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  • unity 两点间生成随机的贝塞尔曲线

    效果如图:

    思路:

      主要问题在于生成随机控制点。

      1. 以start -> end 为z轴建立坐标系,获得 x_Axis, y_Axis, z_Axis

      2.以z轴方向随机长度, x轴y轴随机所在位置

      3.用贝塞尔曲线公式生成曲线

    public class Test : MonoBehaviour
    {
        public Transform m_Start;
        public Transform m_End;
    
        Vector3 m_AxisX;
        Vector3 m_AxisY;
        Vector3 m_AxisZ;
        float m_OriLength;
    
        List<Vector3> m_PointList;
        Vector3[] m_BLine;
    
        void Start()
        {
            m_PointList = new List<Vector3>();
            var start = m_Start.position;
            var end = m_End.position;
            m_OriLength = Vector3.Distance(start, end);
    
            BuildMoveMatrix(start, end);
    
            m_PointList.Add(start);
            RandomPoint(start, end, m_OriLength);
            m_PointList.Add(end);
    
            m_BLine = Bezeir.draw_bezier_curves(m_PointList.ToArray(), m_PointList.Count, 0.005f);
        }
    
        void BuildMoveMatrix(Vector3 start, Vector3 end)
        {
            var z = (end - start).normalized;
            var y = new Vector3(1.0f, 1.0f, -(z.x + z.y) / z.z).normalized;
            var x = Vector3.Cross(z, y).normalized;
    
            m_AxisX = x;
            m_AxisY = y;
            m_AxisZ = z;
        }
    
        void RandomPoint(Vector3 start, Vector3 end, float length)
        {
            if (length < m_OriLength * 0.3f)
                return;
    
            var len = UnityEngine.Random.Range(length * 0.2f, length * 0.8f);
            var point = (end - start).normalized * len + start;
            point = point + 
                    m_AxisX * UnityEngine.Random.Range(-0.3f * length, 0.3f * length) + 
                    m_AxisY * UnityEngine.Random.Range(-0.3f * length, 0.3f * length);
    
            RandomPoint(start, point, Vector3.Distance(start, point));
            m_PointList.Add(point);
            RandomPoint(point, end, Vector3.Distance(point, end));
        }
    }
    
    public class Bezeir
    {
        public static Vector3[] draw_bezier_curves(Vector3[] points, int count, float step)
        {
            List<Vector3> bezier_curves_points = new List<Vector3>();
            float t = 0F;
            do
            {
                var temp_point = bezier_interpolation_func(t, points, count);    // 计算插值点
                t += step;
                bezier_curves_points.Add(temp_point);
            }
            while (t <= 1 && count > 1);    // 一个点的情况直接跳出.
            return bezier_curves_points.ToArray();  // 曲线轨迹上的所有坐标点
        }
    
        /// <summary>
        /// n阶贝塞尔曲线插值计算函数
        /// 根据起点,n个控制点,终点 计算贝塞尔曲线插值
        /// </summary>
        /// <param name="t">当前插值位置0~1 ,0为起点,1为终点</param>
        /// <param name="points">起点,n-1个控制点,终点</param>
        /// <param name="count">n+1个点</param>
        /// <returns></returns>
        private static Vector3 bezier_interpolation_func(float t, Vector3[] points, int count)
        {
            var PointF = new Vector3();
            var part = new float[count];
            float sum_x = 0, sum_y = 0, sum_z = 0;
            for (int i = 0; i < count; i++)
            {
                ulong tmp;
                int n_order = count - 1;    // 阶数
                tmp = calc_combination_number(n_order, i);
                sum_x += (float)(tmp * points[i].x * Math.Pow((1 - t), n_order - i) * Math.Pow(t, i));
                sum_y += (float)(tmp * points[i].y * Math.Pow((1 - t), n_order - i) * Math.Pow(t, i));
                sum_z += (float)(tmp * points[i].z * Math.Pow((1 - t), n_order - i) * Math.Pow(t, i));
            }
            PointF.x = sum_x;
            PointF.y = sum_y;
            PointF.z = sum_z;
            return PointF;
        }
    
        /// <summary>
        /// 计算组合数公式
        /// </summary>
        /// <param name="n"></param>
        /// <param name="k"></param>
        /// <returns></returns>
        private static ulong calc_combination_number(int n, int k)
        {
            ulong[] result = new ulong[n + 1];
            for (int i = 1; i <= n; i++)
            {
                result[i] = 1;
                for (int j = i - 1; j >= 1; j--)
                    result[j] += result[j - 1];
                result[0] = 1;
            }
            return result[k];
        }
    
    }


    参考链接 : https://blog.csdn.net/qq_32688731/article/details/84037072
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  • 原文地址:https://www.cnblogs.com/liucUP/p/10755420.html
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