Difference between revisions of "Inverse transformation"
| Line 22: | Line 22: | ||
The equation above is now solved for <math>\vec{\mathbf{q}}_1</math>:<br/> | The equation above is now solved for <math>\vec{\mathbf{q}}_1</math>:<br/> | ||
:<math>\begin{align} | :<math>\begin{align} | ||
| − | \vec{\mathbf{q}}_1&=\mathbf{R} | + | \vec{\mathbf{q}}_1&=\mathbf{R} \vec{\mathbf{q}}_0 + \vec{\mathbf{p}}\\ |
| − | \vec{\mathbf{q}}_1-\vec{\mathbf{p}}&=\mathbf{R} | + | \vec{\mathbf{q}}_1-\vec{\mathbf{p}}&=\mathbf{R} \vec{\mathbf{q}}_0\\ |
| − | \mathbf{R}^{-1} | + | \mathbf{R}^{-1}(\vec{\mathbf{q}}_1-\vec{\mathbf{p}})&=\mathbf{R}^{-1}\mathbf{R} \vec{\mathbf{q}}_0\\ |
\mathbf{R}^{-1}\vec{\mathbf{q}}_1-\mathbf{R}^{-1}\vec{\mathbf{p}}&=\vec{\mathbf{q}}_0\\ | \mathbf{R}^{-1}\vec{\mathbf{q}}_1-\mathbf{R}^{-1}\vec{\mathbf{p}}&=\vec{\mathbf{q}}_0\\ | ||
\end{align}</math> | \end{align}</math> | ||
Revision as of 16:09, 17 June 2014
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Let 
be a general homogeneous transformation matrix. The inverse transformation 
corresponds to the transformation that reverts the rotation and translation effected by . If a vector is pre-multiplied by and subsequently pre-multiplied by , this results in the original coordinates because 
and multiplication with the identity matrix does not change anything (see transformations).
The general homogeneous transformation matrix for three-dimensional space consists of a 3-by-3 rotation matrix 
and a 3-by-1 translation vector 
combined with the last row of the identity matrix:
![\mathbf{T}=
\left[\begin{array}{ccc|c}
& & & \\
& \mathbf{R} & & \vec{\mathbf{p}}\\
& & & \\ \hline
0 & 0 & 0 & 1
\end{array}\right]](/wiki/robotics/images/math/1/3/2/1327593f05795df92e5c12f1a1a27f84.png)
As stated in the article about homogeneous coordinates, multiplication with is equivalent in cartesian coordinates to applying the rotation matrix first and then translating the coordinates by :
The equation above is now solved for 
:
