# ETOOBUSY đźš€ minimal blogging for the impatient

# Ellipses (for SVG): parameter values

**TL;DR**

After the center (see Ellipses (for SVG): finding the center), itâ€™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 weâ€™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 $\mathbf{Câ€™}$, 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 $Yâ€™$ and $Xâ€™$ direction, in our case) and avoids infinites, so
we can calculate $t_1$ as:

Of course the `atan2`

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

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 $t_{end}$:

- 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. itâ€™s the
*short*arc of ellipse) but $f_A = 1$ (i.e. we need the long one), OR - $\delta > \pi$ (i.e. itâ€™s the
*long*arc of ellipse) but $f_A = 0$ (i.e. we need the short one).

- $\delta \le \pi$ (i.e. itâ€™s the
- 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

penstart 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 solvedâ€¦ theoretically, stay tuned for the implementation!