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503

answers:

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Hi, I've been struggling looking for an understandable way to do this. I have four points, a StartPt, EndPoint, and Intersection points to represent the peak and valley in the bezier.

The BezierSegment in C# requires start, controlPoint 1, controlPoint 2, endpoint - however I don't have any control points I only have these two points that lie along the bezier curves (i'm calling them intersection points above)... how can I calculate the two control points?

Thanks in advance, this has been driving me crazy.

There's some kind of explanation here: http://www.tinaja.com/glib/nubz4pts1.pdf but it's written in postscript and that language makes no sense to me at all - it's over my head.

A: 

There are an infinite number of solutions to a curve passing through 4 points, but the best simple solution is to try to make the curve segment lengths proportional to the chord lengths. The code you link to is the a first order approximation that works well and is pretty fast.

Here's the C# translation of the PostScript code:

static class DrawingUtility
{
    // linear equation solver utility for ai + bj = c and di + ej = f
    static void solvexy(double a, double b, double c, double d, double e, double f, out double i, out double j)
    {
        j = (c - a / d * f) / (b - a * e / d);
        i = (c - (b * j)) / a;
    }

    // basis functions
    static double b0(double t) { return Math.Pow(1 - t, 3); }
    static double b1(double t) { return t * (1 - t) * (1 - t) * 3; }
    static double b2(double t) { return (1 - t) * t * t * 3; }
    static double b3(double t) { return Math.Pow(t, 3); }

    static void bez4pts1(double x0, double y0, double x4, double y4, double x5, double y5, double x3, double y3, out double x1, out double y1, out double x2, out double y2)
    {
        // find chord lengths
        double c1 = Math.Sqrt((x4 - x0) * (x4 - x0) + (y4 - y0) * (y4 - y0));
        double c2 = Math.Sqrt((x5 - x4) * (x5 - x4) + (y5 - y4) * (y5 - y4));
        double c3 = Math.Sqrt((x3 - x5) * (x3 - x5) + (y3 - y5) * (y3 - y5));
        // guess "best" t
        double t1 = c1 / (c1 + c2 + c3);
        double t2 = (c1 + c2) / (c1 + c2 + c3);
        // transform x1 and x2
        solvexy(b1(t1), b2(t1), x4 - (x0 * b0(t1)) - (x3 * b3(t1)), b1(t2), b2(t2), x5 - (x0 * b0(t2)) - (x3 * b3(t2)), out x1, out x2);
        // transform y1 and y2
        solvexy(b1(t1), b2(t1), y4 - (y0 * b0(t1)) - (y3 * b3(t1)), b1(t2), b2(t2), y5 - (y0 * b0(t2)) - (y3 * b3(t2)), out y1, out y2);
    }

    static public PathFigure BezierFromIntersection(Point startPt, Point int1, Point int2, Point endPt)
    {
        double x1, y1, x2, y2;
        bez4pts1(startPt.X, startPt.Y, int1.X, int1.Y, int2.X, int2.Y, endPt.X, endPt.Y, out x1, out y1, out x2, out y2);
        PathFigure p = new PathFigure { StartPoint = startPt };
        p.Segments.Add(new BezierSegment { Point1 = new Point(x1, y1), Point2 = new Point(x2, y2), Point3 = endPt } );
        return p;
    }
}

I haven't tested it, but it compiles. Just call DrawingUtility.BezierFromIntersection with the 4 points you have, and it will return a PathFigure for drawing the curve.

Gabe
Wow! Gabe, Thank you soo much. I have no idea how you can read that postscript code (looks like assembly language MIPS), but this is perfect. Just tried it out and it worked great.Hmmm... Now I've got to solve the same problem but with finding the single control point in a Quadratic Bezier given a start, midpoint, and endpoint. You wouldn't know how Interpolate a QuadraticBezier by any chance?
softwarequestioneer
If you want to know how to do quadratics, make another post and link to it from here.
Gabe
Thanks Gabe! Here's the link to my question about Quadratics: http://stackoverflow.com/questions/2320956/find-control-point-for-quadraticbeziersegment-when-given-start-end-and-1-point
softwarequestioneer
A: 

Here are two good examples:

http://www.codeproject.com/KB/graphics/ClosedBezierSpline.aspx http://www.codeproject.com/KB/graphics/BezierSpline.aspx

Also see this animation to better understand how BezierSplines work http://en.wikipedia.org/wiki/B%C3%A9zier_curve

thomas nn