No Yilmaz ,I don't want to stick a Locator on a vertex, this is what I
am trying to do.
What Eigen Vectors can do is to get the orientation of a volume .
SO basically I could get the polygon orientation of the whole object
based on the volume that will need to broken down
to matrices of the vertex that make it to be fed to Numpy .
Assuming that you have a cube in space and you rotate it to an
arbitrary rotation value.
At this point if you freeze the transformation and center the pivot
you will have the pivot point
oriented to the world and not the actual roattion of the cube.
Eigen Vectors and Eigen values can give yo that rotation to be fed in
to the transform matrix and regain that orientation based on the
object and not the pivot point.
This is a case but many times you can get objects form different
places or different conversion application and need to get a decent
pivot to manipulate the object.

I hope my explanation was clear enough.

Thank you

On Mar 24, 12:04 pm, Ozgur Yılmaz <[email protected]> wrote:
> so you want to stick a locator on to an objects vertex ?
>
> E.Ozgur Yilmaz
> Lead Technical Directorwww.ozgurfx.com
>
> On Mon, Mar 22, 2010 at 1:05 AM, maurizio1974 <[email protected]>wrote:
>
> > Hi
>
> > When I was at ILM there was a cool tool that would calculate the
> > orientation of an object based on the volume
> > thanks to the eigenvector.
> > There they were using some old module called Linear Algebra to get the
> > eigenvector result from the arrays
> > of the object, but I know that Numpy has integrated the Linear Algebra
> > Module so I guess that is the same.
>
> > I haven't been able to find a good way to just get an object parsing
> > on the matrices of the individual vertex and then
> > feed a 4x4 matrix to Numpy so that would give me a matrix that applied
> > to a Locator would result to the right orientation of the object even
> > if the original pivot is completely in a different position like when
> > we freeze the transform on a rotated object in space.
>
> > Anyone got a clue of a reasonably simple way to do that ?
>
> > Thank you in advance.
>
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