计算三次贝塞尔曲线的弧长，曲线长度。为什么不工作？ [英] Calculate the arclength, curve length of a cubic bezier curve. Why is not working?

问题描述

我正在使用此算法计算arclength（立方贝塞尔曲线的长度）

I'm calculating the arclength (length of a cubic bezier curve) with this algorithm

    function getArcLength(path) {

        var STEPS = 1000; // > precision
        var t = 1 / STEPS;
        var aX=0;
        var aY=0;
        var bX=0, bY=0;
        var dX=0, dY=0;
        var dS = 0;
        var sumArc = 0;
        var j = 0;

        for (var i=0; i<STEPS; j = j + t) {
            aX = bezierPoint(j, path[0], path[2], path[4], path[6]);
            aY = bezierPoint(j, path[1], path[3], path[5], path[7]);
            dX = aX - bX;
            dY = aY - bY;
            // deltaS. Pitagora
            dS = Math.sqrt((dX * dX) + (dY * dY));
            sumArc = sumArc + dS;
            bX = aX;
            bY = aY;
            i++;
        }

        return sumArc;
    }

但我得到的是类似于915.但是曲线是480而且没有更多。 （我肯定知道这是因为曲线几乎是一条线）
路径数组有这个值：
498 51 500 52 500 53 500 530

But what I get is something like 915. But the curve is 480 and no more. (I know for sure this because the curve is almost a line) The path array has this values: 498 51 500 52 500 53 500 530

bezierPoint函数是：

The bezierPoint function is:

        function bezierPoint(t, o1, c1, c2, e1) {
        var C1 = (e1 - (3.0 * c2) + (3.0 * c1) - o1);
        var C2 = ((3.0 * c2) - (6.0 * c1) + (3.0 * o1));
        var C3 = ((3.0 * c1) - (3.0 * o1));
        var C4 = (o1);

        return ((C1*t*t*t) + (C2*t*t) + (C3*t) + C4)
    }

我做错了什么？

推荐答案

因为 bX 和 bY 被初始化为0，当i = 0时第一个段测量距离原点的距离到了路径的起点。这会增加额外的sqrt（498 ^ 2 + 51 ^ 2）长度。如果你初始化 bX = path [0] 和 bY = path [1] ，我认为它会起作用。

Because bX and bY are initialized to 0, the first segment when i = 0 measures the distance from the origin to the start of the path. This adds an extra sqrt(498^2+51^2) to the length. If you initialize bX = path[0] and bY = path[1], I think it will work.

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