160 



SITING AND COVERAGE OF GROUND RADARS 



15.6.11 



General Formula 

 for the Reflection Area 



In Figure 59 is shown an image antenna T" sending 

 radiation through a plane of indefinite extent. In 



Figure 59. Fresnel zones on the reflecting surface. 



order to simplify the calculations it will be assumed 

 that the distance from the reflection point to the 

 target is large, so that the rays from the Fresnel 

 zones may be considered parallel. With regard to 

 the transmitter distance, however, no such approxi- 

 mation will be made. 



hi = depth of the image antenna below the 



reflecting surface, in feet. 

 ^ = angle of the lobe considered with reference 



to the tangent plane at the reflection point, 



in radians. 

 m = number of the Fresnel zone. 

 m = for the center of the first zone. 

 m = +1 for the far edge of the first zone. 

 m = — 1 for the near edge of the first zone. 

 m = +2 or —2 for the edge between the second 



and third zones. 

 n = lobe number. For a given radar station n is 



related to the angle SF by the equation 



n = (4/ii/X) sin *. 

 X <= wavelength in feet. 

 d n = distance from the transmitter to the near 



edge for the Fresnel zone and lobe considered, 



in feet. 

 d f = distance from the transmitter to the far edge 



for the Fresnel zone and lobe considered, 



in feet. 

 di = distance from the transmitter to the center 



of the first Fresnel zone for a particular lobe, 



in feet. 



In Figure 59 is shown the first Fresnel zone for an 

 angle ^ with a corresponding value of n. Ray 2 

 passes through the center of the first zone and rays 

 1 and 3 pass through the near and far edges respec- 

 tively. Because of the great distance of the target 

 the rays 1, 2, and 3 are parallel. 



For the first zone the path difference between 1 

 and 2 is X/2. For zone m the path difference is 

 mX/2 (where m = 1, 2, 3, etc.). Since the points 

 di and d n are not equidistant from the target, the 

 distance Xi cos M* must be subtracted from ray 2 to 

 compensate for the increased path length of ray 1 

 above the plane. 



rrik 



= h- 



hi 



sin ^ 



— Xi cos ^ 



In the right triangle 



h 2 = xi* sin 2 * + f-A- 

 \sm ^ 



Xi cos ty 



Eliminating Zi from these equations and solving 

 for Xi 



— m\ cos^> + \/wi 2 X 2 + 4m\/ii sin^ 



2 sin 2 * " • (75) 



For the far point of the zone 



m\ 



= h - 



hi 

 sin ^ 



+ x 2 cos ^ 



also 



lr = x 2 2 sin 2 ^ 



hi 

 sin ^ 



+ X-2 cos ^ 



By a similar process of elimination of U and solving 

 for Xi\ 



m\ cos ^ + s/m-X 1 -+- 4mX/*i sin ^ ,_.. 



Xi = „ ■ ., , — - (7b) 



2 sin 2 * 



For the near point of the zone 



hi — raA cos^ + \/m 2 \ 2 + 4m\hi sin 1 !' 



tan^ 



2 sin 2 * 



Since sin ^ = nX/4/h and SF is small, cos ^ may be 

 taken as unity with the following error: 



