the Stability of Ships . 2*91 
the sections ; the volume immersed, by the inclination on the 
side AD W, being equal, in a practical sense, to the volume which 
emerges on the side BDN. Let each or either of these volumes 
be denoted by the letter A = 18509 cubic feet, being the ave- 
rage value between 18432 and 18587. Through the points D, in 
all the sections, draw lines F f perpendicular to the lines NW ; 
the same plane passes through all the lines F /: in the next 
place, the distance between the centres of gravity of the volumes 
immersed, and caused to emerge, in consequence of the vessel's 
inclination, estimated in the direction of a line NW, perpendicu- 
lar to the plane F /, is to be obtained. To effect this, the line 
D/*, in the principal section, is found by mensuration to be 
= 13-8, ZU = 7.03; and the area-^- AW <6 = 2.95. The area 
ADW h has been found = 124.87. £ Wherefore, according to 
the preceding solution, the distance ZL = - 9 - 5 - x 7 '°i ~ 0 17 
124.87 /’ 
which being added to the line D l = 13.8, the sum will be DL 
= 1 3-97> an( * the product arising from multiplying this line 
into the area ADW^/6 will be 13-97 x 124.87 = 1742. Similar 
products being obtained, arising from multiplying the several 
areas ADW h into the perpendicular distances of their centres 
of gravity from the plane F f, also the several products arising 
from multiplying the areas BDNc into the perpendicular dis- 
tances of their centres of gravity from the plane F /, in each of 
the 34 vertical sections, the results will be as expressed in the 
adjacent table. 
The point M is the centre of gravity of the triangle ADW, as in the general 
construction and solution : M / is drawn through M, perpendicular to NW : the line 
IL is found according to the method described in the same general solution, 
t Page 288. j Ibid. 
