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Thread: Trying to find "best" rotation matrix

  1. #1
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    Trying to find "best" rotation matrix

    $\displaystyle
    \text{Let }A_i =
    \left[ \begin{array}{cc}
    R_i & p_i \\
    0 & 1\end{array}\right]
    \text{ where } R_i \in SO[3] \text{ , and } p_i \in R^3.
    $

    $\displaystyle
    \text{Further, let } \overline{R} =
    \left[ \begin{array}{cc}
    R & 0 \\
    0 & 1\end{array}\right]
    \text{ where } R \in SO[3].
    $

    Now, given a set of $\displaystyle A_i$, I'm looking to find a least squares (or some other optimal form) of $\displaystyle R$ such that

    $\displaystyle
    A_i \overline{R}
    \left[\begin{array}{c}0 \\ -L \\ 0 \\ 1 \end{array}\right]
    =
    \left[\begin{array}{c}c_x \\ c_y \\ c_z \\ 1 \end{array}\right]
    $

    I've tried to present the mathematical description of a problem I'm facing. I have a rod that is connected to the end of a 6dof robot. The robot provides position/orientation information in the form of a homogeneous transform ($\displaystyle A_i$). However, the orientation needs to be calibrated (hence the separate $\displaystyle R$ matrix). To do so, I fix the end of the rod and trace out a sphere which provides me the position and orientation vectors that should all intersect at the center point $\displaystyle c$.

    Obviously if there were only 1 data point (i.e. 1 position vector, 1 orientation matrix, and 1 center point), the solution could (rather trivially) be determined in a unique fashion. What I'm looking for is a way to solve this in a least squares (or some other "optimal") sense. Ideas?

    My best has been to use the following (note: slightly different notation here as I'm using inhomogeneous coordinates)...

    $\displaystyle
    \text{Let } A_i \in SO[3], p_i \in R^3, R \in SO^3, c \in R^3.
    $

    $\displaystyle
    A_i R
    \left[ \begin{array}{c} 0 \\ -L \\ 0 \end{array}\right]
    + p_i
    =
    c
    $

    Thus,
    $\displaystyle
    R_2 = \frac{-1}{L} A_i^T (c - p_i) \text{ where } R_2 \text{ is the 2nd column of } R
    $

    I could try to take the mean over $\displaystyle R_2$ and attempt to solve for the entire matrix using the single column, but it will certainly be messy.

    EDIT: On 2nd thought, I can't solve for the matrix with only a single column
    Last edited by anon123; Aug 5th 2010 at 03:13 PM.
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  2. #2
    A Plied Mathematician
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    Warning: this might be a somewhat boneheaded approach. Given the equation

    $\displaystyle A_i \overline{R}
    \left[\begin{array}{c}0 \\ -L \\ 0 \\ 1 \end{array}\right]
    =
    \left[\begin{array}{c}c_x \\ c_y \\ c_z \\ 1 \end{array}\right],$

    isn't it true that $\displaystyle A_{i}, L, c_{x}, c_{y},$ and $\displaystyle c_{z}$ are all known? If that's the case, can't you come up with a general description of the matrix $\displaystyle \overline{R}$ in terms of, say, Euler angles, and combine that matrix with

    $\displaystyle \left[\begin{array}{c}0 \\ -L \\ 0 \\ 1 \end{array}\right],$

    then solve the resulting system, and finally solve the smaller problem? It's kind of like this: Solve ABx = b by first setting y = Bx, then solve Ay = b, and then solve y = Bx. The only difference is that in your second problem, you're solving for the matrix instead of the vector.

    Any reason why this wouldn't work?
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  3. #3
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    If I were trying to solve for R given a single $\displaystyle A_i$ and $\displaystyle p_i$, then that'd be the way to go obviously. However, I have an overdetermined system with a BUNCH of $\displaystyle A_i$ and $\displaystyle p_i$. Therefore, even if I were to solve each one individually, the best I'd be able to do is to average them (which isn't exactly robust to outliers).
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  4. #4
    A Plied Mathematician
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    Well, it's beyond me, I'm afraid. Maybe Captain Black can help you.
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