294 ON NEWTON'S SOLUTION OF KEPLER'S PROBLEM. [38 



drawing a line. For if we lay an edge of the ruler across the points EC 

 and then make a parallel edge to pass through the point M it will cut 

 the curve in the point P required. 



Thus we may always find a first approximate value of the eccentric 

 anomaly, without making repeated trials, whether the eccentricity be large 

 or small. 



I described this graphical method of solving Kepler's problem at the 

 Birmingham meeting of the British Association in 1849. It is referred to 

 in a paper by Mr Proctor in Vol. xxxm. of the Monthly Notices, p. 390. 



The construction is so simple that it has probably been proposed before, 

 though I have nowhere met with it. 



Note on Professor Zenger's solution of the same problem given in Number 9 of 



Vol. XLII. of the " Monthly Notices." 



The only peculiarity in this solution is in the mode of obtaining the 

 first approximate value employed. The subsequent approximations are carried 

 on by means of the first method given above. Professor Zenger's process 

 may be represented in a slightly different form as follows : 



We have x z = e sin x, 



and therefore 



sin (x z) = sin (e sin x} = e sin x -1 1 - e 2 sin 2 x + e 4 sin 4 x etc. [ , 



I O 1 ^U 



or sm(x z)=fsmx; 



where / = e 1 1 - - e 2 sin 2 x + e 4 sin 4 x - etc. . 



ID A i\j 



Hence tan (x-z) = - 



1 / 



cosz 



Now, an approximate value of / is e, and the error in the determi- 

 nation of tan (x z) if we were to put 



, x e sin z 

 tan(cc z)=- - , 

 1 e cos z 



would be of the 3rd order in e. 



