如何在一组点上绘制闭合曲线? [英] How do I draw a closed curve over a set of points?

问题描述

基本上，我想绘制一个多边形，但我希望边缘显示为软而不是硬.由于多边形的形状很重要，因此边缘必须越过这些点.

Basically I want to draw a polygon, but I want the edges to appear soft rather than hard. Since the shape of the polygon is important, the edges have to go over the points.

我发现单调三次样条曲线对于打开的曲线(即，不会自行缠绕的曲线)是准确的，但是我发现的算法会预先计算点0和N.闭合的曲线?

I've found monotone cubic splines to be accurate for open curves (i.e., curves that don't wrap around on themselves), but the algorithms I've found precalculate points 0 and N. Can I somehow change them to work with a closed curve?

我正在用JavaScript实现此功能，但是伪代码也一样.

I am implementing this in JavaScript, but pseudo-code would just as well.

推荐答案

那里是一种简单的方法(由Maxim Shemanarev开发)，用于构造(通常)在一组点上设置的美观的闭合Bezier曲线.示例:

There is an easy method (developed by Maxim Shemanarev) to construct (usually) good-looking closed Bezier curves set over a set of points. Example:

算法的关键时刻:

和示例代码:

  // Assume we need to calculate the control
    // points between (x1,y1) and (x2,y2).
    // Then x0,y0 - the previous vertex,
    //      x3,y3 - the next one.

    double xc1 = (x0 + x1) / 2.0;
    double yc1 = (y0 + y1) / 2.0;
    double xc2 = (x1 + x2) / 2.0;
    double yc2 = (y1 + y2) / 2.0;
    double xc3 = (x2 + x3) / 2.0;
    double yc3 = (y2 + y3) / 2.0;

    double len1 = sqrt((x1-x0) * (x1-x0) + (y1-y0) * (y1-y0));
    double len2 = sqrt((x2-x1) * (x2-x1) + (y2-y1) * (y2-y1));
    double len3 = sqrt((x3-x2) * (x3-x2) + (y3-y2) * (y3-y2));

    double k1 = len1 / (len1 + len2);
    double k2 = len2 / (len2 + len3);

    double xm1 = xc1 + (xc2 - xc1) * k1;
    double ym1 = yc1 + (yc2 - yc1) * k1;

    double xm2 = xc2 + (xc3 - xc2) * k2;
    double ym2 = yc2 + (yc3 - yc2) * k2;

    // Resulting control points. Here smooth_value is mentioned
    // above coefficient K whose value should be in range [0...1].
    ctrl1_x = xm1 + (xc2 - xm1) * smooth_value + x1 - xm1;
    ctrl1_y = ym1 + (yc2 - ym1) * smooth_value + y1 - ym1;

    ctrl2_x = xm2 + (xc2 - xm2) * smooth_value + x2 - xm2;
    ctrl2_y = ym2 + (yc2 - ym2) * smooth_value + y2 - ym2;

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