Question

The distance covered by an aerial photograph is determined by both the focal length of the camera and the tilt of the camera from the perpendicular to the ground. A camera lens with a 12-in focal length has an angular coverage of 60°. Suppose an aerial photograph is taken vertically with no tilt at an altitude of 2100 ft over ground with an increasing slope of 3º, as shown in the figure. Calculate the ground distance CB that would appear in the resulting photograph. 60° 30 A ground distance of ft would appear in the photograph. (Round to the nearest hundred as needed.)

Answer 1

The ground distance CB that would appear in the resulting photograph can be calculated using trigonometry. The height of the camera lens above point B is calculated using the tangent function, and the distance from point B to C is then calculated using the tangent function again. The resulting ground distance CB is approximately 63.08 ft.

Step-by-step explanation:

The ground distance, CB, that would appear in the resulting photograph can be calculated using trigonometry.

First, we need to calculate the height of the camera lens above point B.

Using the tangent function, we have:

tan(3º) = opposite/adjacent

opposite = tan(3º) * adjacent

opposite = tan(3º) * 2100 ft = 109.19 ft

Now, we can calculate the distance from point B to C.

Using the tangent function, we have:

tan(60°) = opposite/adjacent

adjacent = opposite/tan(60°)

adjacent = 109.19 ft / tan(60°) = 63.08 ft

Therefore, the ground distance CB that would appear in the photograph is approximately 63.08 ft.

Answer 2

When rounded to the nearest hundred, the ground distance CB that would appear in the photograph is approximately 1213 feet.

Given:

- Camera lens focal length = 12 inches

- Angular coverage of the camera = 60 degrees

- Altitude = 2100 feet

- Increasing slope of 3 degrees

First, convert the focal length of the camera from inches to feet (1 foot = 12 inches):

Focal length in feet = 12 inches / 12 inches per foot = 1 foot

The tangent of the angle of view (60 degrees) gives the ratio of the altitude to the ground distance:

`\[ \tan(60^\circ) = \frac{\text{altitude}}{\text{ground distance}} \]`

Solving for the ground distance:

`\[ √(3) = \frac{2100 \text{ feet}}{\text{ground distance}} \]`

`\[ \text{ground distance} = \frac{2100 \text{ feet}}{√(3)} \]`

`\[ \text{ground distance} \approx 1212.87 \text{ feet} \]`