示例项目源代码
如果您对它的派生方式不感兴趣,请跳至答案部分。
背景资料
贝塞尔曲线很有趣,使用起来很有趣。此动画显示了如何在两个锚点(P0 和 P2)之间相对于控制点 (P1) 随着时间的推移绘制二次曲线。
您需要做的不是绘制控制点 (P1),而是在 t=0.5 处绘制曲线上的点:

(来源:whilenotnull.com)
幸运的是,这很容易使用维基百科页面上给出的公式:http ://en.wikipedia.org/wiki/Bezier_Curve#Quadratic_B.C3.A9zier_curves
这是 Actionscript 中的公式:
public function calculatePoint(p0:Point, p1:Point, p2:Point, t:Number):Point
{
var p:Point = new Point(calculateTerm(p0.x, p1.x, p2.x, t), calculateTerm(p0.y, p1.y, p2.y, t));
return p;
}
public function calculateTerm(p0:Number, p1:Number, p2:Number, t:Number):Number
{
var negT:Number = 1 - t;
var a0:Number = Math.pow(negT, 2) * p0;
var a1:Number = 2 * negT * t * p1;
var a2:Number = Math.pow(t, 2) * p2;
var pos:Number = a0 + a1 + a2;
return pos;
}
因此,如果您插入三个点:var t0:Point = calculatePoint(p0, p1, p2, 0.5);
您将获得曲线上要绘制“控制点”的点。
回答
现在我们可以编写一个函数,假设第二个参数是曲线上的一个点,并确定控制点的坐标:
public function derivePoint(p0:Point, b1:Point, p2:Point, t:Number = 0.5):Point
{
var p:Point = new Point(deriveTerm(p0.x, b1.x, p2.x, t), deriveTerm(p0.y, b1.y, p2.y, t));
return p;
}
public function deriveTerm(p0:Number, bt:Number, p2:Number, t:Number):Number
{
var negT:Number = 1 - t;
var a0:Number = Math.pow(negT, 2) * p0;
var a1:Number = 2 * negT * t;
var a2:Number = Math.pow(t, 2) * p2;
var p1:Number = (bt - a0 - a2) / a1;
return p1;
}
从此,我更新了您的代码片段,以(希望)通过您的“控制点”绘制曲线:
// clear old line and draw new / begin fill
var g:Graphics = graphics;
g.clear();
g.lineStyle(2, 0, 1);
g.beginFill(0x0099FF,.1);
//move to starting anchor point
var startX:Number = anchorPoints[0].x;
var startY:Number = anchorPoints[0].y;
g.moveTo(startX, startY);
// Connect the dots
var p0:Point = new Point(startX, startY);
var p2:Point;
var numAnchors:Number = anchorPoints.length;
for (var i:Number=1; i<numAnchors; i++) {
p2 = new Point(anchorPoints[i].x, anchorPoints[i].y);
// curve to next anchor through control
var b1:Point = new Point(controlPoints[i].x,controlPoints[i].y);
var p1:Point = derivePoint(p0, b1, p2);
g.curveTo(p1.x, p1.y, p2.x, p2.y);
p0 = p2;
}
// Close the loop
g.curveTo(controlPoints[0].x,controlPoints[0].y,startX,startY);
示例项目源代码