Corrected Image Coordinates: 18-Geom-A4 Photogrammetry, May 2014, Q1

Question 1: Corrected Image Coordinates (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A measured image point and the two calibration quantities needed to correct it:

Quantity	Value
Measured coordinates (fiducial axes) $x_a,\,y_a$	$62.579,\ -80.916$ mm
Calibrated principal point $x_p,\,y_p$	$0.008,\ -0.001$ mm
Radial (lens) distortion at the point $\Delta r$	$-0.0021$ mm

Find. The corrected photo-coordinates $(x_c, y_c)$ — i.e. coordinates referred to the principal point of autocollimation and freed of symmetric radial lens distortion.

Two corrections in sequence: shift the measured point from the fiducial origin to the principal point ($\bar x=x_a-x_p$), then move it radially by $-\Delta r$ along the radius $r$ to remove symmetric lens distortion.

Approach. Reduce the measured coordinates to the principal point, form the radial distance $r$, then remove the symmetric radial distortion by scaling each component by $\Delta r/r$.

Reduce to the principal point. Corrections are always referred to the principal point, so subtract it from the measured coordinates: $$\bar x = x_a - x_p = 62.579 - 0.008 = 62.571\ \text{mm},$$ $$\bar y = y_a - y_p = -80.916 - (-0.001) = -80.915\ \text{mm}.$$
Radial distance from the principal point. The distortion is radial, so we need $r=\sqrt{\bar x^{2}+\bar y^{2}}$: $$r = \sqrt{62.571^{2} + 80.915^{2}} = \boxed{102.286\ \text{mm}}.$$
Resolve the distortion into $x$ and $y$. A symmetric radial distortion $\Delta r$ acts along the radius; its components are proportional to $\bar x/r$ and $\bar y/r$. Because $\Delta r$ is the distortion present in the image, the correction subtracts it: $$x_c = \bar x - \Delta r\,\frac{\bar x}{r}, \qquad y_c = \bar y - \Delta r\,\frac{\bar y}{r}.$$ With $\Delta r/r = -0.0021/102.286 = -2.053\times10^{-5}$, the component corrections are $-\Delta r\,\bar x/r = +0.00128$ mm and $-\Delta r\,\bar y/r = -0.00166$ mm.
Apply the corrections. $$x_c = 62.571 + 0.00128 = 62.5723\ \text{mm},$$ $$y_c = -80.915 - 0.00166 = -80.9167\ \text{mm}.$$

The corrections are only micrometres because the distortion itself is only $-2.1\ \mu$m, but they are exactly the kind of systematic error that must be removed before the coordinates enter an analytical solution.

Quantity	Value
Radial distance $r$	102.286 mm
Corrected $x_c$	62.5723 mm
Corrected $y_c$	−80.9167 mm

Concept & Theory

Topic. Interior-orientation refinement of measured image coordinates: principal-point reduction and symmetric radial lens distortion.

Key relations. Reduction $\bar x=x_a-x_p$, $\bar y=y_a-y_p$; radius $r=\sqrt{\bar x^2+\bar y^2}$; radial-distortion components $\delta x=\Delta r\,(\bar x/r)$, $\delta y=\Delta r\,(\bar y/r)$; corrected $x_c=\bar x-\delta x$, $y_c=\bar y-\delta y$. In full calibration $\Delta r=k_1 r^3+k_2 r^5+\dots$ (odd-power Seidel series).

Why this works. Symmetric radial distortion displaces every point along its own radius from the principal point by an amount that depends only on $r$. Decomposing $\Delta r$ with the direction cosines $\bar x/r$ and $\bar y/r$ therefore gives the exact $x$ and $y$ shifts, and subtracting them restores the straight-line (collinear) geometry the later resection/aerotriangulation assumes.

Common pitfall. Computing $r$ from the raw fiducial coordinates instead of first reducing to the principal point, and getting the sign of the correction backwards — a negative $\Delta r$ (barrel-type, inward) means the true point lies slightly farther out, so the correction moves the point outward in $x$ here.

Source. Wolf, Dewitt & Wilkinson, Elements of Photogrammetry, ch. 4 (camera calibration & image refinement); Mikhail, Bethel & McGlone, Introduction to Modern Photogrammetry, ch. 3.

Question 1 of 9: Corrected Image Coordinates

Question 1: Corrected Image Coordinates (10 marks)