Question

我正在尝试学习3D编程，现在我正在尝试了解如何使用四元数来围绕轴旋转矢量。

据我所知，为了将矢量v绕轴a旋转，在将两个矢量转换为四元数之后，我们将v乘以a，然后将乘积乘以a的共轭。

我想围绕（1,0,0）旋转v（0,1,0）90度，我应该得到一个结果向量v（0,0,1）（或0,0， - 1，取决于旋转方向。）

我没有得到我期待的输出。这是代码：

    int main()
    {
        //I want to rotate this vector about the x axis by PI/2 radians:
        Quaternion v(0, 1, 0, 0);
        v.normalize();


        float angle = PI / 2.0f;
        float cos = math::cos(angle / 2.0f);
        float sin = math::sin(angle / 2.0f);

        Quaternion q(1.0f*sin, 0.0f*sin, 0.0f*sin, cos);

        std::cout << "q not normalized = " <<"\t"<< q.x << " " << q.y << " "      << q.z << " " << q.w << std::endl;

        q.normalize();

        std::cout << "q normalized = " <<"\t\t"<< q.x << " " << q.y << " " << q.z << " " << q.w << std::endl;
        std::cout << std::endl;

        Quaternion r;


       //I multiply the vector v by the quaternion v, then I multiply by the    conjugate.
        r = q * v;
        //do I need to normalize here?
        r = r * q.conjugate();
        //and here?


        //shouldn't the resulting vector be 0,0,1? 

        std::cout << "r not normalized = " << "\t" << r.x << " " << r.y << " " << r.z << " " << r.w << std::endl;
        r.normalize();

        std::cout << "r normalized = " << "\t\t" << r.x << " " << r.y << " " << r.z << " " << r.w << std::endl;
        std::cout << std::endl;

        system("pause");
        return 0;
    }

这是输出：

q未标准化，与q标准化相同： x = 0.707107，y = 0，z = 0，w = 0.707107

未正常化： x = 0.707107，y = 0，z = 1，w = -2.12132 r标准化： x = 0.288675，y = 0，z = 0.408248，w = -0.866025

我做错了什么？我是否从这个过程中了解了什么？

Answer 1

基本上沿着x轴（1,0,0）以90度角旋转矢量，使用下面的方法，这适用于Euler和四元数

| 1    0           0 |   | 0 |     | 0 |
| 0   cos90   -sin90 | * | 1 |  =  | 0 |
| 0   sin90    cos90 |   | 0 |     | 1 |

了解旋转矩阵http://en.wikipedia.org/wiki/Rotation_matrix

使用四元数旋转轴周围的矢量

1 个答案: