TL;DR

After the center (see Ellipses (for SVG): finding the center), it’s time to find the values for $t$ that represent our arc of ellipse.

Last post was a bit of a roller-coaster to find the center of the ellipse. Now we’re going to find the parameters values, but this is definitely easier at this point.

If we go back to the translated-then-rotated representation, i.e. the one centered in $\mathbf{C’}$, we can easily translate the origin on the center and obtain:

$$\mathbf{P''_1} = (x''_1, y''_1) = (x'_1 - C'_x, y'_1 - C'_y) \\ \mathbf{P''_2} = (x''_2, y''_2) = (-x'_1 - C'_x, -y'_1 - C'_y) \\$$

Remember: in the first translation, the origin was put in the midpoint of $\mathbf{P}_1$ and $\mathbf{P}_2$, so they happen to have opposite coordinate values.

These are all known quantities at this point. To find the respective values of $t$, we remember that this is the angle of the point corresponding to re-normalizing the ellipse back to a unitary circle, so for $t_1$ we have:

$$cos(t_1) = \frac{x''_1}{r_x} = \frac{x'_1 - C'_x}{r_x} \\ sin(t_1) = \frac{y''_1}{r_y} = \frac{y'_1 - C'_y}{r_y}$$

This will allow us to find the right value of $t_1$. Many programming languages provide the atan2 function, which takes two parameters (in the $Y’$ and $X’$ direction, in our case) and avoids infinites, so we can calculate $t_1$ as:

$$t_1 = atan2(\frac{y'_1 - C'_y}{r_y}, \frac{x'_1 - C'_x}{r_x})$$

Of course the atan2 function does not complain if we scale both arguments by the same factor, so we can use the equivalent expression:

$$t_1 = atan2(r_x \cdot (y'_1 - C'_y), r_y \cdot (x'_1 - C'_x))$$

The same goes for $t_2$, of course:

$$t_2 = atan2(r_x \cdot (-y'_1 - C'_y), r_y \cdot (-x'_1 - C'_x))$$

Now we have that $t_1$ and $t_2$ are values in the interval $]-\pi, \pi]$, and we have to:

• establish which comes first and which second (based on which arc of the ellipse we are interested into);
• find an equivalent contiguous range.

This is basically finding $t_{begin}$ and $t_{end}$. We can do like this:

• initialize $t_{begin} = t_1$ and $t_{end} = t_2$;
• make sure that $t_{begin}$ is not greater $t_{end}$. To do this, we subtract $2\pi$ from $t_{begin}$ if it is greater than $t_{end}$:

$$t_{begin} > t_{end} \Rightarrow t_{begin} \leftarrow t_{begin} - 2\pi$$

• at this point, we can campute $\delta = t_{end} - t_{begin}$ and assign to $t_{begin}$ the value of $t_{end}$ in either of the following cases:
  • $\delta \le \pi$ (i.e. it’s the short arc of ellipse) but $f_A = 1$ (i.e. we need the long one), OR
  • $\delta > \pi$ (i.e. it’s the long arc of ellipse) but $f_A = 0$ (i.e. we need the short one).
• shift $t_{begin}$ in a range we like (e.g. $[0, 2\pi[$), if we want;
• last, re-calculate $t_{end} = t_{begin} + \delta$

Now we have a contiguos interval $[t_{begin}, t_{end}]$ that allows us sweep the whole arc.

The sweeping in this interval might make your pen start from the end point $\mathbf{P}_2$ and go back to $\mathbf{P}_1$. As we are eventually interested into the bounding box, anyway, this is not an issue here.

Problem solved… theoretically, stay tuned for the implementation!