5.1.8. Example 4.8

A rectangular area 130 km x 120 km is to be mapped from aerial photographs taken to a scale of 1/20000. The focal length of the lens of the camera to be used is 152 mm and each print is to be 230 mm square. Provision is to be made for a 60% overlap between successive exposures and a 25% lateral overlap.

In this chunk, we are tasked with mapping a rectangular area of 130 km by 120 km using aerial photographs. The key details are the scale of the photographs, which is 1/20000, and the camera lens focal length of 152 mm. The scale indicates that 1 unit on the photograph represents 20000 units on the ground. Thus, the aerial photographs must adequately cover the given area while accounting for overlaps to ensure complete coverage.

Imagine you are taking a series of photographs of a large park to create a complete map. You need to ensure each photo overlaps with the previous one—like pieces of a puzzle—so that no sections of the park are missed.

a) Average height above ground level: H = f / S = 152 mm / (1/20000) = 152 * 20000 * 10^-3 = 3040 m.

To calculate the average height at which the aircraft must operate, we use the formula H = f/S, where H is the height, f is the focal length of the camera, and S is the scale of the photograph. With the focal length of 152 mm and a scale of 1/20000, we convert the scale to its reciprocal and compute the height needed to take photographs that accurately reflect the specified area.

Think of using a telescope where the lens size corresponds to how far you can see. Here, the focal length of the camera lens determines how high up we need to be to capture a wide enough view of the area being mapped.

b) Let the flight line be parallel to the 130 km length. Since there is a 60% overlap between successive exposures, the effective length of each photograph is 40% of 230 mm, i.e., 0.4*230 = 92 mm. The ground distance covered by this photo length is 92 mm * 20000 * 10^-3 = 1840 m. Number of photographs per strip = 130,000 / 1840 = 70.65 ≈ 71 photos. The operating speed of the aircraft is 200 km/h. To cover a length of 130 km, the aircraft needs 130 / 200 = 0.65 hour. Since the exposures are at regular intervals, time interval between exposures = 0.65 hour / 71 ≈ 33.12 sec.

This chunk focuses on determining the time needed between each photograph taken. A 60% overlap means that each photo covers only 40% of its full width when capturing the next section. Calculating the effective photo length allows us to determine how many photos are needed to cover the entire flight path. Dividing the total coverage time by the number of photographs gives the time interval between exposures, ensuring a systematic approach to photographing the area.

Imagine taking quick snapshots of a parade. You can only capture sections that overlap with each previous shot; if you time your shots improperly, you may miss capturing important details in the parade.

c) The width of the area to be photographed is 120 km. A 25% lateral overlap results in an effective photo length of 230 mm, which is 0.75 * 230 = 172.5 mm. The ground distance covered by this width is 172.5 mm * 20000 * 10^-3 = 3450 m. Number of strips = 120,000 / 3450 ≈ 35 strips. Minimum number of photographs required = 71 * 35 = 2485.

Next, we calculate how many strips of photographs are needed to cover the width of the rectangular area, accounting for a 25% lateral overlap. By reducing the effective photo length due to overlap, we find the total ground coverage per photograph. Then, dividing the total area width by the ground coverage gives the number of strips required. Finally, multiplying the number of photographs per strip by the number of strips gives us the minimum number of photographs needed.

Think of layering sheets of colorful cellophane to create a large artwork. Each sheet needs to overlap with neighboring sheets for a coherent final piece. Similar strategies apply when taking aerial photographs, ensuring all parts of the area are covered.

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