根据两点,获取贝塞尔曲线.

[1dit]

小弟数学从来都是59分,所有请各位多多关照....

如图
http://hi.csdn.net/attachment/201105/27/148499_13064671916HT0.png

[gomoku]

static public PointF[] GetBezier(PointF s, PointF e)
{
    const float distance = 3f;
    MyPoint p1 = s, p4 = e;

    MyPoint normal = (p4 - p1).Pert;
    MyPoint t2 = p1 + (p4 - p1) * 0.25f + normal * distance;
    MyPoint t3 = p1 + (p4 - p1) * 0.75f - normal * distance;

    // t2 = 0.75^3*p1 + 3*0.75^2*0.25*p2 + 3*0.75*0.25^2*p3 + 0.25^3p4
    // t3 = 0.25^3*p1 + 3*0.25^2*0.75*p2 + 3*0.25*0.75^2*p3 + 0.75^3p4
    float a = 3 * 0.75f * 0.75f * 0.25f;
    float b = 3 * 0.25f * 0.25f * 0.75f;
    MyPoint m = t2 - p1 * (0.75f * 0.75f * 0.75f) - p4 * (0.25f * 0.25f * 0.25f);
    MyPoint n = t3 - p1 * (0.25f * 0.25f * 0.25f) - p4 * (0.75f * 0.75f * 0.75f);
    // a*p2 + b*p3 = m
    // b*p2 + a*p3 = n
    // (a+b)*(p2+p3) = m+n
    // (a-b)*(p2-p3) = m-n
    // 2*p2 = (m+n)/(a+b) + (m-n)/(a-b);
    // 2*p3 = (m+n)/(a+b) - (m-n)/(a-b);
    MyPoint p2 = ((m + n) / (a + b) + (m - n) / (a - b)) / 2;
    MyPoint p3 = ((m + n) / (a + b) - (m - n) / (a - b)) / 2;

    return new PointF[] { p1, p2, p3, p4 };
}

public class MyPoint
{
    public float X { get; set; }
    public float Y { get; set; }
    public float Length { get { return (float)Math.Sqrt(this * this); } }
    public MyPoint Pert { get { MyPoint r = this / this.Length; return new MyPoint() { X = r.Y, Y = -r.X }; } }
    
    public static MyPoint operator +(MyPoint p1, MyPoint p2) { return new MyPoint() { X = p1.X + p2.X, Y = p1.Y + p2.Y }; }
    public static MyPoint operator -(MyPoint p1, MyPoint p2) { return new MyPoint() { X = p1.X - p2.X, Y = p1.Y - p2.Y }; }
    public static MyPoint operator *(MyPoint p1, float f) { return new MyPoint() { X = p1.X * f, Y = p1.Y * f }; }
    public static MyPoint operator /(MyPoint p1, float f) { return new MyPoint() { X = p1.X / f, Y = p1.Y / f }; }
    public static implicit operator PointF(MyPoint p1) { return new PointF(p1.X, p1.Y); }
    public static implicit operator MyPoint(PointF p1) { return new MyPoint() { X = p1.X, Y = p1.Y }; }
    public static float operator *(MyPoint p1, MyPoint p2) { return p1.X * p2.X + p1.Y * p2.Y; }

[gomoku]

把图贴出来后才发现是要约束曲线跑在半径为3的通道内:)
1楼的方法不能达成该要求，它只是强迫Bezier经过1/4和3/4处的两点，该两点距离直线的距离为3。

[gomoku]

Bezier曲线为：
B(t) = (1-t)^3*p1 + 3*(1-t)^2*t*p2 + 3*(1-t)*t^2*p3 + p4

假设p1和p4在x轴上，那么对于控制点p2(x2,y),p3(x3,-y)，最高最低点将发生在以下函数的拐点上：
f(t) = 3*(1-t)^2*t + 3*(1-t)*t^2

求导：
f'(t) = 6t*t - 6t + 1

解得：
t2 = (6-Sqrt(12)/12
t3 = (6+Sqrt(12)/12

把一楼方法中的0.25（1/4处）和0.75（3/4处）换成t2和t3即可。

[1dit]

[Quote=引用 2 楼 gomoku 的回复:]

把图贴出来后才发现是要约束曲线跑在半径为3的通道内:)
1楼的方法不能达成该要求，它只是强迫Bezier经过1/4和3/4处的两点，该两点距离直线的距离为3。
[/Quote]

public Point[] GetBezierPoint(Point p1, Point p4)
{
    Point[] result = new Point[4];
    result[0] = p1;
    result[3] = p4;

    //  TODO...
    //我这里需要的是p2,p3的坐标代数式.
    //麻烦明示.
    //result[1] = ?
    //result[2] = ?
    
    
    return result ;
}

[1dit]

@gomoku
3Q,我知道了.