Question

有没有人看到过针对cornu螺旋（又名，回旋体）或花键的正确绘制算法？对于弧线和线条，我们有像Bresenham算法这样的东西。这适用于回旋体吗？

维基百科页面有这个Sage代码：

p = integral(taylor(cos(L^2), L, 0, 12), L)
q = integral(taylor(sin(L^2), L, 0, 12), L)
r1 = parametric_plot([p, q], (L, 0, 1), color = 'red')

是否有参数图的示例代码？我在网上搜索时看不到多少。

Answer 1

我没有看到任何现有的高速算法。但是，我确实理解了绘制这样东西的常用方法。基本上，您递归地分割L，直到计算出的左，中，右点足够接近您可以绘制该直线的直线。我能够使用MathNet.Numerics.dll进行集成。一些代码：

    public static void DrawClothoidAtOrigin(List<Point> lineEndpoints, Point left, Point right, double a, double lengthToMidpoint, double offsetToMidpoint = 0.0)
    {
        // the start point and end point are passed in; calculate the midpoint
        // then, if we're close enough to a straight line, add that line (aka, the right end point) to the list
        // otherwise split left and right

        var midpoint = new Point(a * C(lengthToMidpoint + offsetToMidpoint), a * S(lengthToMidpoint + offsetToMidpoint));
        var nearest = NearestPointOnLine(left, right, midpoint, false);
        if (Distance(midpoint, nearest) < 0.4)
        {
            lineEndpoints.Add(right);
            return;
        }
        DrawClothoidAtOrigin(lineEndpoints, left, midpoint, a, lengthToMidpoint * 0.5, offsetToMidpoint);
        DrawClothoidAtOrigin(lineEndpoints, midpoint, right, a, lengthToMidpoint * 0.5, offsetToMidpoint + lengthToMidpoint);
    }

    private static double Distance(Point a, Point b)
    {
        var x = a.X - b.X;
        var y = a.Y - b.Y;
        return Math.Sqrt(x * x + y * y);
    }

    private static readonly double PI_N2 = Math.Pow(Math.PI * 2.0, -0.5);

    public static double C(double theta)
    {
        return Integrate.OnClosedInterval(d => Math.Cos(d) / Math.Sqrt(d), 0.0, theta) / PI_N2;
    }

    public static double S(double theta)
    {
        return Integrate.OnClosedInterval(d => Math.Sin(d) / Math.Sqrt(d), 0.0, theta) / PI_N2;
    }

cornu螺旋/样条的高速绘图算法？

1 个答案: