Question

On a paper-print positive, the measured x distance between fiducials (1 and 3) was 224.9 mm and y between fiducials (2 and 4) was 223.5 mm. These x and y distances determined in camera calibration were 225.433 and 225.693 mm, respectively. Using the method of example 4-1, calculate shrinkage-corrected coordinates of points a,b, and c whose coordinates were measured on the paper print as follows:

Point x, mm y, mm 20.3 92.1 85.8 102.5 48.6

Answer 1

Given data rightarrow On a paper - print positive focal length of a 'x' direction (f_1) = 224.9 mm focal length of a 'y' direction (f_2) = 223.5 mm The distance measured from X & Y directions are H_1 = 225.433 mm H_2 = 225.693 mm coordinates of a, b, c are for point 'a' X_1 = (H/f) x_1 rightarrow 225.433/224.9 x_1 = (1.0028) x_1 \r\n = (1.0028) (20.3) X_1 rightarrow 20.358m Y_1 = (H/f) y_1 = 225.693/223.5 times (-92.1) y_1 rightarrow -93.003m for point 'b' X_2 = (H/f) X_2 rightarrow 225.433/224.9 times (48.6) rightarrow 48.715m Y_2 = (H/f)y_2 rightarrow 225.693/223.5 times 85.8 = 86.641m \r\n for point 'c' X_3 = (H_x/f_x) times x_3 rightarrow 225.433/224.9 times (-111.1) rightarrow -111.363m Y_3 = (H_y/f_y) times y_3 = 225.693/223.5 times (-102.5) rightarrow -103.505m So, the calculated shrinkage corrected coordinate a = (x, y) rightarrow (20.358m, -93.003m) b = (x_2, y_2) rightarrow (48.715m, 86.641) c = (x_3, y_3) rightarrow (-111.363m, -103.505m)