I'm trying to implement two JavaScript functions to calculate
- A 2D coordinate transformation from the reference frame $O$ to $O'$;
- The inverse transformation from $O'$ to $O$;
The first step it's very simple for me and it works. My mathematical process is this:
$$
\begin{pmatrix}
x'\\
y'\\
1
\end{pmatrix}
=
\begin{pmatrix}
R & u_0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
u\\
0
\end{pmatrix}
=
\begin{pmatrix}
cos\theta & -sin\theta & x_0\\
sin\theta & cos\theta & y_0\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
1
\end{pmatrix}
$$
Where $R$ is the rotation matrix, $u_0$ is the $O'$ origin in the $O$ frame reference and $u$ is a point in $O$. At the end I get this two equations:
$$
\left\{
\begin{array}{c}
x' = xcos\theta – ysin\theta + x_0\\
y' = xsin\theta + ycos\theta + y_0\\
\end{array}
\right.
$$
The problem is how to calculate the inverse transformation. I'm doing something like this but it doesn't work:
$$
\begin{pmatrix}
x\\
y\\
1
\end{pmatrix}
=
\begin{pmatrix}
R^{-1} & -R^{-1}u_0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
u'\\
0
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{cos\theta} & sin\theta & -\frac{x_0}{cos\theta}-y_0sin\theta\\
-sin\theta & \frac{1}{cos\theta} & {x_0}{sin\theta}-\frac{y_0}{cos\theta}\\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x'\\
y'\\
1
\end{pmatrix}
$$
The final equations are:
$$
\left\{
\begin{array}{c}
x = x'\frac{1}{cos\theta} + y'sin\theta -\frac{x_0}{cos\theta}-y_0cos\theta\\
y = -x'sin\theta + y'\frac{1}{cos\theta} + {x_0}{sin\theta}-\frac{y_0}{cos\theta}\\
\end{array}
\right.
$$
Can you tell me where I wrong?
Best Answer
Pay attention to what transformation you want to represent, whether active or passive.
Anyway, if $R$ is the matrix that represents a rotation of angle $\theta$ in a plane, its inverse is the matrix that represents a rotation of the opposite angle $- \theta$, around the same origin, and it is easy to see that it is represented by the transpose matrix: $R^{-1}=R^T$ (it seems that this is your mistake, when you use $\frac{1}{\cos \theta}$ in the inverse matrix.).
From this it is also easy to see that the inverse of a rigid tranformation $T$ that , in homogeneous coordinates, is represented by the matrix $$ T=\begin{pmatrix} R&\vec u\\ 0&1 \end{pmatrix} $$
is represented by the matrix $$ T^{-1}=\begin{pmatrix} R^T&-R^T\vec u\\ 0&1 \end{pmatrix} $$