38] ON NEWTON'S SOLUTION OF KEPLER'S PROBLEM. 293 



This curve is commonly called " the curve of sines." It will be suf- 

 ficient to trace the portion of the curve for which the ordinates are positive. 



Let AOB be the line of abscissae, and let AO be taken equal to OB, 

 and let each of them be divided into 90 equal parts representing degrees 

 of angle. Let A N be any abscissa representing the angle x, and let the 

 corresponding ordinate NP = c sin x ; then the greatest ordinate will be 

 OC=c, corresponding to the abscissa AO. 



Suppose the curve line APCB to be divided into 180 parts which 

 correspond to equal divisions on the line of abscissae A NOB. 



Then if E be taken in AO so that .#0 = ^x57 '296 divisions, or if 

 AE = 90 e x 57 '296 divisions, and if CE be joined and PM be drawn 

 parallel to it through P meeting the line of abscissae in M, then AM will 

 represent the mean anomaly corresponding to the eccentric anomaly repre- 

 sented by AN. 



For, since the triangles PMN, CEO are similar, 



MN _PN _ 



EO ~ CO = X> 



and therefore MN = EO sin x = 57'296 (e sin x). 



Hence MN represents the number of degrees in x z, and therefore AM 

 represents the mean anomaly z. 



Conversely, if A M represents any given mean anomaly, then if MP be 

 drawn parallel to EC, it will cut the curve in the point P corresponding 

 to the eccentric anomaly. 



By the employment of a parallel ruler we may find the eccentric anomaly 

 corresponding to any given mean anomaly, or conversely, without actually 



