CALCULATIONS FOR OPTICAL-INTERFERENCE REGION 



89 



The value of p by successive approximations in 

 equation (136) is found to be 

 p = 0.08127 

 V = 47.35, d = 1,065 km 

 D = 0.9851 

 s = 0.00184 

 h2 



h 



= 2,772 



7!.2 = 83,160 meters 

 3. Vertical polarization, p ^ 1, (f) ^ 180° {sea 

 water). The procedure given here is first to find 

 the position of the point for a given n assuming the 

 reflection coefficient to equal —1 and then find the 

 shift caused by the change in the reflection coeffi- 

 cient. For the most important case, n = 1, the new 

 maximum distance is given by 



d = do (1 + pD), (138) 



where do is the free-space distance corresponding to 

 the given value of A. p is found from the value of •■{/, 

 the grazing angle at the reflection point given by p 

 found above in the calculation for p = 1 and 

 (j) = 180°. For the new contour tip p changes, but 

 the angle ^ does not change sharply, so that p and (f) 

 as found for the p obtained by the simplified calcula- 

 tion is a close approximation. 



For p ^ 1,4> ^ 180°, a = 5 + cj) - T [see equa- 

 tion (29)] consists of two parts. 5 = irR/r = (A/X) 2x 

 = 7iT and (j)' = 4> — IT. Writing U = A''7r, N the lobe 

 number for the case p 7^ 1, ({> ^ 180°, the require- 

 ment for the new lobe tip is A^ = 1; for the first 

 maximum 9. = ir = R-jr/r -]- ip — ir where R is the 

 path difference A'ariable at the new lobe tip. Hence 

 the value of R at the new tip is given by 



R = r (2 - 0/7r). (139) 



Using the results in (2), 



p = 0.10385, 

 D = 0.9788, 

 r = 9.4, 

 do = 566, 

 we find from Figure 24 or equation (107), ;/- = 0.725°; 

 from Figures 14 and 15 in Chapter 4, p = 0.76, 

 = 170°. Substituting in equation (138), 



d = 566(1.744) = 987 km. 

 From equation (139), 



R = (9.4) (1.056) = 9.92. 

 Also 



The above value of p can now be improved by 

 substitution in equation (137) with D replaced by 

 pD which gives 



p = 0.0985. 

 In the calculation just made, p has been assumed 

 constant. This can be checked by finding \p as de- 

 termined by the new value of p. The result is 

 \l/ = 0.766° and the corresponding values are p = 0.74 

 (as against 0.76 previously) and 4> = 169° (as against 

 170°), Avhich is good enough. We now find 



^ ^ p ^ (10985 ^ 0.00224 



43.9 





987 

 22^5 



= 43.9. 



- = 2,359 



h-i = 70,770 meters. 

 Hence the new maximum point of the lobe is at a 

 distance of 987 km and at an elevation of 70,700 

 meters, as compared with a distance of 1,110 km 

 and height 86,900 meters for perfect reflection. 



5.5.6 Maximum Range Versus Receiver 

 (or Target) Height 



If the value of A has been determined by using 

 the minimum detectable power in equation (3) 

 or (5), the corresponding contour is a curve of 

 maximum range versus receiver height for com- 

 munication or maximum range versus target height 

 for radar. Generally, the lower part of the lowest 

 curve (n < 1) is of greatest interest. If A is suffi- 

 ciently small (i.e., 20 log A numerically large and 

 negative), the complete contour has points below 

 the line of sight. If the transmitter antenna is low 

 (hi < 30X^'^^ meters), the lower points are likewdse 

 given by the diffraction formula, discussed in Sec- 

 tions 5.1.7 and 5.7.3. Several such curves, the lower 

 part of the lowest lobe corresponding to various 

 transmitter heights for X = 1 meter and 20 log A 

 = — 130, are given in Figure 27. 



Consider, for example, the curve for hi — 2 meters. 

 The uppermost points of the curve correspond to 

 the tip of the lowest lobe and were found by the 

 procedure used in Section 5.6.5, putting n = 1. 

 The lowest points were found from the diffraction 

 formula by the method of Section 5.7.3 •with the 

 aid of Figures 31 to 36. It has been pointed out 

 that for n < 1, the optical interference formula is 

 inadequate. 



To locate a point between the upper extreme 

 (n = 1) and the diffraction points, a curve of A 



