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Answer:
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