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Robotic Kinematics – the Inverse Kinematic Solution

Robotic Kinematics – the Inverse Kinematic Solution. ME 3230 Kinematics and Mechatronics Dr. R. Lindeke. Cartesian Space. Joint Space. Joint Space. FKS vs. IKS. In FKS we built a tool for finding end frame geometry from Given Joint data:

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Robotic Kinematics – the Inverse Kinematic Solution

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  1. Robotic Kinematics – the Inverse Kinematic Solution ME 3230 Kinematics and Mechatronics Dr. R. Lindeke ME 3230

  2. Cartesian Space Joint Space Joint Space 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: Cartesian Space ME 3230

  3. So this IKS Problem is Nasty (in Mathematics this is considered a Hard Modeling Issue) • It a more difficult problem because: • The Equation set is “Over-Specified”: • 12 equations in 6 unknowns • Space can be “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 ME 3230

  4. But the IKS is VERY Useful – some Uses of IKS include: • Building Workspace Maps • Allow “Off-Line Programming” solutions • The IKS allows the engineer to equate Workspace capabilities with Programming realities to assure that execution is feasible(as done in ROBCAD & IGRIP) • The IKS Aids in Workplace Design and Operational Simulations ME 3230

  5. Doing a Pure IKS solution: the R Manipulator R Frame Skeleton (as DH Suggest!) Same Origin Point! ME 3230

  6. LP Table and Ai’s ME 3230

  7. FKS is A1*A2: ME 3230

  8. Forming The IKS: In the Inverse Problem, The RHS MATRIX is completely known (perhaps from a robot mapped solution)! And we use these values to find a solution to the joint equations that populate the LHS MATRIX ME 3230

  9. If 2 Matrices are Equal then EACH and EVERY term is also Uniquely equal as well! Forming The IKS: • Examining these two matrices • n, o, a and d are “givens” in the inverse sense!!! • (But typically we want to build general models leaving these terms unspecified) • Term (1, 4) & (2,4) on both sides allow us to find an equation for : • (1,4): C1*(d2+cl2) = dx • (2,4): S1*(d2+cl2) = dy • Form a ratio to build Tan(): • S1/C1 = dy/ dx • Tan  = dy/dx •  = Atan2(dx, dy) ME 3230

  10. Forming The IKS: • After  is found, back substitute and solve for d2: • (1,4): C1*(d2+cl2) = dx • Isolating d2: d2 = [dx/Cos1] - cl2 ME 3230

  11. Alternative Method – “doing a pure inverse approach” • Form A1-1 then pre-multiply both side by this ‘inverse’ • Leads to: A2 = A1-1*T0ngiven ME 3230

  12. Simplifying & Solving: • Selecting and Equating (1,4) • 0 = -S1*dx + C1*dy • Solving: S1*dx = C1*dy • Tan() = (S1/C1) = (dy/dx) •  = Atan2(dx, dy) – the same as before • Selecting and Equating (3,4) -- after back substituting  solution • d2 + cl2 = C1*dx + S1*dy • d2 = C1*dx + S1*dy - cl2 ME 3230

  13. Performing IKS For Industrial Robots: (a more involved problem) • First let’s consider the concept of the Spherical Wrist Simplification: • All Wrist joint Z’s intersect at a point • The ‘n Frame’ is offset from this Z’s intersection point at a distance dn (the hand span) along the a vector of the desired solution (3rd column of desired orientation sub-matrix!) which is the z direction of the 3rd wrist joint • This follows the DH Algorithm design tools as we have learned them! ME 3230

  14. Performing IKS – with a Spherical Wrist • We can now separate the POSE effects: • ARM joints • Joints 1 to 3 in a full function manipulator (without redundant joints) • They function to maneuver the spherical wrist to a target POSITIONrelated to the desired target POSE • WRIST Joints • Joints 4 to 6 in a full functioning spherical wrist • Wrist Joints function as a primary tool to ORIENT the end frame as required by the desired target POSE ME 3230

  15. Performing IKS: Focus on Positioning • We will define a point (called the WRIST CENTER) where all 3 z’s intersect as: • Pc = [Px, Py, Pz] • We find that this position is exactly: Pc = dtarget - dn*a • Px = dtarget,x - dn*ax • Py = dtarget,y - dn*ay • Pz = dtarget,z - dn*az Note: dn is the so called ‘Handspan’ (a CONSTANT) ME 3230

  16. Treating the Arm Types (as separate robot entities): • Cartesian (2 types) • Cantilevered • Gantry • Cylindrical • Spherical (w/o d2 offset) • 2 Link Articulating (w/o d2 offset) ME 3230

  17. Cantilevered Cartesian Robot J2 J3 X0 P-P-P Configuration Y0 Z0 J1 ME 3230

  18. Gantry Cartesian Robot P-P-P Configuration X0 Z0 Y0 J3 J2 J1 ME 3230

  19. Cylindrical Robot • Doing IKS – given a value for: • X, Y and Z of End • Compute , Z and R Z0 R Y0 X0 ME 3230

  20. Spherical Robot • Developing a IKS (model): • Given Xe, Ye, & Ze • Compute ,  & R  R Z0 Y0 X0 ME 3230

  21. 2-Link Articulating Arm Manipulator Z0 3 L2 L1 X0 Y0 2 1 ME 3230

  22. Focusing on the ARM Manipulatorsin terms of Pc: • Prismatic: • q1 = d1= Pz (its along Z0!) – cl1 • q2 = d2 = Px or Py - cl2 • q3 = d3= Py or Px - cl3 • Cylindrical: • 1 = Atan2(Px, Py) • d2 = Pz – cl2 • d3 = Px/C1 – cl3 {or +(Px2 + Py2).5 – cl3} ME 3230

  23. Focusing on the ARM Manipulatorsin terms of Pc: • Spherical: • 1 = Atan2(Px, Py) • 2 = Atan2( (Px2 + Py2).5 , Pz) • D3 = (Px2 + Py2 + Pz2).5 – cl3 ME 3230

  24. Focusing on the ARM Manipulatorsin terms of Pc: • Articulating: • 1 = Atan2(Px, Py) • 3 = Atan2(D, (1 – D2).5) • Where D = • 2 =  -  •  is: Atan2((Px2 + Py2).5, Pz) •  is: ME 3230

  25. Focusing on the ARM Manipulators in terms of Pc: • 2 = Where D = Atan2((Px2 + Py2).5, Pz) - ME 3230

  26. One Further Complication Must Be Considered: • This is called the d2 offset problem • A d2 offset is a problem that states that the nth frame has a non-zero offset along the Y0 axis as observed in the solution of the T0n with all joints at home • (like the 5 dof articulating robots in our lab!) • This leads to two solutions for 1 the So-Called “Shoulder Left” and “Shoulder Right” solutions ME 3230

  27. Defining the d2 Offset issue Here: ‘The ARM’ might contain a prismatic joint (as in the Stanford Arm – discussed in text) or it might be the a2 & a3 links in an Articulating Arm as it 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 1is at Home, the wrist center is not on the X0 axis!) ME 3230

  28. Lets look at this Device “From the Top” – a ‘plan’ view of the structure projected to the X0 Y0 plane F1  11 ME 3230

  29. Solving For 1: • We will have a Choice of two poses for 1: ME 3230

  30. This device (like the S110 robots in lab) is called a “Hard Arm” Solution • We have two 1’s • These lead to two 2’s (Spherical) • One for Shoulder Right & one for Shoulder Left • Or four 2’s and 3’s in the Articulating Arm: • Shoulder Right  Elbow Up & Down • Shoulder Left  Elbow Up & Down ME 3230

  31. Now, lets look at Orientation IKS’s • Orientation IKS again relies on separation of joint effects • We (now) know the first 3 joints control positions • they would have been solved by using the appropriate set of equations developed above • The last three (wrist joints) will control the achievement of our desired Orientation ME 3230

  32. These Ideas Lead to an Orientation Model for a Device that is Given by: • This ‘model’ separates Arm Joint and Wrist Joint Contribution to the desired Target Orientation • Note: target orientation is a ‘given’ for the IKS model! ME 3230

  33. Focusing on Orientation Issues • Lets begin by considering ‘Euler Angles’ (they are a model that is almost identical to a full functioning Spherical Wrist as defined using the D-H algorithm!): • Step 1, Form a Product: • Rz1*Ry2*Rz3 • This product becomesR36 in the model on the previous slide ME 3230

  34. Euler Wrist Simplified: this matrix, which contains the joint control angles, is then set equal to a ‘U matrix’ prepared by multiplying the inverse of the ARM joint orientation sub-matrices and the Desired (given) target orientation sub-matrix: NOTE: R03 is Manipulator dependent! (Inverse required a Transpose) ME 3230

  35. Continuing: to define the U-Matrix Rij are terms from the product of the first 3 Ai’s (rotational sub-matrix) ni, oi & ai are from given target orientation – we develop our models in a general way to allow computation of specific angles for specific cases ME 3230

  36. Now Set: LHS (euler angles) = RHS (U-Matrix) U as defined on the previous slide! (a function of arm-joint POSE and Desired End-Orientation) ME 3230

  37. Solving for Individual Orientation Angles (1st solve for ‘’ the middle one): • Selecting (3,3) C= U33 • With C we “know” S = (1 - C2).5 • Hence:  = Atan2(U33, (1-U332).5) • NOTE: leads to 2 solutions for ! ME 3230

  38. Re-examining the Matrices: • To solve for : Select terms: (1,3) & (2,3) • CS = U13 • SS = U23 • Dividing the 2nd by the 1st: S /C = U23/U13 • Tan() = U23/U13 •  = Atan2(U13, U23) ME 3230

  39. Continuing our Solution: • To solve for : Select terms: (3,1) & (3,2) • -SC = U31 • SS = U32 (and dividing this by previous) • Tan() = U32/-U31 (note sign migrates with term!) •  = Atan2(-U31, U32) ME 3230

  40. Summarizing: •  = Atan2(U33, (1-U332).5) •  = Atan2(U13, U23) •  = Atan2(-U31, U32) Uij’s as defined on the earlier slide! – and – U is manipulator and desired orientation dependent ME 3230

  41. Let consider a Spherical Wrist: Same Origin Point Here drawn in ‘Good Kinematic Home’ – for attachment to an Articulating Arm ME 3230

  42. IKSing the Spherical Wrist ME 3230

  43. Writing The Solution: Uij’s as defined on the earlier slide! – and – U is manipulator and desired orientation dependent ME 3230

  44. Let’s Solve(here I try the ‘Pure Inverse’ Technique): Note: R4’s Inverse ME 3230

  45. Simplifying ME 3230

  46. Solving: • Let’s select Term(3,3) on both sides: • 0 = C4U23 – S4U13 • S4U13 = C4U23 • Tan(4) = S4/C4 = U23/U13 • 4 = Atan2(U13, U23) • With the givens and back-substituted values (from the arm joints) we have a value for 4 andthe RHS is completely known! ME 3230

  47. Solving for 5 & 6 • For 5: Select (1,3) & (2,3) terms • S5 = C4U13 + S4U23 • C5 = U33 • Tan(5) = S5/C5 = (C4U13 + S4U23)/U33 • 5 = Atan2(U33, C4U13 + S4U23) • For 6: Select (3,1) & (3, 2) terms • S6 = C4U21 – S4U11 • C6 = C4U22 – S4U12 • Tan(6) = S6/C6 = ([C4U21 – S4U11]/[C4U22 – S4U12]) • 6 = Atan2 ([C4U22 – S4U12], [C4U21 – S4U11]) ME 3230

  48. Using the Pure Inversing: • We removed ambiguity from the solution • We were able to solve for joints “In Order” • Without noting – we see the obvious relationship between Spherical wrist and Euler Orientation! ME 3230

  49. Summarizing: • 4 = Atan2(U13, U23) • 5 = Atan2(U33, C4U13 + S4U23) • 6 = Atan2 ([C4U22 – S4U12], [C4U21 – S4U11]) ME 3230

  50. Lets Try One: • Cylindrical Robot w/ Spherical Wrist • Given would be a Target matrix from Robot Mapping! (it’s an IKS after all!) • The d3 “constant” is 400mm; the d6 offset (the ‘Hand Span’) is 150 mm. • 1 = Atan2((dx – ax*150),(dy-ay*150)) • d2 = (dz – az*150) • d3 = ((dx – ax*150)2+p(dy-ay*150)2).5 - 400 ME 3230

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