With this analysis, we find that there can be at most two solution sets.
Case 9: al=a3=0. The first two joint axes intersect and the last two joint axes intersect. The reduced system is
a2a3 S3 a2a3d2 C3 = r1 (5.68)
aitx S1 alty C1 + a2a3 C3 = r2 (5.69)
alpx S1l alPy C1 = r3 (5.70)
0 = r4. (5.71)
Here, r4 = 0 poses a constraint on pose parameter pz. Equation (5.70) yields two distinct values of el, then Eqs. (5.68) and (5.69) will form a linear system in S3 and C3 which can be solved for a unique value of e3 for each value of e1. Two solution sets are thus obtained.
Case 10: a2=a3=0. The intermediate two joint axes intersect and the last two joint axes intersect. The reduced system becomes
alty S1 + altx C1 a263d2 C3 = r1 (5.72)
itx S1 lty C1 + 0203 C3 = r2 (5.73)
alpx S1l 0lPy C1 = r3 (5.74)
alPy S1 + alPx C1 = r4. (5.75)
Equations (5.74) and (5.75) yield a unique solution for el. The value of el obtained-can be substituted in Eq. (5.72) or (5.73) to solve for C3 -which provides twop possible- values
..