TOPOGRAPH! OF SITING 



l«) 



The application <>l these formulas will be shown by 

 a number of examples. 



Example J. Intervening Obstruction* — (Iraphical 

 Solution. In Figure 3 is shown a profile as may be 



Figure 3. Intervening obstruction of radiation between 

 two points. 



obtained from a topographical map. It is desired to 

 ascertain whether P 2 will receive radiation from Pi 

 without being obstructed by the intervening hill P. 

 Owing to the curvature of the earth the line marked 

 "datum level" is actually curved instead of being 

 straight as shown. To compensate for this distortion 

 the line of sight of the radar is taken as the parabola 

 P1XP2 (shown dashed) instead of the straight line 

 PiQPz. If X lies above the top of the intervening 

 hill P, the ray is not obstructed. The distance QX 

 is from equation (5). 



QY = ^ 2 = 2^X_30 = 3()()ft 



Scaling this distance down from Q, it will be found 

 that X lies above P and there is no obstruction to 

 the radiation. 



It will be noted that QX is a maximum midway 

 between Pi and P 2 . 



QX t 



d l 



(6) 



Where there are several obstructions to be con- 

 sidered the work may be speeded by drawing a line 

 S1S2 (Figure 4) parallel to PiP 2 at a vertical distance 



ISOOy 

 FT'Sj, 



Figure 4. Several obstructions of radiation between 

 two points. 



below it equal to the maximum dip. Then intervening 

 hills which do not rise above <SiS 2 will not have to 

 be considered, and those that do cut the line may 



be checked for obstruction by equation (5) as before. 



Example J. Remote Shielding— Graphical Solution. 

 It is frequently desired to know from a position as 

 Pi what degree of shielding will be obtained from a 

 given profile. In Figure 5 the rise Q'X' is computed 



Figure 5. Remote shielding obtainable from a given 

 profile. 



by equation (5) . Since P' lies below X' it is shielded 

 from Pi and the minimum height of radiation is 

 indicated by the dotted lines 



Q'X' = 



di'ds' 90 X 80 



= 3,600 ft 



Example 4- Visibility Determined by Computation. 

 In many cases it is not necessary to construct profiles, 

 as visibility may be determined by a simple compu- 

 tation. The critical height, h c , which an intervening 

 hill must equal or exceed to obstruct the line of 

 sight, may be computed from Figure 4. Let the 

 height of the line PiP 2 at <& be h'. Then 



In - h! V - h 2 



di 



or 



V 



h\di 



d 2 

 h 2 d} 



di 



rf 2 



But the height h' exceeds h c by the dip (did 2 )/2; 

 therefore 



hidi + hidx d]di 

 ~2~ 



T- 2 - W 



di + d 2 



For the values given in Figure 4 the critical height 

 at di is 



_ 3,000 X 35 + 1,500 X 15 _ 15 X 35 

 c - " 15 + 35 2 



= 2,287.5 ft . 



Since the hill P is 2,300 ft high it will interfere with 

 radiation to P 2 . 



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 P» shields the distant mountainous 

 island. The height of the line PiQ' at a distance of 



