376 CHARTS, GRAPHS, TABLES, AND COMPUTATIONS 



should a surface duct have been present, it would have been necessary to 

 calculate the angle of penetration, 



dp = V'2 [N, - Nh - 156.9 (Ah) (in km)], 



to find the smallest initial elevation angle that yields a non-trapped ray. 

 Any initial elevation angle less than 9p cannot be used in bending calcula- 

 tions. 



Schulkin's approach (a) of (3.13) yields the results shown in table 9.19 

 for mrad, table 9.20 for 10 mrad, table 9.21 for 52.4 mrad, and table 

 9.22 for 261.8 mrad, where dk+i is determined from (3.58) using 

 Tk = a -\- hk, and a is the radius of the earth, 



a = 6370 km. 



It should be remembered that dk = 0, 10, 52.4, or 261.8 mrad only for the 

 first-level calculation, and that thereafter 6k is equal to the dk+i computed 

 for the preceding layer; e.g., for the second layer of table 9.19 

 (^0 = Omrad), dk = 6.15 mrad,which is the ^t+icalculatedforthe first layer. 

 The exponential model solution (b) may be found by using tables 

 9.10 through 9.17. Interpolation will usually be necessary for A''^, do, and 

 height; this interpolation may be done linearly. In practice, one of these 

 three variables will often be close enough to a tabulated value that inter- 

 polation will not be necessary, thus reducing from 7 to 3 the number of 

 interpolations necessary. Since in the problem forA'^s = 404.9, h = 10.0 km 

 and 00 = 10 mrad 



To. 10. 0(10 mrad) = 15.084 uirad 



and at A = 20.00 km, do = 10 mrad. 



To, 20.00(10 mrad) = 15.946 mrad, 



and thus by linear interpolation for h = 10.870 km, do = 10 mrad, 



TO, 10. 870, (10 mrad) = 15.084 + (15.946 - 15.084) 20 00 - 10 00 

 = 15.159 mrad. 



