SITING AND COVERAGE OF GROUND RADARS avi 
For the values given in Figure 4 the critical height 
at d; is 
h _ 3,000 X 35 + 1,500 X 15 _ 15 X 35 
: 15 + 35 2 
= 2:287.5 ft . 
Since the hill P is 2,300 ft high it will interfere with 
radiation to Ps. 
Example 5. Remote Shielding—Computation. An 
important problem is illustrated in Figure 5. A trans- 
mitter is located at Pi, and it is desired to know if 
the nearby hill Ps shields the distant mountainous 
island. The height of the line P,Q’ at a distance of 
d,' from P; is denoted by h’ and may be obtained 
from the relation: 
he—-h’ hy —h’ 
d.’ dy,’ ? 
giving re dy'he — do'hi 
dy’ —d’ ~ 
For this case the rise Q’X’ due to earth curvature 
must be added to give the elevation X’ of the line 
of sight. The height of the lowest ray is therefore 
dy'he — do'hi, , di'd2! 
p= Shoo, aa @ 
For the values given in Figure 5 
_ 90 X 2,200 — 80 X 2,400 
=. 90 — 80 a 
= 4,200 ft . 
It is apparent that the hill P’ is shielded from Pi 
by the nearby hill P. except by diffraction. 
90 X 80 
2 
50 MILES 
Ficure 6. Vertical angle computation. 
Example 6. Vertical Angles. The slope of the line 
of sight at the near end (Figure 6) is given by 
_hz—-h__d 
O11 = “5 980d ~ 10,560 (9) 
6, is in radians and is measured from the horizontal 
at h,. The angle with respect to the horizontal at the 
far end is given by 
hy = he d 
92 = “5 280d 10,560" 
Thus for the values shown 
4, = ———____ — —— = — (0.0142 radian , 
5,280 X 50 
and 02 = 0.00474 radian . 
Example 7. Angle of Diffraction. One of the principal 
problems in connection with intervening obstacles is 
the computation of the angle of diffraction. In Figure 
7 the angle of diffraction is 0,. The line P,P is the 
geometrical shadow line; h, is the height of the 
Ficur: 7. Angle of diffraction computation. 
direct line P:P2 at the distance d, and is given by 
equation (7). 
h. — H 
Oa = 5 980d ° 
(10) 
For the values shown in Figure 7 
4,000 X 30 + 1,500 X 20-20 X 30 
Ne 20 + 30 2 
= 2,700ft , 
2,700 — 2,500 _ : 
6, = “5,280 X 20> 0.00189 radian . 
P, is therefore in the illuminated region. Had hz 
been 100 ft instead of 1,500 ft, h, would be 2,140 
ft and 
_ 2.140 — 2,500 
0a = 5280 X20 = —0.00341 radian , 
and P» would then have been in the shadow region. 
DIFFRACTION OF RADIO WAVES 
Introduction 
Whenever interference effects are important, the 
reflecting surface must be examined to determine to 
what extent the assumption of an ideal plane or 
spherical surface with uniform values for the ground 
constants is valid. This uniformity holds when the 
reflecting surface is the sea; but it is often not true 
over land areas, and especially at the coastline. 
More important deviations from the ideal case are 
roughness of the reflecting surface, such as waves on 
the sea, or irregularities of the land, such as hilly 
or broken terrain. This frequently causes diffuse 
reflection and virtual elimination of useful reinforce- 
ment of the direct ray. To deal with these practical 
terrain problems the methods of physical optics are 
employed. 
