Bezier曲线与Catmull-Rom曲线

一、N次贝塞尔

1.一般参数公式

$$B(t)=\sum _{i=0}^n{C}_n^i\ast P_i\ast (1-t{)}^{(n-i)}\ast t^i(t\in [0,1])$$

2.代码

(1)求$n!$ 方法

/// 

/// 求阶乘
/// 
/// 
/// 
public static int Factorial(this int n)
{
    if (n == 0)
    {
        return 1;
    }
    return n * (n - 1).Factorial();
}

(2)求组合$C_n^m$ 方法

/// 

/// 求组合
/// 
/// 
/// 
/// 
public static int Combination(this int n, int m)
{
	return n.Factorial() / (m.Factorial() * (n - m).Factorial());
}

(3)获取Bezier曲线的点

/// 

/// 
/// 
/// 
/// 
/// 
private static Vector3 GetBezierPoint(float t, params Vector3[] p)
{
    int n = p.Length - 1;
    float u = 1 - t;

    Vector3 result = Vector3.zero;

    for (int i = 0; i < p.Length; i++)
    {
        result += p[i] * Mathf.Pow(u, n - i) * Mathf.Pow(t, i) * Combination(n, i);
    }

    return result;
}

(4) 创建曲线

/// 

/// 
 /// 
 /// 
 /// 
 /// 
 public static void DrawBezier(this LineRenderer line, List<Vector3> pos, params Vector3[] target)
 {
     if (target == null || target.Length <= 0) return;

     if (pos == null)
     {
         pos = new List<Vector3>();
     }
     pos.Clear();

     float resolution = 0.01f;
     int loops = Mathf.FloorToInt(1f / resolution);

     for (int i = 0; i <= loops; i++)
     {
         pos.Add(GetBezierPoint(i * resolution, target));
     }

     line.positionCount = pos.Count;
     line.SetPositions(pos.ToArray());
 }

Catmull-Rom曲线

1、推导公式

公式推导过程
由上述公式可推出经过四个点($P_0$,$P_1$,$P_2$,$P_3$)的曲线方程为
$$B(t)=a+b\ast t+c\ast t^2+d\ast t^3$$
其切线方程为
$$C(t)=b+2\ast c\ast t+3\ast d\ast t^2$$
满足如下几个条件

(1) $B(0)=P_1$,故a = $P_1$
(2) $C(0)=\frac{P_2-P_1}{2},故b=\frac{P_2-P_1}{2}$
(3) $B(1)=P_2,故a+b+c+d=P_2$
(4) $C(1)=\frac{P_3-P_1}{2},故b+2c+3d=\frac{P_3-P_1}{2}$
由上诉(1)(2)(3)(4)可推出
$a=P_1$
$b=\frac{P_2-P_1}{2}$
$c=\frac{2P_0-5P_1+4P_2-P3}{2}$
$d=\frac{-P_0+3P_1-3P_2+P_3}{2}$

2、代码

/// 

/// 绘制曲线
/// 每次取四个点:当前点的前一个点，当前点，当前点之后的第一个点，当前点之后的第二个点
/// 
/// 
/// 是否循环
/// 经过的点
public static void DrawCatmullRomSpline(this LineRenderer line, bool loop, params Vector3[] target)
{
    if (target == null || target.Length <= 0) return;

    List<Vector3> pos = new List<Vector3>();

    int current = 0;

    //绘制0~n-1
    for (; current < target.Length - 1; current++)
    {
        if (current == 0) // 第一个点时，其相关的四个点为 0,0,1,2
        {
            GetCatmullRomSplinePos(pos, target[0], target[0], target[1], target[2]);
        }
        else if (current == target.Length - 2) // 为倒数第二个点: n-3,n-2,n-1,n-1
        {
            GetCatmullRomSplinePos(pos, target[current - 1], target[current], target[current + 1], target[current + 1]);
        }
        else
        {
            GetCatmullRomSplinePos(pos, target[current - 1], target[current], target[current + 1], target[current + 2]);
        }
    }

    if (loop) // 循环就是绘制最后一个点与第一个点之间的线段 ： n - 2,n-1,0,1
    {
        current = target.Length - 1;
        GetCatmullRomSplinePos(pos, target[current - 1], target[current], target[0], target[1]);
    }

    line.positionCount = pos.Count;
    line.SetPositions(pos.ToArray());
}

/// 

/// 绘制两点之间的曲线
/// 
/// 
/// 
/// 
/// 
/// 
private static void GetCatmullRomSplinePos(List<Vector3> pos, Vector3 p0, Vector3 p1, Vector3 p2, Vector3 p3)
{
    float resolution = 0.01f;

    int loops = Mathf.FloorToInt(1f / resolution);

    for (int i = 0; i <= loops; i++)
    {
        pos.Add(GetCatmullRomPosition(i * resolution, p0, p1, p2, p3));
    }
}

/// 

/// 会的对应t值的点
/// 
/// 
/// 
/// 
/// 
/// 
/// 
private static Vector3 GetCatmullRomPosition(float t, Vector3 p0, Vector3 p1, Vector3 p2, Vector3 p3)
{
    Vector3 a = 2 * p1;
    Vector3 b = p2 - p0;
    Vector3 c = 2 * p0 - 5 * p1 + 4 * p2 - p3;
    Vector3 d = -p0 + 3 * p1 - 3 * p2 + p3;

    Vector3 pos = 0.5f * (a + (b * t) + (c * t * t) + (d * t * t * t));

    return pos;
}