Bezier curve arc length

Question

See also: same question on Math.SE

How can I find the arclength of a Bezier curve? For example, a linear Bezier curve has the length:

length = sqrt(pow(x[1] - x[0], 2) + pow(y[1] - y[0], 2));

But what about quadratic, cubic, or n-degree Bezier curves?

(My goal was to estimate a sampling resolution beforehand, so I don't have to waste time checking if the next point is touching the previous point.)

Answer 1

A simple way for cubic Beziers is to split the curve into N segments and sum the segments' lengths.

However, as soon as you need the length of only part of the curve (e.g. up to a point 30% of the length along), arc-length parameterization will come into play. I posted a fairly long answer on one of my own questions about Béziers, with simple sample code.

Answer 2

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 But do not quote me on this, I do not have references at hand at the moment.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]);
}

Answer 3

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

(a^2·(q^2 - 2·q·r + r^2) + 2·a·(r - q)·(b·(p - r) + c·(q - p)) + (b·(p - r) + c·(q - p))^2)·LN((√(a^2 - 2·a·b + b^2 + p^2 - 2·p·q + q^2)·√(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2) + a^2 + a·(c - 3·b) + 2·b^2 - b·c + (p - q)·(p - 2·q + r))/(√(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2)·√(b^2 - 2·b·c + c^2 + q^2 - 2·q·r + r^2) + a·(b - c) - 2·b^2 + 3·b·c - c^2 + (p - 2·q + r)·(q - r)))/(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2)^(3/2) + (√(a^2 - 2·a·b + b^2 + p^2 - 2·p·q + q^2)·(a^2 + a·(c - 3·b) + 2·b^2 - b·c + (p - q)·(p - 2·q + r)) - √(b^2 - 2·b·c + c^2 + q^2 - 2·q·r + r^2)·(a·(b - c) - 2·b^2 + 3·b·c - c^2 + (p - 2·q + r)·(q - r)))/(a^2 + 2·a·(c - 2·b) + 4·b^2 - 4·b·c + c^2 + (p - 2·q + r)^2)

Where LN is the natural logarithm, and ^ denotes power and √ the square root.

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.

Answer 4

I worked out the closed form expression of length for a 3 point Bezier (below). 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)));