Theoretical background for Calibration method

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    Is there any literature specific to this sandbox for the calibration methods used in the sandbox ?
    I understand that tie points can be used to calculate homography matrix for transformation from camera space to projection space.
    But I would like to get a deeper understanding of how both base plane equations and projector matrix is calculated and used in the sandbox.
    I am a Python coder so simply looking at the code isn’t helping much.

    For full disclosure, I have built a sandbox for wind simulation as a part of my masters thesis (not as aesthetically pleasing and efficient as this one) and have borrowed heavily from Gravbox at U of Iowa.

    Oliver Kreylos

    Base plane equation:

    1. Collect a bunch of 3D points d_i in depth image space from the camera’s depth image.

    2. Convert depth image-space points to metric camera space using camera’s intrinsic depth un-projection matrix (read from camera firmware): c_i = DP * d_i.

    3. Find the plane equation (nx, ny, nz) * c_i = o that best fits all points c_i using a standard least-squares linear system solver A^T * A * x = A^T * b.

    Projector calibration matrix:

    1. Collect a set of tie points that associate a 2D point p_i in projector image space with a 3D point c_i in metric camera space.

    2. Create an over-determined linear system for (p_ix * w_i, p_iy * w_i) = H_2 * c_i where H = (h0, …, h3, h4, …, h7, h8, …, h11) is an unknown 3×4 matrix, H_2 is the 2×4 matrix consisting of the upper two rows of H, and w_i = h8*c_ix + h9*c_iy + h10*c_iz + h11 is the homogeneous weight of H * (c_ix, c_iy, c_iz, 1).

    3. Solve the system A^T * A * x = 0 by finding the eigenvector of A^T * A that has the smallest eigenvalue.

    4. Extend the resulting matrix H to an OpenGL projection matrix by inserting a third row (0, 0, 0, -1) and pre-multiplying it with the inverse of a viewport matrix that maps the projector’s screen rectangle to [-1, 1] in both x and y, and the camera-space tie points’ z range, after multiplication with the extended H, to [-0.5, 0.5] in z. Z maps to [-0.5, 0.5] instead of [-1, 1] to avoid clipping away geometry that is lower or higher than the lowest or highest tie point, respectively.


    It took me a couple of days to understand these few lines. It definitely helped fill up a lot of knowledge gaps I had.
    Thanks a lot !

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