396 : PROPAGATION THROUGH THE STANDARD ATMOSPHERE 
which, solved for p, becomes 
pale NAS 
123 
7 (123) 
IfkR>2, 
1 
~=. 124 
pee (124) 
If R <2, 
3-—R 
~~ + 125 
D i (125) 
For R > 3, 
Dw~1l. (126) 
The calculations will be divided into four types. 
Type I. The direct calculation of the radio gain 
(or field) when the heights and distance apart of the 
antennas and the wavelength are given. 
Type II. The calculation of the radio gain as a 
function of the receiver height he when the trans- 
mitter antenna height h;, distance d, and wavelength 
d are given. 
Type Ill. The calculation of the radio gain as a 
function of the distance d when the transmitter 
antenna height h;, receiver height he, and wavelength 
are given. 
Type lV. The calculation of the possible positions 
of the receiver in space (h2,d) when the radio gain 
will have the given value for given values of the 
gain factor A, the transmitter antenna height h,, and 
the wavelength . Special cases, such as the re- 
ceiver antenna height, hz for given d, or d for given ho, 
can be solved by use of the curves in Type II and 
Type III. 
This type of problem is of importance in estimat- 
ing the range of a set when the minimum detectable 
power of the receiver and power output of the trans- 
mitter are known. 
Problem of Type I. Radio Gain 
for Fixed Heights and Distance 
The radio gain at a given receiver is to be found, 
their heights, as well as the wavelength being given. 
The polarization is assumed horizontal, the 
effective earth’s radius 4/8. 
The following data are assumed. 
Transmitter height: h; = 50 meters 
Receiver height: he = 1,500 meters 
Distance apart: d = 100 kilometers 
Wavelength: \ = 1 meter (f = 300 mc) 
Gains (over doublet): Gi: = Gz = 100 (20 db) 
Onr-Way TRANSMISSION 
1. d;, = 188 kilometers (Figure 2). d <d,, so 
that the receiver is in the optical region. 
2. The u,v coordinates of the receiver are 
dp = V2kah, = 29,100 meters 
pees A are 
dr 29.1 
3. Referring to Figure 19, s (= p/v) is estimated to 
be 0.05. Since the result is very sensitive to slight 
changes in s, it is desirable to improve the value of s. 
In Newton’s method, the next approximation, using 
equations (101) or (119), is 
ae — 3 — 
o=3— Lia) ee SSC (127) 
Fs) a¢—gs—t[ 144 _ 4] 
2 ve 
s’ = 0.04794. 
The next approximation gives the same result. 
4. Using the above value of s’ and the relation 
p = sv (v = 3.48), pis equal to 
p = 0.1645. 
With the value of p = 0.1645, equation (117) and 
Figure 22 give the value 
D = 0.95. 
5. The path difference variable R is obtained from 
equation (112) or Figures 21, 22, 23, as 
R = 5.477. 
6. The number 7, from equation (115), is 2.91, so 
that the lobe number 7 is 
= 1.88. 
n= 
i) 
Hence the receiver is on the upper part of the first 
lobe close to a null. 
7 Q = nm = 5.9 [see equation (116)], 
© _ 995, 
y 
sin? 2 = 0.0353. 
2 
8. To use equation (110), the value of the free- 
space gain factor Apis needed. Figure 3 in Chapter 2 
gives 
20 log Ay = — 118. 
Substituting in equation (110), 
log A = 20log Ao + 10 log (0.0025 + 0.1342) 
= —118 —8.642 —127 
By equation (8), using gains of 20 db, 
10 log = 2 pyar 4 27, 
1 
Accordingly the radio gain is —87 db and the received 
power is given by 
: P, — P,10°°”. 
Suppose the receiver has a minimum detectable 
power of 10° watt, then the required minimum 
