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;
 lineRenderer.SetPosition(i - 1, pixel);
 //存储Point
 path[i - 1] = pixel;
 Debug.Log(path[i - 1]);
}
return path;
}

来源：https://blog.csdn.net/BBoyLXiang/article/details/88815637