1. Slide Set 3 – Review of Matrix Methods Applicable to Robot Control
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
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
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 Pt in 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
T01 =
The solution of where A1 is in Frame0 is the product of these two matrices!
T1A =

11. Solution is given by: T01· T1A = 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 Zaxis B =

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
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 a kinematic chain for mapping the Tool (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 the T0n 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

51. Now its your turn: Homework
Do Text problems from Chapter 2 in Spong & Vidyasagar: #5, 7, 8, 15, and 18
Approach problem 18 as a Mapping Problem