CHAP. VI.] MOMENT OF INERTIA. 77 



Hence &* = 5 2 = radius of gyration, or k / & x ^b 

 That is, the radius of gyration is a mean proportional between 



1 7 i * r 



-b and -6. 



The centre of gravity of the parallelogram is at O the inter- 

 section of the diagonals, and this is therefore the centre of the 

 central curve. 



If we suppose the parallelogram divided into laminae parallel 

 to D C, and suppose each lamina divided by G H parallel to 

 BC, the centres of gravity of each will lie upon GH. Right 

 and left of G H we then have a group of forces whose points of 

 application lie in lines parallel to G H, and the lines joining 

 any pair, one on each side of G H, are parallel. By (1) of the 

 preceding Art., therefore, GH and EF are conjugate axes of 

 the central curve. For the lengths of the half diameters, we 



find the mean proportional between-^ b and -% 5, -~ a and -^ , 



respectively, by the half circles B F and B H. We thus find k 

 and /', and can then construct the central ellipse directly, or 

 find the principal axes, and then construct it. The centre of 

 action of the moments of the parallelogram, with reference to 

 any axis parallel to A B, is as we have seen, Art. 60, the pole of 

 a line parallel and equally distant from O on the other side. If 

 we draw this line then, as D C, then from G draw two tangents 

 to the central ellipse, and unite the points of tangency by a 

 line ; the intersection of this line with O G is the centre of 

 gravity of the moments of the forces themselves considered as 

 forces, or area of the parallelogram, with reference to A B. 



2d. Triangle. PI. 11, Fig. 37. 



The moment of inertia of a triangle for the axis B C is 



j- a A 3 * whence J& = -~ A 2 , and for an axis E F distant i = 

 K A, which passes through fhe centre of gravity, 



? = tf-# = A 2 . (Art. 59.) 



* rh 



I a 

 Jo 



a? d x = ~ a A 3 , A being the line A D, a = B O. 



