Key property on a tilted/vertical photograph – distances of plumb and horizon points For a (nominally vertical) aerial photograph considering tilt geometry, the product of the distances from the principal point to the plumb (nadir) point and to the horizon point equals which quantity?

Introduction / Context:
Understanding special points on an aerial photograph—principal point, plumb (nadir) point, isocentre, and horizon point—is central to interpreting tilt and relief effects. Certain elegant geometric relations connect their distances and the camera focal length.

Given Data / Assumptions:

• Principal point is the intersection of the principal ray with the photo plane.
• Plumb (nadir) point is the perspective center's vertical projection onto the photo plane.
• Horizon point is the intersection of the tilt-induced horizon line with the principal line.
• f is the camera focal length (principal distance).

Concept / Approach:
For a tilted photograph, geometry yields a classic result: the product of the distances from the principal point to the plumb point (PP–N) and from the principal point to the horizon point (PP–H) equals f^2. This follows from similar triangles formed by the optical axis and the tilted image plane, linking conjugate points along the principal line.

Step-by-Step Solution:
Set up principal line with points PP, N (plumb), and H (horizon) defined by tilt.Apply similar triangles in central projection between optical axis and horizon construction.Derive invariant: (PP–N) * (PP–H) = f^2.Thus, the required product equals f^2.

Verification / Alternative check:
In the limit of zero tilt, PP ≈ N and H moves to infinity, keeping the product finite as f^2 (distance tends to infinity while the other tends to zero such that their product remains f^2), consistent with the invariant.

Why Other Options Are Wrong:
Multiples such as 2 f^2 or 3 f^2 do not arise from the projective similarity; values f or duplicated f are dimensionally inconsistent.

Common Pitfalls:
Mixing up isocentre–principal point relations; while related, the invariant specifically involves the plumb and horizon points with the principal point along the principal line.

Final Answer:
f^2