70 MOMENT OF INERTIA. [CHAP. 



to act in different directions, parallel to X X at the point 

 A t A,, etc. 



As before we have the force polygon 001234 for an arbi- 

 trary axis as X X, and from the corresponding equilibriur 

 polygon, we determine the statical moments with reference to 

 X X, 01' 1'2', etc., to the basis C 0. These moments we agaii 

 consider as parallel forces acting at A l A 2 , etc., for which we 

 have C 1' 2' 3' 4' and corresponding equilibrium polygor 

 C I IT HI',. etc. "We then determine the centre of action S, by 

 a second polygon 0" I II" Ht", etc., the sides of which are 

 respectively perpendicular to the first, according to the process 

 for finding the centre of gravity, Art. 30. The line joining O 

 with S gives the direction 0/"Y Y, the diameter of the curve 

 conj ugate to X X. To find the length of the semi-diameters 

 O b and O a, we must find the moments of inertia of the forces 

 with reference to X X and Y Y, taking the distances of the 

 points of application as measured parallel to these lines. 



Therefore instead of C 0, we must take C y as basis or pole 

 distance, and then find the radii of gyration as already indi- 

 cated in Art. 55, viz., O b' and O a'. These distances laid off 

 along Y Y and X X give the semi-conjugate diameters of the 

 curve of inertia. 



From the Fig. we see that the force P L whose direction from 

 left to right we shall always consider positive, and 2! P = 4 

 have the same sign. On the other hand the total moment of 

 inertia y and the moment of inertia of P t , viz., 01" have dif- 



C?/ 2 xO~y 

 ferent signs. The square of radius of gyration ^ = ^ p ' 



is therefore negative, the radius itself or the semi-diameter O b 

 is imaginary. 



In similar manner, we see thatOa the radius of gyration for 

 Y Y is real, since the total moment of inertia O x and P = 0^, 

 have the same signs. The curve is then a double hyperbola 

 with the conjugate semi-diameters O a and O b. 



It is then easy to find the assymptotes K K and J J, and by 

 bisecting the angle which they make, the principal axes A A 

 and B B. In order to find the length of these axes, we have 

 the well-known principle that for any point as a, the product 

 of a k and k O (a k being parallel to the assymptote J J) is equal 



1 /A 2 4-B 2 \ 



to -T the sum of the squares of the semi-axes ( j ). If then 



