Question

Photographs at a scale of 1:18,000 are required to cover an area 12 miles x 8 miles. The camera has a focal length of 6 inches and focal plane dimensions of 9 inch x 9 inch. The endlap is 60% and sidelap 30%. The average terrain in the area is 1,200 ft above datum. This is a second-order aerial survey.

a. What is the flying elevation of the plane? The flying elevation is the flying height of the plane above datum.

b. What is the contour interval? The answer must be a whole number.

c. How many photos are required to cover the area?

Answer #1

Given,

Scale = 1:18,000

Covered area: 12 miles * 8 miles

Camera focal length = 6 in

Focal plane dimensions: 9 in * 9 in

End lap = 60%

Side lap = 30%

The average terrain in the area is 1,200 ft above datum

This is a second-order aerial survey

a) The flying elevation of the plane:

H - h = 9000 ft

H = 10,200 ft above datum

Flying height of the plane above datum = 10,200 ft

b) contour interval:

Flying Height = Contour Interval * C factor

Assume C factor = 750 (It is a value between 500 to 1500)

Contour Interval = 10,200/750 = 13.6 ~ 14

c) Number of photos required to cover the area:

Ground area, A = 12 * 8 = 96 sq. miles = 96 * 5280 * 5280 =
2676326400 ft^{2}

Ground area covered by each photograph, a = (1-P_{l}) *
l/s * (1-P_{w}) * w/s

= (1-0.6) * (9/12)/(1/18000) * (1-0.3) * (9/12)/((1/18000) =
51030000 ft^{2}

Number of photos = A/a = 2676326400/51030000 = 52.45 ~ 53 photos

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