unity绘制一条流动的弧线(贝塞尔线)
本文实例为大家分享了unity绘制一条流动弧线的具体代码,供大家参考,具体内容如下
最终效果
把下面脚本复制,直接拖上脚本,设置两个点(物体)的位置
GameObject1是开始点的位置,GameObject2是结束点的位置
public Transform[] controlPoints; public LineRenderer lineRenderer; public float centerPoint =0.1f; private int layerOrder = 0; //生成弧线中间的点数 private int _segmentNum = 20; //偏移 float m_offset; float m_speed = 0.5f; void Start() { if (!lineRenderer) { lineRenderer = GetComponent<LineRenderer>(); } lineRenderer.sortingLayerID = layerOrder; //调用画贝斯尔线 GetBeizerList(controlPoints[0].position, (controlPoints[0].position + controlPoints[1].position) * 0.5f + new Vector3(0, centerPoint, 0), controlPoints[1].transform.position, _segmentNum); } private void Update() { m_offset = m_offset - m_speed * Time.deltaTime; //控制offset使材质移动 GetComponent<LineRenderer>().material.mainTextureOffset = new Vector2(m_offset, 0); } /// <summary> /// 根据T值,计算贝塞尔曲线上面相对应的点 /// </summary> /// <param name="t"></param>T值 /// <param name="p0"></param>起始点 /// <param name="p1"></param>控制点 /// <param name="p2"></param>目标点 /// <returns></returns>根据T值计算出来的贝赛尔曲线点 private static Vector3 CalculateCubicBezierPoint(float t, Vector3 p0, Vector3 p1, Vector3 p2) { float u = 1 - t; float tt = t * t; float uu = u * u; Vector3 p = uu * p0; p += 2 * u * t * p1; p += tt * p2; return p; } /// <summary> /// 获取存储贝塞尔曲线点的数组 /// </summary> /// <param name="startPoint"></param>起始点 /// <param name="controlPoint"></param>控制点 /// <param name="endPoint"></param>目标点 /// <param name="segmentNum"></param>采样点的数量 /// <returns></returns>存储贝塞尔曲线点的数组 public Vector3[] GetBeizerList(Vector3 startPoint, Vector3 controlPoint, Vector3 endPoint, int segmentNum) { Vector3[] path = new Vector3[segmentNum]; for (int i = 1; i <= segmentNum; i++) { float t = i / (float)segmentNum; Vector3 pixel = CalculateCubicBezierPoint(t, startPoint, controlPoint, endPoint); //设置lineRenderer的control Points lineRenderer.positionCount = i; lineR【原创作者:http://www.1234xp.com/tw.html网络转载请说明出处】enderer.SetPosition(i - 1, pixel); //存储Point path[i - 1] = pixel; Debug.Log(path[i - 1]); } return path; }
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