76 TECHNICAL SURVEY 
the earth curvature and the visibility directly from 
the map by methods given below. 
In Figure 2 is shown the relations between various 
heights on the earth’s surface. In considering the 
reference line (sea level) flat as on a map or ordinary 
profile diagram, use is made of the line H,TH2T’ 
instead of the curve H,HH2H’. This will be com- 
pensated for by using a fictitious ray path P,PP2P’ 
instead of the line P,QP2Q’. The deviation of this 
fictitious path from PiQP2Q’ at P is QP = HT and 
is called the dip. The deviation at P’ is Q'P’ = H'T’ 
and is called the rise. 
In the figure on the left the triangles HH2T and 
H,KT are similar and 
TH, HT 
THe Teigae° 
or approximately (right-hand figure) 
HT X 2ka = did2. 
Therefore the dip, 
5,280 X did2 X 3 _ dide 
QP = “3 x 3,960 x 4 2 o) 
Similarly for the rise 
Oia? dy'd.! 
(QI = 2° 
The application of these formulas will be shown by 
a number of examples. 
Example 2. Intervening Obstructions — Graphical 
Solution. In Figure 3 is shown a profile as may be 
Ficure 3. Intervening obstruction of radiation between 
two points. 
obtained from a topographical map. It is desired to 
ascertain whether P» will receive. radiation from P; 
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 
P,XP, (shown dashed) instead of the straight line 
P,QP>. If X lies above the top of the intervening 
hill P, the ray is not obstructed. The distance QX 
is from equation (5). 
did2 _ 20 X 30 
Qx = = 
y= = 800 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 P; and Po». 
2 
5 1 
QXx max — 8 (a oP as) . (6) 
Where there are several obstructions to be con- 
sidered the work may be speeded by drawing a line 
S1S2 (Figure 4) parallel to PiP2 at a vertical distance 
WILL NOT 1 
OBSTRUCT 
Figure 4. Several obstructions of radiation between 
two points. 
below it equal to the maximum dip. Then intervening 
hills which do not rise above S,S2 will not have to 
be considered, and those that do cut the line may 
be checked for obstruction by equation (5) as before. 
Example 3. Remote Shielding—Graphical Solution. 
It is frequently desired to know from a position as 
P, 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 P; and the minimum height of radiation is 
indicated by the dotted lines 
dds’ _ 90 X 80 
Dalam 2 
Example 4. Visibility Determined by Computation. 
In many case. it is not necessary to construct profiles, 
as visibility may be determined by a simple compu- 
tation. The critical height, h,, 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 P;P2 at d, be h’. Then 
Dap st ld 0 1 
QXx' = = 3,600 ft . 
dy dime 
oh ay me hid + hed 
di + de 
But the height h’ exceeds h, by the dip (did2)/2; 
therefore 
— Iide + hedi _ dide 
a mrresyir | ou (7) 
