I'm not really sure why you're doing it the way you're doing it...
This is how I solved it quickly (There's better ways but I'm lazy and I
didn't want to keep working on it.) Just a simple recursive function.
def dist( objA, objB ):
Ax, Ay, Az = objA.getTranslation(space="world")
Bx, By, Bz = objB.getTranslation(space="world")
return ( (Ax-Bx)**2 + (Ay-By)**2 + (Az-Bz)**2 )**0.5
def lenJointChain(obj,len=0):
if obj.getChildren()==[]:
return len
else:
child=obj.getChildren(type='transform')[0]
return lenJointChain(child, len=len+dist(obj,child))
lenJointChain(pm.ls(sl=True)[0])
On Wednesday, May 28, 2014 3:18:13 PM UTC-4, flaye wrote:
>
>
>
> # Find the full length of the chain
> sel=[]
> sel=mc.ls(sl=True)
> length = 0
> for i in range(1, len(sel)):
> length += abs(mc.getAttr(sel[i]+'.tx'))
>
> print length
>
>
> The above code will provide the correct answer, on the condition that the
> primary axis of the joints (in this case, X) is known.
>
> I've also tried using the MVector function, but unfortunately the numbers
> are off
>
> import maya.api.OpenMaya as om #for api 2.0
> sel=[]
> sel=mc.ls(sl=True)
>
> startjnt=sel[0]
> numjnts=len(sel)
> length=0
> for i in range(0,numjnts-1):
>
> currentjnt=sel[i]
>
> nextjnt=mc.listRelatives(currentjnt,children=1)[0]
>
> print currentjnt,nextjnt
>
> t1=mc.xform(currentjnt,q=1,t=1)
> t2=mc.xform(nextjnt,q=1,t=1)
>
> v1=om.MVector(t1)
> v2=om.MVector(t2)
>
> dist=om.MVector(v2-v1).length() #equivalent to MEL's mag?
> length+=dist
> print length
>
>
> any ideas? Thanks.
>
>
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