the excentricfrom the mean anomaly of a planet. 12 g 
into the proceeding, and therefore no correction by trial and 
error is requisite. 
Let ALP be the orbit of a planet, C the centre of the ellipse, 
S that focus in which the sun is placed, and AMP a circle 
described on the greater axis AP as a diameter. Let L be 
the true place of the planet, and AM the corresponding mean 
anomaly. Through L let the straight line ER be drawn, 
perpendicular to AP, and let it meet AP in R and the circle in 
E. Let EC, CM, SM be drawn, and let ST be perpendicular 
to EC, SG to CM and MN to ST. Then MN is parallel to 
ET, and NT is equal to the sine of the arc EM. It is easily 
proved, as in almost every writer on the subject, that ST is 
equal to the arc EM. 
In this Problem it is supposed that AC, CS, AM are given, 
and it is required to find AE the excentric anomaly, for AE 
being found, the true anomaly ASL is easily obtained. 
In each of the three following methods the angles CMS, 
CSM are used, and their difference is found by this pro- 
portion. 
MDCCCXVI. 
S 
