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Lecture 24: Controlling the UncontrollableBuilding 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 answeralthough I think the answer is yes.

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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

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τ 2 = −G ⋅x

and the closed reduced problem would become

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˙ x = A − B ⋅G( ) ⋅x

We would write a characteristic polynomial incorporating the gainsand match the coefficients to the coefficients of a desired polynomial

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Here the characteristic polynomial we obtain has no constant term

No matter what the gains, there is always a zero eigenvalue!

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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.)

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Here is a plot of the gains vs. w0

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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)

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The gains associated with these look not that different from the previous ones

the glitches are gains swapping roles

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I would love to substitute this into a simulationbut 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

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x = c iVi exp λ it( )i=1

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∑

eigenvalues

eigenvectors

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The constants are determined by the initial conditions

Recall that the state is

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x =

q4

q5

q16

q17

u1

u2

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

=

φ1

θ1

φ3

θ3

˙ θ 1˙ ψ 2

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

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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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12

13

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

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The Spherical Bicycle: Extended Method of Zs

Let’s review the method of Zs as described in Lecture 19

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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 weirdBear 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

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We note that pi is linear in uj, which we can write as

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pi = Z ijuj

We can obtain the Zs from the usual momentum expression

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pi = M ij ˙ q j = M ik ˙ q k = M ikS jku j ⇒ Z ij = M ikS j

k

or

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Zij = ∂pi

∂u j

The latter is often easier

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€

Z ij = M ikS jk

It is clear from the expression

that Z does not depend on u

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We already have the equations for the evolution of q:

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˙ q i = S ji u j

Hamilton’s equations become

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˙ p i = Z ij ˙ u j + ˙ Z ijuj = ∂L

∂qi + Qi

where any explicit

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˙ q i = S ji u j

We need to replace

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˙ Z ij =∂Z ij

∂qk ˙ q k =∂Z ij

∂qk Smk um

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So that we have Hamilton’s equations in the form

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Zij ˙ u j +∂Z ij

∂qk Smk umu j = ∂L

∂qi + Qi

Remember that these are not actually correct because I have not written the Lagrange multipliers

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Z ij ˙ u j +∂Z ij

∂qk Smk umu j = ∂L

∂qi + Qi + λ kCik

is the correct form

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We have learned to multiply by S to obtain the correct formthe reduced Hamilton’s equations

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Zij ˙ u jSni +

∂Zij

∂qk Smk umu jSn

i = ∂L∂qi Sn

i + QiSni

We need to solve these simultaneously with the q equations

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˙ q i = S ji u j

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€

∂L∂qi is, formally, a terribly complicated expression

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∂L∂qi = 1

2˙ q p

∂M pq

∂qi ˙ q q − ∂V∂qi

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∂L∂qi = 1

2Sm

p um ∂M pq

∂qi Snqun − ∂V

∂qi

However, if we relegate the connectivity constraints to the pseudononholonomic worldmost of the derivatives of M are zero, and it is not at all bad

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€

Z ij ˙ u jSni +

∂Z ij

∂qk Smk umu jSn

i = 12

Smp um ∂M pq

∂qi Snqun − ∂V

∂qi

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sn

i + QiSni

The full formal expression for the reduced Hamilton’s equations

This is a set of nonlinear first order ordinary differential equations in uthe 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

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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 on

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∂L∂qi and Si

j actually being pretty simple functions of q

(and also the generalized forces)

This is where we shall have to extend our thought processes

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€

ZijSni ˙ u j +

∂Zij

∂qk Smk Sn

i umu j = ∂L∂qi Sn

i + QiSni

Write Hamilton’s equations as follows

We replace the Z variables by constants

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Z ij → Tij , ∂Z ij

∂qk Smk → Tijm

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TijSni ˙ u j + TijmSn

i umu j = ∂L∂qi Sn

i + QiSni

This choice of substitutions aligns with Hamilton’s equations, not the reduced equations

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Of course, they are not constants, so we have to add algebraic equationsto our system to allow us to calculate them as we go forward

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Tij = Zij qk( ), Tijm =∂Z ij qp( )

∂qk Smk qq( )

Let me try to summarize this

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the simple holonomic constraints

generalized coordinates

Lagrangian

nonholonomic and pseudononholonomic constraints

Hamilton’s equations

Start from square one

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˙ q i = S ji u j

introduce the Zs

reduced Hamilton’s equations with Zs

generalized forces

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

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L = 12

M ij ˙ q i ˙ q j −V qk( )

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pi = ∂L∂˙ q i

= M ij ˙ q j

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C ji ˙ q j = 0

€

˙ q j = Skjuk

€

SkjCr

k = 0[ ]r

j

€

pi = M ijSkjuk

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€

˙ p i = ∂L∂qi + Qi

€

Qi = ∂ ˙ W ∂˙ q i

€

∂L∂qi = 1

2∂Mmn

∂qi SrmSs

numun − ∂V∂qi

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˙ p i = M ijSkj ˙ u k + M ij

˙ S kj ˙ u k + ˙ M ijSk

j ˙ u k

€

˙ p i = M ijSkj ˙ u k + M ij

∂Skj

∂qm Srmuruk +

∂M ij

∂qm SkjSr

muruk

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€

M ijSkj ˙ u k + M ij

∂Skj

∂qm Srmuruk +

∂M ij

∂qm SkjSr

muruk = 12

∂Mmn

∂qi SrmSk

nuruk − ∂V∂qi + Qi

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M ijSkj ˙ u k = −M ij

∂Skj

∂qm −∂M ij

∂qm Skj + 1

2∂Mmn

∂qi Skn

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sr

muruk − ∂V∂qi + Qi

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

€

M ijSkjSp

i ˙ u k = −M ij∂Sk

j

∂qm −∂M ij

∂qm Skj + 1

2∂Mmn

∂qi Skn

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sr

mSpi uruk − ∂V

∂qi Spi + QiSp

i

Finally the (correct) reduced Hamilton equations

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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

€

M ijSkjSp

i ˙ u k = −M ij∂Sk

j

∂qm −∂M ij

∂qm Skj + 1

2∂Mmn

∂qi Skn

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sr

mSpi uruk − ∂V

∂qi Spi + QiSp

i

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But first, note that we’ll want to do this for qdot as wellbecause the qdot equations are also pretty messy

We have S and we want its symbolic equivalent

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So we’ll have symbolic qdot equations

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We have 22 simple odesaccompanied by 52 not very simple algebraic equations

The reduced Hamilton equations are more complicated.

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€

M ijSkjSp

i ˙ u k = −M ij∂Sk

j

∂qm −∂M ij

∂qm Skj + 1

2∂Mmn

∂qi Skn

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sr

mSpi uruk − ∂V

∂qi Spi + QiSp

i

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

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Spi , ∂Sk

j

∂qm , M ij , ∂Mmn

∂qi

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We know S and M and can calculate the gradients directly

The symbolic version (both of them in one block)

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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

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€

M ijSkjSp

i ˙ u k = −M ij∂Sk

j

∂qm −∂M ij

∂qm Skj + 1

2∂Mmn

∂qi Skn

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sr

mSpi uruk − ∂V

∂qi Spi + QiSp

i

Here’s the equation we are building