SITING AND COVERAGE OF GROUND RADARS 107 
When Y and W, are small 
¥+% = Foy, 
d 
and 
Yowo 2 ou, 
Ta 
and hence by subtraction 
ee ae 
Ta 
Since W is a small quantity rg may be taken to equal 
A + B,and A = d,, that is 
_ BS th 
1 B+ a 
This is the angle of the target with respect to the 
tangent plane CH as seen from the antenna. The 
angle. desired however is y, which is measured with 
respect to the horizontal at the antenna, GH. As 
shown in Figure 54 
y¥=Wa+t 0.: 
From equation (61) 
v, Vv. (70) 
~ REED ° 
The line of minimum path difference (A = 0) is 
along the earth’s surface from the transmitter to 
the horizon, and beyond it is along the line of sight 
tangential to the horizon since the direct and indirect 
waves are equal in that case. Maximum path differ- 
ence occurs directly below the antenna and is equal 
to 2h;. Since the path difference is also n\/2, the 
maximum value of 7 is 4h;/X. In practice the vertical 
directivity of the antenna limits n to a much smaller 
value. 
Consider a wave which is reflected from directly 
under the antenna, and let hy denote the height 
above the reflector at which the path difference is 
n\/2. Then 
(7) 
Ta + ho — (hs — hy) = =, 
or 
md 
4 . 
Thus if \ is 10 ft the center of the first lobe will be 
2.5 ft high at zero range. For most purposes the lobes 
and nulls may therefore be considered to start at 
the origin. 
To use this method it is best to arrange the calcu- 
lations in a tabular form. Points along the lobe center 
are selected by using various values of d, for the 
value of n desired. Next W is obtained from equation 
ho a 
(71) 
‘(68) and substituted in equation (69), and B and 
W are substituted in equation (70) yielding Ya, which 
is combined with 0 to obtain y. The curve of constant 
path difference is then plotted from y and ra, which 
are now known. 
Example 13. The General Lobe Angle Formula. To 
illustrate this method a radar 3,000 ft high and 
operating at 100 me will be used. A trial value of 
60 miles is arbitrarily selected for d, and substituted 
in equation (68), giving 
3,000 — 3 X (60)? 
v= 5,280 X 60 = 0.003788 radian . 
In equation (69) using n = 1 and d = 9.84 ft, 
1 X 9.84. 
10,560 SOD d 
B= TSO. = 70.85 miles . 
Dy ea 
2 X 60(0.003788) 10,560 
70.85 — 60 : 
Wa = 70.85 + 60 x 0.003788 = 0.000314 radian . 
60 ‘ 
¢= — 5,280 7 — 0.01136 radian . 
y = 0.000314 — 0.01136 = —0.01105 radian . 
60 + 70.85 = 130.85 miles . 
ll 
Ta 
Laying out the angle y from the antenna and 
marking off the distance ra gives one point on the 
curve of constant path difference. Enough other 
points are computed to enable one to draw a smooth 
curve. The computations may be arranged as shown 
TasLE 8, General lobe angle formula. (Example 13.) 
dh, vy, B, 
miles radians miles 
(m = 1) 65 .002800 696.0 
62 .003290 141.2 
60 003788 70.85 
58 .004300 44.60 
55 .005118 26.32 
50 -006628 13.45 
30 .016090 1.91 
(n = 2) 60 .003788 36 
58 -004300 386.0 
56 -004840 137.5 
55 .005118 101.0 
53 -005700 62.6 
50 .006628 36.78 
30 .016090 4.09 
radians 
.0023220 
0012810 
.0003140 
—.0005618 
—.0018050 
—.0038380 
—.0141600 
0031770 
-0020390 
-0015090 
0004733 
—.0010115 
—.0122300 
Wa, =) =, Td, 
radians radians miles 
.0123105 .00999 ‘761.0 
.0117456 .01046 203.2 
.0113636 .01105 130.85 
.0109850 .01155 102.60 
.0104166 .01222 81.32 
.0094698 .01331 63.45 
-0056820 .01984 31.91 
.0113636 36 bo 
-0109850 .007808 440.0 
.0106060 .008567 193.5 
-0104166 .008908 156.0 
-0100381 .009565 115.6 
.0094698 .010480 86.78 
.0056820 .017910 34.09 
