Question

The flying height of an aircraft above the base of a building is 500 m. In the nominally vertical aerial photograph, the radial distance from the principal point to the top of the building is 90 mm. The photo size is 230 mm × 230 mm. The relief displacement of the building is 6 mm.
The estimated height of the building is _______ m (Rounded off to the nearest integer).

Hint:

Use the relief displacement relation d = h r / H, from similar triangles between the camera, principal point, and the top of the object, and solve for h.

Answer 1

Correct Answer: 33

Step 1: Recall the relief displacement formula.
In a nominally vertical (near-vertical) aerial photograph, an object that stands above the ground datum, such as the top of a building, is imaged at a position radially displaced outward from the position it would occupy if it were exactly at datum level. This radial shift is called relief displacement, and for an object of height \(h\) above the datum whose top is imaged at radial distance \(r\) from the photo's principal point (nadir point), when the camera is at flying height \(H\) above the same datum, the relief displacement \(d\) is given by \( d = \dfrac{h\,r}{H} \).
Step 2: Identify the known quantities from the question.
Here the flying height above the base of the building (the chosen datum) is \(H = 500\ \text{m}\), the radial distance from the principal point to the image of the top of the building is \(r = 90\ \text{mm}\), and the relief displacement of the building's top is given directly as \(d = 6\ \text{mm}\). The stated photo format, \(230\ \text{mm} \times 230\ \text{mm}\), is not needed in this formula since \(r\) is already given, it is only supplementary information confirming that \(r=90\ \text{mm}\) is a physically reasonable radial distance on a photo of this size.
Step 3: Rearrange the relief displacement formula for h and substitute.
Rearranging \( d = \dfrac{h\,r}{H} \) for the unknown height gives \( h = \dfrac{d\,H}{r} \). Substituting the given values, \( h = \dfrac{6\ \text{mm} \times 500\ \text{m}}{90\ \text{mm}} = \dfrac{3000}{90}\ \text{m} = 33.33\ \text{m} \).
Step 4: Round to the required precision.
Rounding \(33.33\ \text{m}\) off to the nearest integer, as instructed, gives \(33\ \text{m}\).
\[ \boxed{h \approx 33\ \text{m} \; (\text{accepted range } 33 \text{ to } 34)} \]