xI nL (P PL)
x2 bL (P P L)
x3 bL (P PL)
xe = = (7.42)
x4 (tL.b t.bL)/2
x5 (nL-t n.tL)/2
x6 (bL.n b.nL)/2
where n, b, t, and p are the vectors that describe the desired end-effector pose P as defined in (2.12) and vectors nL, bL, tL, and PL are their corresponding vectors from the left hand side of equation (7.1). Two functions can be
defined as
f(81,2) = x12 + x22 + x32 (7.43)
222 ( .4
g(e1,2) = x42 + x52 + x6 (7.44)
A pair (81e,2) that is a zero of both f and g is guaranteed to correspond to a solution set of Eq. (7.1) so that the iterative method described above will only converge to a true solution. However, now, the forward kinematics must be computed at each iteration since the components of vectors nL, bL, tL, and PL are all needed to evaluate functions f and g as defined by Eqs. (7.42) and (7.43).
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