244 On finding the Length of a perpendicular Degree, 



tity far below the errors of observation. In this instance, and 

 in all those where this method has been actually used, or where 

 it can be supposed to apply, we may conclude that the angle 

 of the spherical triangle opposite to the base /3, is equal to a>, 

 the difference of longitude of the two stations. But in every 

 spherical triangle, the sines of the angles are proportional to 

 the sines of the opposite sides ; and hence, on account of the 

 equations (x), we learn that m is the angle of the same triangle 

 opposite to the side 90 — \J/, and m' the angle opposite to the 

 side 90— vp. 



Having obtained a knowledge of all the parts of the sphe- 

 rical triangle, if we apply to it one of the analogies of Napier, 

 we shall get, 



*-*' 



cos — - — . , 



rp. « 1 m-\-m 



Tan T = . w x cotan -y-. 

 sin-^- 



Now this formula is different from the method in the Trigo- 

 nometrical Survey in no other respect, except that the arcs \J/ 

 and \J/, which are what are called the reduced latitudes, come 

 in place of the true latitudes k and X\ But as the reduced 

 latitudes depend upon the excentricity of the spheroid, it fol- 

 lows that the difference of longitude is no more independent 

 of the figure of the earth in this mode of computation, than in 

 any other. Because the difference of latitude is very small, 



we may write cos ~~ , or even the radius of the tables, in- 

 stead of cos ■ ~ : and I have found, by reducing properly 

 and putting e == — , 



e cos 8 — — I. 



Sin^p- = sin-^- x(l- 



The foregoing expression will now become 



x— x' 

 cos / 



Tan T = , x+x > x cotan-^— x(l+ e cos*-±-), 



sin-y- 



or, in logarithms, 



Logtan. - = log! x+x , X cotan -T-J + Mecos 8 — . 



This, then, is the formula by which we must compute the dif- 

 ference of longitude on a spheroid of which s is the compres- 

 sion ; and if we make e = 0, it will coincide with the method 



in 



