r2=0 and r3=0. These two constraints and Eqs. (6.44) and (6.45) yield a system of equations in S5 and C5,


 r21 $5 + r22 C5 = -r23 r31 S5 + r32 C5 = -r33,


which can be solved for 05 when the determinant given by (r21r32 r31r22) is not zero, otherwise there is no solution. With 85 known, the remaining angles can be obtained in closed-form.

 If the last three axes are parallel, a similar result is obtained by exchanging the roles of base and end-effector frames. When the intermediate axes are parallel, Eq. (6.2) should be used. The constraints r2 = r3 = 0 then yield a value of 01 and the inverse kinematic problem can be solved in closed form as well.

 Case 2: two consecutive sets of two parallel axes. If axes 1 and 2 are parallel and axes 3 and 4 are parallel, Eq. (6.30) and Chapter 5, case 2 yield r2 = 0 which can be used to solve for 95 from Eq. (6.44). If axes 2 and 3 are

parallel and axes 4 and 5 are parallel, then using Eq. (6.2) and Eq. (6.40) will yield a value of 91.

 Case 3: Two parallel axes followed by two intersecting axes. When this special geometry concerns the first four joint axes of the 5-DOF arm, using Eq. (6.30) and Chapter 5, case 3 yields r3 0. This constraint applied to Eq. (6.45) gives a value of 95. If the upper part of the 5-DOF robot

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