416 PROPAGATION THROUGH THE STANDARD ATMOSPHERE 
The radio gain is then given by: 
One-way: 
10 log = = 20 log A + 13.5 + 0; 
1 
The radar gain is then given by: 
Radar: 
10 log = = 4log A +.7.5-+27+10—3.5 
1 
= 40 log A + 41. 
These equations are evaluated in Table 6 and the 
one-way values are plotted in Figure 26. 
From Figure 41, 
20 log gi: = 18.5, 
20 log h; = 40. 
Therefore 
20 log A 
— 229 — 40 — 18.5, 
= —287.5 db. 
Referring to Figure 37, the pairs of values of sd and 
eh, along the contour 20 log A = —287 are given 
in Table 7. By means of Figures 40 and 41, sd and 
eh, are changed to d and he. The points found are 
to the right of the curve for eh; = 7.3, so that they 
correspond to points in the diffraction region. 
TABLE 6 TABLE 7 
Radio Gain | Radar Gain diem sd cha hz meters 
20 log A] sd d |20log A in db in db —— 
—170 | 6.2] 159 | —140 —127 —239 tae i ge ie 
—180 | 6.9] 177 | —150 —137 —259 140 1B 70 96 
- —190 | 7.6 | 193 —160 —147 —279 151 14 115 158 
—200 8.3 | 213 —170 —157 —299 161 15 16.5 226 
—210 | 9.0] 231 | —180 —167 —319 
—220 | 9.7 | 249 | —190 —177 —339 
aan ine ae san ee ms Radar: The value of 10 log (P2/P:) is the same as 
—250 |11.8| 305 | —220 —207 —399 for the one-way calculation, —159 db. This must 
—260 | 12.5 | 320 | —230 —217 —A419 be changed to 20 log A by equation (5), 
—270 | 13.2 | 340 | —240 —227 —439 
—280 | 13.8 | 357 | —250 —237 —459 20 log A = —141 db, 
and, as above, 
20 log A = —141 — 40 — 18.5, 
Type1V. The determination of contours along 
which the radio gain (or A) is constant (the coverage 
problem): A radar has a wavelength of 0.107 meter 
and a power output of 750 kw. Assume a receiver in 
space with a minimum detectable power of 107° 
watt. The maximum possible distance between the 
radar transmitter whose elevation is 100 meters and 
the receiver for varying heights of the receiver is to 
be found. The gain of the radar antenna is 10,000 
(or 40 decibels), the gain of the receiver will be 
assumed to be 30 decibels. 
For the radar problem, a target of radar cross 
section « = 50 square meters is assumed to take the 
place of the receiver. The minimum detectable 
power of the radar is taken as 10°’ watt; the range 
of the set for varying altitudes of the target will be 
calculated. 
One-way: The radio gain sought is the ratio 
of the minimum detectable power to the power 
output, or 
Pai Om ei 4 
P, 750X108? 3 
or 
10 log (P2/P:) = —160 + 1 = —159 db. 
From equation (8), 
20 log A = — 159 — 40 — 30, 
= — 229 db. 
— 199.5 db. 
Referring to Figure 37, we see that the contour 
20 log A = —199.5 is to the left of em = 7.3 [see 
caution in equation (181)]. Therefore it is not possible 
to get the necessary power return P: from the given 
target so long as it is below the line of sight. The 
desired contour would lie above the line of sight. 
The determination of the contour is discussed on 
page 400. 
Sea Water, VHF, 
Vertical Polarization 
1. Graphical Aids. Graphical aids are given in 
this section, which, as in text on p.413 are valid for 
all practical distances when both antennas are low 
[h < h, (see Figure 35)] and 2hihz << dd. If one 
or both antennas are elevated, they will give the 
value of the radio gain for the first mode, which is a 
good approximation for the result found by summing 
all the modes when the receiver is well within the 
diffraction region, i.e., when d > d;. [If one antenna 
is low (hk </h,) and one well elevated (h > 40h,), 
the result found by using several modes is given in 
Referring to equation (162) and Section 5.7.1, 
A = 2A,A iF .(H199'):(H199')2, (184) 
with gg’ = 1 forh < h,. 
