Aerial photography motion blur: if v is the ground speed of the aircraft (m/s), t is the camera shutter time (s), and the photograph scale is 1:S, what is the linear image displacement due to motion during exposure?

Introduction / Context:
In vertical aerial photography, forward motion of the aircraft during the finite shutter opening time causes linear image displacement (motion blur) on the photograph. Controlling this blur is crucial for sharp imagery and accurate photogrammetric measurements.

Given Data / Assumptions:

• Ground speed of aircraft = v (m/s).
• Shutter time (exposure) = t (s).
• Photo scale = 1:S (dimensionless; S = ground distance / photo distance).
• Nominal flying height is such that scale is constant across the small exposure time.

Concept / Approach:
During exposure, the aircraft advances a ground distance g = v * t. On the photo, this ground distance is reduced by the scale factor S. Hence the corresponding smear length on the film/sensor is d = g / S.

Step-by-Step Solution:
Compute ground travel during exposure: g = v * t.Convert to photo displacement using scale 1:S: d = g / S.Therefore: d = v * t / S.This is the standard first-order relation for motion blur in vertical photography.

Verification / Alternative check:
From lens geometry, f/H = 1/S for vertical photos; using image-space expression d = (v * t) * (f/H) also reduces to d = v * t / S, confirming the relationship.

Why Other Options Are Wrong:

• d = v * t * S: inverts the scale; would imply larger blur for larger S, contrary to reality.
• d = v / (t * S): wrong units and dependence.
• d = S / (v * t): dimensionally incorrect.
• d = (v * t)^2 / S: introduces a non-physical square term.

Common Pitfalls:
Confusing scale 1:S with its reciprocal; ignoring along-track versus across-track components; neglecting image motion compensation (IMC) mechanisms in modern cameras.

Final Answer:
d = v * t / S