CODEWITHSUNDEEP
c++mathbezier
cdk
cdk

Reputation: 6778

Finding points of inflection on a 2d Bezier curve

I need to determine points of inflection (points where the curvature changes) on a 2d Bezier curve, parameterized by t, 0 <= t <= 1, if they exist. My original method was to sample along the curve, evaluating second derivatives and finding the point where the derivative's sign changes.

2DVector curvature1, curvature2;
for (double t = 0, t <= 1.0; t += STEP) {
    curvature1 = bezier.CurvatureAt(t);
    curvature2 = bezier.CurvatureAt(t + (STEP/2.0 >= 1.0 ? 0 : t + STEP/2.0));
    if (isNegative(curvature1) ? isPositive(curvature2) : isNegative(curvature2)) {
        inflection_point = t;
    }
 }

where CurvatureAt() is a method that evaluates the second derivative of the bezier at t, but as the bezier curve is a vector valued function the derivative is returned as a 2D vector (not std::vector, a 2D vector class). I dont know how to interpret "where the sign changes" for vectors. Basically i dont know how to write isNegative or isPositive in the above snippet.

are there any other ways to find points of inflection on a 2d Bezier curve?

I dont think its possible to determine a closed form solution to this problem because the Bezier can be of arbitrary degree, however please correct me if I'm wrong here.

Upvotes: 1

Views: 2400

Answers (3)

jamadagni
jamadagni

Reputation: 1264

To determine the points of inflection on a bezier, find the time or times in the interval (0, 1) [excluding the endpoints of course] for which the cross product of the first and second derivatives of the parametric equation of the bezier is zero i.e. f' X f'' = 0.

This is noted in various sources like this page and p 4 of this paper.

Upvotes: 1

Petar Ivanov
Petar Ivanov

Reputation: 93040

Curvature is related to but not the same as second derivative.

The signed curvature of a parametric curve P(t) = (x(t), y(t)) is actually a number and is defined as:

k(t) = (x'y'' - x''y') / (x' * x' + y' * y')^(3/2)

If you use this formula your original idea should work.

Upvotes: 2

Bentoy13
Bentoy13

Reputation: 4976

I think you don't need such a loop. According to this page, you can compute at any point the curvature of a Bezier curve. As a Bezier curve has a polynomial expression, you can easily compute when the sign of the curvature changes.

Upvotes: 0

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