Lecture 20: Putting it all together An actual mechanism A physical abstraction: links, springs, wheels, motors, etc. A mathematical model of the physical abstraction Stability and control of the mathematical model A simulation built on the mathematical model 1

The physical abstraction all the elements (links) must be rigid if something is flexible in the actual mechanism approximate the motion by a spring or springs (this is akin to finite element modeling) Low inertia elements (like drive belts) can be replaced by constraints 2

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4 W HAT DO WE DO WITH THIS SYSTEM ? How many links? What are they? What do we have to keep and what can we let go? How are they related?

5 The big one is how to deal with the rider We can neglect the chain and the derailleur details The simple choice is to neglect all motion of the rider and lump him in with the frame More complicated would be to model each leg/pedal set as a planar linkage For now let’s do the simple thing

6 Simple thing = four links: frame with rider, fork assembly, front wheel, rear wheel How do we connect them? There’re both orientations and connections to be found/defined I haven’t talked about constraints yet today but it’s worth taking a look at them in a (partially) abstract sense What can you tell me about this?

7 For orientation I need body axes Simple conventions: Align the two wheel Ks with the axles their I and J axes rotate Align the I frame axis with the top tube Let the J frame axis point down Align the fork K axis with its long axis Its K axis then points into the screen Let its I axis point “forward” 3 4

8 Orientation constraints The fork K axis points up and is tilted back an angle  The front wheel K axis is aligned with the fork J axis The rear wheel K axis is aligned with the frame K axis

9 We learned how to align K axes with arbitrary directions (chapter 2 pp ) Our target vectors are unit vectors and can be written directly as We see that cos  = v 3 and tan  = v 2 /v 1.  lies on (0, 2π) and  lies on (0, π)

10 The K 4 condition is trivial:  4 =  1 and  4 =  1 The others require us to use the formalism, which is sufficiently complicated that it warrants Mathematica I will address this after I have done some more general comments

11 Connectivity constraints which are best written in terms of vectors in the body frames and we know all the necessary body axis vectors at this point Connect 1 to 2 and 1 to 4 in the frame frame Connect 2 to 3 in the fork frame (The 1 and 2 CMs are nontrivial)

12 Simple block model containing most of the dynamics Frame is green, the fork is blue and the two wheels are red

13 There will also be rolling constraints, and these will also attach the bike to the ground

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The mathematical model is built on the foundation of a Lagrangian T denotes the kinetic energy and V the potential energy Each link has six degrees of freedom before it is constrained and its kinetic energy is 15

Mechanisms operating in a gravitational field have a potential energy for each link, generally Springs connect links, and so their potential energy involves more than one link 16 My standard boilerplate does not include springs; if you have them, you need to add them Look at parts of the standard boilerplate and see what the pieces mean

17 The kinetic energy is the sum of the individual kinetic energies and the potential energy is the sum of the gravitational energies

18 A big piece of the boilerplate is devoted to finding the angular velocity in body coordinates (and in inertial coordinates, not needed for this set up)

19 The angles in these formulas are the Euler angles, and we spent a lot of time on them They a set of axes fixed in the body to the inertial frame — there’s boilerplate for that, too

20 The rotations get defined at the top

21 At this point I have built a candidate Lagrangian containing six times as many variables as links There is as yet no mechanism model because the links are not connected We connect them (and otherwise constrain their motion) with constraints We consider three types of constraints: simple holonomic, nonsimple holonomic and nonholonomic constraints The only nonholonomic constraints we have considered seriously are rolling constraints: If your mechanism has wheels, you will have to deal with nonholonomic constraints

22 Holonomic constraints can also be divided into orientation and connectivity constraints Orientation constraints are generally simple connectivity constraints are generally nonsimple Simple holonomic constraints are linear relations among the coordinates You should always apply these as soon as you have figured them out Nonsimple holonomic constraints are nonlinear relations among the coordinates whether or not to apply them immediately or convert them to pseudononholonomic constraints is a matter of judgment

23 After you have applied whatever holonomic constraints you wish you can assign the remaining coordinates to a set of generalized coordinates You now have a Lagrangian in terms of the coordinates you will use for the rest of the analysis If there are springs or effective springs, their contributions are supposed to have been included

24 You have still to deal with the nonholonomic constraints and any external forces I include in the nonholonomic constraints any pseudononholonomic constraints I have added If there are no nonholonomic constraints, then the Euler-Lagrange equations are a practical way of approaching the problem (and they work even for systems constrained nonholonomically) The Euler-Lagrange equations are excellent for analyzing stability and designing controls For brute force calculations and simulations I prefer Hamilton’s equations and their variants

25 Recall the rolling constraint In our most usual convention, where the wheel axle is the body axis K and

26 The z component of this is actually integrable although we generally carry it along

27 We learned how to handle nonholonomic constraints using Lagrange multipliers but this was quite clumsy and nonintuitive, so I suggested what I consider a better method The nonholonomic constraints are a set of homogeneous linear equations in the derivatives of the coordinates As such, they can be written in matrix form This says that any vector must be orthogonal to all the rows of C In other words: it must belong to the null space of C

28 There are some technical issues The null space must be complete, which is taken care of if the constraints are independent You should check this by checking the rank of C. If it is not of full rank, you need to perform a row reduction and go forward with a reduced set of nonholonomic constraints (we have seen this) Once you have the appropriate null space vectors, put them in a matrix and write where u is a new vector

29 This sort of locks you into the Hamilton approach We know how q evolves and we can replace its derivative in terms of u in Hamilton’s equations This is easier to do operationally than symbolically, but let’s review the symbolism from Lecture 12

30 Let’s move on and see where this can take us. is a function only of the qs the momentum equations the momentum combine in two steps

31 And finally we can get rid of the Lagrange multipliers 0! combine the momentum equationwith what we just did momentum equation in terms of u

32 Before multiplying by S the free index was i, which runs from 1 to N After multiplying by S, the free index becomes p which runs from 1 to K We have reduced the number of momentum equations to the number of degrees of freedom and gotten rid of the Lagrange multipliers. As usual, it is not as ghastly in practice as it looks in its full generality

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