Question

2- A camera attached to a robot arm shown below. The transformation X E SE(3) is constant. The robot arm moves from posture one to posture two. The transformations A E SE(3 and BESE(3) are measured and can be assumed to be known. Supposed that transformations X and A are given as follows, determine transformation B. (8 points) (Problem 3.25 of the textbook) 1 pose 1 camera X = [ 1 0 1 0 | 0 0 1 0 0 0 0 1 0 1 1 0 0 , 1 A= 0 0 -1 0 0 1 0 0 1 0 1 0 0 1 1 0 1 1 pose 2 {c}

Answer 1

Let us assume camera axis are cartesian co-ordinate axis itself.

So for a particular point (x,y,z,1) transformation to pose {t'} will result in:

$\\$Let T be the transformed co-ordinates.$ \\\therefore T = X*A*(x,y,z,1)^{T} \\\therefore T = \begin{bmatrix} 1 &0 &0 &1 \\ 0 & 1 & 0& 0\\ 0&0 &1 &0 \\ 0& 0& 0& 1 \end{bmatrix}\begin{bmatrix} 0&0 &1 &0 \\ 0 & 1 & 0& 1\\ -1&0 &1 &0 \\ 0& 0& 0& 1 \end{bmatrix}\begin{bmatrix} x \\ y\\ z \\ 1 \end{bmatrix} \\\therefore T = \begin{bmatrix} 0&0 &1 &1 \\ 0 & 1 & 0& 1\\ -1&0 &1 &0 \\ 0& 0& 0& 1 \end{bmatrix}\begin{bmatrix} x \\ y\\ z \\ 1 \end{bmatrix} \\\therefore T = (z+1,y+1,z-x,1)^{T}$

Thus above is the transformation done in pose {t'}

Same should be obtained by transformation X and B i.e. {c} -> {t} -> {t'}

$\\\therefore T = B*X*(x,y,z,1)^{T} \\\therefore T = B\begin{bmatrix} 1 &0 &0 &1 \\ 0 & 1 & 0& 0\\ 0&0 &1 &0 \\ 0& 0& 0& 1 \end{bmatrix}\begin{bmatrix} x \\ y\\ z \\ 1 \end{bmatrix} \\\therefore T =B \begin{bmatrix} x+1 \\ y\\ z \\ 1 \end{bmatrix}$

$\\\therefore \begin{bmatrix} z+1 \\ y+1\\ z-x \\ 1 \end{bmatrix} =B \begin{bmatrix} x+1 \\ y\\ z \\ 1 \end{bmatrix} \\$Back calculating:$ \\\therefore B = \begin{bmatrix} 0& 0& 1&1 \\ 0 & 1 & 0 & 1\\ -1 &0 & 1 & 1\\ 0& 0 &0 &1 \end{bmatrix}$

Hence above is the required transformation B.