(a) Elevation normal to
principal plane
FIGURE 34
0, which is angle oLK\. Then 6 = d + a. Let H = flying height above sea level in meters,
and d' = angle of dip in minutes. Then d' = 1 .775 Vn 9 Eq. a. This relationship is based
on the mean radius of the earth, and it includes allowance for atmospheric refraction.
From Fig. 34a, tan a = oK 2 /f, Eq. b. Then d^1 = 1.775 V2925 = 96.0', or d = 1°36'. Also,
tan a = 86.85/152.7 = 0.5688, giving a = 29°38'. Thus, O = 1°36' + 29°38' = 31°14'.
From Fig. 34a, OK 1 =/tan O, or oK} = 152.7(0.6064) = 92.60 mm. This dimension serves
to establish the true horizon.
- Write the equation for the scale of a constant-scale line
Since the optical axis is inclined, the scale S of the photograph is constant only along a
line that is normal to the principal line, and so such a line is called a constant-scale line.
As we shall find, every constant-scale line has a unique value of S.
Refer to Fig. 35, where A is a point on the ground and a is its image. Line AQ is nor-
mal to the principal plane, Q lies in that plane, and q is the image of Q. Line Rq is a hori-
zontal line in the principal plane. If the terrain is truly level and curvature of the earth may
be disregarded, the vertical projection of the distance from A to L is H. Let e = distance in
photograph from true horizon to line qa. Along this line, S = qa/QA = LqILQ = LRILN.
But LR = e cos 6 and LN=H. Thus, S = (e cos S)IH, Eq. c.
Vertical
