17

我一直在努力寻找一种可以理解的方法来做到这一点。我有四个点，一个 StartPt、EndPoint 和 Intersection 点来表示贝塞尔曲线中的峰值和谷值。

C# 中的 BezierSegment 需要 start、controlPoint 1、controlPoint 2、endpoint - 但是我没有任何控制点我只有沿着贝塞尔曲线的这两个点（我称它们为上面的交点）......如何我可以计算两个控制点吗？

在此先感谢，这让我发疯了。

这里有某种解释：http: //www.tinaja.com/glib/nubz4pts1.pdf但它是用后记写的，这种语言对我来说根本没有意义 - 它在我的脑海中。

4

4 回答 4

18

一条经过 4 个点的曲线有无数种解法，但最好的简单解法是尽量使曲线段的长度与弦长成正比。您链接到的代码是一阶近似值，运行良好且速度非常快。

这是 PostScript 代码的 C# 翻译：

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;
    }
}

我没有测试过它，但它可以编译。只需调用DrawingUtility.BezierFromIntersection您拥有的 4 个点，它就会返回PathFigure用于绘制曲线的 a。

于 2010-02-23T07:33:40.820 回答
1

这里有两个很好的例子：

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

另请参阅此动画以更好地了解 BezierSplines 的工作原理 http://en.wikipedia.org/wiki/B%C3%A9zier_curve

于 2010-04-21T12:16:21.967 回答
1

您应该考虑使用基数（规范）样条曲线，它使用路径上存在的一组点，加上一个“张力”参数，该参数控制拐角与拐角切线的平滑程度。

在 Windows 窗体中，可以使用DrawCurve和DrawClosedCurve方法。在 WPF 中没有直接的等价物。这里有两篇文章描述了在 C# 中使用 WPF 中的基数样条。

Floris - WPF Cardinal Spline 的 AddCurve

Petzold - WPF 和 Silverlight 中的规范样条

于 2011-07-03T06:30:30.273 回答
0

as3 版本：

package 
{
    import flash.geom.Vector3D;

    public class DrawingUtility 
    {
        private var x1:Number; 
        private var y1:Number; 
        private var x2:Number;
        private var y2:Number;

        // linear equation solver utility for ai + bj = c and di + ej = f
        private function solvexy(a:Number, b:Number, c:Number, d:Number, e:Number, f:Number):Vector.<Number>
        {
            var returnVal:Vector.<Number> = new Vector.<Number>();
            var j:Number = (c - a / d * f) / (b - a * e / d);
            var i:Number = (c - (b * j)) / a;
            returnVal[0] = i;
            returnVal[1] = j;
            return returnVal;
        }

        // basis functions
        private function b0(t:Number):Number { 
            return Math.pow(1 - t, 3);
        }
        private function b1(t:Number):Number {
            return t * (1 - t) * (1 - t) * 3;
        }
        private function b2(t:Number):Number {
            return (1 - t) * t * t * 3;
        }
        private function b3(t:Number):Number {
            return Math.pow(t, 3);
        }

        private function bez4pts1(x0:Number, y0:Number, x4:Number, y4:Number, x5:Number, y5:Number, x3:Number, y3:Number):void
        {
            // find chord lengths
            var c1:Number = Math.sqrt((x4 - x0) * (x4 - x0) + (y4 - y0) * (y4 - y0));
            var c2:Number = Math.sqrt((x5 - x4) * (x5 - x4) + (y5 - y4) * (y5 - y4));
            var c3:Number = Math.sqrt((x3 - x5) * (x3 - x5) + (y3 - y5) * (y3 - y5));
            // guess "best" t
            var t1:Number = c1 / (c1 + c2 + c3);
            var t2:Number = (c1 + c2) / (c1 + c2 + c3);
            // transform x1 and x2
            var x1x2:Vector.<Number> = solvexy(b1(t1), b2(t1), x4 - (x0 * b0(t1)) - (x3 * b3(t1)), b1(t2), b2(t2), x5 - (x0 * b0(t2)) - (x3 * b3(t2)));
            x1 = x1x2[0];
            x2 = x1x2[1];
            // transform y1 and y2
            var y1y2:Vector.<Number> = solvexy(b1(t1), b2(t1), y4 - (y0 * b0(t1)) - (y3 * b3(t1)), b1(t2), b2(t2), y5 - (y0 * b0(t2)) - (y3 * b3(t2)));
            y1 = y1y2[0];
            y2 = y1y2[1];
        }

        public function BezierFromIntersection(startPt:Vector3D, int1:Vector3D, int2:Vector3D, endPt:Vector3D):Vector.<Vector3D>
        {
            var returnVec:Vector.<Vector3D> = new Vector.<Vector3D>();
            bez4pts1(startPt.x, startPt.y, int1.x, int1.y, int2.x, int2.y, endPt.x, endPt.y);

            returnVec.push(startPt);
            returnVec.push(new Vector3D(x1, y1));
            returnVec.push(new Vector3D(x2, y2));
            returnVec.push(endPt);
            return returnVec;
        }
    }
}
于 2013-07-29T11:10:09.267 回答