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Let's assume I have a cubic bezier curve that is provided with A, B, C, D points, where

  • A is the start of the curve

  • B is the first control point

  • C is the second control point

  • D is the end of the curve.

The curve's parametric equation is given like this:

$x(t) = A_x + 3(B_x - A_x)t + 3(A_x - 2B_x + C_x)t ^ 2 + (3(B_x - C_x) + D_x - A_x) t ^ 3, 0\leqslant t\leqslant 1$ $y(t) = A_y + 3(B_y - A_y)t + 3(A_y - 2B_y + C_y)t ^ 2 + (3(B_y - C_y) + D_y - A_y) t ^ 3, 0\leqslant t\leqslant 1$

Let's assume this curve is (UPD: almost) identical to the arc of the ellipse with the center in the point $O$.

Question:

What is the best strategy to find the center $O$, radii and rotation of (UPD: the approximated) ellipse that represents the corresponding arc?

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Unfortunately the hypothesis that the curve is "identical to the arc of the ellipse" is impossible. Bezier curves can come close enough to circular or elliptic arcs to be visually indistniguishable from them; but a bezier curve cannot exactly coincide with an arc of an ellipse. If it did, for some ellipse $E$ given by an equation $Q(x,y)=0$, then $x(t)$ and $y(t)$ would satisfy $Q(x(t),y(t))=0$ for all $t$ with $0 \leq t \leq 1$, and thus for all real $t$. But then $E$ would contain points whose coordinates $x(t)$ and $y(t)$ are arbitrarily large (by taking $t \to \infty$), which is impossible $-$ unless both $x$ and $y$ are constants, in which case the "curve" is just a single point.

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  • $\begingroup$ 0≀𝑑≀1 corresponds to the limits of $x$ and $y$ for the elliptic arc, so this puts the limits onto $x$ and $y$ that can be supplied into the ellipse equation 𝑄(π‘₯,𝑦)=0. Are you stating that even within those limited ranges x(0)≀x≀x(1) and y(0)≀y≀y(1) for the ellipse equation 𝑄(π‘₯,𝑦)=0 they are still not identical? $\endgroup$
    – Eugene Dudnyk
    Commented Apr 11, 2020 at 20:23
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    $\begingroup$ The point is that $Q(x(t),y(t))$ is a polynomial in $t$, so if it's exactly zero for all $t$ in an interval such as $0 \leq t \leq 1$ then it must be identically zero $-$ whence $Q(x(t),y(t)) = 0$ holds for all real $t$, whether inside or outside the interval. $\endgroup$
    – Noam D. Elkies
    Commented Apr 11, 2020 at 20:46
  • $\begingroup$ Thanks, I'll note that. Still, I am looking for the best approximation into the elliptic arc. For example, in this paper, the method for the inverse problem is described. The error of approximation of elliptic arc to bezier curve is negligible. I am looking for the same kind of method, but inverse to the one described in that paper. $\endgroup$
    – Eugene Dudnyk
    Commented Apr 11, 2020 at 21:04
  • $\begingroup$ I'm also interested in this. There seems to be many ways (posts and websites) for approximating a Bezier curve from an ellipse. So why not the other way? $\endgroup$
    – DrDress
    Commented Feb 3, 2021 at 17:10

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