82 MOMENT OP INERTIA. [CHAP. VI. 



principle of Art. 60. A little consideration will show that these 

 centres of gravity will coincide approximately with the centres 

 of gravity of the areas themselves, except for areas (3) (4) (5) 

 and (6). Finding then for these areas the centres of action of 

 the moments considered as forces, we construct the equilibrium 

 polygon O' I' II'. . . . VIII'. The distance 0" 8" cut off by the 

 first and last sides of this polygon gives the moment of inertia 

 to the pole distances H and H' and the reduction base a. Thus 

 0" 8" measured to scale of force and multiplied by a H H' is 

 the moment of inertia of the cross-section with reference to o S. 



The radius of gvratio i is then k \ = . 



a08 



The division will be performed if we take H' = 8 = 2 P. 

 This we can easily do now without drawing a new polygon, 

 since what is required is the intersection of the outer sides only. 

 Thus take a new pole C/ distant from o S, H' = 8. Now we 

 know that each side of the new polygon for this pole distance 

 will intersect the corresponding side of the first in a line paral- 

 lel to o CY [Art. 27]. Since the new polygon may start from 

 any point, we may take the first side to coincide with O VIII'. 

 Then the line of intersection of any two sides is O VHE' 8". 

 Produce any side as IV 7 V to intersection e with this line ; from 

 e draw e a^ parallel to C t ' 4'. 



Through a' the intersection of o' 1' and V IV, the resultant 

 of (1) (2) (3) and (4), must pass. The change of pole cannot 

 affect this resultant, which must therefore pass through a/, the 

 intersection of e a/ with the vertical through a' parallel to o S. 

 .Hence <V a/ is the direction of the last side of the new poly- 

 gon, and 8"Qi" is the moment of inertia for the new pole dis- 

 tance o Ci = 8. The radius of gyration then is k = V H O/'S". 

 In other words, k is a mean proportional between H and 0/'8". 

 The construction of k is given by the serai-circle described upon 

 0^'S" + H. The ordinate to this semi-circle through O/' per- 

 pendicular to oS gives k. We thus find the semi-diameter 

 S a = S a' of the central ellipse. 



In order to find the other semi-diameter S b = S b', we might 

 divide the cross-section into areas by lines parallel to S X, and 

 then proceed as above. This is, however, unnecessary. "With 

 the same areas as before, we can find the central curve for that 

 area on each side of XX, and then the centre of application of 



