Hi, Igor,
1. plainAngle is between 0~180, you want to know when you need to invert it?
I assumed m_transform is for leaf, then swap branch and leaf in
QVector3D::crossProduct.
No need lessZero() function.
2. In this case, be careful when branch and leaf are in the same line (same or
opposite directions) , crossProduct produces null vector3d.
Regards,
Bin
On 1 Apr 2018, at 8:56 pm, Igor Mironchik
<igor.mironc...@gmail.com<mailto:igor.mironc...@gmail.com>> wrote:
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://urldefense.proofpoint.com/v2/url?u=https-3A__en.wikipedia.org_wiki_Euler-5Fangles&d=DwMDaQ&c=-0XTxx5JZxtPyuSXdvX8qQ&r=_JxpcpJpSMrVwuVMK05qMw&m=JDoIqjY-zYljVJcp8DYX1co3l1ElN2hO1_68-23VIAU&s=tezn68iaGXO6oGne6GHcWtS-CoDkU-Uj0fztStttyik&e=>
I hope that helps.
Konstantin.
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