68 Statics. 



EG = / x( 



+ T# 4- KM + ./VT -|- &c. . . + | .tffl 



This formula, expressed in common language, furnishes the 

 following rule ; 



To find the distance of the centre of gravity G, from one of 

 the extreme ordinates DF, 



(l). Take a sixth of the first ordinate DF ; a sixth of the last 

 ordinate AB, multiplied by triple the number of ordinates less 4 ; 

 then the second ordinate, double the third, triple the fourth, and so on ; 

 which may be called thejirst sum. 



(2). To half the entire sum of the two extreme ordinates, add all 

 the intermediate ordinates, for a second sum ; 



' (3). Divide the first sum by the second, and multiply the quotient 

 by the common interval between two adjacent ordinates* 



For example, if there were 7 perpendiculars, whose values 

 were 18,23,28, 30,30,21, 0, feet; and each interval were 

 20 feet; I should take a sixth of 18, which is 3; and since 

 the last term is 0, I should add to 3, the second ordinate 23, 

 double of 28, triple of 30, 4 times 30, and so on, which would 

 give 397. To the half of 18, 1 should next add, 23, 28, &c. the 

 result of which would be 141; now dividing 397 by 141, and 

 multiplying by 20, 1 should have 



* 397 * 20 



H or = 56 feet 4 inches nearly.* 



When we once know how to determine the centre of gravity 



of any section of a solid, that of the solid itself is easily found. 



Hence, by means of what is above laid down, we can determine 



the centre of gravity of the hold of a ship, or of the space em- 



braced by the outer surface of a ship's bottom. Let it be pro- 



posed to find the distance of the centre of gravity of this space 



from the keel. We imagine it composed of several laminae par- 



*" 61, allel to the section at the water's edge. The bulk of each lam- 



ina will be equal to half the sum of the two opposite surfaces of 



>ra - this lamina, multiplied by their perpendicular distance, and the 



*See Bouguer, Trail* du Navire, p. 213. 



