120 



SITING AND COVERAGE OF GROUND RADARS 



di from Pi is denoted by h' and may be obtained 

 from the relation: 



hj ~ 

 rfl' 



giving 



h' = 



di'k, 



hi - h! 



di' - ds' ' 



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 



di'hi 



h = 



di'hi _ dVdY 



(8) 



rfY - di' ' 2 



For the values given in Figure 5 



, = 90 X 2,200 - 80 X 2,400 90 X 80 

 90-80 ' + " 2 



= 4,200 ft . 



It is apparent that the hill P' is shielded from Pi 

 by the nearby hill P 2 except by diffraction. 



4000FT 



1500 



50 MILES 



Figure 6. Vertical angle computation. 



Example 6. Vertical Angles. The slope of the line 

 of sight at the near end (Figure 6) is given by 



fci 



d 



5,280d 10,560 ' 



0i is in radians and is measured from the horizontal 

 at hi. The angle with respect to the horizontal at the 

 far end is given by 



hi — }n d 



5,280d 

 Thus for the values shown 



10,560 



1,500 - 4,000 



50 



and 



5,280 X 50 10,560 



2 = 0.00474 radian . 



= -0.0142 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 B d . The line PiP is the 

 geometrical shadow line; h c is the height of the 



Figure 7. Angle of diffraction computation. 



direct line P1P2 at the distance d\ and is given by 

 equation (7). 



_ hf -H 



d 5,280di ' 



For the values shown in Figure 7 



4,000 X 30 4- 1,500 X 20 20 X 30 



(10) 



K = 



20 + 30 

 = 2,700 ft , 

 2,700 - 2,500 



5,280 X 20 



= 0.00189 radian 



Pi is therefore in the illuminated region. Had A 2 

 been 100 ft instead of 1,500 ft, h c would be 2,140 

 ft and 



2,140 - 2,500 



= -0.00341 radian , 



5,280 X 20 

 and P 2 would then have been in the shadow region. 



(9) 1S * DIFFRACTION OF RADIO WAVES 



1541 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. 



