Representing rational quadratic Bezier with non-rational cubic Is it possible to exactly represent a rational quadratic Bezier curve with a non-rational cubic Bezier curve? I know that it is always possible to exactly represent an n-degree Bezier with an (n+1)-degree Bezier by carefully choosing the control points, so I'm wondering if a similar technique can be applied to the rational-to-non-rational question.
My feeling is that it should be possible, since we are given an extra control point to work with. But, I also understand that the rational representation effectively lies in a higher dimensional space, which may not be the same thing.
 A: As you probably know, a quarter-circular arc can be represented exactly with a quadratic rational Bézier curve. A cubic Bézier curve can only approximate that arc (quite well, but not exactly).
This shows that there are quadratic rational Bézier curves that cannot be re-parameterized as cubic Bézier curves.
Details:
Here is a parametrization of the quarter-circular arc around the origin
in the first quadrant:
$$\begin{align}
    x(t) &= \frac{1-t^2}{1+t^2}
&   y(t) &= \frac{2t}{1+t^2}
&   t&\in[0,1]
\end{align}$$
The same parametrization can be achieved with a rational quadratic Bézier curve
using the following control points and (non-normalized) weights:
$$\begin{align}
    P_0 &= (1,0)
&   P_1 &= (1,1)
&   P_2 &= (0,1)
\\  w_0 &= 1
&   w_1 &= 1
&   w_2 &= 2
\end{align}$$
If instead you want a circular arc around the origin given by real
polynomials $x(t)$ and $y(t)$, those would need to fulfill
$$\forall t\in[0,1]\colon\ x^2(t) + y^2(t) = 1\tag{*}$$
So clearly, those polynomials cannot be both zero.
Let $d$ be the maximum of the degrees of both polynomials.
Then the left-hand side of $(*)$ is a polynomial in $t$ of degree $2d$
with strictly positive leading coefficient.
Making the left-hand side a nonzero constant would therefore require $d=0$,
but that would yield only a point, not a curve segment.
By employing affine transformations, this result can be extended to
any elliptical arc anywhere in the plane: Polynomial Bézier curves cannot
parameterize it exactly, no matter how large the polynomial degrees involved.
A: A rational quadratic Bezier curve is, in each coordinate, the quotient of two unrelated quadratic polynomials. An unweighted cubic Bezier curve is in each coordinate a cubic polynomial. It is generally not possible to express quotients of arbitrary quadratic polynomials as cubic polynomials.
To help see the issue, try using some negative weights so that the denominator has a zero in $(0, 1)$. The resulting quadratic rational Bezier curve will have a pole, and regular Bezier curves cannot have poles.
