Of KEPLER'S PROBLEM. 217 



o rnC A 



Also, fince fin 2.v = , f ^ ; we have 1 — fin 2v = 



r A 

 -cofA fm2 ~ 2 „A 



— tan 2 — : now let 2v zz ^ zz 90 — 2X 



1 -f- coi A A 2 



col 2 — 



2 



tnen ! — fin 2y =: t — cof 2X =: 2 fin 2 X : therefore fin X- ~ 



\ 1 A r 



tan - X -= = tan -~ x fm 45 . 



Inverting this analyfis, we derive the following rule for 

 computing the eccentric anomaly, when the mean anomaly 

 is a right angle : Take tan A == e X fee 45 ° ; then fin X 35 



tan — x fin 45% and p zz 90 ° — 2X. This rule would be ri- 



2 



gorous and. exact, if we could give to e the value it has in the 



m — p 

 fin — - — 



formula e zz S X- — '• buC as this cannot be- done, we 



mud be content with approximating to the arch fought as near- 

 ly as our purpofe may require. We will, therefore, by means 

 of the rule, compute a feries of arches, p, p', p", &c. by fucceffively 



r m — p m — p' 



fin — — £- fin L 



2 2 

 fubftituting for^, the values £, 1 X 5 s X , &c 



~{m—p) ~{m—p') 



and the feries p, p' 9 p", &c will converge very quickly to the. 

 exact value of the arch of eccentric anomaly, erring alternately 

 in defect and in excefs. Eor the arches here denoted by p, p\ 

 />", &c. are manifeftly no other than the double of the arches 

 formerly denoted by t, t, it", &c. refpectively. 



The cafe of the general problem that we have here refolved, 

 may be thus enunciated : " To draw a ftraight line from aa 



" eccentric 



