has the special structure, Eq. (6.2) can be used and the constraint r3 = 0 applies to Eq. (6.41). Angle aI can be directly computed.
Case 4: Two intersecting axes followed by two parallel axes. This structure corresponds to Chapter 5, case 6. Here, the constraint is rI = 0 and, as in the preceding cases, 81 or 85 can be directly computed from Eq. (6.39) or (6.43), respectively.
Case 5: Three intersecting axes. Pieper (1968) has
shown that a six-DOF manipulator with three intersecting axes can always be solved in closed form. This result
applies to the simpler case of five-DOF robots. This
structure corresponds to case 8 of Chapter 5 where the constraints are rl=0 and r4=0. If the three intersecting axes are the first three, Equation (6.30) should be used. With Eqs. (15) and (6.46), a value of a5 can be obtained directly. This same method can be used when the last three axes intersect by first exchanging the roles of end-effector and base frames. When the intermediate three axes are intersecting, use of Eqs. (6.2), (6.39) and (6.42) will yield a value of al.
Case 6: Two consecutive sets of two intersecting axes. This structure is analyzed in case 9 of Chapter 5. The
constraint equation is r4 = 0. Depending on where this
special structure is located along the five-DOF arm, 01 or
..