如何在C＃中找到開始，結束和2個交叉點的BezierSegment的控制點 - AKA Cubic Bezier 4點插值

Question

我一直在努力尋找一種可以理解的方法來做到這一點。 我有四個點，一個StartPt，EndPoint和Intersection點來表示貝塞爾曲線中的峰值和谷值。

C＃中的BezierSegment需要start，controlPoint 1，controlPoint 2，endpoint - 但是我沒有任何控制點我只有這兩個點位於貝塞爾曲線上（我稱之為上面的交點）...我可以計算出兩個控制點嗎？

在此先感謝，這一直讓我發瘋。

這里有一些解釋： http ： //www.tinaja.com/glib/nubz4pts1.pdf，但它是用postscript寫的，而且這種語言對我來說毫無意義 - 這是我的頭腦。

Answer 1

通過4個點的曲線有無數個解，但最好的簡單解決方案是嘗試使曲線段長度與弦長成比例。 您鏈接的代碼是一階近似，效果很好並且非常快。

這是PostScript代碼的C＃轉換：

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;
    }
}

我沒有測試它，但它編譯。 只需使用您擁有的4個點調用DrawingUtility.BezierFromIntersection ，它將返回一個用於繪制曲線的PathFigure 。

Answer 2

這是兩個很好的例子：

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

另請參閱此動畫以更好地了解BezierSplines如何工作http://en.wikipedia.org/wiki/B%C3%A9zier_curve

Answer 3

您應該考慮使用Cardinal（Canonical）Splines，它使用路徑上存在的一組點，加上一個“張力”參數，用於控制角落平滑到角落切線的程度。

在Windows窗體中，可以使用DrawCurve和DrawClosedCurve方法。 在WPF中沒有直接的等價物。 這里有兩篇文章描述了使用C＃在WPF中使用基數樣條線。

Floris - 用於WPF Cardinal Spline的AddCurve

Petzold - WPF和Silverlight中的典型樣條

Answer 4

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;
        }
    }
}