How can I compute rotation angle of an image?

Question

An image is translated with $x_0$ and $y_0$ (vertical and horizontal displacements), rotated with an angle $θ$, and then is scaled on both vertical and horizontal directions with the scaling parameters $α$ and $β$. The resulting composed geometric transform in homogeneous matrix form is $$\begin{pmatrix}\sqrt 3 & 1 & 20 \\ -1 & \sqrt 3 & 10 \\ 0 & 0 & 1\end{pmatrix}$$

I need to compute the rotation angle $θ$, the scaling parameters $α$ and $β$, and the vertical and horizontal displacements $x_0$ and $y_0$.

Answer 1

Let $F\colon \mathbb P^2\to\mathbb P^2$ be your transformation on the projective plane. It can be decomposed as $F = S_{(\alpha,\beta)}\, R_{\theta}\, T_{(x_0,y_0)}$ as explained by AugSB. To find the angle $\theta$ you can take any two points and see how the line segment between them gets rotated. The line segment from $(0,0)$ to $(1,0)$ get's mapped to the line segment from $(20,10)$ to $(\sqrt 3+20,-1+10)$, where I use $(x,y)$ to denote the point $(x:y:1)$ in $\mathbb P^2$. Hence $$ \theta = \measuredangle (1,0), (\sqrt 3,-1) = \cos^{-1}\left(\frac{\sqrt 3}{2}\right) = \frac{\pi}{6} = 30^\circ. $$ Now consider the line segment of length $1$ from $(0,0)$ to $(\sqrt{3}/2,-1/2)$, this has an angle of $-30^\circ$ to the $x$-axis, so after translation and rotation it will be parallel to the $x$-axis. Hence the length of its image under $F$ determines $\alpha$. We have $F(0,0)=(20,10)$ and $F(\sqrt{3}/2,-1/2)=(21, -\sqrt 3+10)$ so that $$ \alpha = \|(1, -\sqrt 3)\| = 2. $$ The same reasoning applied to the line segment from $(0,0)$ to $(1/2,\sqrt{3}/2)$ yields $$ \beta = \|(\sqrt 3,1)\| = 2. $$ Finally, we can find $(x_0,y_0)$ by applying the inverses of $S_{(\alpha,\beta)}$ and $R_\theta$ to $F(0,0)=(20,10)$: \begin{align*} S_{(2,2)}\, R_{30^\circ}\, T_{(x_0,y_0)}(0,0) &= (20,10) \\ R_{30^\circ}\, T_{(x_0,y_0)}(0,0) &= (10,5) \\ \end{align*} Note that $R_\theta = \begin{pmatrix} \sqrt 3/2 & -1/2 & 0 \\ 1/2 & \sqrt 3/2 & 0 \\ 0 & 0 & 1\end{pmatrix}$, so that $R_\theta^{-1} = \begin{pmatrix} \sqrt 3/2 & 1/2 & 0 \\ -1/2 & \sqrt 3/2 & 0 \\ 0 & 0 & 1\end{pmatrix}$ and we get $$ (x_0, y_0) = T_{(x_0,y_0)}(0,0) = R_\theta^{-1} (10,5) = \left(\frac 5 2(2\sqrt 3 - 1), \frac 5 2(\sqrt 3 + 2)\right). $$

Answer 2

You are using three matrices: $$\text{Traslating matrix: } T_{(x_0,y_0)} = \begin{pmatrix} 1 & 0 & x_0 \\\ 0 & 1 & y_0 \\\ 0 & 0 & 1 \end{pmatrix},$$ $$\text{Rotating matrix: } R_\theta = \begin{pmatrix} \cos\theta & \sin\theta & 0 \\\ -\sin\theta & \cos\theta & 0 \\\ 0 & 0 & 1 \end{pmatrix},$$ $$\text{Scaling matrix: } S_{(\alpha,\beta)} = \begin{pmatrix} \alpha & 0 & 0 \\\ 0 & \beta & 0 \\\ 0 & 0 & 1 \end{pmatrix}.$$

You just need to solve the system given by: $$S_{(\alpha,\beta)} R_\theta T_{(x_0,y_0)} = \begin{pmatrix} \sqrt{3} & 1 & 20 \\\ -1 & \sqrt{3} & 10 \\\ 0 & 0 & 1 \end{pmatrix}.$$