380 CHARTS, GRAPHS, TABLES, AND COMPUTATIONS 



and at 20.0 km (from table 9.8) 



To.20.0 = (0.1165) (400.0) - 17.9573 ± 7.5131 

 = 28.6427 ± 7.5131 mrad. 



Thus, by linear interpolation 



Ti.2 = To.10.87 = 27.3973 + (28.6427 - 27.3973) 9q qq I |q qq ± 7.5227 



^ .„ _<^„„ „_.„.. 10.87 - 10.00 , 

 T (7.5227 - 7.5131) 20.00 - 10.00 



Ti,2 = 27.5056 ± 7.5218 mrad. 



Similarly, for the remaining 0's, 



ri,2(io mrad) = 13.9548 ± 0.9701 mrad, 



Ti,2(53.4 mrad) = 5.2186 ± 0.0817 mrad, 



Ti, 2(261. 8 mrad) = 1.2695 ± 0.0158 mrad. 



Determination of the bending by means of the graphical method (f) 

 of Weisbrod and Anderson yields, from figure 3.18 for 500 tan 6, for the 

 first layer. 



At do At ^0 At ^0 At do 



h(m.) = mrad = 10 mrad = 52.4 mrad = 261.8 mrad 



0.000 0.0 5.0 26.2 134.0 



0.340 3.0 5.8 27.0 134.0 



which yields for the bending in the first layer, 



At do At ^0 At ^0 At do 



= mrad = 10 mrad = 52.4 mrad = 261.8 mrad. 



11.67 3.24 0.66 0.13 



Similarly, the bending for the entire profile may be obtained, and shown 

 to be 



At do At do At ^0 At do 



= mrad =10 mrad =52.4 mrad = 261.8 mrad. 



24.42 14.00 5^32 iTlT 



