on the chosen equation. With el known, two values of 83 can be computed from the value of S3 derived from Eq. (5.52).

 Case 6: a3=al=0. The last two joint axes are parallel and the first two axes intersect. The reduced system is 0 = r1 (5.56)

 altx S1 at y C1 = r2 (5.57)

 alpx S1 alPy C1 a2a3 S3 = r3 (5.58)

 a2a3d2 S3 + a2a3 C3 = r4. (5.59)

Equation (5.56) is a realizability constraint on pose

parameter tz (Eq. (5.17)). For an end-effector pose that satisfies this constraint, Eq. (5.57) yields two values for e1. Equations. (5.58) and (5.59) can then be solved for 83 uniquely.

 Case 7: a3=a2=0. The last two joint axes are parallel and the intermediate two axes intersect. The reduced system is


 alty S1 + altx C1 = rI (5.60)

 altx S1 1ty C1 = r2 (5.61)

aPpx S1 alPy Ci a2a3 S3 = r3 (5.62)

alPy S1 + alPx C1 + a2a3d2 S3 = r4. (5.63)


Here, Eqs. (5.60) and (5.61) yield a unique value for e1, then one of Eqs. (5.62) or (5.63)-can be used to solve. for

..