Lecture 27. Two Big Examples

1. Lecture 27. Two Big Examples The eleven link diver The Stanford Arm (Kane & Levinson)

2. The Eleven Link Diver Parts of the diver The joints neck C1 head shoulder wrist hand C7 upper arm T6 thorax elbow lower arm abdomen L1 pelvis hip thigh knee calf ankle foot

3. The torques can only control the differential motions, so (The first three denote the position and orientation of one reference link, here the abdomen.) Or, as we wrote them last time

5. Characterizing a dive CM path and rotation determined by initial conditions Figure of the diver is determined by the diver — the control torques The diver must find torques that will put the internal angles on specific paths in time

6. Control by feedback linearization, which we’ve explored and can review quickly Consider a second order system: We want q to follow a specific path: The equation for the error is Choose t such that rhs = (you can’t always do this) The error then obeys

7. To apply to our problem note: There are 13 equations for you don’t actually have to do this as we saw for the three link diver The torques appear only in equations 4-13 One can use equations 1-3 to eliminate the first three If then the generalized coordinates will go to their desired paths AND YOU CAN DO THIS

8. The next issue is computational I have 26 first order odes, some of which are very messy I like to see what I’m doing using Mathematica, but the big set is very very slow. I borrow an idea from Kane and Levinson (1983), which swaps complexity for size — the so-called method of Zs

9. The idea is to write the odes symbolically in terms of coefficients to be evaluated at each time step 26 odes PLUS 642 algebraic equations

10. Dives 10 meter platform

11. Recall the model 11 link diver Parts of the diver The joints neck C1 head shoulder wrist hand C7 upper arm T6 thorax elbow lower arm abdomen L1 pelvis hip thigh knee calf ankle foot (not to scale)

12. Cobbled together from various sources — he’s big

13. All dives start from an erect position diver on toes with arms fully extended upwards. Diver leaps up and to the left with positive rotation. Analysis starts once the diver has cleared the platform. t = dt t = 0

14. Initial conditions: position and velocity of the center of mass determines the path of the diver

15. initial angles and rotations yi = 0 qi = 0 determine the initial posture and angular momentum Desired qi, qdi determine the posture during the dive We want them to go to a specified value and then return to the initial value of p to ensure straight entry into the water. initial uniform rotation

16. Let’s look at a couple of examples, and then think about how to do this

17. Mock jackknife

20. Mock tuck

23. We can look at the Mathematica code and at “movies” of these dives

24. The Stanford Arm: Kane & Levinson Revisited This is a difficult problem The difficulty lies in figuring out how to represent it Once you have a decent model, it becomes easy (if possibly tedious) What are the angular velocities of the links? What are the generalized forces? We will roll up our sleeves and tackle it.

25. Note that blindly following the recipes I have given you won’t work or at least won’t work easily You need to go back to the ideas that lie behind the recipes So, here we go . . .

26. There are six links, hence 36 possible coordinates. There are only six degrees of freedom so we have to get rid of 30 coordinates I don’t like their coordinate system, so let me change the notation

27. They letter their links, and I want to number mine so link A is link 1, link B is link 2, etc I want to arrange things such that all the rotations are about K axes This is a holonomic system, but some of the connectivity constraints are awful Finally we have one prismatic (sliding) joint, which is something we haven’t seen

28. I have a seven simple holonomic constraints Choose the origin at KL A*, then The third link, C does not rotate about its own K axis

29. Link A rotates about the vertical so make that K1 and let I1 be parallel to Link B rotates about I1, so let that be K2 q1 will be y1 q2 will be y2

30. Link C moves parallel to Link D rotates about that direction Let K4 = K3 q3 = y3 Let that be J2 and equal to K3

31. Finally, Link F rotates about which we can call I5 and set equal to K6 q5 = y6 Link E rotates about which we can call I4 and set it equal to K5 q4 = y5

32. We can summarize all of this as follows functions of y1 functions of y1, y2 functions of y1, y2 functions of y1, y2, y4 functions of y1, y2, y4, y5 functions of y1, y2, y4, y5, y6

33. Now, how do we incorporate all of this into our usual systems/thoughts? First note that we can finesse the usual f and q variables out of the picture We can calculate the angular velocities and momenta in body coordinates without them Thus we can write the rotational kinetic energy without them This is a little bit tricky, and we need to think as we go forward

34. Each link can be represented by a vector referred to its body axes For all but link 3 (link C) the XX, YY and ZZ coefficients are constant (and their values are irrelevant to the argument!) The angular velocity is represented by the rates of change of the base vectors where the last term is zero except for link 3

35. We can write any vector as The angular velocity term is contained in the change of base vectors In our case, then

36. Our job is then to extract the components of wB from this equation Write Take the body components of the right hand vector The ZZ component of the first one gives me wBY TheXX component of the second one gives me wBZ TheYYcomponent of the third one gives me wBX The values of XX and YY and ZZ are irrelevant!

37. Here it is in Mathematica

38. We need the angular velocities in inertial coordinates for the generalized forces The base vectors are, of course, defined with respect to the inertial system, so these are indeed exactly what we want

39. Where do we now stand on variables? We need a kinetic energy and a potential energy The rotational kinetic energy depends only on the five ys we have defined The translational kinetic energy depends on the eighteen center of mass coordinates But we’ve already eliminated four of those So we have fourteen CM coordinates, five ys and the extension coordinate of link 3

40. So we have twenty coordinates and only six degrees of freedom The CM coordinates are all connected by fourteen connectivity constraints They are pretty immense, so I put them into a constraint matrix and go with pseudononholonomic constraints Before I do that, let us look at the generalized coordinates

41. the extension of link 3 The variables I care about are the last six

42. The fourteen constraints can be written

43. Differentiate the constraints, form the usual constraint matrix, and find a null space matrix The u vector will have six components, and with the proper choice of S these will be the rates of change of the important variables: yi and q6 The only thing we need now to complete the problem is the generalized forces

44. acts on link 1 acts on links 2 and 1 (-) acts on links 3 and 2 (-) acts on links 4 and 3 (-) acts on links 5 and 4 (-) acts on links 6 and 5 (-)

45. So we have where All the angular velocities here are expressed in the inertial system

46. This leads us to a set of generalized forces, the first fourteen of which are zero The nonzero terms are

47. Now we have everything we need to set up the equations and simulate the robot So, let’s go do it!