# [Interest] Rotating leaves

Bin Chen bchen at topcon.com
Sun Apr 1 23:27:19 CEST 2018

```Should not need to invert plainAngle, a code segment for you:

QVector3D leaf(1, 1, 0);

leaf.normalize(); //need to be normalised otherwise the angle could be wrong.

qDebug()<<leaf;

QVector3D branch(-1, 0, 0);

branch.normalize(); //need to be normalised otherwise the angle could be wrong.

QVector3D v3 = QVector3D::crossProduct(leaf, branch).normalized();

if(v3.isNull()) {

//leaf and branch are in the same line.

v3 = QVector3D::crossProduct(leaf, QVector3D(1, 0, 0)).normalized();

if(v3.isNull()) {

v3 = QVector3D::crossProduct(leaf, QVector3D(0, 1, 0)).normalized();

if(v3.isNull()) {

v3 = QVector3D::crossProduct(leaf, QVector3D(0, 0, 1)).normalized();

}

}

}

qDebug()<<v3;

qDebug()<<angle;

QQuaternion qut =  QQuaternion::fromAxisAndAngle( v3, angle );

qDebug()<<qut*leaf;

On 2 Apr 2018, at 6:56 am, Bin Chen <bchen at topcon.com<mailto:bchen at topcon.com>> wrote:

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.mironchik at gmail.com<mailto:igor.mironchik at 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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