我有一个字符只能在Y轴上转动，所以基本上左右，如果up是Vector3.up。角色有一个孩子，这是手榴弹产生的位置（他的右手），现在我试图弄清楚如何为父母计算正确的旋转，以便孩子准确地看着目标，但只能“水平”。父级的 x 和 z 旋转必须为 0。有没有办法做到这一点？如果孩子只在“水平面”上看正确的方向，而不是向上或向下看以完美地面对目标，这很好。
这个答案几乎是我想要的：http://answers.unity.com/comments/1339850/view.html
唯一的问题是产生的旋转是所有三个轴的混合，而不仅仅是一个 y 旋转。
【问题讨论】：
标签： c# unity3d quaternions
我会使用三角函数来解决这个问题。 cmets中的解释：
Vector3 rotAxis = Vector3.up;
// flatten child pos, target pos, char pos, and child forward.
Vector3 flatChildPos = Vector3.ProjectOnPlane(childTransform.position, rotAxis);
Vector3 flatTargetPos = Vector3.ProjectOnPlane(targetTransform.position, rotAxis);
Vector3 flatCharPos = Vector3.ProjectOnPlane(charTransform.position, rotAxis);
Vector3 flatChildForward = Vector3.ProjectOnPlane(childTransform.forward, rotAxis);
Vector3 flatCharToTarget = flatTargetPos-flatCharPos;
Vector3 flatChildToChar = flatCharPos-flatChildPos;
// Find the angle going from the direction of child to parent to the child's forward
// The sign will come in handy later
float childAngle = Vector3.SignedAngle(flatChildToChar, flatChildForward,
rotAxis);
float absChildAngle = Mathf.Abs(childAngle);
// Find the distance between child and character:
float childDist = flatChildToChar.magnitude;
// Find the distance between character and target:
float targetDist = flatCharToTarget.magnitude;
// Consider the triangle made by character position, position of target, and
// desired child position. Use sin rule to find angle of target's corner
float targetAngle = Mathf.Rad2Deg * Mathf.Asin(
Mathf.Sin(absChildAngle * Mathf.Deg2Rad)
* childDist/targetDist);
// determine angle in parent's corner:
float desiredParentAngle = 180f - absChildAngle - targetAngle;
// Determine sign of angle at character from target to child.
// It's the same as the sign of angle at child from char to target.
float sign = Mathf.Sign(childAngle);
// Consider the triangle made by character position, position of target, and
// current child position. Determine the current signed angle in character corner.
float currentParentAngle = Vector3.SignedAngle(flatCharToTarget, -flatChildToChar,
rotAxis);
// Calculate the diff in angle needed to make the current triangle the desired one.
float diffAngle = desiredParentAngle * sign - currentParentAngle;
// Apply the diff in world space
Quaternion newRot = Quaternion.AngleAxis(diffAngle, rotAxis) * charTransform.rotation;
有趣的是，在某些情况下可能有两种可能的轮换。确定何时是这种情况以及第二次轮换留给读者的练习。提示：targetAngle 和 desiredParentAngle 有时也有多个有效值；）
【讨论】：
Afaik 你只需要反转孩子的localEulerAngles.y
localEulerAngles 是子代相对于其父代的欧拉角偏移旋转。因此，仅采用 y 组件已经告诉您子代与其父代的偏移量。
所以我想为了相应地旋转父级，你会这样做，例如
// as usual get the direction from child to target
var targetDirection = targetObject.transform.position - childObject.transform.position;
// flatten the direction by erasing any difference in the Y-axis
var targetDirectionFlat = Vector3.Scale(new Vector3(1,0,1), targetDirection).normalized;
// get the Y angle offset between parent and child
var offset = -childObject.transform.localEulerAngles.y;
// rotate the parent to face the flattened direction minus the offset
// both are rotations around the Y axis only
parent.transform.rotation = Quaternion.LookDirection(targetDirectionFlat) * Quaternion.Euler(Vector3.up * offset);
注意：在智能手机上输入，但我希望 ides 清楚
【讨论】：