Hi,
On 01.04.2018 23:56, Bin Chen 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.
Big thanks. Great, swapping helped. So the rule is when you need to
rotate something around some axis then first argument of
QVector3D::crossProduct and QVector3D::dotProduct should be that vector
that should be rotated?
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.
_______________________________________________
Interest mailing list
Interest@qt-project.org <mailto:Interest@qt-project.org>
https://urldefense.proofpoint.com/v2/url?u=http-3A__lists.qt-2Dproject.org_mailman_listinfo_interest&d=DwICAg&c=-0XTxx5JZxtPyuSXdvX8qQ&r=_JxpcpJpSMrVwuVMK05qMw&m=JDoIqjY-zYljVJcp8DYX1co3l1ElN2hO1_68-23VIAU&s=iR7_jc5oqZY41bU9FD_ergWeOPoFC0aUG4SyhsDkHIY&e=
Confidentiality Notice: This message (including attachments) is a
private communication solely for use of the intended recipient(s). If
you are not the intended recipient(s) or believe you received this
message in error, notify the sender immediately and then delete this
message. Any other use, retention, dissemination or copying is
prohibited and may be a violation of law, including the Electronic
Communication Privacy Act of 1986.
_______________________________________________
Interest mailing list
Interest@qt-project.org
http://lists.qt-project.org/mailman/listinfo/interest
