ri = ril S5 ri2 C5 + ri3, i=l, . 4.
Indeed, if we replace uz, qz, u.q, and q.q by their expressions in terms of S5 and C5, as given by Eqs. (6.34)(6.38) and substitute in Eqs. (5.17)-(5.20) (substitute for tz, pz, t.p, and p.p and let dl=0 ), we obtain rI = a4(n.p) S5 + a4(b.p) C5 + r4(t.p)
r3d3 7'2d2r3 d4, (6.43)
r2 = -rla4nz S5 -Tla4bz C5 TT4tz + 23, (6.44)
r3 = (Tla4d4nz -Tla4bz) S5 + (rla4nz + r1a4d4bz) C5
+ (Tlr 4d4tz -Tlpz) + d2 + T2d3, (6.45)
and
r4 =-2[a4d4 (n.p) a4 (b.p)] S5
2[a4 (n.p) + a4d4 (b.p)] C5 2r4d4 (t.p) r2d2d3
+ (p.p + al2 a -d22 a3 2
d32 + a42 + d42)/2. (6.46)
In the analysis of special four-DOF geometries in Chapter 5, we found cases where the reduced system of equations included a constraint of the form ri = 0. Such a constraint applied to one of Eqs. (6.39)-(6.46) will usually yield a value of 81 or e5 which in turn makes the 5-DOF inverse kinematics problem solvable in closed form.
Case 1: Three joint axes are parallel. When the
parallel axes are the first three (i.e. axes 1, 2 and 3), Eq. (6.30) can be used. Case 1 of Chapter 5 shows that
..