Question

A vertical aerial photograph is obtained over flat terrain with a 30 cm focal-length camera lens from an altitude of 18288 m. If the width of a dolerite dyke on this vertical photograph is 2 mm, its actual width on the terrain is $\underline{\hspace{1cm}}$ m. [round off to 2 decimal places]

Hint:

For vertical airphotos: Scale $=f/H$. Always put $f$ and $H$ in the same units first. Then $\text{ground}=\text{photo}\times(H/f)$. A fast check: at 1:$n$, \,1 mm on photo $=n$ mm on ground.

Answer 1

Step 1: State the governing relation (vertical photo over flat terrain).
For a true vertical photograph with negligible relief, the photo scale is \[ S \;=\; \frac{\text{photo distance}}{\text{ground distance}} \;=\; \frac{f}{H}, \] where $f$ is the camera focal length and $H$ is the camera height above ground (flying height). Hence, $\displaystyle \text{ground distance}=\text{photo distance}\times \frac{H}{f}$.

Step 2: Convert all quantities to consistent SI units.
\[ f = 30~\text{cm} = 0.30~\text{m}, \qquad H = 18288~\text{m}, \qquad \text{photo width} = 2~\text{mm} = 0.002~\text{m}. \]

Step 3: Compute the photo scale and its inverse explicitly.
\[ S=\frac{f}{H}=\frac{0.30}{18288}=1.642\ldots\times 10^{-5}\ \Rightarrow\ \text{inverse scale}=\frac{H}{f}=\frac{18288}{0.30}=60960. \] Thus the scale is \(1:60960\) (a very common mapping scale).

Step 4: Convert the measured width on the photo to the ground width.
\[ W_{\text{ground}}=(0.002~\text{m})\times 60960 = 121.92~\text{m}. \]

Step 5: Rounding and reasonableness check.
Rounded to two decimals, $121.92$ m already has two decimals. Sanity check: $2$ mm $\approx 1/500$ of an A4 width; at 1:60,960, every $1$ mm represents $\approx 60.96$ m. Hence $2$ mm $\approx 121.92$ m — consistent.

Final Answer:\quad \[ \boxed{121.92~\text{m}} \]