COMPUTATIONS OF ATMOSPHERIC REFRACTION 



379 



layer, one makes use of (3.46) or figure 3.17. First, however, one must 

 determine the value of c in (3.46) to be used. Usually interpolation will 

 be necessary in table 9.18, but in the N s = 450.0 case it is not possible, 

 and thus the straight N s — 450.0 exponential atmosphere values are used. 

 From (3.46) 



^(A^„ h) = N{h) + .¥, [1 - exp (-ch)], 



and for the layer running from h = io h = 0.340 km, figure 3.17 yields 



377.2 (1 - exp (cO)] = 0.0 



450.0 [1 - exp (-C X 0.340)] = 32.8, 



A (450.0, 0) = 400.0 + = 400.0 



A (450.0., 0.340) = 365.0 + 32.8 = 397.8, 

 whence 



AA = A (450.0, 0.340) - A (450.0, 0) = 397.8 - 400.0 = -2.2 A^ units. 

 Therefore, the departure term of (3.23), 



and 



and, therefore, 



and 



+ 



7/c+l L 



AAiNs) 





becomes 



+ 6.388 L 



2.2 



= +0.689 mrad. 



The remaining calculations are tabulated in table 9.23 for the ^o = mrad 

 case, in table 9.24 for the ^o = 10 mrad case, in table 9.25 for the ^o = 52.4 

 mrad case, and in table 9.26 for the ^o = 261.8 mrad case. The sum of 

 the departures for the mrad case is 





+ 



/fc+i L 



iN 



AA (N.) 



k+i 



N, 



= —5.335 mrad. 



Determination of the bending is required in part (e) of the problem by 

 using regression lines. By (3.10), using table 9.7 and 9.8, it is found for 

 the 00 = mrad case that at 10.0 km (from table 9.7) 



TO, ,0.0 = (0.1149) (400.0) - 18.5627 ± 7.5227 

 = 27.3973 ± 7.5227 mrad 



