as follows. 



Let IK be the surface of the fluid, and HECD the immersed portion 

 of the section, as in the general dia- 

 gram preceding. Bisect D H and c E, ~ \'- " 

 the parallel sides of the figure, in the 

 points t and z, and draw tz;* then, 

 by the principles of mechanics, the l =^^-^ 

 straight line tz passes through the 

 centre of gravity of the figure HECD, 

 and divides it into two parts such, that 

 gz : gt : : SDH-J-CE : DH + 2cE. 



Through the point E draw E* parallel to DC, and bisect DC and 

 HE in the points i and h; draw ih, bisecting E the side of the 

 triangle HE* in the point q, and join th and uq, intersecting one 

 another in the point o ; then is the point o thus determined, the centre 

 of gravity of the triangular space H E t. 



Draw the diagonal D E, bisecting q i in p the centre of gravity of 

 the rectangular space ECD t, and join po intersecting tz in g ; then is 

 g the centre of gravity of the quadrilateral space HECD which falls 

 below the plane of floatation. Through the point g thus determined, 

 draw the straight line gv parallel to DC, the lowest immersed side of 

 the section, and meeting CE in the point v; then is gv the quantity 

 to be assigned by the construction. 



From the point z and parallel to CD, draw zf meeting DH perpen- 



* It is a circumstance entirely accidental, that the lines tz and <E terminate in 

 the same point t, and consequently, has nothing to do with the conditions of the 

 problem j it only happens when D H is double of c E. 



z 2 



