Question

对于绘图应用程序，我将鼠标移动坐标保存到数组中，然后使用lineTo绘制它们。生成的线条不平滑。如何在所有聚集点之间生成单条曲线？

我用google搜索但我只找到3个绘制线的函数：对于2个样本点，只需使用lineTo。对于3个样本点quadraticCurveTo，对于4个样本点，bezierCurveTo。

（我尝试在阵列中每4个点绘制一个bezierCurveTo，但这会导致每4个采样点扭结，而不是连续的平滑曲线。）

如何编写一个函数来绘制一个包含5个样本点的平滑曲线？

Answer 1

您可以使用基数样条曲线来执行此操作：

这个函数是这样的，点数组排列为[x1, y1, x2, y2, ... xn, yn]，张力在[0.0,1.0]之间，可选的段数决定了每个点之间的分辨率。

<强> Here's an online demo of this in action

UPDATE 发布了我的基本实现的错误版本，这是正确的 -

结果将是一个新的数组，其中包含您迭代的平滑线 -

function getCurvePoints(ptsa, tension, numOfSegments) {

    // use input value if provided, or use a default value   
    tension         =   (tension != 'undefined') ? tension : 0.5;
    numOfSegments   =   numOfSegments   ? numOfSegments : 16;

    var _pts = [], res = [],    // clone array
        x, y,                   // our x,y coords
        t1x, t2x, t1y, t2y,     // tension vectors
        c1, c2, c3, c4,         // cardinal points
        st, t, i;               // steps based on num. of segments

    // clone array so we don't change the original
    _pts = ptsa.slice(0);

    _pts.unshift(pts[1]);           //copy 1. point and insert at beginning
    _pts.unshift(pts[0]);
    _pts.push(pts[pts.length - 2]); //copy last point and append
    _pts.push(pts[pts.length - 1]);

    // ok, lets start..

    // 1. loop goes through point array
    // 2. loop goes through each segment between the two points + one point before and after
    for (i=2; i < (_pts.length - 4); i+=2) {

        // calc tension vectors
        t1x = (_pts[i+2] - _pts[i-2]) * tension;
        t2x = (_pts[i+4] - _pts[i]) * tension;

        t1y = (_pts[i+3] - _pts[i-1]) * tension;
        t2y = (_pts[i+5] - _pts[i+1]) * tension;

        for (t=0; t <= numOfSegments; t++) {

            // calc step
            st = t / numOfSegments;

            // calc cardinals
            c1 =   2 * Math.pow(st, 3)  - 3 * Math.pow(st, 2) + 1; 
            c2 = -(2 * Math.pow(st, 3)) + 3 * Math.pow(st, 2); 
            c3 =       Math.pow(st, 3)  - 2 * Math.pow(st, 2) + st; 
            c4 =       Math.pow(st, 3)  -     Math.pow(st, 2);

            // calc x and y cords with common control vectors
            x = c1 * _pts[i]    + c2 * _pts[i+2] + c3 * t1x + c4 * t2x;
            y = c1 * _pts[i+1]  + c2 * _pts[i+3] + c3 * t1y + c4 * t2y;

            //store points in array
            res.push(x);
            res.push(y);

        }
    }

    return res;
}

如何使用javascript HTML5画布通过N点绘制曲线？

1 个答案: