Now sin^ ^ = K (1 - cos 20) 



sin' (f) = 2,/Q- Vi cos 2<;6 + (1/8) cos 40 



sin' = 5/16 - (15/32) cos 20 + (3/16) cos 40 - (1/32) cos 60 



sin' = 35/128 - (7/16) cos 20 + (7/32) cos 40 - (1/16) cos 60 + (1/128) cos 80 



and these values placed in (46) give 



h = a (di - dj cos 20 + d, cos 40 - d^ cos 60 + d; cos 80) 



d=eV4-eV64- (3/256)6*^ -(233/16,384)e% 



< h < a - b (47) 



d, = eV4 + eVl6 + 7eV512 + 3eV2048 , 



dj = 5eV64 + lleV256 + 115eV4096 

 d, = 9eV512 + 37eV2048, d^ = 53eVl6,384 

 a, e are the semimajor axis, eccentricity of the reference ellipsoid. 



We now check (47) using the values of a and e for the Clarke 1866 spheroid. From (34) 

 and (47) with a = 6,378,206.4 meters one has h(meters) = 10,788.3852 -10,811.2646 cos 20 

 + 22.9147 cos 40 - 0.0350 cos 60. ^'^^ 



As a check, equation (48) should give 



h = a - b = 6,378,206.4 - 6,356,583.8 = ■l\fiTl.^ meters 

 when = 90°. Placing = 90° in (48) gives 



h = 10,788.3852 + 10,811.2646 + 22.9147 + 0.0350 = 21,622.5995 meters. 



Since we have the values of Q and for '^0f[,ax ^^om (6) we now check the value given by 

 (48) against the closed formula (43), 



cos Q ,, , , 

 h = a N(0). 



cos 



= 45° 02' 55';i06, cos = 0.70650624, cos 20 = - 0.00169788 

 cos 40 = - 0.99999423, cos 60 = + 0.00509360. 

 Q = 44° 51' 15':851, cos Q = 0.70890136, N(0) = 6,389,045.266. 

 cos Q 



- N(0) = (6,378,206.4) (0.70890136) / (0.70650624) - 6,389,045.266 



cos 



= 6,399,829.094 - 6,389,045.266 = 10,783.828 meters 

 Equation (48) gives 



h = 10,788.3852 + 18.3562 - 22.9146 - 0.0002 = 10,783.827 meters, 

 when = 0, h = and (48) gives 



h = 10,788.3852 - 10,811.2646 + 22.9147 - 0.0350 = + 0.0003 meter. 



Unless h were required to very high precision it is clear from the above checks that the 

 formula (48) is adequate. 



20 



