Camera Calibration

For this activity we would transform the camera coordinates into world coordinates. We used a checkerboard as the calibration object.
Since the board is placed in a 90 degree angle, I assigned the left face as the x-axis, the right face as the y-axis and the vertical axis as the z-axis. Doing this would simplify most of the equations. The origin is place bottom center of the board.
25 cornerpoints were randomly assigned. So that I would not get lost in using the locate function in Scilab, I placed markers in the image. (This would have no effect since we don't need the color information of the image) (NOTE: see code for the coordinates of the points). Each square in the board measures 1x1 inch.
In lecture 2, the camera axis is labeled as x and the image plane as yz. Since the xyz axis can be permutated without affecting the location of a point in the plane, the same derivation can be used but this time change all x with z, all y with x, and all z with y.
To solve for the camera parameters, we used the following equation:
Labeling the first matrix as Q, the second as a (camera parameters), and the third as d (image plane coordinates), we get:
//Scilab code
image = imread("cboard-marked-gs.bmp") - 1;
imshow(image, [])
d = locate(25, flag = 1);
coords = [0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 2 3 3 4 5 5 6 6 6; //x
1 1 2 2 3 4 4 4 4 5 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0; //y
2 6 4 7 5 1 3 6 9 7 3 5 1 3 5 8 6 2 8 4 2 7 1 4 9]; //z
Q = [];
for i = 1:25
x = coords(1, i);
y = coords(2, i);
z = coords(3, i);
yi = d(1, i);
zi = d(2, i);
Qi = [x y z 1 0 0 0 0 -(yi*x) -(yi*y) -(yi*z);
0 0 0 0 x y z 1 -(zi*x) -(zi*y) -(zi*z)];
Q = cat(1, Q, Qi);
end
d = matrix(d, [length(d), 1]);
a = inv(Q'*Q)*Q'*d
//end of code

The computed camera parameters are:
-19.387751
9.411270
-0.564041
172.318473
-2.744767
-3.956096
20.590540
36.630648
-0.008846
-0.013345
-0.001769

To this if this calibration is correct, I first tried to convert the object coordinate to image coordinate using the camera calibration above and the following equation:
//Scilab code
a = [-19.387751 ;9.411270 ;-0.564041 ;172.318473 ;-2.744767 ;-3.956096 ;
20.590540 ;36.630648 ;-0.008846 ;-0.013345 ;-0.001769 ]
coords = [0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 2 3 3 4 5 5 6 6 6;
1 1 2 2 3 4 4 4 4 5 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0;
2 6 4 7 5 1 3 6 9 7 3 5 1 3 5 8 6 2 8 4 2 7 1 4 9];
yi = []
zi = []
for i = 1:6
x = coords(1, i);
y = coords(2, i);
z = coords(3, i);
yi(i) = (a(1)*x + a(2)*y + a(3)*z + a(4))/(a(9)*x + a(10)*y + a(11)*z + 1);
zi(i) = (a(5)*x + a(6)*y + a(7)*z + a(8))/(a(9)*x + a(10)*y + a(11)*z + 1);
end
d = cat(2, yi, zi);
//end of code

After processing all the coordinates used, the deviations from using the locate function of scilab and this equation are:

yi	zi
0.52826	0.87672
0.87410	0.22682
0.15622	0.77860
0.83752	0.86413
0.84163	0.82441
0.64452	0.62763
0.89180	1.10098
0.33847	0.09812
0.41275	0.15379
0.37815	0.40735
0.66961	0.21872
0.87220	0.29042
0.83183	0.46501
0.87415	1.30520
0.98954	0.23742
0.01032	0.98742
1.06601	0.00136
0.50450	0.61056
0.42662	0.28194
0.77159	0.11263
0.87657	0.02865
0.11396	1.14443
1.33988	0.06806
0.99433	1.24125
0.22386	0.0234

The mean is 0.659 and 0.519 for yi and zi respectively with standard deviation equal to 0.335 and 0.425. This translates to less than 2% error for all the values computed.
Using the camera parameters obtained, I tried predicting the image coordinates of some unused cornerpoints. The results are as follows:

Predicted			Located			Difference
171.537	98.9273		172.187	98.9699		0.650	0.043
183.460	96.2413		186.133	95.1664		2.674	1.075
208.713	47.3315		209.588	47.6228		0.875	0.291
153.706	76.0082		155.705	76.4342		1.999	0.426
135.287	73.8909		135.42	75.5151		0.133	1.624
115.791	114.611		115.135	113.550		0.657	1.061

The predictions were quite accurate.

For this activity, I give myself a grade of 10 because the results I obtained were precise and accurate.
Thanks to Cole Fabros for some clarifications on the needed equations
Collaborators: Raf Jaculbia