454 Mr. A. Lodge on a New Geometrical Representation of 



and twice the triangle KPL is the product-coefficient ahout 

 the pair OK, OL. 



For, draw PM perpendicular to AK ; then PM = a cos 0, 

 MK = 5 sin ; hence 



PK 2 = a 2 cos 2 + b 2 sin 2 = k 2 . 



Similarly, PL 2 = Z 2 . 



Also LK (the base of the triangle KPL) = a + 6 ; and, if S 

 is the centre of the circle, SP = i(a— b), and the angle 

 KSA=20, therefore the height of the triangle 



= i(a — b) sin 20= (a — b) sin 0cos ; 



.*. twice area of triangle KPL = (a 2 — b 2 ) sin 0cos = h. 

 g ' Q.E.D. 



The sign of the product h is positive if the positive direc- 

 tions of OK, OL include between them the axis of minimum 

 moment (as in the figure), for in that case is positive and 

 less than a right angle, and a 2 —b 2 is positive ; or is nega- 

 tive and less than a right angle, and a 2 — b 2 is negative. 

 This condition is the same as the following : — If the centre S is 

 in the positive quadrant KOL, the product of inertia is positive 

 if 0, P are on opposite sides of KSL. If S is taken on one 

 of the axes OK, OL, when the product is positive P is in the 

 positive quadrant. 



If Q be taken on AB so that SQ = SP, the figure PKQL 

 is a parallelogram with sides equal to the radii of gyration 

 about K, OL, whose area equals the product-coefficient 

 about OK, OL, and one of whose diagonals is the sum of the 

 principal radii of gyration, the other being the difference ; 

 P, Q being the ends of the shorter diagonal. 



Hence, if the radii PK, PL, and the product-coefficient 

 about OK, OL are given, it is easy to find the radius SK of 

 the gyration-circle at 0, and to construct for the principal 

 axes. 



It is worth noticing that if PK is drawn perpendicular to 

 OK, LQ lies along LO, and P is on the positive or negative 

 side of LO, according as the product of inertia is positive or 

 negative. 



It is not difficult, being given the position of the centre of 

 area G, to construct for the central principal axes. For, 

 suppose G lies on OK ; draw GL' parallel to OL. Then the 

 radius about GK is known, the product of inertia about GK, 

 GL' is equal to that about OK, OL, and they are both there- 

 fore represented by parallelograms of the same height, and 

 with one pair of sides of the same length. The other sides of 

 the new parallelogram are of length equal to v'PL 2 — OG 2 , 



