476 PROPAGATION THROUGH THE STANDARD ATMOSPHERE 
Tf d is in statute miles and h in feet, 
. (7) 
Cole 
dy, = V2h, + V3h2 fork = 
The relation between hy, h, and d, is graphically 
presented in Chapter 5, Figure 2 as a nomogram. 
Height of Obstacle 
As a first case, consider a smooth earth and two 
terminals at the ground. The earth itself forms an 
obstacle which reaches its maximum height h,, in the 
middle of the path (Figure 4). By equation (2) 
@) 
9 2 
he 2 d 
~ Qka ~ 8ka" 8) 
A point P on the ground at distances d’ and d’’ from 
the two terminals (see Figure 4) has an elevation 
d+d di-d 
2 ale e i 
: < u d ae 
Ficure 4. Height of earth as an obstacle. 
above the straight line connecting the terminals 
given by 
(e +: a @ = oy 
a Ee 
9 
2ka .2ka (9) 
or, after a simple reduction, 
h Oe = 56.10 Saal’ (10) 
7 oe ; 
where h, d’, and d”’ are given in meters. If h is in 
feet and d’ in statute miles, 
yee 
nae? 
fork = 4/3. (11) 
Secondly, assume that the terminals are elevated 
(Figure 5). The elevation of the straight line con- 
necting the terminals, for a flat earth, is equal to 
d! he site d! hy 
where d’, d’’ are again the distances to the terminals, 
and hy, he are the corresponding elevations. 
In order to account for the effect of the earth’s 
curvature, Figure 5 may be considered as a plane 
earth diagram on which a ray will appear curved, the 
deviation from a straight line being downward and 
given by equation (10).See dashed line in Figure 5. 
h= (12) 
h, 
| 
| 
Ra aS sede a8 
Ficure 5. Height for elevated terminals. 
Hence, the total height above the theoretical 
ground is 
d'he Aes dh, d'd”’ 
bam. ~- Ge, ? 
When the heights are expressed in feet and the dis- 
tances in miles, the first term remains unaltered, 
while the second term again becomes d’d’’/2 for 
k = 473. 
Equation (13) is used to decide whether and by 
how much an obstacle such as a hill will obstruct a 
given transmission path. 
Extended Obstacle 
When the obstacle is of appreciable horizontal 
extension, it may not possess a single peak to which 
equation (13) can be applied without ambiguity. 
The case of twin mountains is shown in Figure 6 for 
straight rays (earth’s radius ka). 
(13) 
FreureE 6. Height of equivalent diffracting edge. 
The optical peak P of the obstacle for radio or radar 
transmission is the point from which both terminals 
are just visible. For a given profile, the limiting rays 
to the terminals may be found by trial and error by 
applying equation (13) to those points of the profile 
which are most likely to represent limiting elevations. 
In the theory of diffraction (Chapter 8)P marks 
the position of the equivalent diffracting edge. 
Degree of Shielding 
As a measure of the degree of shielding, the angle 
between the tio limiting rays drawn from the 
terminals to the (actual or equivalent) peak of the 
obstacle of height h, may be used. 
Since all angles considered are small, the sine or 
tangent of the angle may be replaced by the angle 
in radians: Consider first the ray going from the 
first terminal to P (Figure 7). The angle of the ray 
with the horizontal at the terminal is 
Ligeti 
= 14 
- d’ ka o 
and its angle with the horizontal at P is 
byt yn 
= ane iy 15 
By = a + is a + ae (15) 
The angle of the ray going from the second terminal 
to P is determined correspondingly. 
The angle between the two 1ays is tren equal to 
pit b= {J+ ur) - nh) (16) 
2 OCC aC 
where d = d’ + d’’, and (ka) = 1.18- 107 (meter) '. 
