444 Lord Kelvin on Graphic Solution 



University, suggests a corresponding method for the solution 

 of dynamical problems. 



In dynamical problems regarding the motion of a single 

 particle in a plane, it gives the following plan for drawing 

 any possible path under the influence of a force of which the 

 potential is given for every point of the plane. Suppose, 

 for example, it is required to find the path of a particle 

 projected, with any given velocity, in any given direction 

 through any given point P (fig. 1). Calculate the normal 

 component force at this point; and divide the square of the 



Fig. 1. 



velocity by this value, to find the radius of curvature 

 of the path at that point. Taking this radius on the com- 

 passes, find the centre of curvature, C , in the line, P U K, 

 perpendicular to the given direction through P , and describe 

 a small arc, P PiQi, making P^ equal to about half the 

 leno-th intended for the second arc. Calculate the altered 

 velocity for the position Q l5 according to the potential law ; 

 and, as before for P , calculate a fresh radius of curvature 

 for Q! by finding the normal component force for the altered 

 direction of normal and for the velocity corresponding to the 

 position of Qi- With this radius, find the position of the 

 centre of curvature, Ci, in PiC L, the line of the radius 

 through Px- With this centre of curvature, and the fresh 

 radius of curvature, describe an arc PjP 2 Q 2 making P 2 Q 2 equal 

 to about half the length intended for the third arc ; calculate 

 radius of curvature for position Q 2 ; draw an arc P 2 P 3 Q 3 ; and 

 continue the procedure. This process is well adapted for 

 findino' orbits by the ( trial and error* method described in 

 my article " On Some i Test Cases ' of the Maxwell- 

 Boltzmann Doctrine regarding Distribution of Energy/' sect. 

 13; Proc. Royal Soc, June 11, 1891. 



