我复制了一个代码,用于将3D旋转矩阵转换为四元数和后退。 jMonkey中使用了相同的代码(我刚刚将其重写为我的C ++类)。但是,它不能正常工作(至少不像我期望的那样。)
e.g。我做了这个测试:
matrix (a,b,c):
a : 0.707107 0.000000 0.707107
b : 0.000000 -1.000000 0.000000
c : -0.707107 0.000000 0.707107
>>> ortonormality:
a.a b.b c.c 1.000000 1.000000 1.000000
a.b a.c b.c 0.000000 0.000000 0.000000
>>> matrix -> quat
quat: 0.000000 0.594604 0.000000 0.594604 norm(quat) 0.707107
>>> quat -> matrix
matrix (a,b,c):
a: 0.000000 0.000000 1.000000
b: 0.000000 1.000000 0.000000
c: -1.000000 0.000000 0.000000
我认为问题出在 matrix -> quat
,因为之前我使用过quat -> matrix
程序,并且工作正常。由正交矩阵构成的四元数不是单一的,这很奇怪。
matrix -> quat
程序
inline void fromMatrix( TYPE m00, TYPE m01, TYPE m02, TYPE m10, TYPE m11, TYPE m12, TYPE m20, TYPE m21, TYPE m22) {
// Use the Graphics Gems code, from
// ftp://ftp.cis.upenn.edu/pub/graphics/shoemake/quatut.ps.Z
TYPE t = m00 + m11 + m22;
// we protect the division by s by ensuring that s>=1
if (t >= 0) { // by w
TYPE s = sqrt(t + 1);
w = 0.5 * s;
s = 0.5 / s;
x = (m21 - m12) * s;
y = (m02 - m20) * s;
z = (m10 - m01) * s;
} else if ((m00 > m11) && (m00 > m22)) { // by x
TYPE s = sqrt(1 + m00 - m11 - m22);
x = s * 0.5;
s = 0.5 / s;
y = (m10 + m01) * s;
z = (m02 + m20) * s;
w = (m21 - m12) * s;
} else if (m11 > m22) { // by y
TYPE s = sqrt(1 + m11 - m00 - m22);
y = s * 0.5;
s = 0.5 / s;
x = (m10 + m01) * s;
z = (m21 + m12) * s;
w = (m02 - m20) * s;
} else { // by z
TYPE s = sqrt(1 + m22 - m00 - m11);
z = s * 0.5;
s = 0.5 / s;
x = (m02 + m20) * s;
y = (m21 + m12) * s;
w = (m10 - m01) * s;
}
}
quat -> matrix
程序
inline void toMatrix( MAT& result) const {
TYPE r2 = w*w + x*x + y*y + z*z;
//TYPE s = (r2 > 0) ? 2d / r2 : 0;
TYPE s = 2 / r2;
// compute xs/ys/zs first to save 6 multiplications, since xs/ys/zs
// will be used 2-4 times each.
TYPE xs = x * s; TYPE ys = y * s; TYPE zs = z * s;
TYPE xx = x * xs; TYPE xy = x * ys; TYPE xz = x * zs;
TYPE xw = w * xs; TYPE yy = y * ys; TYPE yz = y * zs;
TYPE yw = w * ys; TYPE zz = z * zs; TYPE zw = w * zs;
// using s=2/norm (instead of 1/norm) saves 9 multiplications by 2 here
result.xx = 1 - (yy + zz);
result.xy = (xy - zw);
result.xz = (xz + yw);
result.yx = (xy + zw);
result.yy = 1 - (xx + zz);
result.yz = (yz - xw);
result.zx = (xz - yw);
result.zy = (yz + xw);
result.zz = 1 - (xx + yy);
};
抱歉TYPE, VEC, MAT, QUAT
这是课堂版的一部分......应该由double, Vec3d, Mat3d, Quat3d
或float, Vec3f, Mat3f, Quat3f
替换。
编辑:
我还检查了我是否直接使用jMonkey获得了相同的行为(如果我在Java到C ++转换中犯了一个错误)。我这样做 - 使用此代码:
Matrix3f Min = new Matrix3f( 0.707107f, 0.000000f, 0.707107f, 0.000000f, -1.000000f, 0.000000f, -0.707107f, 0.000000f, 0.707107f );
Matrix3f Mout = new Matrix3f( );
Quaternion q = new Quaternion();
q.fromRotationMatrix(Min);
System.out.println( q.getX()+" "+q.getY()+" "+q.getZ()+" "+q.getW() );
q.toRotationMatrix(Mout);
System.out.println( Mout.get(0,0) +" "+Mout.get(0,1)+" "+Mout.get(0,2) );
System.out.println( Mout.get(1,0) +" "+Mout.get(1,1)+" "+Mout.get(1,2) );
System.out.println( Mout.get(2,0) +" "+Mout.get(2,1)+" "+Mout.get(2,2) );
答案 0 :(得分:2)
你的矩阵:
matrix (a,b,c):
a : 0.707107 0.000000 0.707107
b : 0.000000 -1.000000 0.000000
c : -0.707107 0.000000 0.707107
是正交的,但它不是旋转矩阵。旋转矩阵具有行列式1;你的矩阵有行列式-1,因此是improper rotation。
我认为您的代码可能是正确的,问题出在您的数据中。尝试使用真正的旋转矩阵。