204 A NEW and UNIVERSAL 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 

 pafling through the point E : Then, fince the fector ADE is 

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

 both, the triangle EDC will be equal to the circular feclor 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 

 m = arch AM, [a zz arch AE, and s zz eccentricity DC : Then, 

 fince DF rr e fin ^, we fhall have this equation, exprefling the 

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

 centric anomaly : 



m — [a, — g fin p. 



3. In the equation juft found, let us put m zz in and /i-2v: 

 and, remarking that fin p = fin 2v zz 2 fin v X cof >-, we fhall 

 readily obtain, 



n — v n e fin v X cof v : 



And, if we further fuppofe ezztX m ^ " — — , and, by means 



of this formula, exterminate g, we fhall find 



fin (n — v) — e fin v X cof v. 

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

 from the nature of the problem, the arches, m and p, never ex- 

 ceed 180 ; % and, of confequence, the arches n and p, never ex- 

 ceed 90 . 



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

 dent, that the equation laft found will 110 longer have the form 

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

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

 able by known methods. 



In flriclnefs, indeed, we cannot confider e as a given quan- 

 tity; for the exact: value of e depends upon the arch v, and 

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



in 



