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).
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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.
Step 1: Set up similar triangles for relief displacement.
Picture the camera lens at the top of a vertical line of height $H$ above the datum plane (here the base of the building), looking straight down at the principal point (nadir). The actual top of the building, of unknown height $h$ above the same datum, lies horizontally offset from the vertical line through the camera by some ground distance, and this creates two similar right triangles: a small one formed by the lens, the principal point on the photo, and the image of the building top on the photo, and a large one formed by the lens, the nadir point on the ground, and the true ground position directly below the building top.
Step 2: Write the similar-triangles ratio.
Because these two triangles are similar, the ratio of the relief displacement $d$ on the photo (the shift between where the building top is imaged and where its ground base would be imaged) to the radial image distance $r$ of the building top from the principal point equals the ratio of the building's height $h$ to the flying height $H$ above the same datum: $\dfrac{d}{r} = \dfrac{h}{H}$, which rearranges to $d = \dfrac{h\,r}{H}$, the same relief displacement relation used above, arrived at here through geometric similarity instead of by quoting the formula directly.
Step 3: Solve for the building height.
From $\dfrac{d}{r} = \dfrac{h}{H}$, $h = \dfrac{d}{r} \times H = \dfrac{6\ \text{mm}}{90\ \text{mm}} \times 500\ \text{m}$. The ratio $\dfrac{6}{90} = \dfrac{1}{15} = 0.0667$, so $h = 0.0667 \times 500\ \text{m} = 33.33\ \text{m}$.
Step 4: Note the irrelevance of the photo format and round the result.
The $230\ \text{mm} \times 230\ \text{mm}$ photo format only tells us the frame is large enough to contain a point at $r = 90\ \text{mm}$ from the centre (since the maximum possible radial distance, to a corner, is about $\sqrt{115^2+115^2} \approx 163\ \text{mm}$), it does not enter the height computation. Rounding $33.33\ \text{m}$ to the nearest whole metre, as the question asks, gives $33\ \text{m}$.
\[ \boxed{h \approx 33\ \text{m}} \]
