05) Substituting (111) into (110) we have: 
  
  
  
  
  
  
  
t 
a-l 
tan (z a T 
06) ^. (a, o ‘a 1-0 t ,a-l | 
a € 1 - (1 +7) dt (112) 
a 
9 0 
a-l | 
tan 
A kz), o T° A t 2 Di 
07) a T pê-1 a TT toe (113) 
> a | 
a a 
o -T 
0 &-1l a p 
Ato nla), [00 AL (1s) | 
To e(T,-T,) 
98) or ' q^ - mé 
n &-l a p 
A, =A Lan Giu = a(T -T ) (115) 
& p 
where 
% 
A = p" T T. * LH 
)9) mé a 9 8 
O 
a=- à T T t4 LH 
y | p p 
From Figure 12 or 14 we have: 
0) TRL, 10, | (116) 
where tv 1s angular refraction, and A, is the refraction of an aerial observer, 
From 95 follows: 
m, rin (2), : 
n ” r, sin (z) (117) 
r ; 
with (116), setting + A 1, end with 7 being a small angle, 
a 
1) we may write (117) as: 
n 
='=1 +17 ctn (z), (118) 
P 
6T 
1 cor 3 i * > I ; 2 ie it. ; f : i ne non 3