1. ME 4135 Differential Motion and the Robot Jacobian
Slide Series 6
R. R. Lindeke, Ph.D.

2. Lets develop the differential Operator – bringing calculus to Robots
The Differential Operator is a way to account for “Tiny Motions” (T)
It can be used to study movement of the End Frame over a short time intervals (t)
It is a way to track and explain motion for different points of view

3. Considering motion: We can define a General Rotation of a vector K:
By a general matrix defined as:

4. These Rotation are given as:
But lets remember for our purposes that this angle is very small (a ‘tiny rotation’) in radians
If the angle is small we can have use some simplifications:
Cos small  1
Sin small  small

5. Substituting the Small angle Approximation:
Similarly for Y and Z:

6. Simplifying the Rotation Matrices (form their product):
Note here: we have neglected higher order products of the  terms!

7. What about Small (general) Translations?
We define it as a matrix:
General ‘Tiny Motion’ is then (including both Rot. and Translation):

8. So using this idea:
Let’s define a motion which is due to a robot’s joint(s) moving during a small time interval:
T+T = {Rot(K,d)*Trans(dx,dy,dz)}T
Consider Here: T is the original end frame pose
Substituting for the matrices:

9. Solving for the differential motion (T)

10. Factoring T (on the RHS)

11. Further Simplifying:
We will call this matrix the del operator: 

12. Thus, the Change in POSE (T or dT) is:
dT (T) = T
Where:  = {[Trans(dx,dy,dz)*Rot(K,d)] – I}
Thus we see that this  operator is analogous to the derivative operator d( )/dx but now taken with respect to HTM’s!

13. Lets look into an application:
Given:
Subject it to 2 simultaneous movements:
Along X0 (dx) by units (/unit time)
About Z0 a Rotation of 0.001rad (/unit time)

14. Graphically:  Here: init = Atan2(3,5) = 1.0304 rad R
Rinit = ( ) .5 = units
init = Atan2(3,5) = rad R 

15. Where is the Frame n after one time step?
Considering Position:
Effect of “Translation”:
X= and Y = 5.000
New Rf = ( ).5 = u
Effect of Rotation
fin = = rad
Therefore: Xf = Cos(fin) * Rf =
And: Yf = Sin(fin) * Rf =

16. Where is the Frame n after one time step?
Considering Orientation:

17. After 1 time step, Exact Pose is:

18. Lets Approximate it using this  operator
Tnew = Tinit + dT = Tinit + Tinit – the 1st law of differential calculus
Where:

19. Thus, Tnew is Approximately:

20. Comparing: “Exact”: Approximate:
Realistically these are all but equal but using the ‘del’ approximation, but finding it was much easier!

21. We can (might!) use the ‘del’ approach to move a robot in space:
Take a starting POSE (Torig) and a starting motion set (deltas in rotation and translation as function of unit times)
Form  operator for motion
Compute dT (Torig)
Form Tnew = Torig + dT
Repeat as time moves forward over n time steps

22. Taking Motion W.R.T. other Spaces (another use for this del operator idea)
Original Model (the motion we seek is defined in an inertial space):
dT = T (1)
However, if the motion is taken w.r.t. another (non-inertia) space:
dT = TT (2)
Here T implies motion w.r.t. itself – a moving frame – but could be motion w.r.t. any other non-inertia space (robot or camera, etc.)
Consider as well: the pose change (motion that is happening) itself (dT) is independent of point of view so, by equating (1) and (2) we can isolate T
T = (T)-1T

23. Solving for the specific Terms in T
Positional Change Vector w.r.t. (any) Tspace:
Angular effects wrt Tspace:
d, n, o & a vectors are extracts from the T Matrix
dp is the translation vector in 
 is the rotational effects in 

24. Subbing into a ‘del’ Form:

25. An Application of this issue:
TRpart
TCamPart
TWCR
If the Part is moving along a conveyor and we “measure” its motion in the Camera Frame (let the camera measure it at various times) and we would need to pick the part with the robot, we must track wrt to the robot, so we need part motion “del” in the robot’s space.

26. This is a Motion “Mapping” Issue:
Pa R WC Ca Pa Pa R  C Pa
Pa R WC Ca Pa
Knowns: C
Robot in WC
Camera in WC
And of course Part in Camera (But we don’t need it for now!)

27. Lets Isolate the “Middle”
R WC Ca R  C
R WC Ca
To solve for R we make progress from “R to R” directly (R) and “The long way around”:

28. Rewriting into a Standard Form:
It can be shown for 2 Matrices (A & B):
A-1*B = (B-1*A)-1 (1)
Or B-1*A = (A-1*B)-1 (2)
If, on the previous page we consider:
TWCCam as “A” and TWCR as “B”,
and define the form: (TwcCam)-1*TWCR as “T”
Then, Using the theorem (from matrix math) stated as (2) above “T”-1 is: (TWCR)-1* TWCCam

29. Continuing: Rewriting, we find that R = “T”-1(Cam) “T”
R is now shown in the “standard form” for non-inertial space motion
the terms: d, n, o & a vectors come from our ‘complex T’ matrix
the dp and  vectors can be extracted from the Cam
These term are required to define motion in the robot space
Of course the “T” is really: (TwcCam)-1*TWCR here!
Its from this “T” product that we extract n, o, a, d vectors

30. R is given by simplifying:
1st: (TwcCam)-1*TWCR = “T”
Then these Scalars:
WHERE:
d, n, o & a vectors are extracts from the “T” Matrix above
dp is the ‘translation’ vector in Cam
 is the vector of ‘rotational effects’ in Cam

31. Lets Examine the Jacobian Ideas
Fundamentally:

32. In This Model, Ddot & Dq,dot are:
We state, then, that the Jacobian is a mapping tool that relates Cartesian Velocities (of the n frame) to the movement of the individual Robot Joints

33. Lets build one from ‘1st Principles’ – Here is a Spherical Arm:
Lets start with only linear motion equations are straight forward!  

34. Writing the Position Models:
Z = R*Sin()
X = R*Cos()*Cos()
Y = R*Cos()*Sin()
To find velocity, differentiate these as seen here:

35. Writing it as a Matrix:
This is the Jacobian; It is built as the Matrix of partial joint contributions (coefficients of the velocity equations) to Velocity of the End Frame

36. Here we could develop an Inverse Jacobian:
It was formed by taking the partial derivatives of the IKS equations

37. We say the Jacobian is a 6 by n (6xn) matrix
The process we just did is limited to finding Linear Velocity! … and We need both linear and angular velocities for full functioned robots!
We can approach the problem by separations as we did in the General case of Inverse Kinematics –
Here we separate Velocity (Linear from Rotational), not Joints (Arms from Wrists)
Generally speaking, in the Jacobian we will obtain one Column for each Joint and 6 rows for a full velocity effect
We say the Jacobian is a 6 by n (6xn) matrix

38. Separation Leads to: A Cartesian Velocity Term V0n:
An Angular Velocity Term 0n:
Each of these “Ji’s” are 3 Row by n Columned Matrices

39. Building the “Sub-Jacobians”:
We follow 3 stipulations:
Velocities can only be added if they are defined in the same space – as we know from dynamics
Motion of the end effector (n frame) is taken w.r.t. the base space (0 frame)
Linear Velocity effects are physically separable from angular velocity effects
To address the problem we will consider moving a single joint at a time (using DH separation ideas!)

40. Lets start with the Angular Velocity (!)
Considering any joint i, its Axis of motion is: Zi-1 (Z in Frame i-1)
The (modeling) effect of a joint is to drive the very next frame (frame i)
If Joint i is revolute:
here k(i-1) is the Zi-1 direction (by definition)
This model is applied to each of the joints (revolute) in the machine (as it rotates the next frame, all subsequent fames, move similarly!)
But if the Joint is Prismatic, it has no angular effect on its “controlled” frame and thus no rotoation from it on all subsequent frames

41. Developing the (base) J
We need to add up each of the joint effects
Thus we need to “normalize” them to base space to do the sum
DH methods allow us to do this!
Since Zi-1 is the active direction in a Frame of the model, we really need only to extract the 3rd column of the product of A1 * …*Ai-1 to have a definition of Zi-1 in base space. Then, this Ai’s products 3rd column is the effect of Joint i as defined in the (common) base space (note, the ‘qdot’ term is the rate of rotation of the given joint)

42. So the Angular Velocity then is:
Note Zi-1 for Jointi – per DH algorithm!
As stated previously, Zi-1 is the 3rd col. of A1*…Ai-1 (rows 1,2 & 3). And we will have a term in the sum for each joint

43. Going Back to the Spherical Device:
Notice: 3 columns since we have 3 joints!
Here, Z1 depends on the Frame Skeleton drawn!

44. Building the Linear Jacobian
It too will depend on the motion associated with Zi-1
It too will require that we normalize each joints linear motion contribution to the base space
We will find that revolute and prismatic joints will make functionally different contributions to the solution (as if we would think otherwise!)
Prismatic joints are “Easy,” Revolutes are not!

45. Building the Linear Jacobian

46. Building the Linear Jacobian – for Prismatic Joints
When a prismatic jointi moves, all subsequent links move (linearly) at the same rate and in the same direction

47. Building the Linear Jacobian – Prismatic Joints
Therefore, for each prismatic joint of a machine, the contribution to the Jacobian (after normalizing) is:
Zi-1 which is the 3rd column of the matrix given by: A0 * … * Ai-1
This is as expected based on the model on the previous slide (and our “move only one and then normalize it” method)

48. Building the Linear Jacobian for Revolute Joints
This is a dicer problem,
but then, remembering the idea of prismatic joints on angular velocity …
But no that won’t work here – just because its a rotation, and it changes orientation of the end – revolute motion also does have a linear contribution effect to the motion of the end
This is a “levering effect” which moves the origin of the n-frame as we saw when discussing the del-operator on the -R structure.
We must compute and account for this effect and then normalize it too.

49. Building the Linear Jacobian – Revolute Joints
Using this model we would expect that a rotation i would ‘lever the end’ by an amount that is equivalent (in direction) to the CROSS product of the ‘driver’ vector and the ‘connector’ vector and with a magnitude equal to Joint velocity

50. Building the Linear Jacobian – Revolute Joints
This is the directional resultant (DR) vector given by:
Zi-1 X di-1n
[with Magnitude equal to joint speed!]
Note the “Green” Vector is equal to the ‘red’ DR vector!

51. Building the Linear Jacobian – Revolute Joints
Zi-1 X di-1n is the direction of the linear motion of the revolute joint i on n-Frame motion
It too must be normalized
Notice: di-1n = d0n – d0i-1 (call it eq. 3)
This “normalizes” the vector di-1n to the base space
But the d-vectors on the r.h.s. are really origin position of the various frames (Framei-1 and Framen) – i.e. the positions of frame “Origins”
So let’s rewrite equation 3 as: di-1n = On – Oi-1

52. Building the Linear Jacobian – Revolute Joints
The contribution to the Jv due to a revolute joint is then:
Zi-1 X (On – Oi-1)
Where:
Zi-1 is the 3rd col. of the T0i-1 (A1*… *Ai-1)
Oi-1 is 4th col. of the T0i-1 (A1*… *Ai-1)
On is 4th col. Of T0n (the FKS!)
NOTE when we pull the columns we only need the first 3 rows!

53. Building the Linear Jacobian
Summarizing:
The Jv is a 3-row by n columned matrix
Each column is given by joint type:
Revolute Joint: Zi-1 X (On – Oi-1)
Prismatic Joint: Zi-1
And notice: select Zi-1 and Oi-1 for the frame before the current joint column

54. Combining Both Halves of the Jacobian:
For Revolute Joints:
For Prismatic Joints:

55. What is the Form of the Jacobian?
Robot is: (PPRRRR) – a cylindrical machine with a spherical wrist:
Z0 is (0,0,1)T; O0 = (0,0,0)T always, always, always!
Zi-1’s and Oi-1’s are per the frame skeleton

56. Lets try this on the Spherical ARM we did earlier:2 1
The robot indicates this frame skeleton:

57. Lets try this on the Spherical ARM we did earlier:
LP Table: Fr
Link
Var  d a 
C 
S 
C 
S 
0→1 1 R 1 90 C1 S1
1→2 2
2+90
-S2 C2
2→n 3 P 0 d3

58. Lets try this on the Spherical ARM we did earlier:
Ai’s:

59. Lets try this on the Spherical ARM we did earlier:
T1 = A1!
T2 = A1 * A2
T0n = T3 = A1*A2*A3

60. Lets try this on the Spherical ARM we did earlier: THE JACOBIAN
The Jacobian is Of This Form:

61. Lets try this on the Spherical ARM we did earlier: THE JACOBIAN
Here:

62. After total Simplification, THE Full JACOBIAN is: