osg中四元数学习

首先几个四元数相乘是几个旋转动作叠加

其次构造四元数方法osg::Quat quat(float radians, const Vec3f& axis),其中radians是旋转弧度, 后面的axis是旋转轴向量;

osg::Quat rotation(osg::Quat(roll,osg::Vec3(0.0,1.0,0.0))*osg::Quat(-(yaw+osg::inDegrees(90.0f)),osg::Vec3(0.0,0.0,1.0)));

   

其中的旋转分量, x轴是俯仰(pitch), y轴是横滚(roll), z轴是航向角度(yaw),.

   

旋转表示:

   

四元数和旋转矩阵之间转换代码:

#include<iostream>

#include<cmath>

using namespace std;

   

typedef double ValType;

   

struct Quat;

struct Matrix;

   

struct Quat {

ValType _v[4];//x, y, z, w

   

/// Length of the quaternion = sqrt( vec . vec )

ValType length() const

{

    return sqrt( _v[0]*_v[0] + _v[1]*_v[1] + _v[2]*_v[2] + _v[3]*_v[3]);

}

   

/// Length of the quaternion = vec . vec

ValType length2() const

{

    return _v[0]*_v[0] + _v[1]*_v[1] + _v[2]*_v[2] + _v[3]*_v[3];

}

};

   

struct Matrix {

ValType _mat[3][3];

};

   

#define QX q._v[0]

#define QY q._v[1]

#define QZ q._v[2]

#define QW q._v[3]

   

void Quat2Matrix(const Quat& q, Matrix& m)

{

    double length2 = q.length2();

    if (fabs(length2) <= std::numeric_limits<double>::min())

    {

        m._mat[0][0] = 0.0; m._mat[1][0] = 0.0; m._mat[2][0] = 0.0;

        m._mat[0][1] = 0.0; m._mat[1][1] = 0.0; m._mat[2][1] = 0.0;

        m._mat[0][2] = 0.0; m._mat[1][2] = 0.0; m._mat[2][2] = 0.0;

    }

    else

    {

        double rlength2;

        // normalize quat if required.

        // We can avoid the expensive sqrt in this case since all 'coefficients' below are products of two q components.

        // That is a square of a square root, so it is possible to avoid that

        if (length2 != 1.0)

        {

            rlength2 = 2.0/length2;

        }

        else

        {

            rlength2 = 2.0;

        }

         

        // Source: Gamasutra, Rotating Objects Using Quaternions

        //

        //http://www.gamasutra.com/features/19980703/quaternions_01.htm

         

        double wx, wy, wz, xx, yy, yz, xy, xz, zz, x2, y2, z2;

         

        // calculate coefficients

        x2 = rlength2*QX;

        y2 = rlength2*QY;

        z2 = rlength2*QZ;

         

        xx = QX * x2;

        xy = QX * y2;

        xz = QX * z2;

         

        yy = QY * y2;

        yz = QY * z2;

        zz = QZ * z2;

         

        wx = QW * x2;

        wy = QW * y2;

        wz = QW * z2;

         

        // Note. Gamasutra gets the matrix assignments inverted, resulting

        // in left-handed rotations, which is contrary to OpenGL and OSG's

        // methodology. The matrix assignment has been altered in the next

        // few lines of code to do the right thing.

        // Don Burns - Oct 13, 2001

        m._mat[0][0] = 1.0 - (yy + zz);

        m._mat[1][0] = xy - wz;

        m._mat[2][0] = xz + wy;

         

         

        m._mat[0][1] = xy + wz;

        m._mat[1][1] = 1.0 - (xx + zz);

        m._mat[2][1] = yz - wx;

         

        m._mat[0][2] = xz - wy;

        m._mat[1][2] = yz + wx;

        m._mat[2][2] = 1.0 - (xx + yy);

    }

}

   

void Matrix2Quat(const Matrix& m, Quat& q)

{

    ValType s;

    ValType tq[4];

    int    i, j;

   

    // Use tq to store the largest trace

    tq[0] = 1 + m._mat[0][0]+m._mat[1][1]+m._mat[2][2];

    tq[1] = 1 + m._mat[0][0]-m._mat[1][1]-m._mat[2][2];

    tq[2] = 1 - m._mat[0][0]+m._mat[1][1]-m._mat[2][2];

    tq[3] = 1 - m._mat[0][0]-m._mat[1][1]+m._mat[2][2];

   

    // Find the maximum (could also use stacked if's later)

    j = 0;

    for(i=1;i<4;i++) j = (tq[i]>tq[j])? i : j;

   

    // check the diagonal

    if (j==0)

    {

        /* perform instant calculation */

        QW = tq[0];

        QX = m._mat[1][2]-m._mat[2][1];

        QY = m._mat[2][0]-m._mat[0][2];

        QZ = m._mat[0][1]-m._mat[1][0];

    }

    else if (j==1)

    {

        QW = m._mat[1][2]-m._mat[2][1];

        QX = tq[1];

        QY = m._mat[0][1]+m._mat[1][0];

        QZ = m._mat[2][0]+m._mat[0][2];

    }

    else if (j==2)

    {

        QW = m._mat[2][0]-m._mat[0][2];

        QX = m._mat[0][1]+m._mat[1][0];

        QY = tq[2];

        QZ = m._mat[1][2]+m._mat[2][1];

    }

    else /* if (j==3) */

    {

        QW = m._mat[0][1]-m._mat[1][0];

        QX = m._mat[2][0]+m._mat[0][2];

        QY = m._mat[1][2]+m._mat[2][1];

        QZ = tq[3];

    }

   

    s = sqrt(0.25/tq[j]);

    QW *= s;

    QX *= s;

    QY *= s;

    QZ *= s;

}

   

void printMatrix(const Matrix& r, string name)

{

cout<<"RotMat "<<name<<" = "<<endl;

cout<<"\t"<<r._mat[0][0]<<" "<<r._mat[0][1]<<" "<<r._mat[0][2]<<endl;

cout<<"\t"<<r._mat[1][0]<<" "<<r._mat[1][1]<<" "<<r._mat[1][2]<<endl;

cout<<"\t"<<r._mat[2][0]<<" "<<r._mat[2][1]<<" "<<r._mat[2][2]<<endl;

cout<<endl;

}

   

void printQuat(const Quat& q, string name)

{

cout<<"Quat "<<name<<" = "<<endl;

cout<<"\t"<<q._v[0]<<" "<<q._v[1]<<" "<<q._v[2]<<" "<<q._v[3]<<endl;

cout<<endl;

}

   

int main()

{

ValType phi, omiga, kappa;

   

phi = 1.32148229302237 ; omiga = 0.626224465189316 ; kappa = -1.4092143985971;

   

    ValType a1,a2,a3,b1,b2,b3,c1,c2,c3;

   

    a1 = cos(phi)*cos(kappa) - sin(phi)*sin(omiga)*sin(kappa);

    a2 = -cos(phi)*sin(kappa) - sin(phi)*sin(omiga)*cos(kappa);

    a3 = -sin(phi)*cos(omiga);

   

    b1 = cos(omiga)*sin(kappa);

    b2 = cos(omiga)*cos(kappa);

    b3 = -sin(omiga);

   

    c1 = sin(phi)*cos(kappa) + cos(phi)*sin(omiga)*sin(kappa);

    c2 = -sin(phi)*sin(kappa) + cos(phi)*sin(omiga)*cos(kappa);

    c3 = cos(phi)*cos(omiga);

     

    Matrix r;

    r._mat[0][0] = a1;

r._mat[0][1] = a2;

r._mat[0][2] = a3;

   

r._mat[1][0] = b1;

r._mat[1][1] = b2;

r._mat[1][2] = b3;

   

r._mat[2][0] = c1;

r._mat[2][1] = c2;

r._mat[2][2] = c3;

   

printMatrix(r, "r");

   

     //////////////////////////////////////////////////////////

   

Quat q;

Matrix2Quat(r, q);

   

printQuat(q, "q");

   

Matrix _r;

Quat2Matrix(q, _r);

   

printMatrix(_r, "_r");

     

system("pause");

return 0;

}

   

源文档 <http://hi.baidu.com/simbaforrest/blog/item/328d7fb4a4c0ac7a8ad4b24b.html>

posted on 2012-02-06 13:14  sunliming  阅读(4517)  评论(0编辑  收藏  举报