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View Full Version : Xpresso - Link to clone with step effector data


Kwartjuh
02-13-2013, 10:36 AM
Hey Guys!

I am struggling with this project i am working on.
Basicly i have a set of tiles that first move out and then rotate one by one. I want to link other objects to each clone.

I basicly have one tile-object and placed it in a cloner. Withing the cloner, i simply increase the grid array size so they all move neatly out.
To get the rotation done, i rotate the base-object and then add a step effector to the cloner.

And here is where the problem comes in: To link other objects to each clone i used expresso. After a bit of searching i got it working pretty well. But the data is only based on the cloner, not on the step effector data. I thought by simply doing the exact same xpresso workflow for the step effector it would work, but it doesnt.

Here is a picture of the current xpresso setup:

http://dev.shifthorizon.nl/xpresso_problem.png

Can anyone point me in the right direction?

littledevil
02-13-2013, 06:07 PM
You should provide a scene or more pictures. It is very difficult to tell what
you are heading for. Terms like 'To get the rotation done' do not not help either.

there are basically two mistakes in your posted screenshot:

1. A Data Node does not work in conjunction with effector node. It is meant to
read and write the MoData arrays of Cloner/Matrix/... objects. To read effector
data/influence you have to use the effector node.

2. Your setup does not really make any sense, even if you ignore the effector part.
I guess Targetswitch is an iterator, so you just iterate through all the MoData and
read the position vector of each element and write it into your ClonerChild position.
The result will be your ClonerChild to snap to the position vector of the last particle in
your MoData array. Which won't have any effect unless you uncheck fix clones in your
cloner. If you do uncheck it, your whole clone array will be offset by its own size on
each update. If you press play the array should shoot out of side due to the exponential
nature of this calculation.

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02-13-2013, 06:07 PM
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