40 Mr Larmor, On Rigidly connected Points, [Jan. 27, 



Draw 0/3, Oy to represent the velocities of 

 B and in magnitude and direction; divide 

 the fixed line /3y in v so that ftv is to vy as 

 BP to CP. Then Ov represents the velocity 

 of P in magnitude and direction, and 



R = k^- (15), 



pv . yv 



where k represents the constant factor 



(be — cb) sin w 



It is worthy of notice that altering the accelerations of B or G 

 only alters the value of the factor k ; so that the curvatures of 

 the trajectories of all points on BO are altered in the same ratio. 



We have thus from Fig. 2 a complete specification of the 

 velocity and curvature for any point P on BC; for the plane of 

 the trajectory is that parallel to the directions of motion of B 

 and 0. The acceleration of P may be constructed from those of 

 B and C in the same way as the velocity. The value of the 

 curvature involves the accelerations of B and C only as entering- 

 into k. A known value of the curvature at any one point de- 

 termines k once for all, and the values of these accelerations are 

 no longer necessary. A geometrical form for k is given by (21). 



The construction of Fig. 2 applies to any line in the moving 

 solids in so far as the determination of velocities only is con- 

 cerned ; for the determination of accelerations and curvatures it 

 applies only to a chord of the curve of inflexions, as above. 



If however the planes of the curvatures at any two points 

 on the line are parallel, it must be such a chord. This may 

 be established by an easy extension of the method of Fig. 1. 

 For taking B, to represent these two points, we have now 

 BB' and also CO' and BA curved instead of straight lines, and 

 we obtain a quadratic equation giving two positions of N on 

 AB' for which the curvature of the path of P is zero ; this gives 

 two points, real or imaginary, on BO, which are also on the curve 

 of inflexions. 



The formula (10) for the curvature should be reducible to 

 a form depending only on the geometry of the diagram. In 



fact, if BB' = h, CC' = k, B'BC = /3, C'OB = y, 



we have 



B'C' 2 =a? + h 2 + k* — 2ah cos /3 - 2ak cos y — 2hk cos m, 

 so that, as B'C = a, h is determined in terms of k by the equation 

 h 2 — 2h (a cos /3 + k cos &>) — 2ak cos 7 + k 2 = (16), 



