80 MOMENT OF LNEKTIA. [CHAP. TL 



circle upon F U = -~ F N, and make FW^^FN, FVis the 



2i o 



semi-diameter sought. We thus find 3 and 4, and can now 

 construct the central ellipse. This being constructed we can 

 find the centre of gravity of the moments with reference to any 

 axis parallel to A B or E F, according to Art. 60, or the moment 

 of inertia for any axis through 0, by drawing a parallel tangent 

 to the ellipse. The distance from to the point of tangeucy 

 gives then the radius of gyration for that axis. 



th. Segment of Parabola. PL 11, Fig. 39. 



Let the segment be limited by B C = 2 A, and A D = ?. 

 Then it is evident that these two axes are conjugate (Art. 61), 

 and the centre of the central curve is 0, the ratio of A to 

 OD being as 3 to 2. Hence AD and E'F', parallel to CD 

 through 0, are conjugate axes of the central curve. To find the 

 length of the semi-diameters along these axes we find first the 

 moment of inertia of the segment with reference to an axis Y Y 

 parallel to E' F' and tangent to the parabola at A. We have 

 then for this moment of inertia 







where p is the parameter of the parabola, and I = A D. Since 

 the area of the seg 

 radius of gyration 



4 

 the area of the segment is -^ h I, we have for the square of the 



_ 

 r c 



The square of the radius of gyration then foi E' F' whose 

 g 



distance from A is i = -= I is 

 5 



2 _ J5_ 3 73 9 72_ 12 72 



"7 "25 -175*' 



a being the semi-diameter along AD. It is easier here to com- 

 pute a, viz., a = 0.26186 I, and lay it off from O, thus finding 

 3 and 4. 



For the other semi-diameter we find the moment of inertia 

 for A D and the direction E' F'. Thus 



