1. The L-E (Torque) Dynamical Model: Inertial Forces Coriolis & Centrifugal Forces Gravitational Forces Frictional Forces

2. Lets Apply the Technique -- Lets do it for a 2- Link “Manipulator” Link 1 has a Mass of m1; Link 2 a mass of m2

3. Before Starting lets define an L-E Algorithm: Step 1 Apply D-H Algorithm to build Ai matrices and find F i the “link frame” Step 2Set T 0 0 = I ; i=1; D(q)=0 Step 3 Find  c i the Homogenous coordinate of the center of mass of link i WRT F i Step 4 Set F c as the translation of Frame F i to C M of i Compute Inertia Tensor D i bar about C M wrt F c Step 5Compute: z i-1 (q); T 0 i ; c i bar (q); D i (q)

4. Before Starting lets define a L-E Algorithm: Step 6Compute Special Case of J i (q) Step 7 Partition J i and compute D(q) = D(q) i-1 + {A T m i A + B T D i B} Step 8 Set i = i+1 go to step 3 or if i=n+1 then set i=1 & continue Step 9 Compute C i (q) -- Vel. Coupling matrix; h i (q) – gravity loading vector; and friction i Step 10Formulate Torque i equation Step 11Advance “i” go to step 9 until i>n

5. We Start with Ai’s Not Exactly D-H Legal (unless there is more to the robot than these 2 links!)

6. So Lets find T 0 2 T 0 2 = A 1 *A 2

7. I’ll Compute Similar Terms back – to – back rather than by the Algorithm

8. c 2(bar) Computation:

9. Finding D i Considers each link a thin cylindrical bar These are Inertial Tensors with respect to a F c aligned with the link Frames i at the C m

10. Continuing for Link 1

11. Simplifying:

12. Continuing for Link 2

13. Now let’s compute the Jacobians

14. Finishing J 1 Note the 2nd column is all zeros – even though Joint 2 is revolute – this is the special case!

15. Jumping into J 2

16. Continuing:

17. And Again:

18. Summarizing J 2 is:

19. Developing the D(q) Contributions D(q) I = (A i ) T m i A i + (B i ) T D i B i A i is the “Upper half” of the J i matrix B i is the “Lower Half” of the J i matrix D i is the Inertial Tensor of Link i defined in the Base space

20. Building D 1 D(q) 1 = (A 1 ) T m 1 A 1 + (B 1 ) T D 1 B 1 Here:

21. Looking at the 1 st Term (Linear Velocity term)

22. Looking at the 2 nd Term (Angular Velocity term) Recall that D 1 is: Then:

23. Putting the 2 terms together, D(q) 1 is: 2x2 because it is a 2-jointed robot!!

24. Building the Full Manipulator D(q) D(q) 2 = D(q) 1 +(A 2 ) T m 2 A 2 + (B 2 ) T D 2 B 2 Recalling (from J 2 ):

25. Building the ‘2 nd’ Term (the Linear Vel. part from the second link):

26. Specifics on term (1,1) ~ details!

27. Building the 3 rd Term: Recall D 2 : Then:

28. Combining the 3 Terms to construct the Full D(q) Manipulator Inertial Tensor “Solution”: 2x2 because it is 2 Joints (still)!

29. Simplifying then, the Manipulator inertial Tensor D(q) is: NOTE: D(q) man is Square in number of Joints! 2x2 here because this is a 2-jointed robot!!

30. This Completes the Fundamental Steps: Now we compute the Velocity Coupling Matrix and Gravitation terms:

31. For the 1 st Link

32. Plugging ‘n Chugging (1 st link) From Earlier:

33. P & C cont – 1 st Link:

34. Finding h 1 : Given: gravity vector points in –Y 0 direction g k =(0, -g 0, 0) T g 0 is gravitational constant In the h model A ki j is extracted from the relevant Jacobian matrix Here: 3-directions of Cartesian base space! Here, i=1

35. Continuing (i =1): From Jacobian j Upper half – specified components Gravitational effect: Would be 3 terms (BUT) only K = 2 survives since gravitational vector points in –Y 0 direction

36. Stepping to Link 2 (i = 2)

37. Computing h 2 Would be 3 terms (BUT) only K = 2 survives as gravitational vector points in –Y 0 direction

38. Building “Torque” Models for each Link In General:

39. For Link 1 (i=1): The 1 st terms: 2 nd Terms:

40. Writing the Complete Link 1 Model

41. And, Finally, For Link 2 (i=2):

42. Initial Link 2 terms: ! st Terms: 2 nd Terms:

43. Finalizing Link 2 Torque Model: