116 TECHNICAL SURVEY 
effects of particular terrain features will suggest 
methods of combining them to analyze a particular 
site. 
A large, flat land area will in general produce lobes 
and nulls at angles given py equation (57) with an 
envelope twice as large as the free space pattern. 
If the land area is not level, the lobe pattern will be 
tilted by the angle of the land. However, the problem 
is essentially a matter of diffraction since the land is 
of limited extent. Equation (16) should be used to 
determine whether the area is sufficiently flat to act 
as a regular reflector. 
If the land is flat from the antenna out to a 
distance d, the relative intensity of the reflected ray 
is 4% when d, = h; X cot y. This assumes the land 
beyond d; to be nonreflecting and that the distant 
boundary acts as a diffracting edge. As d; increases 
further, the relative intensity increases to about 1.18 
and then decreases again and oscillates about unity 
in gradually decreasing swings. This is accompanied 
by a variation of phase. 
Several typical terrain problems will be solved in 
detail to illustrate the methods. 
Example 17. Limited Reflecting Area. A 200-mc 
radar, Figure 64, with an antenna as described in 
he5O FEET #2 200 MG 
LAND REFLECTING FROM 600 TO 3000 FEET 
C F(¥)> 14° 
400 
ANTENNA 
Y 
200 
HEIGHT hy IN FEET 
IMAGE t= REFLECTING 
SURFACE 
ROUGH LAND 
OIFFUSE REFLECTOR 
OEeeeee eee ee eed 
Ce) 4000 8000 12900 
DISTANCE d, IN FEET 
Figure 64. Lobes from a limited reflecting area. 
Example 16, is 50 ft above a smooth reflecting surface 
(a lake) which extends from 600 to 3,000 ft. From 0 
to 600 ft and from 3,000 ft on is rough land. The 
shore line diffraction method will be used to deter- 
mine the effect of the réflection from the limited — 
area upon the antenna pattern f,. The vertical 
pattern is plotted from Figure 63 and shown dotted. 
To obtain the pattern for the reflected wave the 
shore at 600 ft is taken as a diffracting edge, and the 
relative intensity computed as a function of y as 
though the surface from 600 ft on were a perfect 
reflector. This is then repeated using the shore at 
3,000 ft. The difference between these two functions 
is then the effect of the area between 600 and 3,000 
ft. From equation (83) for n = 1 
600 XK 4.92 
ADS TSS GOP 
= 1.18; T 3,000 = 5.9 ; 
1 + 75. = 0.918 ; 
1.18 
Ye00 = x l= 
3,000 = 0.279 . 
From equation (78) 
4 X (50)? 
Neo0 = sarees = 3.39 > 3,000 = 0.678 . 
From Figure 27, using the plus sign for vgo0 and the 
minus sign for 3 99, is obtained the relative intensity 
2600 = 1.073; 23000 = 0.375. 
The reflection factor for n = 1 is given by 
= 1.073 — 0.375 = 0.698. 
From equation (57) 
_i xa 
TSZED 0.0246 radian = 1.41 
Other values are given in Table 11. 
TaBLE 11. Limited reflecting area. (Example 17.) 
n Y V600 U3,000 Z600 = 23,000 2 fr f(y) 
0 Oo +1.30 -+0.583 1.177 0.870 0.307 0.693 0.693 
0.1 0.14 +1.26 +0.498 1.181 0.810 0.371 0.658 0.658 
0.4 0.56 +1.15 +0.241 1.164 0.645 0.519 0.973 0.973 
0.5 0.70 +1.11 +0.155 1.153 0.592 0.561 1.147 1.147 
0.6 0.85 +1.071 +0.077 1.142 0.544 0.598 1.314 1.314 
1.0 1.41 +0.918 —0.279 1.073 0.375 0.698 1.700 1.650 
1.5 2.11 +0.727 —0.707 0.960 0.257 0.703 1.223 1.137 
2.0 2.82 +0.535 —1.136 0.838 0.186 0.652 0.349 0.314 
3.0 4.28 +0.155 —1.995 0.592 0.116 0.476 1.475 1.090 
4.0 5.64 —0.237 —2.853 0.393 0.082 0.311 0.689 0.386 
5.0 7.05 —0.619 —3.710 0.277 0.067 0.210 1.210 0.436 
The values of z multiplied by f, from Figure 63 
are plotted in Figure 65 as the reflected pattern. The 
DIRECT PATTERN- f, 
-_--= 
~ 
i REFLECTING 
a 
ee pees eS PLANE 
—— BLE -—— =S" —_—_— — — — ee 
naeeae 
NE Ass PATTERN - zf, 
° 0.2 0.4 0.6 0.8 1.0 
RELATIVE INTENSITY 
Figure 65. Components of the modified antenna for a 
limited reflecting area. 
resultant of the two vectors, f, and z2f4, in terms of 
n is given by the cosine law: 
fr= V1 + 2 = 2 cos (nm) . (94) 
Thus for n = 0.1 
fr = V 1+ (0.371)? — 2 X 0.371 cos (0.1m) = 0.658 
