Finding the Correct Pole Vector Position

Hi all!

With the help of some more math-inclined friends, I was able to script up a way to find the correct pole vector position in Maya.

Solving the problem required a little linear algebra and lots of simplification. Follow my script comments for a walk through of what I did.

Here’s the script:

import maya.cmds as cmds

Finding pole vector position D given points A B and C 

1. Find vectorAB and vectorAC
2. 0 = (vAC-vAB(k))(vAB)
3. k= ((vABx*vACx) + (vABy*vACy) +(vABz*vACz)) / (vABx^2 +vABy^2 + vABz^2)
4. Find the difference between vectorAC and vectorAK. That's the normal vector of plane ABC.
5. The location of point K is point A plus vectorAK
6. The location of point D is point C plus the normal vector. You can always scale the normal vector if that's too small or large a distance.

def findPoleVector(objectA,objectB,objectC):
    # our three given points
    A = cmds.xform(objectA,q=1,t=1,ws=1)
    B = cmds.xform(objectB,q=1,t=1,ws=1)
    C = cmds.xform(objectC,q=1,t=1,ws=1)

    # find vectorAB and vectorAC
    vecAB = [B[0]-A[0],B[1]-A[1],B[2]-A[2]]
    vecAC = [C[0]-A[0],C[1]-A[1],C[2]-A[2]]

    # find K, which is the scalar (constant) for vectorAB that will give us the point where vector KC is perpendicular to vectorAB
    # the equation for that is: k=(vecABx*vecACx + vecABy*vecACy + vecABz*vecABz)/(vecABx^2,vecABy^2,vecABz^2)
    # we need to use float division in this case to find K or else Python will return 0
    k =  float(((vecAB[0]*vecAC[0]) + (vecAB[1]*vecAC[1]) + (vecAB[2]*vecAC[2]))) / float((vecAB[0]*vecAB[0] + vecAB[1]*vecAB[1] + vecAB[2]*vecAB[2]))
    # now use k to find vector AK, which we will use later on to find where K is
    vecAK = [vecAB[0]*k,vecAB[1]*k,vecAB[2]*k]
    # use vectorAK to find the normal vector of the plane, which is found by subtracting vecAC from vecAK
    nVector = [vecAC[0]-vecAK[0],vecAC[1]-vecAK[1],vecAC[2]-vecAK[2]]

    # find point D by taking the nVector (like a 3d slope) from point C
    D = [C[0] + .5*nVector[0],C[1] + .5*nVector[1],C[2] + .5*nVector[2]]
    # do the same thing we did for point D, now with point K
    K = [A[0] + vecAK[0],A[1] + vecAK[1],A[2] + vecAK[2]]

    # done!
    # these are just visual markers that are in the location of the points we calculated
    dLoc = cmds.spaceLocator(n="D")
    kLoc = cmds.spaceLocator(n="K")
    print "A possible pole vector location is: " + str(D)
    print "The normal vector is equal to: " + str(nVector)
    return D

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s