Hi,
Sure, I know this...
const QVector3D branch(...);
const QVector3D leaf( 0.0f, 1.0f, 0.0f );
const QVector3D axis = QVector3D::crossProduct( branch, leaf );
const float cosPlainAngle = QVector3D::dotProduct( branch, leaf );
const float plainAngle = qRadiansToDegrees( std::acos( cosPlainAngle ) );
const QQuaternion quat = Qt3DCore::QTransform::fromAxisAndAngle( axis,
plainAngle );
m_transform->setRotation( quat );
But in a view of Qt 3D this is only a half of the solution. In a half of
cases this works, but in another cases I need -plainAngle.
So at this point I found the next solution:
static inline bool lessZero( const QVector3D & v )
{
return ( v.x() < 0.0f || v.y() < 0.0f || v.z() < 0.0f );
}
if( lessZero( branch ) )
plainAngle = -plainAngle;
So I actually asked not for the math as it is but for checking of my
solution for correctness.
On 01.04.2018 12:48, Konstantin Shegunov wrote:
Hi Igor,
What Bin Chen wrote is probably the most painless way of achieving
what you want. If you are however interested in the math, here goes my
stab:
If I understand you correctly, you know the leaf normal, and the
branch direction vector, then you're searching for the matrix that
transforms the former to the latter.
Basically you need to find the matrix that satisfies: b = A * n (b is
the branch direction, n is the leaf normal).
This equation however is underdetemined, meaning you can have several
rotations done in sequence that give you the same result, so you'd
need to do some "trickery". One of the usual ways to solve such a
problem is to use Euler angles[1], where the idea is to make elemental
rotations with respect to the principle axes of the (global)
coordinate systems. To that end you'd need to calculate the
projections (i.e. dot products) of b and n to the principal axes and
extract the angles of rotation from there, then construct each
rotation matrix around a principal axis of the coordinate system and
finally multiply them to obtain the final transformation.
[1]: https://en.wikipedia.org/wiki/Euler_angles
<https://en.wikipedia.org/wiki/Euler_angles>
I hope that helps.
Konstantin.
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