CALCULATION OF RADIO GAIN 401 
In Chapter 6, purely graphical methods of de- 
termining the contours (lobes) are given. Here we 
are concerned with finding individual: points on the 
contour. Thus, for example, n = 1 if the tip of the 
lowest contour is wanted (as in range determination). 
Points near the tip require values of n near 1, such as 
n = 1.2 or n = 0.9. For the next higher lobe, the 
tip of the lobe corresponds to n = 3 and so on. The 
nulls are at n = 0, 2, 4,.... However, while the 
optical formula gives a null at n = 0, that is, near 
the line of sight, the true value of the field, or of A, 
is greater. To obtain the correct result, the contours 
for the diffraction field (pp 413- 416) and those 
obtained for the optical region should be merged 
into smooth overall curves. 
For values of r > 3, R = nr> 38, 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 
Amom sin 5. (132) 
If n is given, sin 22/2 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 Figure 3 in Chapter 
2 and is equal to 
a edasin - (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 dp corresponding to 20 log A = —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, 0/2 = 0.67 [from equation (116)], giving 
d = 2.(566) sin (0.67), 
= 1,076 km. 
To find the corresponding height he, a curve of the 
type on paga 397 is needed for —20 log A versus 
height. From this, the height corresponding to 
20 log A = —180 can be read. Alternatively, the 
calculation given later in (2) will determine both 
d and he. 
If instead of A, either the radio gain or radar gain 
is given, A is first found from equations (8) or (5). 
2. Calewation 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 
Newton’s method p; = p — Sp) to the equation 
J'(p) 
eS ee Mi ep 22 
S(p) = ae ql (1 — D)? + 4Dsin Be (134) 
remembering that D is a function of p as given in 
equation (117), where % 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 n = 1, equation (134) reduces to 
ea)! (135) 
Vo v 
The correction formula corresponding to equation 
(134), denoting by p; the improved value of 7p, is 
v— Uo 
Ja — D)? + 4D sin? (2/2) 
oe 3(1 — | a +f) 
pp?! 4pD ee ales AG —p? 
(136) 
Pi =p 
The correction formula corresponding to equation 
(135) (n = 1) is 
b_@£5) 
Vo Vv 
Pi=?p ee (ee a (137) 
p?D? 2p \1 — p? 
Taking n = 1, d) from A = 3d/87d) (or Figure 3 
in Chapter 2) is 566 km, and 
Taking p = 0.1 as an initial value as explained above, 
then D is found to be 0.9788 [equation (117)]. 
Using R = 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 = —130db, p = 0.10385. The corresponding 
values of D and v are 
D = 0.9789 
y = 49.38, d = 1,110 km. 
In (1) d was found to be 1,182 km. Now s = p/v 
= 0.0021. From equation (121), 
u = — 2.898 
h 
1 
he = 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. 
The value of p by successive approximations in 
