bezier-curvepolynomials
How do I convert a segment of parabola to a cubic Bezier curve?
The parabola segment is given as a polynomial with two x values for the edges.
My target is to convert a quadratic piecewise polynomial to a Bezier path (a set of concatenated Bezier curves).
The simplest heuristic approximation method is the one proposed by Tiller and Hanson. They just offset the legs of the control polygon in perpendicular directions:
The blue curve is the original one, and the three blue lines are the legs of its control polygon. We offset these three lines, and intersect/trim them to get the points A and B. The red curve is the offset. It's only an approximation of the true offset, of course, but it's often adequate.
If the approximation is not good enough for your purposes, you split the curve into two, and approximate the two halves individually. Keep splitting until you're happy. You will certainly have to split the original curves at inflexion points, if any, for example.
Here is an example where the approximation is not very good, and splitting would probably be needed:
There is a long discussion of the Tiller-Hanson algorithm plus possible improvements on this web page.
The Tiller-Hanson approach is compared with several others (most of which are more complex) in this paper.
Another good reference, with more up-to-date materials, is this section from the Patrikalakis-Maekawa-Cho book.
For even more references, you can search for "offset" in this bibliography.
In Theorem $3.4$ of this paper, a more general algorithm for finding the the conic associated with a rational quadatic spline is derived. Specialized for non-rational quadratic splines and parabolas, we get the equation for the parabola $$ \begin{bmatrix}x&y&1\end{bmatrix}Q\begin{bmatrix}x\\y\\1\end{bmatrix}=0 $$ where $$ Q=2\left(uw^T+wu^T\right)-vv^T $$ and, using $a=p_0$, $b=p_1$, and $c=p_2$, $$ u=\begin{bmatrix}b_y-c_y\\c_x-b_x\\b_xc_y-b_yc_x\end{bmatrix}\\ v=\begin{bmatrix}c_y-a_y\\a_x-c_x\\c_xa_y-c_ya_x\end{bmatrix}\\ w=\begin{bmatrix}a_y-b_y\\b_x-a_x\\a_xb_y-a_yb_x\end{bmatrix} $$
Best Answer
You can do this in two steps, first convert the parabola segment to a quadratic Bezier curve (with a single control point), then convert it to a cubic Bezier curve (with two control points).
Let $f(x)=Ax^2+Bx+C$ be the parabola and let $x_1$ and $x_2$ be the edges of the segment on which the parabola is defined.
Then $P_1=(x_1,f(x_1))$ and $P_2=(x_2,f(x_2))$ are the Bezier curve start and end points and $C=(\frac{x_1+x_2}{2},f(x_1)+f'(x_1)\cdot \frac{x_2-x_1}{2})$ is the control point for the quadratic Bezier curve.
Now you can convert this quadratic Bezier curve to a cubic Bezier curve by define the two control points as: $C_1=\frac{2}{3}C+\frac{1}{3}P_1$ and $C_2=\frac{2}{3}C+\frac{1}{3}P_2$.