View Full Version : Newbie skinning and partitioning question
05 May 2003, 12:19 AM
I am trying to rig a hand, and am having to experiment with placement and therefore am re-skinning often..... AND then adjusting the CV set memeberships over and over again. (Rigid skin BTW)
Can someone give me the basics (and maybe some of the dos and don'ts) of creating a partition that can be used to skin?
Here is what I assume the process would be:
1 - create sets that contain the proper CVs and name them the same name as the corresponding joints (ie. "joint1" would dictate a set called "joint1Set")
2 - select all the sets and create a partition
3- select the skeleton and bind to the partition
.... Is this correct?
1 - Does it parse the set name to find out which joint to apply so use of the word "Set" must come after the Exact name of the joint?
05 May 2003, 02:08 AM
OK.... I am hoping to put a flag on this post again cause this has to be straight forward and I am just missing it.
It appears that using a partition and doing rigid binding to it still has some sort of vicinity measurement going on when assigning the sets to the joints. I changed the set names and it doesn't seem to phase it and the sets that aren't right next to their joint don't get picked up at all.(but instead get lumped into another set??)
All I want to do is create sets of vtxs that will be assigned to the relevant joint when skinning????
05 May 2003, 02:21 AM
Sorry... wanted to include a breakdown of what is happening.
I have a polyMesh hand and joints running throught it.... simple.
I select the verticies that I want together and create sets for them to correspond to the joints. So I have.
And I have joints for each of these sets. I create a partition by selecting all of these sets and Create>Sets>Partition. I check the partition created and it matches the sets. I then Skin>Bind Skin>Rigid Bind... and select the partition.... but I get.
... so you can see that the index_baseSet was lost even though it exists in the partition and there is a joint for it.... and all the set members of that set are no inside the :
01 January 2006, 06:00 AM
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