and (5.47). The two solution sets are then completed as shown previously.

 Case 4: a2=a3=0. The last three joint axes are

parallel. The reduced system simplifies to


 alty S1 + altx C1 = r1 (5.48)

 altx S1 alty C1 = r2 (5.49)

 1px S1 alPy C1 = r3 (5.50)
alPy S1 + alPx C1 + a2a3 C3 = r4. (5.51)


Two out of the first three equations (Eqs. (5.48)-(5.50)) can be used to solve uniquely for el. The third (unused

equation) becomes a realizability constraint on the pose. With 81 known, Eq. (5.51) yields a value for C3 which in turn gives two possible values for e3. Two solution sets can be obtained after computing e2 and e4.

 Case 5: a2=a3=0. The intermediate joint axes are

parallel and the last two axes intersect. The reduced sytem becomes


 alty S1 + altx C1 + a2a3 S3 = rl (5.52)

 altx S1 7lty C1 = r2 (5.53)

 0lpx S1 alPy C1 = r3 (5.54)

 aly S1 + alPx C1 r4. (5.55)

Two out of the last three equations (5.53)-(5.55) can be solved uniquely for el. The third equation becomes a

realizability constraint on pose element tz or pz depending

..