86 TROPOSPHERIC REFRACTION 



angle CQP = 90° + 6, we have, from the law of sines, 



r sin (90° + 6) = r sin (90° - 0) = (r + dr) sin xj^. (3.64) 



Eliminating sin i/^ from (3.63) and (3.64) we then have 



(n + dn)(r + dr) sin [90° - (^ + dd)] = nr sin (90° - 6), 

 or 



(n + dn)ir + rfr) cos (0 + dd) = nr cos 6. (3.65) 



Multiplication of the (n + (in) and (r + dr) terms, ignoring differential 

 products, yields 



{nr + ndr + rdn) cos (0 + dd) = nr cos 0, 

 or 



{nr + n(/r + rdn) [cos ^ cos ((/0) — sin 6 sin (c?0)] 



= nr cos 0. (3.66) 



Since cos {dd) = 1, and sin {dd) ^ c?^, another multiplication, again ignor- 

 ing products of differentials, yields: 



ndr cos d + rdn cos d — nr sin d dd = 0, 



or, dividing all terms by nrcos 0, 



- + — -ta^nddd = 0. (3.67) 



r n 



Now if (3.67) is integrated between any two thin layers of refractive 

 indices ni and n2, whose radial distances from the earth's center are ri and 

 r2, and the initial elevation angles of a radio ray entering the layers are di 

 and 02: 



dr . dn / . «7^ . ^2 , , n2 , , cos ^2 ^ 



[- / — — / tan ddd = (n j- ^n — + ^n = 



, r Jn, n Je, ri ni cos ^i 



or, taking antilogs of both sides, 



r2n2 cos 02 

 rini cos di 



= 1, 



