Constraint Relaxation IK in 2D - Resolving Constraints

Written by Ryan Juckett

Monday, 13 April 2009 00:00

Article Index
Constraint Relaxation IK in 2D
Reaching the Target
Resolving Constraints
Implementing for Higher Dimensions
Local Minima
Writing the Code
All Pages

Page 3 of 6

Resolving Constraints

When resolving a distance constraint for a bone, we have two end points and the constrained length between them. End point $\mathbf{b_1}$ will be base of the current bone. End point $\mathbf{b_2}$ will be the base of the child bone. Length $l_1$ will be the desired distance between $\mathbf{b_1}$ and $\mathbf{b_2}$ . We will also support positive relaxation weights, $w_1$ and $w_2$ , corresponding to the end points. The higher an endpoint is weighted, the more that endpoint will move in the solution. If an endpoint is given a zero weight, it will not move at all. If both end points are given the same weight, they will move equal distances.

As a first step, we can check the bone weights for a cases that implies we shouldn't do any work. If the total weight, $w_1 + w_2$ , is zero neither joint should move and the constraint is skipped.

Next we check if the squared distance between the end points is zero. Let $\mathbf{v} = \mathbf{b_2}- \mathbf{b_1}$ and then use the dot product $\mathbf{v} \cdot\mathbf{v}$ to get the squared distance. If the result is zero, it implies that the end points are directly on top of each other. This makes it a bit hard to choose the best direction for stretching the bones. We will instead skip constraints in this situation with the hope that either later in this iteration or on a following iteration the points will be stretched apart by one of the neighboring constraints being resolved.

Now that we have found a constraint to modify, we need to compare the distance between the end points with the given bone length $l_1$ . The distance between $\mathbf{b_1}$ and $\mathbf{b_2}$ can be calculated as $d = \sqrt{ \mathbf{v} \cdot \mathbf{v}}$ .

We need to adjust the current distance by $l_1-d$ . We can now compute a vector that would move $\mathbf{b_2}$ the appropriate distance from $\mathbf{b_1}$ by normalizing $\mathbf{v}$ , and rescaling it to the desired change in distance, $l_1-d$ . This gives us a new vector $\mathbf{u} = (l_1-d)\frac{\mathbf{v}}{d}$ .

Rather than directly adding $\mathbf{u}$ to $\mathbf{b_1}$ , we want to distribute it across $\mathbf{b_1}$ and $\mathbf{b_2}$ based on the relaxation weights. Note that when modifying $\mathbf{b_1}$ we will be subtracting $\mathbf{u}$ to move the endpoint in the opposite direction.