88 



CALCULATION OF RADIO GAIN 



is greater. To obtain tlae correct result, the contours 

 for the diffraction field (Section 5.7.3) and those 

 obtained for the optical region should be merged 

 into smooth overall curves. 



For values oi r > 3, R = nr > 3, the tip and upper 

 part of the lowest lobe, and the higher lobes, by 

 equation (126), have a value of D close to 1. Con- 

 sequently, equation (110) reduces to 



n 



A = 24osin- 



(132) 



If n is given, sin 0/2 is determined and the calcula- 

 tion can be performed as a free-space calculation. 

 In terms of d, the equivalent free-space distance 

 do, corresponding to A is given by Figiu-e 3 in Chapter 

 2 and is equal to 



d = 2rfo sin — . 



(133) 



In the stated problem r > 3, so that the free-space 

 calculation suffices. This is given in paragraph (1). 

 In paragraph (2), the method including the diver- 

 gence is given. 



1. Free space. In the given problem, r = 9.4, so 

 that the simplified calculations should be sufficient. 

 The value of do corresponding to 20 log .4 = — 130 

 is found from Figure 3 in Chapter 2 to be 566 km. 

 Hence by equation (133) for n = 1, the true distance 

 d for complete reinforcement by the reflected wave 

 is twice as much and 



d = 2- (566) = 1,132 km. 

 For n = 1.2, 12/2 = 0.6Tr [from equation (116)], giving 



d = 2-(566)sin(0.67r), 

 = 1,076 km. 

 To find the corresponding height h^, a curve of the 

 type in Section 5.6.3 is needed for —20 log A versus 

 height. From this, the height corresponding to 

 20 log A = — 130 can be read. Alternatively, the 

 calculation given later in (2) will determine both 

 d and /h- 



If instead of A, either the radio gain or radar gain 

 is given, A is first found from equations (3) or (5). 



2. Calculation including divergence, n = 1. As in 

 the previous problem, r = 9.4 (from Figure 15 in 

 Chapter 6). A convenient value of p, approximately 

 equal to 1/r, is selected. In this case, we take 

 p = 0.1. This value is then improved by applying 



f(:p) 



Newton's method pi = p 



f'ip) 



to the ecjuation 



Kp) 



Vq 





(134) 



remembering that D is a function of p as given in 

 equation (117), where Vo is the value of v which corre- 

 sponds to the distance do at which the given value of 

 A would occur in free space. 



If 71 = 1, equation (134) reduces to 



- = - (1 + D). (135) 



Vo V 



The correction formula corresponding to equation 

 (134), denoting by pi the improved value of p, is 



Pi = P -' 



i) - I'd -J 



(1 - Dy + 4:D sin- iQ/2) 



3^^-^^|(l-D'0.v- 



pW- ApD 



•is:) 



(136) 

 The correction formula corresponding to equation 



(135) in = 1) is 



1 (1 + Z>) 



Pi = p 



p-D- 2p. 



2pv\l -p-J 



(137) 



Taking n = 1, do from A = 3X/8irdo (or Figure 3 

 in Chapter 2) is 566 km, and 



566 566 _. , 



Vo = = = 25.1. 



dr 22.5 



Taking p = 0.1 as an initial value as explained above, 

 then D is found to be 0.9788 [equation (117)]. 

 Using K = nr = 9.4 and equation (113), 



V = 165.6. 



Substituting these values in equation (135) gives 



Pi = 0.10385. 



Repetition of this procedure requires no change in 

 these five figures. Accordingly, for n = 1 and 

 A=-lSOdh, 79 = 0.10385. The corresponding 

 values of D and v are 



D = 0.9789 



V = 49.38, d = 1,110 km. 

 In (1) d was found to be 1,132 km. Now s = p/v 

 = 0.0021. From equation (121), 



ii=h= 2,898 



h = 86,940 meters. 



For n = 1.2, the calculation proceeds as for n = 1 

 except that equation (134) rather than equation (135) 

 is used: 



R = 11.283. 



