1. Intuitive Kinematics – Converting Between Forward and Reverse Definitions of Space Lecture Series 2 ME 4135 R. R. Lindeke

2. Intuitive Kinematics for Robot Manipulators  Defining the concept of the Kinematic Solution  Finding Kinematic Solutions for POSITIONAL issues (only!) Cartesian Manipulators Cylindrical Manipulators Spherical Manipulators Articulating Manipulators SCARA and Other Redundant Manipulators

3. Intuitive Kinematics for Robot Manipulators  Forward Kinematic Solutions: Given the settings on each Joint (q 1, q 2, , q i ) of the manipulator Determine the End Position (Xe,Ye, Ze) base achieved for the given structural size  Inverse Kinematic Solutions: Given Structural Size and an End Position (Xe,Ye, Ze) base Determine the values for each Joint (q 1, q 2, , q i ) that can place the Manipulator there Note, this solution MAY NOT BE UNIQUE!

4. Cantilevered Cartesian Robot P-P-P Configuration J1 J3 J2 X0X0 Y0Y0 Z0Z0

5. Gantry Cartesian Robot P-P-P Configuration Z0Z0 X0X0 Y0Y0 J1 J2 J3

6. FKS & IKS for a Cartesian Device  In the Forward Sense: On the Gantry J1 was at 435 cm J2 was at 283 cm J3 was at 199 cm and there is a collapse length of 75, 50, & 50 respectively  Where is the End in the ‘Null (base) Space’?

7. FKS -- Cartesian  Xe is J2 + C l1 = 283 + 50 = 333  Ye is 199 + 50 = 249  Ze is 435 + 75 = 510  Exercise Care in the ordering of the Joints as they relate to the Base Frame definition of Space!!!

8. Doing the IKS  Given you want the End Position of: (333, 249, 510) and the Collapse Lengths for the joints are as seen above (75, 50, & 50 respectively)  If J1 is in the Z 0 direction as before, J1 = 510 – 75 = 435 (as expected!) Similarly for J2 & J3 here they are 283 and 199 respectively – again as expected Here because of the directions the Joint motions and Base Axes are defined:  J1 in Z 0 ; J2 in X 0 ; J3 in Y 0 – but be careful as we move deeper into robotics!

9. Cylindrical Robot Work Envelope P-R-P//R-P-P Configuration

10. Cylindrical Robot  Developing a FKS (model):  Given , Z & R  Compute End Position in terms of X 0, Y 0 and Z 0 (Xe,Ye, Ze) R X0X0 Y0Y0 Z0Z0 

11. FKS for Cylindrical Manipulator

12. IKS for Cylindrical Manipulator  Here, before we go on let me make a statement about angular inverse solutions: Sine and Cosine inverses lead to ambiguous angles (they repeat each semi-circle) since they are built from a ratio of a signed over an unsigned vector (Y/R or X/R )  We MUST use inverse Tangent solutions to remove the ambiguity! The tangent is built from a ratio of signed vectors (Y/X)

13. Cylindrical Robot  Doing IKS – given a value for: X, Y and Z of End Compute , Z and R R X0X0 Y0Y0 Z0Z0

14. IKS Cylindrical Manipulator

15. Computing ATan2 Angles  Atan2(X value,Y value ) is a special form of Tan -1 but computed to retain quadrant identity  Consider the 4-cases of: X =  8 and Y =  12 Atan2(8,12) = Tan -1 (12/8) = 56.31   This is a 1 st Quadrant angle! Atan2(-8,12) = 90 + Tan -1 (8/12) = 90 + 33.69 = 123.69  (alternatively it is 180 - Tan -1 (12/8) = 180 – 56.31  = 123.69  )  This is a 2 nd quadrant angle and not the -56.31  value that you find with your calculator! Atan2(-8,-12) = 180 + Tan -1 (12/8) = 180 + 56.31  = 236.31   This is a 3 rd quadrant angle and not the 56.31  value your calculator gives you! Atan2(8,-12) = 270 + Tan -1 (8/12) = 270 + 33.69 = 303.69  (alternatively it is 360 - Tan -1 (12/8) = 360 – 56.31  = 303.69  )  This is a 4 th quadrant angle which is the same as your calculator gives you (-56.31  )!

16. Spherical Robot Workspace R-R-P Configuration

17. Spherical Robot  Developing a FKS (model):  Given ,  & R  Compute X e, Y e, & Z e R  X0X0 Y0Y0 Z0Z0

18. FKS Spherical Manipulator

19. Spherical Robot  Developing a IKS (model):  Given X e, Y e, & Z e  Compute ,  & R R  X0X0 Y0Y0 Z0Z0

20. IKS, Spherical Manipulator

21. 2-Link Articulating Arm Manipulator 33 11 22 L2L2 L1L1 This Machine Rotates about Z 0 Axis (  1 )  2 is measured RELATVE to the base plane  3 is measured RELATVE to L 1 not the base plane All Joint Angles are Right Hand Rule Based Z0Z0 X0X0 Y0Y0

22. FKS 2-Link A. ARM Manipulator

23. 2-Link Articulating Arm Manipulator 33 11 22 L2L2 L1L1 Z0Z0 X0X0 Y0Y0

24. IKS 2-Link A. Arm  All angles defined as ATan2 (using target End Position & Link Lengths)  Requires Construction Lines!!  First Solve for  1 = ATan2(Xe, Ye)  Then the Tilt Angles ( 3 &  2 ) in reversed Order (as shown)!  Solution Indicates 2 acceptable configurations for the Arm: ‘Elbow UP’ and ‘Elbow DOWN’

25. IKS 2-Link A. Arm

26. IKS     22   φ

27. IKS 2-Link Articulating Arm (solving for  3 )

28. IKS 2-Link Art. Arm ( 2 )

29. IKS 2-Link A. Arm ( 2 ) cont.

31. SCARA Manipulator – an over specified Planer Articulating Arm

32. FKS & IKS for this “Over-specified” Arm – Only 2 “State Equations” Exist for 3 Variables  In a Forward Sense solution is simple  In an Inverse sense the above statement indicates the existence of an infinite number of “good” solutions  FKS: Given  1,  2,  3 find X e and Y e  IKS: Given X e and Y e & L1, L2 and L3 find:  1,  2,  3

33. FKS 3-Link Planer A. Arm y1y1 y2y2 y3y3 x1x1 x2x2 x3x3   

34. Focusing on the FKS  Project each link to the two axes (X & Y)  X e is found by summing X-projected lengths of each link  Y e found by summing Y-projected lengths of the links  Example: Link 1 to X: L1*Cos( 1 ) Link 1 to Y: L1*Sin( 1 )

35. FKS Continuing  X2 is projection of L2 – to the X axis therefore it requires what Projection Factor?  Sure  Cos( 1 +  2 )  Y2 is projection of L2 to the Y axis so it is equal to: L2* Sin( 1 +  2 )  Finally:  Xe = L1*Cos(  1 ) + L2*Cos(  1 +  2 ) + L3*Cos(  1 +  2 +  3 )  Ye = L1*Sin(  1 ) + L2*Sin(  1 +  2 ) + L3*Sin(  1 +  2 +  3 )

36. What about the IKS  It’s a 2 Step Process  Requires a Parameterization of one of the Joint Angles – This step will establish the acceptable limits of the solution space  The Parameterized Joint is said to set the BOUNDS for the solution space  We select Joint 1 as the one to be parameterized

37. 3-Link Planer Arm IKS – Step 1 11 2’2’ Note: L2’ is the sum of L2 + L3 formed by freezing  3 at 0˚

38.  1 is Solved as Above  It is the lower angle of a 2-link Articulating Arm  The 2 solutions found for  1 thus form the Upper and Lower Bounds for the solution space  Now pick one within the range  Using this angular value, “Transform” the solution space up Link L1 and ‘unfreeze’  3

39. IKS Step 2:’Redefine’ Space at end of L1 and Re- apply 2-link method in the Transformed space  1,picked

40. Homework Assignment:  Complete the IKS solution for a 3-link planar articulating arm  Develop the FKS and IKS solution for a Planer P-R-P device