I have an ellipse arc parametrized with its center ($x_0, y_0$); radii ($a, b$); rotation of the X-axis ($\theta$); and start/end arc angles ($\phi_s, \phi_e$) measured from the ellipse semi-major axis. I also have a 3×3 affine transformation matrix $M$ that includes skew/shear and non-uniform scale.
The problem is to get new parameters for the ellipse after it is transformed with $M$.
I’ve converted the original parameters into matrix representation $Q$, computed $M^{-T}QM^{-1}$, and decomposed it back as explained in What is the equation for an ellipse in standard form after an arbitrary matrix transformation?
Works excellent for the ellipse itself. But I cannot find a proper way to compute transformed start/end arc angles $\phi_s', \phi_e'$. Obviously, they are not kept the same in case of skew transform. I’ve tried to transform “tip” vectors pointing to start/end arc points with $M$ and then get the transformed angles back with dot/cross products against ellipse X-axis, but the best thing I’ve got is a mess shifted by $±\pi/2$ or $±\pi$ multiplied by $±1$ depending on the initial parameter values.
Does anyone have an idea about obtaining $\phi_s', \phi_e'$ from the original $\phi_s, \phi_e$ after transforming with $M$?
From the specifications of the problem, the parametric equation of the original ellipse is
$ \mathbf{r}(\phi) = \mathbf{C} + \mathbf{v_1} \cos \phi + \mathbf{v_2} \sin \phi $
where $\mathbf{v_1} = a (\cos \theta, \sin \theta) $ and $\mathbf{v_2} = b (-\sin \theta, \cos \theta) $
Now you apply the affine transformation
$$T(\mathbf{v}) = A \mathbf{v}$$
To the points of the ellipse, then the image points are
$ \mathbf{r'}(\phi) = A \mathbf{C} + A \mathbf{v_1} \cos \phi + A \mathbf{v_2} \sin \phi = \mathbf{C'} + \mathbf{f_1} \cos \phi + \mathbf{f_2} \sin \phi $
Vectors $\mathbf{f_1}$ and $\mathbf{f_2}$ are conjugate radii of the ellipse, and the transformed arc will have the same $\phi_s$ and $\phi_e$ as the original arc.
However, you want the angle relative to the major and minor semi-axes of the transformed ellipse.
For that, we can write
$(\mathbf{r'} - \mathbf{C'} ) = [\mathbf{f_1}, \mathbf{f_2} ] \begin{bmatrix} \cos \phi \\ \sin \phi \end{bmatrix} = F \mathbf{u}(\phi)$
Since $\mathbf{u(\phi)}^T \mathbf{u(\phi)} = 1 $
Then
$(\mathbf{r'} - \mathbf{C'} )^T F^{-T} F^{-1} (\mathbf{r'} - \mathbf{C'}) = 1 $
gives the algebraic equation of the transformed ellipse.
The matrix $Q = F^{-T} F^{-1} $ is positive definite, and symmetric, therefore, it can decomposed into $Q = R D R^T $ where $D$ is diagonal with positive diagonal entries (which are the eigenvalues of $Q$) and $R$ is the matrix of corresponding unit eigenvectors. If we split $D$ into $D = D^{(1/2)}D^{(1/2)} $, then we have
$(\mathbf{r'} - \mathbf{C'} )^T R D^{(1/2)} D^{(1/2)} R^T (\mathbf{r'} - \mathbf{C'}) = 1 $
Now define the new vector $\mathbf{z} = D^{(1/2)} R^T (\mathbf{r'} - \mathbf{C'})$
Then clearly $\mathbf{z}$ is a unit vector, i.e.
$ \mathbf{z} = \begin{bmatrix} \cos \phi' \\ \sin \phi' \end{bmatrix} $
From the above definition, it follows that
$ \mathbf{r'} = \mathbf{C'} + R D^{-(1/2)} \mathbf{z}(\phi') $
Note that the columns of the matrix $R D^{-(1/2)}$ are the two semi-axes of the transformed ellipse, and $\phi'$ is the eccentric angle.
Now recall that,
$ \mathbf{r'} = \mathbf{C'} + F \mathbf{u}(\phi) $
From the equality of the two expressions, it follows that
$ \begin{bmatrix} \cos \phi' \\ \sin \phi' \end{bmatrix} = D^{(1/2)} R^T F \begin{bmatrix} \cos \phi \\ \sin \phi \end{bmatrix} $
Thus given $\phi_s$ and $\phi_e$ we can determine from the above equation, unique values for $\phi'_s$ and $\phi'_e$ (because we have the sine and cosine values for $\phi'$).
As a numerical example, let the original ellipse have $a = 30, b = 18, \theta = \dfrac{\pi}{3}$, and let $\phi_s = \dfrac{\pi}{6}=0.5235987$ and $\phi_e = \dfrac{4 \pi}{3}=4.18879 $, and let the transformation matrix be
$ A = \begin{bmatrix} 2 && - 3 \\ 1 && 5 \end{bmatrix} $
The original elliptical arc in shown in the first figure in black.
The transformed ellipse is drawn in blue in the following figure which was generated using the conjugate radii $\mathbf{f_1}$ and $\mathbf{f_2}$.
Next, to plot the transformed arc using it perpendicular semi-axes, the corresponding transformed angles are computed using the above formula to be
$ \phi'_s = 1.775725 $ and $ \phi'_e = 5.440917 $
Using these values, the transformed ellipse is plotted (in red).
As can be seen, the two arcs are identical.