The various methods we're talking about for figuring out a forward vector are
rotating in different directions - here's a bit of test code to confirm.
Code:
FILE* fp = fopen("printout.txt", "w");
osg::Vec3 f(0., 1., 0.);
osg::Quat fq(0., 1., 0., 0.);
osg::Quat q(osg::DegreesToRadians(45.0), osg::Vec3(0., 0., 1.));
osg::Quat qc = q.conj();
osg::Vec3 r1 = q*f;
osg::Vec3 r2 = (q * fq * qc).asVec3();
osg::Vec3 r3 = q.inverse() * f;
fprintf(fp, "quat * vec: %f %f %f\n", r1.x(), r1.y(), r1.z());
fprintf(fp, "q * fq * qc: %f %f %f\n", r2.x(), r2.y(), r2.z());
fprintf(fp, "q.inverse*f: %f %f %f\n", r3.x(), r3.y(), r3.z());
fclose(fp);
Results:
quat * vec: -0.707107 0.707107 0.000000
q * fq * qc: 0.707107 0.707107 0.000000
q.inverse*f: 0.707107 0.707107 0.000000
Since the test rotation you've been using is close to pi/2, it looks like the
axis is switched... but really, it's just rotated in the opposite direction
than you expect.
When you get a rotation quat out of of a PAT or MatrixTransform, (I guess) it's
giving you the coordinate system transform. When you transform a coordinate
system in one direction, it's equivalent to rotating vectors in the opposite
direction. So if you try and rotate a vector directly by this transform, ie by
doing quat*vec, your forward vector will be rotated an equal amount but in the
opposite direction from how the model appears on the screen.
By doing quat * vec * quat.conj(), or by doing quat.inverse() * vec, you're
applying the coordinate system transform to your vector, and everything works
out such that your forward vector and your on-screen model have rotated
together.
I'm still quite new to these maths, so please, an expert should pitch in! I am
basing this all on my own experiences with moving objects around in osg.
Cheers,
Tom[/code]
------------------
Read this topic online here:
http://forum.openscenegraph.org/viewtopic.php?p=29611#29611
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