S3 thereby providing two values for e3, the remaining equation is a pose constraint.
Case 8: al=a2=d2=0. The first three joint axes
intersect and the reduced system becomes
0 = rI (5.64)
aitx S1 aity C1 + a2a3 C3 = r2 (5.65)
oipx S1 alPy C1 C2a3 S3 = r3 (5.66)
0 = r4. (5.67)
Equations (5.64) and (5.67) impose constraints on pose parameters tz and pz. Here again a solution can be obtained in form of a quartic polynomial equation in tl=tan(e1/2) by eliminating e3 from Eqs. (5.65) and (5.66) as we did earlier in case 1.
With e1 known, e3 can be uniquely obtained from Eqs. (5.65) and (5.66) and the solution set completed as usual. This method puts an upper bound of 4 on the number of solution sets since at most four distinct values of 01 can be obtained from the quartic polynomial equation in tI.
An easier inverse kinematic analysis of this structure can be obtained if the roles of end-effector frame and base frame are reversed and the intermediate link-frames are reassigned accordingly. This will put the three
intersecting axes at the end-effector position instead of at the base. The 4-DOF structure is seen to be equivalent to one that has a2 = a3 = 0 which is discussed in case 10.
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