Bezier Curve – Calculate Control Points of Cubic Bezier Curve Approximating a Circle

bezier-curve

I'm not mathematically inclined, so please be patient with my question.

Given

  • $(x_0, y_0)$ and $(x_1, y_1)$ as the endpoints of a cubic Bezier curve.

  • $(c_x, c_y)$ and r as the centerpoint and the radius of a circle.

  • $(x_0, y_0)$ and $(x_1, y_1)$ are on the circle.

  • if it makes the calculation simpler, it's safe to assume the arc is less than or equal to $\frac{\pi}{2}$.

How do I calculate the two control points of the Bezier curve that best fits the arc of the circle from $(x_0, y_0)$ to $(x_1, y_1)$?

Best Answer

Let $(x,y)^R=(-y,x)$ represent rotation by $\pi/2$ counterclockwise and $$ \gamma(t)=(1-t)^3p_0+3t(1-t)^2p_1+3t^2(1-t)p_2+t^3p_3 $$ define a cubic bezier with control points $\{p_0,p_1,p_2,p_3\}$.

Suppose $p_0=(x_0,y_0)$, $p_3=(x_1,y_1)$, and $c=(c_x,c_y)$ are given so that $|p_0-c|=|p_3-c|=r$ ($p_3$ is counterclockwise from $p_0$). Then $p_1=p_0+\alpha(p_0-c)^R$ and $p_2=p_3-\alpha(p_3-c)^R$ where $$ \alpha=\frac43\tan\left(\frac14\cos^{-1}\left(\frac{(p_0-c)\cdot(p_3-c)}{r^2}\right)\right) $$ For a quarter of a circle, $\alpha=\frac43(\sqrt2-1)$, and $\gamma$ is no more than $0.00027$ of the radius of the circle off.

Here is a plot of $\gamma$ in red over the quarter circle in black. We really don't see the circle since it is no more than $0.1$ pixels off from $\gamma$ when the radius is $400$ pixels.

$\hspace{3.5cm}$enter image description here

Computation of $\boldsymbol{\alpha}$

Looking at an arc with an angle of $\theta=\cos^{-1}\left(\frac{(p_0-c)\cdot(p_3-c)}{r^2}\right)$

$\hspace{1.5cm}$enter image description here

we see that the distance from $c$ to the middle of the arc is $$ r\cos(\theta/2)+\frac34\alpha r\sin(\theta/2) $$ we wish to choose $\alpha$ so that this is equal to $r$. Solving for $\alpha$ gives $$ \begin{align} \alpha &=\frac43\frac{1-\cos(\theta/2)}{\sin(\theta/2)}\\ &=\frac43\tan(\theta/4) \end{align} $$

A Slight Improvement

Using a circle of radius $1$, the maximum error in radius produced using $\alpha=\frac43\tan(\theta/4)$ is approximately $$ 0.0741\cos^4(\theta/4)\tan^6(\theta/4) $$ and the error is always positive; that is, the cubic spline never passes inside the circle. Reducing $\alpha$ reduces the midpoint distance by $\frac34\sin(\theta/2)=\frac32\tan(\theta/4)\cos^2(\theta/4)$ times as much, so to distribute the error evenly between the positive and negative, a first guess, assuming that the amplitude of the radius is unchanged, would be to reduce $\alpha$ by $0.0247\cos^2(\theta/4)\tan^5(\theta/4)$.

A bit of investigation shows that, when equalizing the positive and negative swings of the radius, the amplitude increases and that $$ \alpha=\frac43\tan(\theta/4)-0.03552442\cos^2(\theta/4)\tan^5(\theta/4) $$ gives pretty even distribution of the error between positive and negative for $\theta\le\pi/2$. The maximum error, both positive and negative, is approximately $$ 0.0533\cos^4(\theta/4)\tan^6(\theta/4) $$ When $\theta=\pi/2$, this agrees with the article mentioned by bubba in comments.

Note however, that in minimizing the radial error from the circle, the actual variation in radius is increased. Using the simple formula for $\alpha$, which puts the cubic bezier outside the circle, the radius varies by $0.0741\cos^4(\theta/4)\tan^6(\theta/4)$. However, when we minimize the error, the radial variation increases to $0.1066\cos^4(\theta/4)\tan^6(\theta/4)$.

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