Goon
01-29-2006, 10:51 PM
Heh. Nevermind, there seems to have been an error in my script, though I'm still not sure why the utility node solution that I had worked up wasn't working :D
'allo
Ooops. Probably shouldn't have put this in the mel section. Oh well, you guys are down with the vector math, so it shouldn't matter
i'm trying to rig up an ik fk switching system which will match fk to ik, and vice versa. I believe a have a solution at hand. However, being new to vector math and all, i'm a little surprised by some of the behaviour i'm getting.
I have a 3 joint ik chain
-there is 1 vector running from the base to the middle joint :: vector1
-there is 1 vector running from the base to the end joint :: vector2
I'm trying to create a vector which runs from the mid joint, intersects vector2 (and is perpendicular to it). So basically a vector which lies along the ik solver plane, and is perpendicular to the base of the triangle (vector2).
To do this, I'm using the cross product utility to get a vector which is perpendicular to the ik plane (inputs are vectors 1 and 2). But the cross product does not remain perpendicular as I move the ik handle around. Am I missing something that will ensure that it remains perpendicular?
The system i'm using for vectors is having locators feed their worldspace positions into a +/- node. And I'm normalizing all the vectors before and after feeding them into the cross product node.
Thanks
'allo
Ooops. Probably shouldn't have put this in the mel section. Oh well, you guys are down with the vector math, so it shouldn't matter
i'm trying to rig up an ik fk switching system which will match fk to ik, and vice versa. I believe a have a solution at hand. However, being new to vector math and all, i'm a little surprised by some of the behaviour i'm getting.
I have a 3 joint ik chain
-there is 1 vector running from the base to the middle joint :: vector1
-there is 1 vector running from the base to the end joint :: vector2
I'm trying to create a vector which runs from the mid joint, intersects vector2 (and is perpendicular to it). So basically a vector which lies along the ik solver plane, and is perpendicular to the base of the triangle (vector2).
To do this, I'm using the cross product utility to get a vector which is perpendicular to the ik plane (inputs are vectors 1 and 2). But the cross product does not remain perpendicular as I move the ik handle around. Am I missing something that will ensure that it remains perpendicular?
The system i'm using for vectors is having locators feed their worldspace positions into a +/- node. And I'm normalizing all the vectors before and after feeding them into the cross product node.
Thanks