SAMPLE COMPUTATIONS 157 



4.5.4. Sample Computations 



It can be seen from figure 4.52 that 



hi < loss region < Ha + ho, (4.40) 



where hi can be determined from figures like 4.49 to 4.51, which depend 

 upon the station location and the season of the year; hA is the height of 

 the duct; and /i2 is determined from figures 4.47 and 4.48. The following 

 example will illustrate the determination of the shadow zone. 



As an illustrative problem, assume a 500-m transmitting antenna in the 

 spring of the year at 40°N latitude, and that one desires to place the re- 

 ceiving antenna at (a) 40 km and (b) 90 km from the transmitter. How 

 high should the receiving antenna be to be free from fading loss caused 

 by a ground-based superrefractive layer? 



Since both of the distances required are within the standard horizon 

 distance, dho, which, for a 500-m transmitter is 



dho = 92.181 km, 



this is a case of within-the-horizon propagation. 



Since the 40°N latitude position is within the temperate zone, use will 

 be made of the Washington, D.C., model ducting atmosphere. As pre- 

 viously mentioned, 100 m is the median mean duct thickness at Washing- 

 ton, D.C., and 200 m is the median maximum duct thickness observed at 

 Washington. These are the two duct thicknesses to use in this calcula- 

 tion. Also, since it is the spring of the year, figures 4.49 to 4.51 must be 

 used to determine hi. 



For the 40-km distance, case (a), the subscript of (4.38) becomes 



ht - hA = 500 - 100 = 400 m, 

 which is then entered on the abscissa of figure 4.47, where, 



do = r/400 = 83 km. 

 For a 200-m-thick duct, 



300 m. 



