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Say you have this type of cubic Bézier curve:

enter image description here

The 4 control points A,B,C,D have restrictions:

  • A & B have the same Y-axis coordinate

  • C & D have the same Y-axis coordinate

  • B & C have the same x-axis coordinate

My question is:

How do you express the X-axis coordinate of the inflection point of the red curve in function of the control points?

It is a bit related to this question, but the restrictions of the control points are an issue for me..

Thanks.

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Without loss of generality you can choose your coordinate system such that

\begin{align*} A&=\begin{pmatrix}0\\1\end{pmatrix} & B&=\begin{pmatrix}x\\1\end{pmatrix} & C&=\begin{pmatrix}x\\0\end{pmatrix} & D&=\begin{pmatrix}1\\0\end{pmatrix} \end{align*}

Then the Bézier curve is

$$ P(t) = \begin{pmatrix} 3t(1-t)x + t^3 \\ (2t+1)(1-t)^2 \end{pmatrix} = \begin{pmatrix}t^3 - 3t^2x + 3tx \\ 2t^3 - 3t^2 + 1\end{pmatrix} $$

Now if you differentiate, you get

\begin{align*} P'(t) &= (3t^2 - 6tx + 3x, 6t^2 - 6t) \\ P''(t) &= (6t - 6x, 12t - 6) \end{align*}

Looking for the point where they both point in the same direction, you compute the determinant of these two vectors:

$$ \begin{vmatrix} 3t^2 - 6tx + 3x & 6t - 6x \\ 6t^2 - 6t & 12t - 6 \end{vmatrix} =-18\cdot\bigl((2x - 1)t^2 - 2xt + x\bigr) \overset!=0 $$

Solving this quadratic equation, you get two solutions, but the one which satisfies $0\le t\le 1$ is always the following, as can be seen from the plot below:

$$t_{\text{inflection}} = \frac{\sqrt{x-x^2} - x}{1 - 2x}$$

For $x=\frac12$ this expression is undefined, but you know that in that case the inflection point has to lie in the center, at $t=\frac12$.

If you don't have the coordinates above, you can define

$$ x := \frac{\lvert AB\rvert}{\lvert AB\rvert+\lvert CD\rvert} = \frac{B_x - A_x}{D_x - A_x} $$

and obtain $t$ using the same formula.

Plot of t as a function of x

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