方法一:
鼠标控制物体前后左右移动,上下Y值固定; 限制齿轮的旋转圈数(360*3)先计算两个向量角度(不考虑两个向量转的读数大于180,因为理论上不太可能一帧转这么多,以后再说),然后判断两个向量是顺时针还是逆时针,最后计算上一帧角度和目标角度的差值。最后,用上一帧的角度加上插值得到目标角度。虽然麻烦,但可以根据 差值,累积旋转角度,不受顺时针影响。
/// <summary> /// 最初需要左转还是右转 /// </summary> public bool left = false; public void OnDrag(PointerEventData eventData) {
///获取鼠标位置与初始位置之间的向量 var oldDir = moveBackPos - transform.position; var nowDir = Input.mousePosition - transform.position; var an = Vector3.Angle(oldDir, nowDir); var cross = oldDir.x * nowDir.y - oldDir.y * nowDir.x; an = cross > 0 ? an : -an; //cross>0//逆时针 angeleCount = an; if (left) {
if (angeleCount <= 360 * 3 && angeleCount >= 0) {
transform.localEulerAngles = new Vector3(0, 0, lastAngle an); } } else {
if (angeleCount >= -360 * 3 && angeleCount <= 0)
{
transform.localEulerAngles = new Vector3(0, 0, lastAngle + an);
}
}
moveBackPos = Input.mousePosition;
lastAngle = transform.localEulerAngles.z;
}