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Inverse Kinematics Amirkabir University of Technology Computer Engineering & Information Technology Department  Course site: 

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Presentation on theme: "Inverse Kinematics Amirkabir University of Technology Computer Engineering & Information Technology Department  Course site: "— Presentation transcript:

1 Inverse Kinematics Amirkabir University of Technology Computer Engineering & Information Technology Department  Course site:  http://ce.aut.ac.ir/~shiry/lectures.html

2 Inverse Kinematics Given a desired position (P) & orientation (R) of the end-effector Find the joint variables which can bring the robot to the desired configuration.

3 A Simple Example 11 X Y S Revolute and Prismatic Joints Combined (x, y) Finding  : More Specifically : arctan2() specifies that it’s in the first quadrant Finding S :

4 Solvability Given the numerical value of we attempt to find values of The PUMA 560: Given as 16 numerical values, solve for6 joint angles, 12 equations and 6 unknowns 6 equations and 6 unknowns (nonlinear, transcendental equations)

5 Inverse Kinematics More difficult. The equations to solve are nonlinear thus systematic closed- form solution is not always available. Solution not unique.  Redundant robot.  Elbow-up/elbow-down configuration. Robot dependent. 2 solutions!

6 The 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

7 Two-link manipulator If l 1 = l 2 reachable work space is a disc of radius 2l 1. The dextrous Workspace is a point: origin If l 1 l 2 there is no dextrous workspace. The reachable work space is a ring of outer radius l 1 + l 2 and inner radius l 1 - l 2

8 Existence of Solutions A solution to the IKP exists if the target belongs to the workspace. Workspace computation may be hard. In practice it is made easy by special design of the robot.

9 Multiple Solutions The IKP may have more than one solution. We need to be able to calculate all the possible solutions. The system has to be able to choose one. 2 solutions! The closest solution: the solution which minimizes the amount that each joint is required to move.

10 The Closest Solution in Joint Space Weights might be applied. Selection favors moving smaller joints rather than moving larger joints when a choice exist. The presence of obstacles.

11 Two Possible Solutions In the absence of obstacles the upper configuration is selected

12 Number of Solutions Another 4 solution 4 Solutions of the PUMA 560 Depends upon the number and range of joints and also is a function of link parameters ,a,d 8 solutions exits

13 Number of Solutions vs. Nonzero a i The more the link length parameters are nonzero, the bigger the maximum number of solutions!

14 Solvability All systems with revolute and prismatic joints having total of 6 D.O.F in a single series chain are solvable. But this general solution is a numerical one. Robots with analytic solution: several intersecting joint axes and/or many  i = 0, 90 o.

15 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

16 Numerical Solutions Results in a numerical, iterative solution to system of equations, for example Newton/Raphson techniques. Unknown number of operations to solve. Only returns a single solution. Accuracy is dictated by user. Because of these reasons, this is much less desirable than a closed-form solution. Can be applied to all robots.

17 Closed-form solutions Analytical solution to system of equations Can be solved in a fixed number of operations (therefore, computationally fast/known speed) Results in all possible solutions to the manipulator kinematics Often difficult or impossible to find Most desirable for real-time control Most desirable overall

18 Given a 6 axis robot. It can be proven that there exists a closed form solution for inverse kinematics:  If three adjacent revolute joint axes intersect at a point (PUMA, Stanford).  If three adjacent revolute joint axes are parallel to one another (MINIMOVER). In any case algebraic or geometric intuition is required to obtain closed form solutions. Closed-form solutions

19 Methods of Solutions In general the IK problem can be solved by various methods such as: Inverse Transform ( Paul, 1981) Screw Algebra (Kohli 1975) Dual Matrices (Denavit 1956) Iterative (Uicker 1964) Geometric approach (Lee 1984) Decoupling of position and orientation (Pieper 1968)

20 Closed-form Solutions We are interested in closed-form solutions: 1. Algebraic methods 2. Geometric methods

21 Algebraic solution Consider a 3-link manipulator We can derive kinematic equations:

22 i  i-1 didi ii 1000 11 20L1L1 0 22 30L2L2 0 33 Algebraic solution  D-H transformation

23 Algebraic Solution

24 The kinematics of the example seen before are: Assume goal point is the specification of wrist frame, specified by 3 numbers:

25 Algebraic Solution By comparison, we get the four equations: Summing the square of the last 2 equations: From here we get an expression for c 2 And finally:

26 Algebraic Solution The arc cosine function does not behave well as its accuracy in determining the angle is dependant on the angle ( cos(q)=cos(-q)). When sin(q) approaches zero, division by sin(q) give inaccurate solutions. Therefore an arc tangent function which is more consistent is used.

27 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

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

29 Algebraic Solution 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 = 

30 Geometric Solution x y a1a1 a2a2 Two-link Planar Manipulator

31 Geometric Solution Applying the “ law of cosines ” : We should try to avoid using the arccos function because of inaccuracy.

32 Geometric Solution The second joint angle  2 is then found accurately as: Because of the square root, two solutions result:

33 Geometric Solution  1 is determined uniquely given  2  1 =  -  Note the two different solutions for  1 corresponding to the elbow-down versus elbow-up configurations

34 Examples: IK Solution for PUMA 560 Algebraic Solution for UNIMATION PUMA 560 We wish to Solve For  i when 0 6 T is given as numeric values (4.54)

35 Examples: IK Solution for PUMA 560 Paul (1981) suggest pre multiplying the above matrix (4.54) by its unworn inverse transform successively and determine the unknown angle from the elements of the resultant matrix equation.

36 Examples: IK solution for PUMA 560 By multiplying both sides of (4.54) Inverting 0 1 T we get: (4.56)

37 Examples: IK Solution for PUMA 560 1 6 T from (3.13) is:

38 Examples: IK solution for PUMA 560 By adapting 1 6 T from (3.13) and equating elements we have Using trigonometric substitution: We obtain Using difference of angles:

39 Examples: IK Solution for PUMA 560 And so: Solution for  1 : Note: two possible solution for  1

40 Examples: IK Solution for PUMA 560 By equating elements (1,4) and also (3,4) from Equ. (4.56): By squaring and addition of resulting equations and Equ. (4.57): Where

41 Examples: IK Solution for PUMA 560 The above equation is only dependant on  3 so with similar procedure we get: again two different solution for  3. We may rewrite (4.54) as (4.70)

42 Examples: IK Solution for PUMA 560 By adapting 3 6 T from (3.11) and equating (1,4) and (2,4) we have We solve for  23 as Four possible solutions for  2

43 Examples: IK Solution for PUMA 560 By equating elements (1,3) and also (3,3) from (4.70) :

44 Examples: IK Solution for PUMA 560 We continue similar procedure for rest of the Parameters …

45 Examples: IK Solution for PUMA 560 After all eight solutions have been computed, some of them may have to be discarded because of joint limitations. Usually the one closest to present manipulator configuration is chosen. Other possible solutions :

46 Pieper’s Solution When Three Axes Intersect (E.G, Spherical Wrists) A completely general robot with six degrees of freedom does not have a closed form solution, however special cases can be solved. The technique involves decoupling the position and orientation problems. The position problem positions the wrist center, while the orientation problem completes the desired orientation.

47 Pieper’s Solution This method applies to all six joints revolute, with the last 3 intersection It can be applied to majority of industrial robots

48 {0} Z 4Z5 Z 6 P c d d 6 Pieper’s Solution: Basic Concept First, the location of the wrist center, p c is found from the given tool position (d) and the tool pointing direction (here z 6 ). Since the wrist center location depends on the first three joint variables, this results in three equations and three unknowns which are solved for  1 –  3. Then, the relative wrist orientation R 6 3 which is a function of the last three joint variables,  4 –  6 is found from the arm orientation R 3 0 and the given tool orientation R 6 0. The relative wrist orientation is set equal to the kinematic description of R 6 3 and the last 3 joint variables are solved.

49 Pieper’s Solution Step-by-step Procedure Start with the given tool pose, as. Solve portions of the forward kinematics to find find. Find the location of the wrist center, as (last column of ) - (tool offset length) x (third column of ). Set = last column of and solve for for. Solve as (put into ). Set equal to and solve for. Along the way, keep track of the number of solutions for each joint variable.

50 Standard Frames {B} {W} {T} {G} {S}

51 Repeatability and accuracy Repeatability: how precisely a manipulator can return to a taught point? Accuracy: the precision with which a computed point can be attained.


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