where tI = tan(e1/2), Eq. (5.33) yields a quartic polynomial equation in tI. With t1 computed, a value of e1 is obtained and e3 can be computed uniquely from Eqs. (5.29) and (5.32). The remaining angles (e2 and 84) can be computed as indicated earlier.

 We propose a method that allows better insight without the complexity of a quartic polynomial equation. For

simplicity, the sine and cosine of a sum of angles will be represented according to Cijk=cos(ei+ej) and Sij=sin(ei+ej).

 As described in chapter 4, a set of inverse kinematic equations can be obtained by expressing the components of vectors t and p and the inner products t.p and p.p in terms of the joint variables ei, i=l, . 3. The equations obtained are

 t = a3 S123 (5.34)

 ty = -a3 C123 (5.35),

 Px = a3 C123 + a2 C12 + al C1 (5.36)

 py = a3 S123 + a2 S12 + a1 Sl (5.37)
 t.p= al C23 + a2a3 S3 + T3(dl+d2) + d3 (5.38)
 p.p = 2(ala3 C23 + a2a3 C3 + ala2 C2) + ct (5.39) where

 ct = a12 + a22 + a32 + d12 + d22 + d32

 + 2(d1d2 + d1d3 + d2d3).


 Equations (5.34) and (5.35) yield S123 and C123 directly, soa unique value of 8123 81+82+83 is obtained.

..