alty S1 + altx Cl = rI (5.40)
0 = r2 (5.41)
a2a3 S3 = r3 (5.42)
alPy S1 + alPx C1 + a2a3d2 S3 + a2a3 C3 = r4. (5.43) Equation (5.41) imposes a constraint on pose parameter tz, tz = 2T3/T1. When this constraint is satisfied, Eq. (5.40) can be solved and yields two distinct values for e1. Then Eqs. (5.42) and (5.43) form a linear system in S3 and C3 which can be solved uniquely for e3. With E1 and e3
computed, e2 and e4 can be uniquely obtained as shown earlier. Here again we find at most two solution sets.
Case 3: al1=a3=0. First two joint axes are parallel and last two joint axes intersect. The reduced system becomes alty S1 + altx Cl + a2a3 S3 02o3d2 C3 = r1 (5.44)
a2a3 C3 = r2 (5.45)
0 = r3 (5.46)
alPy S1 + alPx C1 = r4. (5.47)
From Eq. (5.19), the pose constraint r3 = 0 translates to
SPz = di + d2 + T2d3"
For a pose matrix that satisfies this constraint, two possible values of 93 can be obtained from Eq. (5.45). For each of those e3 values, a unique value of el is computed from the linear system in S1 and C1 formed by Eqs. (5.44)
..