378 CHARTS, GRAPHS, TABLES, AND COMPUTATIONS 



Hence by linear interpolation, 



.,...,. ,. „„. = 23.560 + [24.442 - 23.5601 ^oiooO I loioOO 

 = 23.637 mrad. 



The bendings for ^o = 10, 52.4, and 261.8 mrad are as given below: 



do = 10; no, 870 (10) = 15.053 mrad, 

 do = 52.4; 710,870 (52.4) = 5.864 mrad, 

 do = 261.8; Tio, 870 (261.8) = 1.280 mrad. 



To use the departures-from-normal method (d) of determining bending, 

 it is first necessary to know the atmosphere which must be used for the 

 calculation. In the problem, 



dN I 

 _ -^ = 102.9 A^ units/km, 



Cl" I initial 



which is within the range of the N ^ = 450.0 exponential atmosphere, as 

 can be seen from table 9.18. Thus one will use table 9.17 to determine 

 the d's and the r's in the N s = 450.0 exponential atmosphere, or one can 

 use the exponential atmosphere tables [1]. 



For an A^s = 450.0 atmosphere to a height of 10.870 km: 



TNsio mrad) = 30.776 mrad, 



TNsiio mrad) = 19.414 mrad, 



rArs(52.4 mrad) = 7.024 mrad, 



T7Vs(26i.8 mrad) = 1.506 mrad. 



Equation (3.58) should be used for the 6 interpolation in preference to 

 linear interpolation, although, if no tables or other facilities are present 

 at the engineering site for easy acquisition of square roots, linear inter- 

 polation will suffice. Proceeding in table 9.17 with (3.58) for the first 

 layer at /i = 0.340 km and the ^o = mrad case: 



V^2 ^ 2(n - ro) ^ ^q6 _ 2(^^ _ ^^) 



ro 



= 6.388 mrad. The remaining d's for the various layers are shown in 

 table 9.23. To determine the value of A at the bottom and top of the 



