The scale at the image of a well-defined object on an aerial photograph is 1:24,000, and the elevation of the object is 1500 ft. The focal length of the camera lens is 6.5". If the air base (B) is 250 ft, determine the elevation of the two points A and C and the distance between them if the coordinates of A and C are as given below.

Question

What will be an ideal response?

Answer 1

First determine the flying height, H, using Equation 14.1.



(1 / 24000) = (6.5/12) / (H – 1500)

H – 1500 = 13000

H = 14,500 feet

Now determine the elevation of point A, ha, using Equation 14.1.

(1 / 13000) = (6.5/12) / (14500 – ha)

14500 – ha = 7042

ha = 7,458 feet

Now determine the elevation of point C, hc, using Equation 14.1.

(1 / 17,400) = (6.5/12) / (14500 – hc)

14500 – hc = 9425

hc = 5,075 feet

Now determine the distance between points A and C.

Use Equations 14.4 and 14.5 to solve for the X and Y coordinates, respectively.

XA = (5.5 / 12) / (1 / 13000)

XA = 5958 ft

XC = (6.5 / 12) / (1 / 17400)

XC = 9425 ft

YA = (3.5 / 12) / (1 / 13000)

YA = 3792 ft

YC = (5.0 / 12) / (1/ 17400)

YC = 7250 ft

Now use Equation 14.6 to solve for the distance between these points.



Therefore, the elevation at point A is 7,458 ft; point C is 5,075 ft and the distance

between the two points is 4,897 ft.