Photogrammetry — flying height from parallax difference and mean photo base A stereo pair taken from flying height H shows a parallax difference of 0.8 mm for two points whose elevation difference is 100 m. If the mean photo base (distance between principal points on the photographs) is 95.2 mm, estimate the flying height H.

Difficulty: Medium

Correct Answer: 12,000 m

Explanation:


Introduction / Context:
In stereophotogrammetry, the relation between stereoscopic parallax and object height allows height determination. Conversely, if height difference and parallax difference are known, one can infer flying height when the mean photo base is given.


Given Data / Assumptions:

  • Elevation difference Δh = 100 m.
  • Parallax difference dp = 0.8 mm (on photos).
  • Mean photo base b = 95.2 mm.
  • Small-relief approximation and mean terrain level near datum so higher-order terms are negligible.


Concept / Approach:

For small relief, a useful approximation relates parallax difference to height difference via the mean photo base and flying height: dp ≈ b * (Δh / H). This follows from p = (B f)/(H − h) with b = (B f)/H and linearization for small Δh relative to H.


Step-by-Step Solution:

Use dp ≈ b * (Δh / H)Rearrange → H ≈ b * Δh / dpCompute ratio b/dp = 95.2 / 0.8 = 119Hence H ≈ 119 * 100 m = 11,900 m ≈ 12,000 m


Verification / Alternative check:

A more exact relation from p = (B f)/(H − h) and b = (B f)/H gives p = b / (1 − h/H). Differencing and linearizing for small Δh leads to dp ≈ b * Δh / H, confirming the approximation.


Why Other Options Are Wrong:

  • 8,000 m and 10,000 m underpredict H for the given dp.
  • 14,000 m overpredicts H; the computed value is near 12,000 m.


Common Pitfalls:

  • Confusing ground air base B with photo base b; here b is already given in mm on the photographs.


Final Answer:

12,000 m.

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