For a general n degree-of-freedom open chain with forward kinematics T (θ), θ ϵ Rn, the inverse kinematics problem can be stated as follows: given a homogeneous transform X ϵ SE(3), find solutions θ that satisfy T (θ) = X. To highlight the main features of the inverse kinematics problem, let us examine the twolink planar open chain of Figure 6.1(a) as a motivational example. Considering only the position of the end-effector and ignoring its orientation, the forward kinematics can be expressed as

Assuming L1 > L2, the set of reachable points, or the workspace, is an annulus of inner radius L1 −L2 and outer radius L1 +L2. Given some end-effector position (x, y), it is not hard to see that there will be either zero, one, or two solutions depending on whether (x, y) lies in the exterior, boundary, or interior of this annulus, respectively. When there are two solutions, the angle at the second joint (the “elbow” joint) may be positive or negative. These two solutions are sometimes called “lefty” and “righty” solutions, or “elbow-up” and “elbow-down” solutions.

Finding an explicit solution (θ1, θ2) for a given (x, y) is also not difficult. For this purpose, we will find it useful to introduce the two-argument arctangent function atan2(y, x), which returns the angle from the origin to a point (x, y) in the plane. It is similar to the inverse tangent tan−1(y/x), but whereas tan−1(y/x) is equal to tan−1(−y/−x), and therefore tan−1 only returns angles in the range [−π/2, π/2], the atan2 function returns angles in the range (−π, π]. For this reason, atan2 is sometimes called the four-quadrant arctangent.

We also recall the law of cosines,

where a, b, and c are the lengths of the three sides of a triangle and C is the interior angle of the triangle opposite the side of length c.

Referring to Figure 6.1(b), angle β, restricted to lie in the interval [0, π], can be determined from the law of cosines,