Inverse Kinematics - University of Iowa

Inverse Kinematics - University of Iowa

Inverse Kinematics How do I put my hand here? IK: Choose these angles! Example: Planar 3-link robot x l1 cos1 l2 cos1 2 l3 cos1 2 3 y l1 sin 1 l2 sin 1 2 l3 sin 1 2 3 3 What is the reachable space? Take l1, l2 fixed and vary 3 l3 2

1 2 3 Take l1 l2 l3 , l1 l2 l3 l2 l1 1 Now vary 1 Finally, vary 2 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 Example (continued) What is the dextrous workspace in the example? The IK Problem Kinematic Problem: Problem given joint angles and/or displacement, compute location and orientation of End Effector. Inverse Kinematic Problem: Problem given location and orientation of EE, find joint variables. Why is IK hard?

May have more than one solution or none at all Amounts to solving nonlinear trascendental equations (can be hard) 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! 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 Solutions numerical solutions We are interested in closed-form solutions 1. Algebraic Methods 2. Geometric Methods 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, 90o. One major application (and driving force) of IK: animation.

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: c s B T W 0

0 s c 0 0 0 x 0 y 1 0 0 1 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 S compute modified goal GT in Smanipulator SS as near as possible to S T GT G' 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

Algebraic Solution The kinematics of the example seen before are: c123 s 123 B 0 T T W 3 0 0

s123 0 l1c1 l2 c12 c123 0 l1s1 l2 s12 0 1 0 0 0 1 Assume goal point is specified by 3 numbers: c s

B WT 0 0 s c 0 0 0 x 0 y 1 0

0 1 Algebraic Solution (cont.) By comparison, we get the four equations: c c123 s s123 x l1c1 l2 c12 y l1s1 l2 s12 Summing the square of the last 2 equations: x 2 y 2 l12 l22 2l1l2 c2 From here we get an expression for c2 2

2 2 1 x y l l c2 2l1l2 2 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 c12=c1c2-s1s2 and s12= c1s2-c2s1:

x k1c1 k 2 s1 y k1s1 k 2 c1 where k1=l1+l2c2 and k2=l2s2. To solve these eqs, set r=+ k12+k22 and =Atan2(k2,k1). Algebraic Solution (IV) k1 l2 2 k2

l1 Then: k1=r cos , k1=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(k2,k1) Finally, 3 can be solved from: 1+ 2+ 3 = Geometric Solution IDEA: Decompose spatial geometry into several

plane geometry problems y L2 L1 x Applying the law of cosines: x2+y2=l12+l22 - 2l1l2cos(180+2) Geometric Solution (II) Then: 2 2 2 1

x y l l c2 2l1l2 2 2 y The LoC gives: x l22 = x2+y2+l12 - 2l1 (x2+y2) cos

So that cos = (x2+y2+l12 - l22 )/2l1 (x2+y2) We can solve for 0 180, and then 1= 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-u2)/(1+u2) sin =2u /(1+u2)

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