ao4 A NEW and UNIFERSJL SOLUTION 



ced to this : " The mean anomaly being given, to find the ano- 

 " maly of the eccentric." 



2. Draw the ftraight line DF at right angles to the diameter 

 pafTmg through the point E : Then, fince the fedor ADE is 

 equal to the fedor ACM, and the fpace ACE is common to 

 both, the triangle EDC will be equal to the circular fedor ECM : 

 therefore, it is manifeft, that the ftraight line DF is equal to the 

 circular arch EM. 



Suppose that the radius of the circle is unity ; and let 

 w — arch AM, joi. = arch AE, and s r: eccentricity DC : Then, 

 fince DF =: e fin ^a, we fhall have this equation, expreffing the 

 relation between the arch of mean anomaly and the arch of ec- 

 centric anomaly : 



711 — ^ =z £ fin ^. 



3. In the equation juft found, let us put m =. in and ^ — 2 f : 

 and, remarking that fin fx, — fin 2v =: 2 fin v X cof v, we fliall 

 readily obtain, 



« — I' IT £ fin K X cof V : 



And, if we further fuppofe f = s X ^-^^-^ % and, by means 



of this formula, exterminate s, we fliall find 



fin (« — v) =: ^ fin c X cof V. 

 It may be remarked here, for the greater precifion, that, 

 from the nature of the problem, the arches, m and ^o., never ex- 

 ceed 180° ;»and, of confequence, the arches « and v, never ex- 

 ceed 90°. 



4. If we confider ^ as a known or given quantity, it is evi- 

 dent, that the equation laft found will no longer have the form 

 of a tranfcendental equation; and that the arch f will be de- 

 termined, when the arch n is given, by a finite equation, refolv- 

 able by known methods. 



In ftridnefs, indeed, we cannot confider ^ as a given quan- 

 tity ; for the exa(ft value of e depends upon the arch v, and 

 cannot be known unlefs the length of that arch were known ; 



ia 



