1. ME 4135 Robotics & Control R. Lindeke, Ph. D.

2. FKS vs. IKS  In FKS we built a tool for finding end frame geometry from Given Joint data:  In IKS we need Joint models from given End Point Geometry: Joint Space Cartesian Space Joint Space Cartesian Space

3. So this IKS is Nasty (as we know!)  It a more difficult problem because:  The Equation set is “Over-Specified”  12 equations in 6 unknowns  Space is “Under-Specified”  Planer devices with more joints than 2  The Solution set can contain Redundancies  Multiple solutions  The Solution Sets may be un-defined  Unreachable in 1 or many joints

4. Uses of IKS  Builds Workspace  Allows “Off-Line Programming” solutions  Thus, compares Workspace capabilities with Programming realities to assure that execution is feasible  Aids in Workplace design and operation simulations

5. Doing a Pure IKS solution: the  R Manipulator  R Frame Skeleton (as last time)

6. LP Table and A i ’sFramesLinkVarda SSSS CCCC S  C  0 → 1 1R  + 90 009010C1-S1 1 → 2 2P0 d2 + cl2 001010

7. FKS is A1*A2:

8. Forming The IKS:

9.  Examining these two equation (n, o, a and d are ‘givens’ in an inverse sense!!!):  Term (1, 4) & (2,4) both side allow us to find an equation for  :  (1,4): C1*(d 2 +cl 2 ) = d x  (2,4): S1*(d 2 +cl 2 ) = d y  Form a ratio to build Tan(  ):  S1/C1 = d y / d x  Tan  = d y /d x   = Atan2(d x, d y )

10. Forming The IKS:  After  is found, back substitute and solve for d 2 :  (1,4): C1*(d 2 +cl 2 ) = d x  Isolating d 2 : d 2 = [d x /C1] - cl 2

11. Alternative Method – doing a pure inverse approach  Form A 1 -1 then pre-multiply both side by this ‘inverse’  Leads to: A 2 = A 1 -1 *T 0 n given

12. Simplify RHS means:

13. Solving:  Selecting and Equating (1,4)  0 = -S1*d x + C1*d y  Solving: S1*d x = C1*d y  Tan(  ) = (S1/C1) = (d y /d x )   = Atan2(d x, d y )  Selecting and Equating (3,4) -- after back substituting  solution  d 2 + cl 2 = C1*d x + S1*d y  d 2 = C1*d x + S1*d y - cl 2

14. Performing IKS For Industrial Robots:  First lets consider the previously defined Spherical Wrist simplification  All Wrist joint Z’s intersect at a point  The nth Frame is offset from this intersection by a distance d n along the a vector of the desired solution (3 rd column of desired orientation sub- matrix)  This result is as expected by following the DH Algorithm!

15. Performing IKS  We can now separate the effects of the ARM joints  Joints 1 to 3 in a full function manipulator (without redundant joints)  They function to position the spherical wrist at a target POSITION related to the desired target POSE  … From the WRIST Joints  Joints 4 to 6 in a full functioning spherical wrist  Wrist Joints function as the primary tool to ORIENT the end frame as required by the desired target POSE

16. Performing IKS: Focus on Positioning  We will define a point (the WRIST CENTER) as:  P c = [P x, P y, P z ]  Here we define P c = d target - d n *a  P x = d target,x - d n *a x  P y = d target,y - d n *a y  P z = d target,z - d n *a z

17. Focusing on the ARM Manipulators in terms of P c :  Prismatic:  q 1 = P z (its along Z 0 !) – cl1  q 2 = P x or P y - cl2  q 3 = P y or P x - cl3  Cylindrical:   1 = Atan2(P x, P y )  d 2 = P z – cl2  d 3 = P x *C1 – cl3 or +(P x 2 + P y 2 ).5 – cl3

18. Focusing on the ARM Manipulators in terms of P c :  Spherical:   1 = Atan2(Px, Py)   2 = Atan2( (P x 2 + P y 2 ).5, P z )  D3 = (P x 2 + P y 2 + P z 2 ).5 – cl3

19. Focusing on the ARM Manipulators in terms of P c :  Articulating:   1 = Atan2(Px, Py)   3 = Atan2(D,  (1 – D 2 ).5 )  Where D =   2 =  -    is: Atan2((Px 2 + Py 2 ).5, P z )   is:

20. One Further Complication:  This can be considered the d 2 offset problem  A d 2 offset is a design issue that states that the nth frame has a non-zero offset along the Y 0 axis as observed with all joints at home, in the solution of the T 0 n  This leads to two solutions for  1  the So-Called Shoulder Left and Shoulder Right solutions

21. Defining the d 2 Offset issue Here: ‘The ARM’ might be a prismatic joint as in the Stanford Arm or it might be a 2 & a 3 links in an Articulating Arm and rotates out of plane A d2 offset means that there are two places where  1 can be placed to touch a given point (and note, when  1 is at Home, the wrist center is not on the X0 axis!)

22. Lets look at this Device “From the Top” a Q 11 F1F1

23. Solving For  1  We will have a Choice (of two) poses for  :

24. In this so-called “Hard Arm”  We have two  1 ’s  These lead to two  2 ’s (Spherical)  Or to four  2 ’s and  3 ’s in the Articulating Arm  Shoulder Right Elbow Up & Down  Shoulder Left Elbow Up & Down

25. The Orientation Model  Evolves from:  Separates Arm Joint and Wrist Joint Contribution to the desiredTarget (given) orientation

26. Focusing on Orientation Issues  Lets begin by considering Euler Angles (they are a model that is almost identical to a full functioning Spherical Wrist):  Form Product:  R z1 *R y2 *R z3  This becomes R 3 6

27. Euler Wrist Simplified: And this matrix is equal to a U matrix prepared by multiplying the inverse of the ARM joint orientation matrices inverse and the Desired (given) target orientation NOTE: R 0 3 is Manipulator Arm dependent!

28. Simplifying the RHS: (our so- called U Matrix)

29. Continuing:

30. Finally:

31. Solving for Individual Orientation Angles (1 st ):  Selecting (3,3) → C  = U 33  With C  we “know” S  =  (1-C 2  ).5  Hence:  = Atan2(U 33,  (1-U 33 2 ).5  NOTE: 2 solutions for  !

32. Re-examining the Matrices:  To solve for  : Select terms: (1,3) & (2,3)  C  S  = U 13  S  S  = U 23  Dividing the 2 nd by the 1 st : S  /C  = U 23 /U 13  Tan(  ) = U 23 /U 13   = Atan2(U 13, U 23 )

33. Continuing our Solution:  To solve for  : Select terms: (3,1) & (3,2)  -S  C  = U 31  S  S  = U 32  Tan(  ) = U 32 /-U 31   = Atan2(-U 31, U 32 )

34. Summarizing:   = Atan2(U 33,  (1-U 33 2 ).5   = Atan2(U 13, U 23 )   = Atan2(-U 31, U 32 )

35. Let do a (true) Spherical Wrist:

36. IKSing the Spherical Wrist FramesLinkVarda 3 → 4 4R 444400-90 4 → 5 5R 555500+90 5 → 6 6R 6666 d6d6d6d600

37. Writing The Solution:

38. Lets See By Pure Inverse Technique:

39. Simplifying

40. Solving:  Examination here lets select (3,3) both sides:  0 = C4U 23 – S4U 13  S4U 13 = C4U 23  Tan(  4 ) = S4/C4 = U 23 /U 13   4 = Atan2(U 13, U 23 )  With the given desired orientation and values (from the arm joints) we have a value for  4 the RHS is completely known

41. Solving for  5 &  6  For  5 : Select (1,3) & (2,3) terms  S5 = C4U 13 + S4U 23  C5 = U 33  Tan(  5 ) = S5/C5 = (C4U 13 + S4U 23 )/U 33   5 = Atan2(U 33, C4U 13 + S4U 23 )  For  6 : Select (3,1) & (3, 2)  S6 = C4U 21 – S4U 11  C6 = C4U 22 – S4U 12  Tan(  6 ) = S6/C6 = ([C4U 21 – S4U 11 ],[C4U 22 – S4U 12 ])   6 = Atan2 ([C4U 21 – S4U 11 ], [C4U 22 – S4U 12 ])

42. Summarizing:   4 = Atan2(U 13, U 23 )   5 = Atan2(U 33, C4U 13 + S4U 23 )   6 = Atan2 ([C4U 21 – S4U 11 ], [C4U 22 – S4U 12 ])

43. Lets Try One:  Cylindrical Robot w/ Spherical Wrist  Given a Target matrix (it’s an IKS after all!)  The d 3 “constant” is 400mm; the d 6 offset (call it the ‘Hand Span’) is 150 mm.   1 = Atan2((d x – a x *150),(d y -a y *150))  d2 = (dz – a z *150)  d3 = [(d x – a x *150) 2,(d y -a y *150) 2 ].5 - 400

44. The Frame Skeleton: Note “Dummy” Frame to account for Orientation problem with Spherical Wrist modeled earlier

45. Solving for U: NOTE: We needed a “Dummy Frame” to account for the Orientation issue at the end of the Arm

46. Simplifying:

47. Subbing U ij ’s Into Spherical Wrist Joint Models:   4 = Atan2(U 13, U 23 ) = Atan2((C1a x + S1a z ), (S1a x -C1a z ))   5 = Atan2(U 33, C4U 13 + S4U 23 ) = Atan2{a y, [C4(C1a x +S1a z ) + S4 (S1a x -C1a z )]}   6 = Atan2 ([C4U 21 - S4U 11 ], [C4U 22 - S4U 12 ]) = Atan2{[C4(S1n x -C1n z ) - S4(C1n x +S1n z )], [C4(S1o x -C1o z ) - S4(C1o x +S1o z )]}