THE CALCINATION OK VERTICAL CO\ KHAGK 



149 



distances such that the earth curvature drop is 

 comparable to hi, equation (57) does not even give 

 the correct order of magnitude for 7. 



Examples 9 and 10. Flat Earth Lobe Angle Compu- 

 tations. Lobe angles for two cases will be computed, 

 Example 9, a 200-mc set at 15 ft and Example 10, 

 a 500-mc set at 50 ft. 



Example 



X = 30 ° X 3.28 = 4.92 ft 

 200 



For n = 1 (first lobe) 

 1 X 4.92 



7 = 



4 X 15 



X 57.3 = 4.7° 



4J015)* 

 1 X 4.92 



Example ID 

 X = 1.97 ft 



7 = 0.564° 

 </i = 5,080 ft 



Table 5. Lobe angle and distance to reflection point. 



Example 9 

 7, degrees di, ft 



Example 10 

 7, degrees di, ft 



In practice some of the lobes listed for Example 9 

 may be absent because of nulls in the antenna 

 pattern. The angle listed for the first lobe of Example 

 10 is slightly over 1 per cent too large. 



15.6.4 



Lobe Angles Corrected 

 for Standard Earth Curvature 



Equation (57) may be modified to include the 

 effect of earth curvature approximately and to give 

 the lobe angles for the majority of sites with accept- 

 able accuracy. 



For antennas several hundred or more feet high, 

 e?i as given by equation (58) may be large enough 

 so that the earth curvature drop is appreciable. 

 In Figure 49 is shown a transmitter of height hi 

 above the horizontal plane GH. The radius of the 

 standard earth is ka. At D, the center of the reflection 

 area for the lobe considered, is drawn a tangent 

 plane CDE, which intersects hi at a distance h\ 

 below the center of the antenna and which will be 

 considered the equivalent antenna height. This then 

 is the part of hi which determines the angle 7' 

 which the lobe center line CL makes with the tangent 

 plane CDE. Subtracting from 7' the angle 8 which 



TO CENTER OF EARTH 

 Figure 49. Lobe angles corrected for earth curvature. 



the tangent plane CDE makes with the horizontal 

 at the base of the antenna GH, the lobe angle 7 

 referred to the horizontal at the antenna is obtained. 

 From equation (49) it follows that 



In' = hi - 



di 2 



(59) 



Here hi and hi are expressed in feet, di in miles, and 

 fc is assumed to be %. 



This height hi is the portion of hi that is effective 

 in connection with the plane CDE and when substi- 

 tuted in equation (57) gives the angle 7'. 



wX 

 4l? 



nX 



4 hi 



dS 

 2 



(60) 



Since the earth's radius ka is perpendicular to GH and 

 CDE, the tangent angle is 6. It is always negative: 



di _ di 



ka 



6 = 



6 = 



5,280 



n\ 



(61) 



d, 



t(/u 



5,280 



(62) 



n is an odd integer for lobe maxima and an even 

 integer for lobe minima, hi in feet, di in miles. 

 The value of di to substitute in equation (62) 

 must also satisfy equation (58). A convenient method 

 of solving these equations is to plot a curve of 

 equation (59) and also of equation (58) in the form 



4(/l!') 2 



n = 



5,280rfiX 



(63) 



Corresponding values of hi and di for the desired 

 value of n are then substituted in equation (62). 



While equation (62) is subject to the same sort 

 of limitation as equation (57), it will be noted that 



