 | | | How can I get the global position of a single point of curve, which is branc | How can I get the global position of a single point of curve, which is branc 2004-02-21 - By Helge Mathee
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you have to multiply by the global transformation of the object.
jscript:
var geo = obj.activeprimitive.geometry;
var pnts = geo.points;
var transform = obj.kinematics.global.transform;
var pntPos = pnts.item(index).position;
var pntPos.MulByTransformationInPlace(transform);
cheers,
-h
-- -- Original Message -- --
From: "Sebastian Faber" <sebastian.faber@(protected)>
To: <XSI@(protected)>
Sent: Saturday, February 21, 2004 1:37 PM
Subject: How can I get the global position of a single point of curve, which
is branch deformed on an other curve?
> I have a curve deformed on another curve via branch deformation. How can I
> get the GLOBAL position of a single Point of this deformed curve?
>
> Any hints are welcome
>
> Sebastian
>
>
>
>
>
> to get the global position of a point of a undeformed geometry would be
like
> this:
>
> set oTrans = oCube.Kinematics.Local.Transform
> set oPoint = oCube.ActivePrimitive.Geometry.Points(0)
> set oPos = oPoint.Position
> logmessage "The local position is: X "& oPos.X & " Y " & oPos.Y & " Z "
&
> oPos.Z
> set oGlobalPos = XSIMath.MapObjectPositionToWorldSpace( oTrans, oPos)
> logmessage "The world position is: X "& oGlobalPos.X & " Y " &
> oGlobalPos.Y & " Z " & oGlobalPos.Z
>
>
>
> LIGA_01 COMPUTERFILM GmbH
> Sebastian Faber
> Landwehrstrasse 60
> D - 80336 Munich - Germany
>
> sebastian.faber@(protected)
> www.liga01.com
>
>
>
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