unity3d - Unity的四元数旋转矩阵是如何计算的？

Question

我不知道如何从 Unity 中的四元数计算旋转矩阵。

一般旋转矩阵计算如下：

http://www.utdallas.edu/~sxb027100/dock/quaternion.html

下面是 wxmaxima 的代码。

l: matrix([w, z, -y, x], [-z, w, x, y], [y, -x, w, z], [-x, -y, -z, w]);
r: matrix([w, z, -y, -x], [-z, w, x, -y], [y, -x, w, -z], [x, y, z, w]);
lr:  l . r;

以一般旋转矩阵中的z轴(0, 0, 1)为中心，有如下公式。

2xz-2wy
2yz+2wx
w^2-x^2-y^2+z^2=1-2x^2-2y^2

在 Unity 中，它是以下公式。

2xz+2wy
2yz-2wx
1-2x^2+2y^2

上面的区别是什么？

下面是 Unity 中的参考代码。

public static Vector3 operator *(Quaternion rotation, Vector3 point)
{
  float num1 = rotation.x * 2f;
  float num2 = rotation.y * 2f;
  float num3 = rotation.z * 2f;
  float num4 = rotation.x * num1;
  float num5 = rotation.y * num2;
  float num6 = rotation.z * num3;
  float num7 = rotation.x * num2;
  float num8 = rotation.x * num3;
  float num9 = rotation.y * num3;
  float num10 = rotation.w * num1;
  float num11 = rotation.w * num2;
  float num12 = rotation.w * num3;
  Vector3 vector3;
  vector3.x = (float) ((1.0 - ((double) num5 + (double) num6)) * (double) point.x + ((double) num7 - (double) num12) * (double) point.y + ((double) num8 + (double) num11) * (double) point.z);
  vector3.y = (float) (((double) num7 + (double) num12) * (double) point.x + (1.0 - ((double) num4 + (double) num6)) * (double) point.y + ((double) num9 - (double) num10) * (double) point.z);
  vector3.z = (float) (((double) num8 - (double) num11) * (double) point.x + ((double) num9 + (double) num10) * (double) point.y + (1.0 - ((double) num4 + (double) num5)) * (double) point.z);
  return vector3;
}

https://github.com/jamesjlinden/unity-decompiled/blob/96fb16e2eb6fff1acf3d4e25fa713defb3d17999/UnityEngine/UnityEngine/Quaternion.cs

标签： unity3dquaternions

Answer 1

一般旋转用于直接（右手）轴系统，而统一是间接（左手）轴系统。这应该解释 - 符号，以及你得到的差异。