ME 4135 Robotics & Control

1. ME 4135 Robotics & Control Slide Set 3 – Review of Matrix Methods Applicable to Robot Control

2. Creating a Rational Approach to Kinematics – A review of Matrix Methods • As the robots got more and more “Revolute” building Inertial models (FKS & IKS) was increasingly complicated • We would like a more logical approach to this problem • We will define a concept of Homogeneous Matrices in S–O3 Space to aid in this model building effort • Consider, we only built Positional DOF models to this point • We need both Position & Orientation models to drive real robots!

3. Considering Translation and Rotation • Translation, in a simple sense, is just the movement of one point from another without changing the orientation of space. • We can assign space frames (coordinate systems) to any object in space – (or all objects in space!). • If we wish to relate one object (and its space frame) to any other space frame we should be able to write a set of equations that represent each axis of the remote space in another’s systems axes and write a vector that relate the positions of the origins of the ‘systems’ to each other.

4. Defining Transformations – here lets consider translational transformations Lets say that we have a point P1 sitting at the origin of Frame1, and a second point A1 located at (2,7,3)1 tell me, What is the pose of the coordinate frame attached to A1 as described in the space Fame 0?

5. Translational Transformation • In physics we said to just add the two vectors (because the vector numbers are ‘the same’ since the axes point in the same directions) • So if A1 is at (2,7,3) in ‘1th Space’ then it is at: (2,7,3) + (12,35,45) = (2+12, 7+35, 3+45) = (14,42,48) in Null Space • But this (simple vector addition) • But this techniques only works for simple translation where space is not ‘reorientated’! • We must then Generalize the method (to me this ‘general approach’ is easier – but it seems more cumbersome when we start thinking this way!)

6. Defining the Homogeneous Transformation Matrix • It is a 4x4 Matrix that describes “3-Space” with information that relates Orientation and Position (pose) of a remote space to a local space nx ox ax dx ny oy ay dy nz oz az dz 0 0 0 1 This 3x3 ‘Sub-Matrix’ is the information that relates orientation of Framerem to Frame Local (This is called R the rotational Submatrix) D vector is the position of the origin of the remote space in Local Coordinate dimensions A vector projects the Zrem Axis to the Local Coordinate System N vector projects the Xrem Axis to the Local Coordinate System O vector projects the Yrem Axis to the Local Coordinate System

7. Defining the Homogeneous Transformation Matrix nx ox ax dx ny oy ay dy nz oz az dz 0 0 0 1 Scaling Factor Perspective or Projection Vector • This matrix is a transformation tool for space motion!

8. HTM – A Physical Interpretation • A representation of a Coordinate Transformation relating the coordinates of a point ‘P’ between 2 like-geometrid (-- ie SO3 --) different coordinate systems • A representation of the Position and Orientation (POSE) of a transformed coordinate frame in the “space” defined by a fixed coordinate frame • An OPERATION that takes a vector P and rotates and/or translates it to a new vector Ptin the same coordinate frame

9. Lets use it on our ‘Translational’ problem • What is the n vector here • Well, since X1 points in X0 direction, it is simply: (1,0,0) • Using the same reasoning: • The o vector is: (0,1,0) • And the a vector is: (0,0,1) • Here the d vector is: • The definition of the origin of Frame1 in Frame0 coordinates: (12,35,45)

10. Solving: • H. Transformation Matrix is: • Point A1 to ‘1 space’: 1 0 0 12 0 1 0 35 0 0 1 45 0 0 0 1 T01 = The solution of where A1 is in Frame0 is the product of these two matrices! 1 0 0 2 0 1 0 7 0 0 1 3 0 0 0 1 T1A =

11. Solution is given by: 1 0 0 14 0 1 0 42 0 0 1 48 0 0 0 1 T01· T1A = Hey, This works! (we got the same answer) -at least for this translational stuff!

12. What about Rotational Transformations? • Lets start with a “Pure Rotation” • A Pure Rotation is one about only 1 axis (a separable rotation) • We will consider this about Z0 (Initially) • This means: Rotate the ‘Remote’ Frame1 by an angle Q about the Z0 axis of ‘Local’ Frame0

13. After Rotation we find this Relationship

14. What is the Representation of P1 in Both Frames? • Assume the (identical) point is at (2,4,6)1 • And we had rotated Frame 1 by 37 degrees about the Z0 Axis • Where is the point as defined in Frame0? • We will employ the Method of Inner Products to find this.

15. By Inner Products: Relating these two definition of the SAME Point:

16. Collecting & Simplifying : Rewriting it in Matrix Form: Psst:This is a R matrix!

17. Converting it to a HTM Form (4x4) Vector of origin1 to orgin0 is (0,0,0) – they are the same point!

18. Lets See?  is 37deg and P1 is (2,4,6) • Cos = 0.799 • Sin = 0.602 • HTM is: • Model is:

19. Solving then: • XP0 = (row1 * P1) = .799*2-.601*4+0*6+0*1=-.806 • YP0 = (row2*P1) = .601*2+.799*4+0*6+0*1 = 4.398 • ZP0 = (row3*P1) = 0*2+0*4+1*6+0*1 = 6 • This is the same as we Observed!

20. What about Pure Rotation about X or Y Axes? • Uses the same Inner Product approach (Cosines of angles between vectors after rotation) • Trotx = • Troty =

21. Lets look at Another Issue! • Since we are in Matrix Math now, We remember that the “order of multiplication” matters • That is A*B  B*A (in general) • When we deal with physical space this is true as well. But it even offers one more added difficulty: • Did we take motion Relatively (space is redefined after an operation) or are all operations taken W.R.T. a fixed geometric space?

22. First Define two simple Operations: • Simple Translation of (4,0,0)A = • Simple Rotation of 90 about ZaxisB =

23. Now Define 2 Cases: • Case 1 is where we “redefine” Space after each operation • Case 2 is where all operations are taken against a fixed (inertia) space frame

24. Check Order issue: • 1st Translate then rotate • Its almost like drawing a cat!

25. Autocad Here! (Case 1:Trans – rot)

26. Given P2 (1,1,0)2 Where is it in Space0? • Lets Guess it is found by applying an (overall) Transformation given by:

27. Is (3,1,0,1)0 Equal to T20*(1,1,0,1)2?

28. Check Order issue: • 2nd – Rotate then Translate • Should be different physically! • Lets See

29. AutoCad Again (Case1: Rot-Trans)

30. Given P2 (1,1,0)2 Where is it in Space0? • Lets Guess it is found by applying an (overall) Transformation given by:

31. Is (-1,5,0,1)0 Equal to T20*(1,1,0,1)2?

32. Checking Case Two • Here we don’t redefine space between operations • That is, all operations are taken WRT a fixed coordinate system

33. Autocad: (Case2 Trans – Rot)

34. Looks Familiar! • The effect is just like the Rotate then Translate operational order when we were in Case 1 • Therefore, To get the Transformation Model, we must write: Trot*Ttrans • Yes, the order of multiplying is reversed from the order of operating!

35. Case 2: Rot - Translate • What about here? • Lets see what we get

36. Autocad: (Case 2, Rot-Trans)

37. Looks Familiar Too • This is just like what happened in Case 1 when we did Translate then Rotate • The overall effect here must be:Ttrans * Trot • Yes the order of multiplying is again reversed from the order of operating!

38. This can be Generalized • For Case 1 operations (space is redefined between each operation), the OVERALL EFFECT is found by taking the product of the operations in the order they are taken • For Case 2 operations (all operations taken W.R.T. a fixed Frame), the OVERALL EFFECT is found by forming the product of the individual operations taken in reverse order

39. This Does Matter! • Robotic Modeling is a Case 1 problem • Euler Orientation is a Case 1 problem • However, RPY Orientation is a Case 2 problem • Finally, Robot Mapping is (typically) a Case 1 problem • Lets look into robot mapping

40. One Last General Idea: Robot Mapping • This is an offline tool for finding Robot Targets (IKS targeting) • Moves Robot programming ‘Upstairs’ – to the engineering office • Relies on CAD models and geometry defined in increasingly complex spaces • Looks at chains of Transforms to define targets and robot tooling in common coordinate frames

41. Robot Mapping: Note: Drill not shown, (the Tool frame is actually located at the drill tip!) n 0 Tool Ho R P Ce Ta

42. Robot Mapping • The idea here is to match up the Tool’s geometric pose with the pose of a Target in our work space. • If we have a part that needs a hole drilled at a certain location, we must get the tool, carried by the robot, to this location (actually a point right above the drilled hole and also at the bottom of the drilled hole will be needed). • Remembering Dynamics, to equate poses, they must be defined in a common coordinate system.

43. Robot Mapping (cont.) • Typically, we would have a lot of geometric information about our working environment ‘just laying around’ • This data would be in the CAD drawings of parts and in CAD facilities plans of our cells and factories • Additionally the information is also provided in equipment drawings (tables, fixtures, even robots to some extent) • But, if we are going to talk about a robot, that machine is a series of adjustable joints and links that can be moved around (in an IKS sense) to put the tool accurately at the working positions we need. • The Necessary Pose is (within the robot): Tno

44. Robot Mapping (cont.) • Thinking about what we know: • We would know where we want a hole in a part (in part coordinates) or: TPHo • We likely want to place the part at a specific location on a table or in a fixture or: TTaP or TFP (known in process documentation) • The Table in the Cell (or fixture on a Table) TCeTa or (TTaF and TCeTa) would be known in our facility designs and/or process documentation

45. Robot Mapping (cont.) • Other thing we would know: • Were our robot is placed in the cell: TCeR from facility drawings • Were the robot Cartesian home frame is located in the Robot space: TR0 from equipment drawings • And finally where a tool is mounted to the end of the robot wrist: Tntool by measuring the tool and holder • What we would like to find is: T0n which contains the information about where the robot needs to ‘pose’ to do the hole drilling operation!

46. Robot Mapping (cont.) • Knowing this stuff, we should be able to generate a kinemetic chain (of HTM’s)– or map – that defines the hole in the cell: TCeTa*TTaP*TPHo • At the same time we can build akinematic chain for mapping theTool (the drill) back to the Cell too: TCER*TR0*T0n*Tntool • Now equate the two – they are defined in a common reference system – and isolate the desired information (T0n) • That is, extract the ‘unknown’ from the ‘known stuff’

47. Robot Mapping (cont.) • Equating the two kinematic chains: TCeTa*TTaP*TPHo = TCER*TR0*T0n*Tntool • Now, to isolate theT0n information (our desired pose for the robot), we must remove the “knowns” step-by-step on the RHS of this equation. • To do this we multiply by their inverses. BUT we must maintain order when we do it!

48. Robot Mapping (cont.) • 1st we move TCER(TCER)-1*TCeTa*TTaP*TPHo = TR0*T0n*Tntool • Here we have to ‘pre-multiply’ both the LHS & RHS by the inverse of the TCER matrix • Then we move TR0 by pre-multiplying by (TR0)-1 -- both sides(TR0)-1*(TCER)-1*TCeTa*TTaP*TPHo =T0n*Tntool

49. Robot Mapping (cont.) • Finally we isolate the T0n matrix by ‘post-multiplying’ both side by: (Tntool)-1 • (TR0)-1*(TCER)-1*TCeTa*TTaP*TPHo* (Tntool)-1 = T0n • This T0n would contain the data we need to solve the IKS equations for any Robot type!

50. Robot Mapping (cont.) • Taking the Inverse of a HTM is “EASY!!!” = • Easy because, physically, it is the same as defining the original ‘close’ frame in the ‘remote’ frame’s space (we have changed – or inversed – our point of view!) This is the Transpose of the R sub-matrix of the original HTM