with the geometric and algebraic approaches taken by the previous authors.

 Without computing the forward kinematics, we will illustrate how Eqs. (4.9)-(4.13) may be easily obtained. For quick reference in the following discussion, we write the equations immediately


tz = S23 C4 S5 + C23 C5 (9.1)

Pz = a2 S2 + a3 S23 d4 C23 (9.2)
p.t = (a3+a2C3) C4S5 + d3 S4S5 C5 (d4 + a2S3) (9.3)
 2 2 2 2-d2
 (p2-a22-a3 -d32d42)/(2a2) = d4 S3 + a3 C3. (9.4)

 To illustrate the simplification obtained by the frame assignment described earlier and inner-product invariance under rotations, we give in detail the development of equation (9.4).


 With 11=15=16=0 and 14=d4z, Eq. (4.4) yields


 p = R1(R2(R314 + 13) + 12)

or

 p = R1R2R3 [14 + R3-113 + R3-1R2-112].


 By orthogonality, the inner-product p.p has the same value as the inner-product of the term in brackets hence


 p.p =[14 + R3-113 + R3-1R2 -112].[14 + R3-113 + R3-1R2-1121

..