CHAP. VI.] MOMENT OF INERTIA. 65 



It is to be remembered that q q z , etc., the distances of the 

 points of application of the forces from the axis, may be meas- 

 ured in any direction, and H is parallel to this direction, and 

 is not therefore necessarily perpendicular to o n. 



The above will be rendered plain by reference to Fig. 34. 

 PI. 10. We suppose four forces applied at the points A t A a .A s 

 A 4 respectively, and acting parallel to X X. Required the mo- 

 ment of inertia of these forces and the radius of gyration, the 

 distances q q^ etc., being measured parallel to Y Y. First we 

 form the force polygon by laying off along X X, 1, 1 2, 2 3, 

 3 4, parallel, and in the direction of action of the forces, choos- 

 ing a pole C, and drawing C 0, C 1, C 2, etc. "We now construct 

 the corresponding equilibrium polygon, C I, III, II III, HI IV, 

 etc. The segments 1', 1' 2', 2' 3', etc.,, represent the statical 

 moments of the forces with reference to X X. That is, these 

 segments to the scale of force multiplied by the pole distance C y 

 parallel to Y Y to the scale of distance, give the statical moments 

 of the forces. Now we take these segments themselves as forces, 

 and suppose them acting at the former points of application. 

 With the same pole as before we draw CO, Cl', C2', etc., and 

 form the corresponding equilibrium polygon CI, III', II III 7 , etc. 

 The sum of the segments of XX cut off by the outer lines of this 

 polygon, or o y, to the scale of force multiplied by H H' or C y* 

 g'ives the moment of inertia of the forces with respect to XX. 

 This moment then is M = y x C y* 

 where Oy = 2P" and "Op = H H'. 



The radius of gyration k is, as we have seen, given by 



/ 



= \l 

 v 



IT XT' 



* SP being equal to 4 in the Fig. Hence 



04 



If, then, we lay off d = 4, and make c = C y, and make 

 the angle dee & right angle, we shall find a point e to the right 



C ?/ H H' 



and Oe will be equal to -^- = -y^p. Upon ey now describe 



a semi-circle, the point of intersection V with the perpendicular 

 through will give (Art. 55) 



/Oy x Cy /HH'^P" 



06 == Y Q^ ' = Y -yp = k = radius of gyration. 



The square of this line, then, multiplied by 2 P or 4, will give 

 5 



