1. Inverse Kinematics How do I put my hand here? IK: Choose these angles!

2. What is the reachable space? Take l 1, l 2 fixed and vary  3 Example: Planar 3-link robot l2l2 l3l3 l1l1 Now vary  1 Finally, vary  2

3. The Workspace Workspace Workspace: volume of space which can be reached by the end effector Dextrous workspace: volume of space where the end effector can be arbitrarily oriented Reachable workspace: volume of space which the robot can reach in at least one orientation

4. Example (continued) What is the dextrous workspace in the example?

5. The IK Problem Kinematic ProblemKinematic Problem: given joint angles and/or displacement, compute location and orientation of End Effector. Inverse Kinematic ProblemInverse Kinematic Problem: given location and orientation of EE, find joint variables. Why is IK hard?Why is IK hard? –May have more than one solution or none at all –Amounts to solving nonlinear trascendental equations (can be hard)

6. Existence of Solutions A solution to the IKP exists if the target belongs to the workspace Workspace computation may be hard. In practice is made easy by special design of the robot The IKP may have more than one solution. How to choose the appropriate one? 2 solutions!

7. Methods of Solutions A manipulator is solvable if the joint variables can be determined by an algorithm. The algorithm should find all possible solutions. closed form solutions numerical solutions Solutions We are interested in closed-form solutions 1. Algebraic Methods 2. Geometric Methods

8. Method of Solution (cont.) Major result: all systems with revolute and prismatic joints having a total of six degrees of freedom in a single series chain are solvable In general, solution is numerical Robots with analytic solution: several intersecting joint axes and/or many  i = 0, 90 o. One major application (and driving force) of IK: animation.

9. Manipulator Subspace when n<6 If n<6, then the workspace will be a portion of an n dimensional subspace To describe the WS: compute direct kinematics, and then vary joint variables On the previous example, the WS has the form:

10. Manipulator SS when n<6 (cont) Usual goal for manipulator with n DoF: use n parameters to specify the goal If 6 DoF are used, n<6 will in general not suffice Possible compromise: reach the goal as “near” as possible to original goal: –1) Given the goal frame compute modified goal in manipulator SS as near as possible to –2) Compute IK. A solution may still not be possible if goal is not in the manipulator workspace For example, place tool frame origin at desired location, then select a feasible orientation

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

12. 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

13. 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 ).

14. 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

15. 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 = 

16. 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 2 2  2l 1 l 2 cos(180  2 ) 22

17. Geometric Solution (II) Then: x y   The LoC gives: l 2 2 = x 2 +y 2 +l 1 2 - 2l 1  (x 2 +y 2 ) cos  So that cos  = (x 2 +y 2 +l 1 2 - l 2 2 )/2l 1  (x 2 +y 2 ) We can solve for 0    180, and then  1 = 

18. Reduction to Polynomial Trascendental equations are difficult to solve since one variable  usually appears as cos  and sin . Can reduce to polynomial in variable u = tan  /2 by using: cos  = (1-u 2 )/(1+u 2 ) sin  =2u /(1+u 2 )