给定 3 个点及其相对位置,计算平面的 3D 旋转
Calculate 3D rotation of a plane given 3 points and their relative position
我正在尝试在 Unity 中旋转 3D 平面,给定三个点。这些点并不总是相同的,但它们始终位于此图像中的红点上
我知道这些点的绝对位置,以及它们相对于平面的位置。例如,我可能知道点 (-5, 5) 位于 (10, -4, 13)。
我知道当三个点在一条线上时(例如(-5, 5),(0, 5)和(5, 5))无法计算完整的旋转,所以我已经抛出出现这种情况时例外。但是当三个点不在一条直线上时,应该可以计算出完整的旋转。
到目前为止,我已经使用下面的代码进行了旋转,但这错过了绕 y 轴的旋转(该示例使用点 (-5, 5)、(5, 5) 和 (-5, -5).
Vector3 p1 = Point1.transform.position;// point 1 absolute position
Vector3 p1Relative = new Vector3(-5, 0, 5);
Vector3 p2 = Point2.transform.position;// point 2 absolute position
Vector3 p2Relative = new Vector3(5, 0, 5);
Vector3 p3 = Point3.transform.position;// point 3 absolute position
Vector3 p3Relative = new Vector3(-5, 0, -5);
Gizmos.DrawSphere(p1, .1f);
Gizmos.DrawSphere(p2, .1f);
Gizmos.DrawSphere(p3, .1f);
Vector3 normal = Vector3.Cross(p2 - p1, p3- p1);
rotator.transform.up = normal;
我如何扩展或更改此代码以包含绕 Y 轴的旋转?提前谢谢你。
此答案在数学堆栈交换中使用 this method described by robjohn。
您可以解决一对 Matrix4x4
s ,您可以使用它们从第一个坐标系转换到第二个坐标系,然后用它来计算原点(中心点)和旋转以对齐轴(使用 Quaternion.LookRotation
);
public class test : MonoBehaviour
{
[SerializeField] Transform point1;
[SerializeField] Transform point2;
[SerializeField] Transform point3;
[SerializeField] Vector2 relativePos1;
[SerializeField] Vector2 relativePos2;
[SerializeField] Vector2 relativePos3;
[SerializeField] Transform demonstrator;
void Update()
{
Vector3 absolutePos1 = point1.position;
Vector3 absolutePos2 = point2.position;
Vector3 absolutePos3 = point3.position;
Vector3 relativePos4 = (Vector3)relativePos1 +
Vector3.Cross(relativePos2 - relativePos1,
relativePos3 - relativePos1);
Vector3 absolutePos4 = absolutePos1 +
Vector3.Cross(absolutePos2 - absolutePos1,
absolutePos3 - absolutePos1);
Vector4 rightColumn = new Vector4(0, 0, 0, 1);
Matrix4x4 P = new Matrix4x4(relativePos2 - relativePos1,
relativePos3 - relativePos1,
relativePos4 - (Vector3)relativePos1, rightColumn);
Matrix4x4 Q = new Matrix4x4(absolutePos2 - absolutePos1,
absolutePos3 - absolutePos1,
absolutePos4 - absolutePos1, rightColumn);
Vector3 originPos = TransformPos(P, Q, Vector3.zero);
// up in source is forward in output
Vector3 forwardPos = TransformPos(P, Q, Vector3.up);
// back in source is up in output
Vector3 upPos = TransformPos(P, Q, Vector3.back);
demonstrator.position = originPos;
demonstrator.rotation = Quaternion.LookRotation(forwardPos
- originPos, upPos - originPos);
}
Vector3 TransformPos(Matrix4x4 P, Matrix4x4 Q, Vector3 input)
{
Matrix4x4 invP = P.inverse;
return Q * invP * input + ((Vector4)point1.position
- Q * invP * relativePos1);
}
}
我正在尝试在 Unity 中旋转 3D 平面,给定三个点。这些点并不总是相同的,但它们始终位于此图像中的红点上
我知道这些点的绝对位置,以及它们相对于平面的位置。例如,我可能知道点 (-5, 5) 位于 (10, -4, 13)。
我知道当三个点在一条线上时(例如(-5, 5),(0, 5)和(5, 5))无法计算完整的旋转,所以我已经抛出出现这种情况时例外。但是当三个点不在一条直线上时,应该可以计算出完整的旋转。
到目前为止,我已经使用下面的代码进行了旋转,但这错过了绕 y 轴的旋转(该示例使用点 (-5, 5)、(5, 5) 和 (-5, -5).
Vector3 p1 = Point1.transform.position;// point 1 absolute position
Vector3 p1Relative = new Vector3(-5, 0, 5);
Vector3 p2 = Point2.transform.position;// point 2 absolute position
Vector3 p2Relative = new Vector3(5, 0, 5);
Vector3 p3 = Point3.transform.position;// point 3 absolute position
Vector3 p3Relative = new Vector3(-5, 0, -5);
Gizmos.DrawSphere(p1, .1f);
Gizmos.DrawSphere(p2, .1f);
Gizmos.DrawSphere(p3, .1f);
Vector3 normal = Vector3.Cross(p2 - p1, p3- p1);
rotator.transform.up = normal;
我如何扩展或更改此代码以包含绕 Y 轴的旋转?提前谢谢你。
此答案在数学堆栈交换中使用 this method described by robjohn。
您可以解决一对 Matrix4x4
s ,您可以使用它们从第一个坐标系转换到第二个坐标系,然后用它来计算原点(中心点)和旋转以对齐轴(使用 Quaternion.LookRotation
);
public class test : MonoBehaviour
{
[SerializeField] Transform point1;
[SerializeField] Transform point2;
[SerializeField] Transform point3;
[SerializeField] Vector2 relativePos1;
[SerializeField] Vector2 relativePos2;
[SerializeField] Vector2 relativePos3;
[SerializeField] Transform demonstrator;
void Update()
{
Vector3 absolutePos1 = point1.position;
Vector3 absolutePos2 = point2.position;
Vector3 absolutePos3 = point3.position;
Vector3 relativePos4 = (Vector3)relativePos1 +
Vector3.Cross(relativePos2 - relativePos1,
relativePos3 - relativePos1);
Vector3 absolutePos4 = absolutePos1 +
Vector3.Cross(absolutePos2 - absolutePos1,
absolutePos3 - absolutePos1);
Vector4 rightColumn = new Vector4(0, 0, 0, 1);
Matrix4x4 P = new Matrix4x4(relativePos2 - relativePos1,
relativePos3 - relativePos1,
relativePos4 - (Vector3)relativePos1, rightColumn);
Matrix4x4 Q = new Matrix4x4(absolutePos2 - absolutePos1,
absolutePos3 - absolutePos1,
absolutePos4 - absolutePos1, rightColumn);
Vector3 originPos = TransformPos(P, Q, Vector3.zero);
// up in source is forward in output
Vector3 forwardPos = TransformPos(P, Q, Vector3.up);
// back in source is up in output
Vector3 upPos = TransformPos(P, Q, Vector3.back);
demonstrator.position = originPos;
demonstrator.rotation = Quaternion.LookRotation(forwardPos
- originPos, upPos - originPos);
}
Vector3 TransformPos(Matrix4x4 P, Matrix4x4 Q, Vector3 input)
{
Matrix4x4 invP = P.inverse;
return Q * invP * input + ((Vector4)point1.position
- Q * invP * relativePos1);
}
}