Arc Length of Bézier Curves

Question

I posted this on gamedev.stackexchange.com earlier, but I didn't get any answers... and only a few views. EDIT: I've got two answers there (as of now), and I'm investigating them now.

How can I find out the distance traveled by a Bézier Curve? For example, the distance traveled by a Linear Bézier Curve is:

$distance = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$

But what of a Quadratic, Cubic, Nth-Degree Bézier Curve?

My goal is to figure out the "resolution" ($resolution = 1-t$) beforehand, so I don't have to waste time checking if the next point is touching the previous point.

From here:

$$\mathbf{B}(t) = \sum_{i=0}^n {n\choose i}(1-t)^{n-i}t^i\mathbf{P}_i$$

Answer 1

See these papers:

• Approximate Arc Length Parametrization, in SIBGRAPI 1996.

• Adaptive sampling of parametric curves, in Graphics Gems V, 1995.

• Computing the arc length of parametric curves, IEEE Computer Graphics and Applications, 1990.

• Point-based methods for estimating the length of a parametric curve, Journal of Computational and Applied Mathematics, 2006.

Answer 2

I showed in the comments how to get Mathematica to generate the arclength function for a quadratic Bézier curve; for this answer I'll give an explicit derivation.

Consider the parametrization

$$\begin{align*}x&=(1-u)^2 x_1+2u(1-u) x_2+u^2 x_3 \\ y&=(1-u)^2 y_1+2u(1-u) y_2+u^2 y_3\end{align*}$$

Letting $\Delta x_i=x_{i+1}-x_i$ and similarly for $\Delta y_i$, the arclength integral corresponding to this parametrization is

$\displaystyle \scriptsize 2\int\sqrt{\Delta x_1^2+\Delta y_1^2+2\left(\Delta x_1\left(\Delta x_2-\Delta x_1\right)+\Delta y_1\left(\Delta y_2-\Delta y_1\right)\right)u+\left(\left(\Delta x_2-\Delta x_1\right)^2+\left(\Delta y_2-\Delta y_1\right)^2\right)u^2}\mathrm du$

We let $c=\Delta x_1^2+\Delta y_1^2$, $b=\Delta x_1\left(\Delta x_2-\Delta x_1\right)+\Delta y_1\left(\Delta y_2-\Delta y_1\right)$, and $a=\left(\Delta x_2-\Delta x_1\right)^2+\left(\Delta y_2-\Delta y_1\right)^2$ to further simplify things.

Consider now the integral

$$2\int \sqrt{c+2bu+au^2}\mathrm du$$

Completing the square yields

$$2\sqrt{a}\int \sqrt{\left(u+\frac{b}{a}\right)^2+\frac{ac-b^2}{a^2}}\mathrm du$$

Skipping the details (but see here for how one might derive the answer), the integral evaluates to

$$\sqrt{a}\left(\left(u+\frac{b}{a}\right)\sqrt{\left(u+\frac{b}{a}\right)^2+\frac{ac-b^2}{a^2}}+\left(\frac{ac-b^2}{a^2}\right)\mathrm{arsinh}\left(\frac{au+b}{\sqrt{ac-b^2}}\right)\right)$$

or

$$\left(u+\frac{b}{a}\right)\sqrt{c+2bu+au^2}+\left(\frac{ac-b^2}{a^{3/2}}\right)\mathrm{arsinh}\left(\frac{au+b}{\sqrt{ac-b^2}}\right)$$

As I've mentioned in the comments, the closed form for the quadratic case is quite complicated (even more so for the cubic case), and you're better off with using numerical quadrature to compute the arclength.