Good call Nando, the Card3D node can be used this way.

Although it does give you skewXYZ params, they are normalized. Instead, it's probably easier to use the "local matrix" which should give you a 1:1 mapping for skew. Just make sure to set the z translation value in the matrix to "-1".

The copy/paste script below shows setting the skewX value to "1" for both a Transform and Card3D matrix node gives identical results.

Kudos again to Nando, Cheers,

Dan.




set cut_paste_input [stack 0]
version 6.2 v4
CheckerBoard2 {
 inputs 0
 name CheckerBoard1
 selected true
 xpos -131
 ypos -178
}
set N2ccf3d40 [stack 0]
Card3D {
 skew {1 1 0}
 useMatrix true
 matrix {
     {1 1 0 0}
     {0 1 0 0}
     {0 0 1 -1}
     {0 0 0 1}
   }
 name Card3D1
 selected true
 xpos -85
 ypos -73
}
push $N2ccf3d40
Transform {
 skew 1
 center {1024 778}
 name Transform1
 selected true
 xpos -179
 ypos -74
}





On 24/10/2011 10:20, [email protected] wrote:
hi.
what about the card3D node? has skew xyz parameters.
nando

On 21.10.2011 11:58, Dan Ring wrote:

Hi Bruce,

I don't know if it helps, but you can directly set the "extra matrix" of a CornerPin node (possibly even expression link it to values extracted from your Transform?)

Dan.



On 20/10/2011 23:20, bjl70 wrote:
Hi All,

I am trying to decompose a standard 2D transformation matrix (Rotation, Scale, Skew & Translation, no Perspective) and apply that information to the parameters in a Transform node.

I can manage to get almost everything in there, but I have recovered 2 values for skew from the matrix: X and Y. The problem is there is only a single skew parameter in a Transform node. Am I correct in thinking that I cannot fully specify my 2D transformation in a Transform node for that reason?

This would seem to be a significant omission if this is the case, but I can't believe that Nuke's engineers would have missed this...

Approaching from a different angle, I have managed to read the matrix associated with a Transform node's parameters, but is there any way to directly set the matrix without having to decompose it into translate, rotate etc?

OK. Thanks in advance,
Bruce


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Dan Ring, Ph.D.
Research Engineer
The Foundry
6th Floor, The Communications Building,
48 Leicester Square,
London, WC2H 7LT, UK
Tel: +44 (0)20 7968 6828 - Fax: +44 (0)20 7930 8906
Web:www.thefoundry.co.uk
Email:[email protected]

The Foundry Visionmongers Ltd.
Registered in England and Wales No: 4642027


_______________________________________________
Nuke-users mailing list
[email protected], http://forums.thefoundry.co.uk/
http://support.thefoundry.co.uk/cgi-bin/mailman/listinfo/nuke-users
_______________________________________________
Nuke-users mailing list
[email protected], http://forums.thefoundry.co.uk/
http://support.thefoundry.co.uk/cgi-bin/mailman/listinfo/nuke-users


--
Dan Ring, Ph.D.
Research Engineer
The Foundry
6th Floor, The Communications Building,
48 Leicester Square,
London, WC2H 7LT, UK
Tel: +44 (0)20 7968 6828 - Fax: +44 (0)20 7930 8906
Web: www.thefoundry.co.uk
Email: [email protected]

The Foundry Visionmongers Ltd.
Registered in England and Wales No: 4642027

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