# 6. Aerial photography is to be taken of a tract of land that is 8 x 8 mi2. Flying height will be 4000 ft above average terrain, and the came

6. Aerial photography is to be taken of a tract of land that is 8 x 8 mi2. Flying height will be 4000 ft above average terrain, and the camera has focal length of 6 inches. If the focal plane opening is 9 x 9 in., and minimum side overlap is 30%, how many flight lines will be needed to cover the tract for the given data

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the number of flight lines needed is approximately 72

Step-by-step explanation:

Given the data in the question;

Aerial photography is to be taken of a tract of land that is 8 x 8 mi²

L × B = 8 x 8 mi²

Flying height H = 4000 ft = ( 4000 × 12 )inches = 48000 in

focal length f = 6 in

[tex]l[/tex] × b = 9 × 9 in²

side overlap = 30% = 0.3

meaning remaining side overlap = 100% – 30% = 70% = 0.7

{ not end to end overlap }

we take 100% { remaining overlap }

[tex]l[/tex]’ = 9 × 100% = 9 in

b’ = 9 × 70% = 6.3 in

Now the scale will be;

Scale = f/H

we substitute

Scale = 6 in /  48000 in = 1 / 8000

so our scale is; 1 : 8000

⇒ 1 in = 8000 in

⇒ 1 in = (8000 / 63360)mi

⇒ 1 in = 0.126 mi

so since

L × B = 8 x 8 mi²

[tex]l[/tex]’ = ( 9 × 0.126 mi ) = 1.134 mi

b’ = ( 6.3 × 0.126 mi ) = 0.7938 mi

Now we get the flight lines;

N = ( L × B ) / ( [tex]l[/tex]’ × b’ )

we substitute

N = ( 8 mi × 8 mi ) / ( 1.134 mi × 0.7938 mi )

N = 64 / 0.9001692

N = 71.0977 ≈ 72

Therefore, the number of flight lines needed is approximately 72