c++ - 我可以从这个矩阵算法中得到旋转吗

Question

我可以从这个矩阵算法中获得旋转(偏航、俯仰、滚动)吗？谢谢

matrix[0][0] = cos(pitch)*cos(yaw);
matrix[0][2] = cos(pitch)*sin(yaw);
matrix[0][1] = sin(pitch);
matrix[0][3] = 0;
matrix[1][0] = -cos(roll)*sin(pitch)*cos(yaw)+sin(roll)*sin(yaw);
matrix[1][1] = cos(roll)*cos(pitch);
matrix[1][2] = -cos(roll)*sin(pitch)*sin(yaw)-sin(roll)*cos(yaw);
matrix[1][3] = 0;
matrix[2][0] = -sin(roll)*sin(pitch)*cos(yaw)-cos(roll)*sin(yaw);
matrix[2][1] = sin(roll)*cos(pitch);
matrix[2][2] = cos(roll)*cos(yaw)-sin(roll)*sin(pitch)*sin(yaw);
matrix[2][3] = 0;
matrix[3][0] = 0;
matrix[3][1] = 0;
matrix[3][2] = 0;
matrix[3][3] = 1;

编辑:我使用了这段代码，但它不起作用:

D3DXMATRIX matrix2 = D3DXMATRIX(matrix);
yaw   = atan2(matrix2._13, matrix2._11);
pitch = asin(matrix2._12);
roll  = atan2(matrix2._32, matrix2._22);

编辑 3:这是我的旧功能，它有效但不是 100%

VECTOR getAnglefromMatrix(float* m)
{
    float pitch, yaw, roll;

    D3DXMATRIX matrix = D3DXMATRIX(m);

    if (matrix._11 == 1.0f)
    {
        yaw = todegree(atan2f(matrix._13, matrix._34));
        pitch = 0;
        roll = 0;
    }
    else if (matrix._11 == -1.0f)
    {
        yaw = todegree(atan2f(matrix._13, matrix._34));
        pitch = 0;
        roll = 0;
    }
    else
    {

        yaw = todegree(atan2(-matrix._31,matrix._11));
        pitch = todegree(asin(matrix._21));
        roll = todegree(atan2(-matrix._23,matrix._22));
    }

    return vector3d(yaw,pitch,roll);
}

我做错了什么？

Answer 1

我们可以浏览这些条目并找到一些对回溯计算有用的条目:

matrix[0][0] = cos(pitch)*cos(yaw);
matrix[0][2] = cos(pitch)*sin(yaw);

这给了我们:

yaw = atan2(matrix[0][2], matrix[0][0])

对于球场:

matrix[0][1] = sin(pitch);

这给了我们:

pitch = asin(matrix[0][1])  //this assumes pitch is between -pi/2 and +pi/2

然后

matrix[1][1] = cos(roll)*cos(pitch);
matrix[2][1] = sin(roll)*cos(pitch);

这给了我们:

roll = atan2(matrix[2][1], matrix[1][1])

我省略了 atan2 的参数变为零的情况。我相信你能弄明白。在这些情况下，欧拉角不是唯一定义的。