THE CALCULATION OF VERTICAL COVERAGE 



153 



sqooo 



401000 



ui 20,000 



eo ioo 



DISTANCE IN MILES 



Figure 53. Lobe angles for Example 12. 



6 = angle between the tangent plane CE and the 

 horizontal at the antenna, in radians. This 

 angle is always negative. 

 ka = radius of the modified earth, 5,280 miles. 

 ^a = angle between the direct ray Td and the 



horizontal plane CE, in radians. 

 ^ = angle between the reflected ray A or B and 



the horizontal plane CE, in radians. 

 n = number of half -wavelengths path difference. 



Substituting 



A = 



n\ 



In the triangle ABrd (cosine law) 



(64) 



(65) 



r d = \/A 2 + B 2 + 2AB cos 2^ . 

 From the definition of path difference: 



A = A + B-rd = 2 X " 5 X 280 , 



A + B - A = VA 2 + B 2 + 2AB cos 2* ; 

 squaring and dropping terms that cancel out gives 



2AB - 2AB cos 2* - 2BA = 2AA - A 2 , 

 or solving with respect to B, 



AA - iA 2 



B = 



A{1 - cos 2*) - A 



(66) 



2 X 5,280 



into equation (66) gives 

 nk 



10,560 



B = 



1 / n\ V 



2 \ 10,560/ 



A(l - cos 2*) - 



nX 



(67) 



10,560 



Several approximations will be introduced to 

 simplify equation (67): 



A will be taken to equal di since ^ is of the order 

 of 3° or less. 



From Figure 54 it follows that sin ^ = hi/ 5, 280 A, 

 or for small angles, V = hi/ 5,280 A. 



Substituting for hi [equation (59)] it follows that 



* = 



hi - hdS 

 5,280di. 



(68) 



Using the approximation 



cos 2* = 1 - 2* 2 



