SOLUTION OF THE EQUATIONS. 615 



Dividing this by 2?r and by 2 sin - = 4 sin ^ cos ^ we have for the 

 weight corresponding to this small area 



1 (!-*) tan J. 



Making a = l, we have at the pole log N= '0014542, log(l -e 2 ) = 9'9970916, 

 and log tan ^ = 8'3388563. 



Hence log (weight) = 8'0407347, and the weight = '01098335. 



To find the weight corresponding to segment of area bounded by a 



small circle round the pole of radius 12 30', we must add the areas of 



the belts for latitudes 85 and 80 to the small circle of radius 2 30'. 

 Thus the weight will be, 



for the pole, '01098335 log (to) = 8*0407347, 



for belt (a), '08773241 log(to) = 8'9431600, 



for belt (b), '17474415 log (to) = 9-2424026, 



weight for 12 = '27345991 log (to) = 9'4368937. 



Next suppose the polar segment to have a radius of 22 30'. Then 

 the areas of the next two belts (c) and (d) must be added to the above, 

 and we get 



27345991 



for belt (c), '26032372 log(to) = 9'4155137, 



for belt (d), '34377762 log (to) = 9'5362776, 



weight for 22 = '87756125 log (to) = 9'9432774. 



From the above equations of condition the coefficients of the final 

 equations for m and m = 1 for all values of n from to 10 have 

 been determined for these polar elements. These have .been added to the 

 coefficients of the corresponding final equations from (X) and (Z) or from 

 (X), (Y) and (Z) for the several belts of latitude so as, in each of the 

 cases taken, to include the whole surface of the Earth. The final equa- 

 tions for values of n from to 6 are given below. 



a a' 



In these final equations as given below, a stands for and a! for 



