64: MOMENT OF INEBTIA. [CIIAP. VT. 



to an axis from which the points of application are distant q l q%, 

 etc., is then 2 P j 8 . This is the product of three quantities, 

 one of which is measured by the scale of force, and the other 

 two by the scale of length. We can therefore regard it as the 

 product of the square of a certain length by the sum of the 

 given forces, or 2 P (f = P S P. We call k the radius of 

 gyration. 



In order to find the moment of inertia of a system of parallel 

 forces then, we must by the preceding Art. construct two force 

 and equilibrium polygons. If the pole distances are H and H', 

 and the segments into which the axis is divided by the produced 

 sides of the polygons are P\ P' 2 and P"! P" 2 etc., respectively, 

 then 



2 P f = H H' P" 

 and the radius of gyration is given by 



7 ,/~H~H 7 ~jiFP 7? 

 or, K = 4/ - * . 



2P 



This expression is easy to construct. Thus for example in. 

 PI. 11, Fig 33, let o n C be the first force polygon, o n the force 

 line, containing the forces P ; C the pole, and H the pole dis- 

 tance. Make o b equal to the second pole distance H,' and draw 

 b G parallel to n c and c t parallel to H. Then 



HH' 

 ct = h = -2P~ 



whence k = y h 2 P" 



If, therefore, in Fig. 33 (J), ra" ra" n is the segment of the 

 axis cut off by the outer sides of the second equilibrium poly- 

 gon, that is, if m" m" n = S P", we have only to prolong m" 

 m" n to L, making in" w" n L = A, and describe a semicircle 

 upon m" L, and erect the perpendicular m" k, which will be 

 equal to k. In general, the pole distance H and H' can be taken 

 arbitrarily, but it is often advantageous to take H (sometimes H' 

 also) equal to 5" P. Then 



= H 



We should then have in Fig. 33 simply to increase m"' m" n 

 by the second pole distance H', and then proceed as above to 



