Outdoor Human Motion Capture using Inverse Kinematics and von Mises-Fisher Sampling Supplemental Material

This is the supplemental material for [5]. It contains a more detailed description of the closed form algorithm to compute inverse kinematics based on the Paden-Kahan subproblems. For an extended and more detailed version of [5] we refer the reader to [7]. 1. Paden-Kahan subproblems We are interested in solving the following problem: exp(θ1ω̂1) exp(θ2ω̂2) exp(θ3ω̂3) = Rj . (1) This problem can be solved by decomposing it into subproblems as proposed in [4]. A comprehensive description of the Paden-Kahan subproblems applied to several inverse kinematic problems can also be found in [3]. The basic technique for simplification is to apply the kinematic equations to specific points. By using the property that the rotation of a point on the rotation axis is the point itself, we can pick a point p on the third axis ω3 and apply it to both sides of Eq. (1) to obtain exp(θ1ω̂1) exp(θ2ω̂2)p = Rjp = q (2) which is known as the Paden-Kahan sub-problem 2. For our problem the 3 rotation axes intersect at the same joint location. Consequently, since we are only interested in the orientations, we can translate the joint location to the origin qj = O = (0, 0, 0) T . In this way, any point p = λω3 with λ ∈ R, λ 6= 0 is a valid choice for p. Eq. (2) can decomposed in two subproblems exp(θ2ω̂2)p = c = exp(−θ1ω̂1)q (3) where c is the intersection point between the circles created by the rotating point p around axis ω2 and the point (a) (b) (c) Figure 1: Inverse Kinematics: (a) decomposition into active (yellow) and passive (green) parameters. Paden-Kahan subproblem 2 (b) and sub-problem 1 (c). q rotating around axis ω1 as shown in Fig. 1 (b). In order for Eq. (3) to have a solution, the points p, c must lie in the same plane perpendicular to ω2, and q, c must lie in the same plane perpendicular to ω1. This implies that the projection of the position vectors 1 p, c,q onto the span of ω1, ω2 respectively must be equal, see Fig. 2 ω 2 p = ω T 2 c and ω T 1 q = ω T 1 c (4) Additionally, the norm of a vector is preserved after rotation and therefore ‖p‖ = ‖c‖ = ‖q‖ (5) Since ω1 and ω2 are not parallel, the vectors ω1, ω2, ω1×ω2 form a basis that span R. Hence, we can write c in the new basis as c = αω1 + βω2 + γ(ω1 × ω2) (6) where α, β, γ are the new coordinates of c. Now, using the fact that ω2 ⊥ ω1×ω2 and ω1 ⊥ ω1×ω2, we can substitute Eq. (6) into Eq. (4) to obtain a system of two equations with 1Since we translated the joint location to the origin we can consider the points as vectors with origin at the joint location qj .