Bezier curve arc length

Arc lengths for Bezier curves are only closed form for linear and quadratic ones. For cubics, it is not guaranteed to have a closed solution. The reason is arc length is defined by a radical integral, for which has a closed for only 2nd degree polynomials.

Just for reference: The length of a quadratic Bezier for the points (a,p) (b,q) and (c,r) is

$$ \frac{(a^2(q^2 - 2qr + r^2) + 2a(r - q)(b(p - r) + c(q - p)) + (b(p- r) + c(q - p))^2)\ln{\frac{\sqrt{a^2 - 2ab + b^2 + p^2 - 2pq + q^2} \sqrt{a^2 + 2a(c - 2b) + 4b^2 - 4bc + c^2 + (p - 2q + r)^2} + a^2 + a(c - 3b) + 2b^2 - bc + (p - q)(p - 2q + r) }{\sqrt{a^2 + 2a(c - 2b) + 4b^2 - 4bc + c^2 + (p - 2q + r)^2} \sqrt{b^2 - 2bc + c^2 + q^2 - 2qr + r^2} + a(b - c) - 2b^2 + 3bc - c^2 + (p - 2q + r)(q - r)}}}{a^2 + 2a(c - 2b) + 4b^2 - 4bc + c^2 + (p - 2q + r)^2}^\frac{3}{2} + \frac{\sqrt{(a^2 - 2ab + b^2 + p^2 - 2pq + q^2) (a^2 + a(c - 3b) + 2b^2 - bc + (p - q)(p - 2q + r)) - \sqrt{b^2 - 2bc + c^2 + q^2 - 2qr + r^2} (a(b - c) - 2b^2 + 3bc - c^2 + (p - 2q + r)(q - r))}}{a^2 + 2a(c - 2b) + 4b^2 - 4bc + c^2 + (p - 2q + r)^2} $$

Hence, it should be easier and cheaper approximate the arc by some other rule, like a polygon or an integration scheme like Simpson's rule, because square roots the LN are expensive operations.

While I am d'accord with the answers you got already, I want to add a simple but powerful approximation mechanism which you can use for any degree Bézier curves: You continually subdivide the curve using de Casteljau subdivision until the maximum distance of the control points of a sub-curve to the sub-curve's baseline is below some constant epsilon. In that case the sub-curve can be approximated by its baseline.

In fact, I believe this is the approach usually taken when a graphics subsystem has to draw a Bézier curve. But do not quote me on this, I do not have references at hand at the moment.

In practice it will look like this: (except the language is irrelevant)

public static Line[] toLineStrip(BezierCurve bezierCurve, double epsilon) { ArrayList<Line> lines = new ArrayList<Line>(); Stack<BezierCurve> parts = new Stack<BezierCurve>(); parts.push(bezierCurve); while (!parts.isEmpty()) { BezierCurve curve = parts.pop(); if (distanceToBaseline(curve) < epsilon) { lines.add(new Line(curve.get(0), curve.get(1))); } else { parts.addAll(curve.split(0.5)); } } return lines.toArray(new Line[0]); } 

I worked out the closed-form expression of length for a quadratic point Bezier curve (consisting of 3 control points). I've not attempted to work out a closed form for 4+ points. This would most likely be difficult or complicated to represent and handle. However, a numerical approximation technique such as a Runge-Kutta integration algorithm would work quite well by integrating using the arc length formula. My Q&A on RK45 on MSE may help with RK45 implementation.

Here is some Java code for the arc length of a 3 point Bezier, with points a,b, and c.

 v.x = 2*(b.x - a.x); v.y = 2*(b.y - a.y); w.x = c.x - 2*b.x + a.x; w.y = c.y - 2*b.y + a.y; uu = 4*(w.x*w.x + w.y*w.y); if(uu < 0.00001) { return (float) Math.sqrt((c.x - a.x)*(c.x - a.x) + (c.y - a.y)*(c.y - a.y)); } vv = 4*(v.x*w.x + v.y*w.y); ww = v.x*v.x + v.y*v.y; t1 = (float) (2*Math.sqrt(uu*(uu + vv + ww))); t2 = 2*uu+vv; t3 = vv*vv - 4*uu*ww; t4 = (float) (2*Math.sqrt(uu*ww)); return (float) ((t1*t2 - t3*Math.log(t2+t1) -(vv*t4 - t3*Math.log(vv+t4))) / (8*Math.pow(uu, 1.5)));