Take the 2-minute tour ×

I'm trying to move an image along Bezier curve. This is how I do it:

- (void)startFly
 [self runAction:[CCSequence actions:
             [CCBezierBy actionWithDuration:timeFlying bezier:[self getPathWithDirection:currentDirection]],
             [CCCallFuncN actionWithTarget:self selector:@selector(endFly)],


My issue is that the image moves not uniformly. In the beginning it's moving slowly and then it accelerates gradually and at the end it's moving really fast. What should I do to get rid of this acceleration?

share|improve this question

3 Answers 3

up vote 14 down vote accepted

It is possible to approximate a solution to this problem for most parametric trajectories. The idea is the following: if you zoom deep enough on a curve, you cannot tell the curve itself from its tangent at that point.

By making this assumption, there is no need to precompute anything more than two vectors (three for cubic Bezier curves, etc.).

So for a curve M(t) we compute its tangent vector dM/dt at point t. The norm of this vector is ||dM/dt|| and thus the distance traveled for a duration Δt can be approximated as ||dM/dt||Δt. It follows that a distance L is traveled for a duration L/||dM/dt||.

Application: quadratic Bezier curve

If the control points of the Bezier curve are A, B and C, the trajectory can be expressed as:

M(t) = (1-t)²A + 2t(1-t)B + t²C
     = t²(A - 2B + C) + t(-2A + 2B) + A

So the derivative is:

dM/dt = t(2A - 4B + 2C) + (-2A + 2B)

You just need to store vectors v1 = 2A - 4B + 2C and v2 = -2A + 2B somewhere. Then, for a given t, if you want to advance of a length L, you do:

t = t + L / length(t * v1 + v2)

Cubic Bezier curves

The same reasoning applies to a curve with four control points A, B, C and D:

M(t) = (1-t)³A + 3t(1-t)²B + 3t²(1-t)C + t³D
     = t³(-A + 3B - 3C + D) + t²(3A - 6B + 3C) + t(-3A + 3B) + A

The derivative is:

dM/dt = t²(-3A + 9B - 9C + 3D) + t(6A - 12B + 6C) + (-3A + 3B)

We precompute v1 = -3A + 9B - 9C + 3D, v2 = 6A - 12B + 6C and v3 = -3A + 3B and the final formula is:

t = t + L / length(t * t * v1 + t * v2 + v3);

Accuracy issues

If you are running at a reasonable framerate, L (which should be computed according to the frame duration) will be sufficiently small for the approximation to work.

However, you may experience inaccuracies in extreme cases. If L is too large, you can do the computation piecewise, for instance using 10 parts:

    t = t + (L / 10) / length(t * v1 + v2);
share|improve this answer

You need to reparamaterize the curve. The easiest way to do this is to calculate the arc lengths of several segments of the curve and use these to figure out where you should sample from. For example, maybe at t=0.5 (halfway through), you should pass s=0.7 to the curve to get the "halfway" position. You need to store a list of arc lengths of various curve segments to do this.

There are probably better ways, but here's some very simple C# code I wrote to do this in my game. It should be easy to port to Objective C:

public sealed class CurveMap<TCurve> where TCurve : struct, ICurve
    private readonly float[] _arcLengths;
    private readonly float _ratio;
    public float length { get; private set; }
    public TCurve curve { get; private set; }
    public bool isSet { get { return !length.isNaN(); } }
    public int resolution { get { return _arcLengths.Length; } }

    public CurveMap(int resolution)
        _arcLengths = new float[resolution];
        _ratio = 1f / resolution;
        length = float.NaN;

    public void set(TCurve c)
        curve = c;
        Vector2 o = c.sample(0);
        float ox = o.X;
        float oy = o.Y;
        float clen = 0;
        int nSamples = _arcLengths.Length;
            float t = (i + 1) * _ratio;
            Vector2 p = c.sample(t);
            float dx = ox - p.X;
            float dy = oy - p.Y;
            clen += (dx * dx + dy * dy).sqrt();
            _arcLengths[i] = clen;
            ox = p.X;
            oy = p.Y;
        length = clen;

    public Vector2 sample(float u)
        if(u <= 0) return curve.sample(0);
        if(u >= 1) return curve.sample(1);

        int index = 0;
        int low = 0;
        int high = resolution - 1;
        float target = u * length;
        float found = float.NaN;

        // Binary search to find largest value <= target
        while(low < high)
            index = (low + high) / 2;
            found = _arcLengths[index];
            if (found < target)
                low = index + 1;
                high = index;

        // If the value we found is greater than the target value, retreat
        if (found > target)

        if(index < 0) return curve.sample(0);
        if(index >= resolution - 1) return curve.sample(1);

        // Linear interpolation for index
        float min = _arcLengths[index];
        float max = _arcLengths[index + 1];
        Debug.Assert(min <= target && max >= target);
        float interp = (target - min) / (max - min);
        Debug.Assert(interp >= 0 && interp <= 1);
        return curve.sample((index + interp) * _ratio);

Edit: It's worth noting that this won't give you the exact arc length, since it's impossible to get the arc length of a cubic curve. All this does is estimate the length of the various segments. Depending on how long the curve is, you may need to increase the resolution to prevent it from changing speeds when it reaches a new segment. I usually use ~100, which I've never had a problem with.

share|improve this answer

I don't know anything about cocos2, but a bezier curve is a kind of parametric equation, so you should be able to get your x and y values in terms of time.

share|improve this answer
Add an example + more explanation and this would be a good answer. –  Byte56 Apr 9 '12 at 22:37

Your Answer


