CPSC 452 Inverse Kinematics Prof. Oussama Khatib, Stanford University Prof. Dezhen Song, Texas A&M University

X

Workspace Reachable Workspace Dexterous Workspace

X Given X Find q=(  1,  2,  3 )

Algebraic Solution The kinematics of the example seen before are: Assume goal point is specified by 3 numbers:

Algebraic Solution (cont.) By comparison, we get the four equations: Summing the square of the last 2 equations: From here we get an expression for c 2

Algebraic Solution (III) When does a solution exist? What is the physical meaning if no solution exists? Two solutions for  2 are possible. Why? Using c 12 =c 1 c 2 -s 1 s 2 and s 12 = c 1 s 2 -c 2 s 1 : where k 1 =l 1 +l 2 c 2 and k 2 =l 2 s 2. To solve these eq s, set r=+  k 1 2 +k 2 2 and  =Atan2(k 2,k 1 ).

Algebraic Solution (IV) k1k1 k2k2 22 l1l1 l2l2  Then: k 1 =r cos , k 1 =r sin , and we can write: x/r= cos  cos  1 - sin  sin  1 y/r= cos  cos  1 - sin  sin  1 or: cos(  +  1 ) = x/r, sin(  +  1 ) =y/r

Algebraic Solution (IV) Therefore:  +  1 = Atan2(y/r,x/r) = Atan2(y,x) and so:  1 = Atan2(y,x) - Atan2(k 2,k 1 ) Finally,  3 can be solved from:  1 +  2 +  3 = 

Geometric Solution IDEA: Decompose spatial geometry into several plane geometry problems x y L1L1 L2L2 Applying the “ law of cosines ” : x 2 +y 2 =l 1 2 +l l 1 l 2 cos(180+  2 )