1. 1 Lecture 24: Controlling the Uncontrollable Building the Unbuildable How do we deal with this uncontrollable system? Can we extend the method of Zs to make it work for the bicycle? We won’t finish the latter, and I’m not sure I know the answer although I think the answer is yes.

2. 2 We already know that the system is not controllable, but we don’t know what happens It is a single input system, and, were it controllable, we’d write and the closed reduced problem would become We would write a characteristic polynomial incorporating the gains and match the coefficients to the coefficients of a desired polynomial

3. 3 Here the characteristic polynomial we obtain has no constant term No matter what the gains, there is always a zero eigenvalue!

4. 4 We can move the other eigenvalues using the gains; choose a simple set of eigenvalues, all equal to -1 I find g[2] through g[6] in terms of g[1], and then set g[1] to zero (This choice fails us later on.)

5. 5 Here is a plot of the gains vs.  0

6. 6 We can now ask the question: what is the consequence of the zero eigenvalue? But first, let’s get a set of eigenvalues that will be useful the poles (as conjugate pairs)

7. 7 The gains associated with these look not that different from the previous ones the glitches are gains swapping roles

8. 8 I would love to substitute this into a simulation but we don’t have one, and are not likely to have one I can look at the linear solutions and see what the zero eigenvalue does there. The solution to the linear problem can be written eigenvalues eigenvectors

9. 9 The constants are determined by the initial conditions Recall that the state is

10. 10 The eigenvector associated with the zero eigenvalue has no 5, 6 components An initial condition that involves u, motion, will not involve this eigenvector-eigenvalue A general initial condition will lead to a long term nonzero result.

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14. 14 We can look at the Mathematica for this, and then come back here

15. The Spherical Bicycle: Extended Method of Zs 15 Let’s review the method of Zs as described in Lecture 19

16. 16 The game is to isolate the us, writing the derivatives as coefficients times us I will denote the coefficients in the momentum (Hamilton and reduced Hamilton) equations by Zs This is going to seem a wee bit weird Bear with me. It’ll make sense eventually We’ll have to do a lot more for the bicycle problem, but let’s start by trying to understand the process once more

17. We note that p i is linear in u j, which we can write as 17 We can obtain the Zs from the usual momentum expression or The latter is often easier

18. 18 It is clear from the expression that Z does not depend on u

19. 19 We already have the equations for the evolution of q: Hamilton’s equations become where any explicit We need to replace

20. 20 So that we have Hamilton’s equations in the form Remember that these are not actually correct because I have not written the Lagrange multipliers is the correct form

21. 21 We have learned to multiply by S to obtain the correct form the reduced Hamilton’s equations We need to solve these simultaneously with the q equations

22. 22 is, formally, a terribly complicated expression However, if we relegate the connectivity constraints to the pseudononholonomic world most of the derivatives of M are zero, and it is not at all bad

23. 23 The full formal expression for the reduced Hamilton’s equations This is a set of nonlinear first order ordinary differential equations in u the coefficients of which are functions of q, often quite complicated functions This would be a snap to integrate numerically were the coefficients to be constant The method of Zs pretends that they are

24. 24 I have adapted the method of Zs from Kane and Levinson’s 1983 paper, cited in the text Their work used Kane’s method, which I explore in the text, but not in class My method of Zs is not as complete as it might be, because I allow some of the coefficients to remain as explicit functions of q This relies onactually being pretty simple functions of q (and also the generalized forces) This is where we shall have to extend our thought processes

25. 25 Write Hamilton’s equations as follows We replace the Z variables by constants This choice of substitutions aligns with Hamilton’s equations, not the reduced equations

26. 26 Of course, they are not constants, so we have to add algebraic equations to our system to allow us to calculate them as we go forward Let me try to summarize this

27. 27 the simple holonomic constraints generalized coordinates Lagrangian nonholonomic and pseudononholonomic constraints Hamilton’s equations Start from square one introduce the Zs reduced Hamilton’s equations with Zs generalized forces

28. 28 Review the development, starting after application of the simple holonomic constraints

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30. 30 and the question arises, how many of these things do you want to consolidate? One option is to keep them all symbolic, for which I will want symbolic S, M, gradS, gradM Finally the (correct) reduced Hamilton equations

31. 31 Rather than introduce Zs at the beginning, let’s extend the idea to all the equations How many of these things do you want to consolidate? Can we work with just M and S and their gradients? S, M, gradS, gradM

32. 32 But first, note that we’ll want to do this for qdot as well because the qdot equations are also pretty messy We have S and we want its symbolic equivalent

33. 33 So we’ll have symbolic qdot equations

34. 34 We have 22 simple odes accompanied by 52 not very simple algebraic equations The reduced Hamilton equations are more complicated.

35. 35 One option is to build all the pieces of this from the four basic arrays

36. 36 We know S and M and can calculate the gradients directly The symbolic version (both of them in one block)

37. 37 We can do the same with S And what we need to do is cobble together the equations of motion, and the accompanying algebraic equations and I think I’ll look at this on the blackboard

38. 38 Here’s the equation we are building