For a scanner viewing off-nadir by an angle θ (theta), the ground distance swept by the sensor's instantaneous field of view (IFOV) is proportional to which trigonometric function of θ?

Difficulty: Medium

Correct Answer: sec2θ

Explanation:


Introduction / Context:
In imaging geometry for whiskbroom and pushbroom sensors, the footprint of an instantaneous field of view (IFOV) projects onto the ground plane. As the scan angle moves away from nadir, geometric foreshortening and projection effects change both the linear dimensions and area of the ground resolution element. Understanding these relationships is crucial for interpreting pixel sizes, radiometry, and map scale across the swath.



Given Data / Assumptions:

  • θ is the scan angle measured from the nadir (vertical).
  • The ground is a flat horizontal plane at mean terrain elevation.
  • Small-angle IFOV in sensor space projects onto the ground by perspective geometry.


Concept / Approach:
For a given angular IFOV, the linear ground dimension along the scan direction increases approximately in proportion to sec θ, because the slant range and projection stretch grow with off-nadir angle. Consequently, the area of the projected IFOV scales roughly with sec^2 θ (one factor from the scan-direction stretch and another from the along-track or range-direction projection, depending on geometry and platform motion). Many remote sensing MCQs refer to the distance 'swept' in the sense of the effective ground extent covered per IFOV across the swath, which aligns with the sec^2 θ dependence for area-like coverage.



Step-by-Step Solution:
Relate ground-projected IFOV to off-nadir angle: linear scale ∝ sec θ.Recognize that total swept ground extent across and along the scan grows with two orthogonal stretches.Conclude that the proportionality for the swept extent (interpreted as composite coverage) is sec^2 θ.Select option: 'sec2θ'.



Verification / Alternative check:
Sensor geometry texts show pixel area increasing more rapidly off-nadir than a single linear factor, consistent with the sec^2 θ rule-of-thumb (ignoring Earth curvature and terrain).



Why Other Options Are Wrong:

  • sin2θ, cos2θ, tan2θ, cot2θ do not match the projection scaling for off-nadir viewing of a small angular IFOV.



Common Pitfalls:
Confusing linear dimension scaling (sec θ) with area scaling (sec^2 θ); ignoring that pixel sizes vary across-track in pushbroom imagery.



Final Answer:
sec2θ

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