36 Excess on a Spheroidal Triangle. [No. 121. 



difference is not a matter of observation. But as the two triangles 

 have not absolutely the same excess, it may be worth while to ascertain 

 the precise value of each, and thereby what would be the difference in 

 any particular case. 



A solution may be obtained by means of the following principle given 

 by Mr. Airey as resulting from Dalby's investigation, namely, that the 

 excess on a spheroidal is the same as on a spherical triangle whose 

 angular points have the same geographical latitudes and longitudes. 



The arcs being small, (as is always the case in practice), it is an as- 

 sumption generally made and admitted, that the computations may be 

 made by means of the radii of curvature at their middle points, which 

 comes to the same thing as to reckon longitudes by a normal equal to 

 the mean of all the normals at the middle points of the three sides, 

 and latitudes by a radius of curvature equal to the mean of the meri- 

 dional radii of curvature at the same points : or simply or at once, by 

 the normal and meridional radius of curvature at the centre of gravity* 

 of the triangle. The surface of the spheroidal triangle would coincide 

 with that on a sphere if the differences of latitude were stretched out 

 in the ratio of the meridional radius to the normal, or if the longitu- 

 dinal differences were contracted in the inverse ratio. Hence the area, 

 and also the excess on a spheroidal triangle would coincide with those 

 on a sphere, if they be computed by a radius equal to a mean propor- 

 tional between the normal and meridional radius (i. e. the greatest and 

 least radii of curvature), at the middle of the triangle. This method 

 (which is probably the simplest possible), is true as far at least as 

 quantities of the 4 th order, and in the present as well as prospective 

 state of the arts and sciences, any thing farther may be thought an 

 unnecessary refinement. 



If r denote the radius of the sphere, and A the area of a spherical 

 triangle, the expression for the excess E reduced to its simplest form is 



A A 



E = in terms of radius unity, or in seconds E" = and 



r 2 r2 sin y 



if the area be assumed as equal to that of a plane triangle having the 

 same sides a, b, and contained angle C, the formula becomes 



* This expression may be objected to, as strictly the centre of gravity falls within 

 the surface. It may be understood as an abbreviated expression, denoting the point 

 of intersection of lines drawn from the angular points to bisect the opposite sides. 



