GEODESY. 



li 



To jret the area of the entire surface of the spheroid, make e^, = — i tt and <^2 



=: -|- ^ TT in (2). The result is 



Surface of spheroid = 2 tt aM i + -^^-^ Nap. log (^j~^j J- (3) 



For numerical applications it is most advantageous to express (3) in a series of 

 powers of e. Thus, by Maclaurin's theorem, 



Surface of spheroid = 47r<r(i — — — — "77 —•••)• (4) 



For the calculation of areas of zones and quadrilaterals it is also most advan- 

 tageous to expand (2) in a series of powers of e sin <^, and e sin ^^ and express 

 the result in terms of multiples of the half sum and half difference of ^1 and <^2- 

 Thus, (2) readily assumes the form 



Z=2iT a-{\ — e"-) [(sin c/>,, — sin </)i) + ^ ^' (sin^ <^2 — sin^ <^j) + . . . J. 

 From this, by substitution and reduction, there results 



wherein 



\ Ci cos <^ sin i A^ — C, cos 3<^ sin | A^ ) 

 Z = 2 77 I _|_ Q COS 5<^ sin f Ac^ — . i ^^' 



A<^ := (j!)2 — <^i , 



c^=2a-\Y- -^ - -g- ~ Ye ~ • ' • r 



a=2«^(f + ^ + o+. ..), (6) 



If Q be the area of a quadrilateral bounded by the parallels whose latitudes are 

 <^i and <f).> and by meridians whose difference of longitude is AA, 



^ 2 TT 



Hence, using the English mile as unit of length, (5) and (6) give for the 



adopted spheroid — 



Area of quadrilateral in square miles. 



( ^1 cos (/) sin I- Acf> — c. cos 2,^ sin | Ac/) ) 

 c^ = AA. (in degrees) i , j • r, ^ , t » 



( + '^r! cos 5^ sm § A^ — . . . ) 



log c,* = 5.7375398. (7) 



logr,^ 2.79173, 

 log 05 = 9.976 — 10. 



•^ = i (<?>. + «^i ). A.J) = <^, — </.,. 

 <f,, </).j = latitudes of bounding parallels, 



AA. = difference of longitude of bounding meridians. 



* ("i, r,, c^ are obtained from C„ C^, C3 respectively by dividing the latter by the number of 

 degrees in the radius, viz : 57.29578. 



