CHAP. VI.] MOMENT OF INERTIA. 79 



For the radius of gyration for G H, at a distance i -5- h 



7 we nave 

 a+b 



1 , 2 

 A 2 r- A 8 I -T- 

 9 \ a + b 



A 2 . 



fl 1 



= I To + o" 

 [_18 9 



This radius a is half the diameter along E F. 



To construct it, put (3 of = -^ h* + ~, r~ r\2 A 2 . 



:s (a + 0) 



Describe a semi-circle upon E F, and at the centre 1? and 

 at the intersection of the diagonals &, erect perpendiculars 



a 



Ol J and K L. Then F J 2 = - A 2 and K L 2 = 7-^ A 2 , since 



2> -- 



EK = ~AandKP= ^ ^ 



a + o a + 6 



K L equal to J M from J, we have 



E K = 1 h and K F = 7 A. If therefore we lay off 



a+b a+b 



and hence the half diameter sought is one-third F M. We 

 thus find 1 and 2. 



To find the other semi-diameter we have the moment of in- 

 ertia for E F and direction G H, ^ (a?+cP b+a 

 hence the square of the radius of gyration is 



1 2 



-^(a+l)h 



This last expression is easily constructed. In the right-angled 



triangle F B N, the hypothenuse F N = A/ (- j 2 -f -^ 

 B N being made equal to C E. If we describe then a semi- 



