lecture 23 the spherical bicycle ii

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Lecture 23 The Spherical Bicycle II

The story so far:

We found the constraint matrix and the null space matrix

We “cheated” a little bit in assessing the rank of the constraint matrixbut I have some confidence that it is actually full rank as found

We’ve put together a modelapplied simple holonomic constraints

and made a Lagrangian

(This is something of a guide to the Mathematica we will look at at the endwhich picks up where we left off last time.)

2

Here’s the picture to remind us

3

A note on scaling

I chose the radius of the wheels to be my length scale,the mass of the wheel to be my mass scale,

and I can choose a time scale such that scaled g = 1

I can do all that without loss of generality, and it is possible to unscale for any real bicycle

The scaled dimensions for the bicycle we have are then

sphere radius and mass are 2 and 40, respectively

the wheel radius and mass are, of course, both unity

The fork is 4 units long, ¼ unit in diameter and has a mass of unity

4

Next we need equations of motion — velocity and momentum

The velocity equations are the usual

€

˙ q i = S ji u j

q has 22 components and u has three components

The components of S (22 x 3) are complicated functions of q

We identify the physical meaning of the components of u by examining the velocity equations

5

The components of q are

€

qi = x1 y1 z1 φ1 θ1 ψ1 L x4 y4 z4 ψ 4{ }T

The components of u are

€

u j =

˙ θ 1˙ ψ 2˙ ψ 4

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

6

The momentum equations are the usual reduced Hamilton equationswhich we have from last time — I’ll spare the details, which are posted

€

M ijSkjSp

i ˙ u k = −∂M ij

∂qm Ssj + 1

2∂Mmn

∂qi Ssn

⎛ ⎝ ⎜

⎞ ⎠ ⎟Sr

m − M ij∂Sr

j

∂qn Ssn

⎛

⎝ ⎜

⎞

⎠ ⎟Sp

i urus − ∂V∂qi Sp

i + QiSpi

€

Apk ˙ u k = Z prsurus − ˜ V p + ˜ Q p

which is of the form

where

€

˜ V p = ∂V∂qi Sp

i

7

We have a total of 25 quite complicated equations and several tasks

Is this system (infinitesimally, linearly) stable?

Can it be controlled?

Questions:

Build a simulation:

Generalized forces

Numerical issues


8


The stability question requires:

an equilibrium position

linearized equations with no generalized forces

I seek an equilibrium for which the bicycle is erectand moving in a straight line at a constant speed

€

u0j =

00

ω0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

9

What does this means for q0?

€

φ1 = φ10, θ1 = − π2

, φ2 = φ10 + π2

, θ2 = γ, φ3 = φ10, θ3 = π2

suffices to satisfy the reduced Hamilton’s equations for

€

u0j =

00

ω0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

We can deduce the rest of the equilibrium variables from the velocity equations

10

€

x i = rWω0t cosφ1, y i = rWω0t sinφ1, i =1L 4ψ 3 = −ω0t, ψ 4 = ω0t

There are five variables that the differential equations don’t care about

€

zi = rWω0t cosφ1, i =1L 4, ψ1

They are determined by the equilibrium, but do not enter the equations

The equilibria contain two constants — the direction of the straightline motion and its speed

11

We have an infinite set of equilibria

€

φ1 = φ10, θ1 = − π2

, ψ1 fixed

x i = rWω0t cosφ10, y i = rWω0t sinφ10, zi fixed

φ2 = φ10 + π2

, θ2 = γ, ψ 2 = π

φ3 = φ10, θ3 = π2

, ψ 3 = −ω0t, ψ 4 = ω0t

u1 = 0 = u2, u3 = ω0

equilibrium and stability condensed from last time

When we get to numbers we’ll be looking at

€

φ10 = π4

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Stability

stability here is a little tricky, but we can follow the idea of stability by expanding around the equilibrium

€

qi = q0i +ε ′ q i, u j = u0

j +ε ′ u j

equilibrium and stability

The velocity equations

€

′ ˙ q i = ω0∂S3

i

∂qk ′ q k + S0 ji ′ u j

13

€

Apk ˙ u k = Z prsurus − ˜ V p

For the u’ equations we can start with the symbolic homogeneous version


After considerable manipulation (see last lecture) we arrive at

€

A0 pj ′ ˙ u j = ω02 ∂ ′ Z p 33

∂qk −∂ ˜ V p∂qk

⎛

⎝ ⎜

⎞

⎠ ⎟ ′ q k + 2ω0Z0 pr3 ′ u r

14


We have a pair of linear vector equations that can be manipulated

€

A0 pj ′ ˙ u j = ω02 ∂ ′ Z p 33


⎛

⎝ ⎜

⎞

⎠ ⎟ ′ q k + 2ω0Z0 pr3 ′ u r€

′ ˙ q i = ω 0∂S3

i

∂qk ′ q k + S0 ji ′ u j

€

′ ˙ u j = A 0 pj ω02 ∂ ′ Z p 33


⎛

⎝ ⎜

⎞

⎠ ⎟ ′ q k + 2ω0A 0 pjZ0 pr3 ′ u r

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Now we can think about a state space picture of this

€

′ ˙ u j = A 0 pj ω02 ∂ ′ Z p33


⎛

⎝ ⎜

⎞

⎠ ⎟ ′ q k + 2ω0A 0 pjZ0 pr3 ′ u r

€

′ ˙ q i = ω0∂S3

i

∂qk ′ q k + S0 ji ′ u j

€

˙ x =′ ˙ q i

′ ˙ u j

⎧ ⎨ ⎩

⎫ ⎬ ⎭=

ω0∂S3

i

∂qk S0 ji

A 0 pj ω02 ∂ ′ Z p 33


⎛

⎝ ⎜

⎞

⎠ ⎟ 2ω0A 0 pjZ0 pr3

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

x

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The block matrix elements have dimensions

€

22 × 22 22 × 33 × 22 3 × 3 ⎧ ⎨ ⎩

⎫ ⎬ ⎭

The whole thing is a 25 x 25 matrix, and its eigenvalues determine the system stability

If everything I have done is correct, the characteristic polynomial is of the form

€

s21 s4 + a1s3 + a2s

2 + a3s + a4( ) = 0

There are 21 zero roots, and four nonzero roots

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The nonzero roots depend on the value of w0, but not f10

For the parameters of the present model there is:

a pair of complex conjugate roots with negative real partswhich grow rapidly with w0

one real negative root

one real positive root

the system is unstable!

18

The latter two roots vs. w0

unreal numerical glitch

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Now we have an interesting question that we need to think about

We have 22 generalized coordinates and nineteen constraints — a three DOF system

How many exponents do we expect for a three DOF system?

SIX

What do the 21 zero roots represent?

This immediately leads us to another question

How can we reconcile this? Will that help with the first question?

20

If we look at the perturbation equations for q, we find that

€

q3 = z1, q6 =ψ1, q

9 = z2, q11 = θ2, q

15 = z3, q21 = z4

do not vary in the linear limit

If we look at the linear equations for the evolution of qwe find that they depend only on q4, q16, u1 and u2

The linear equations for the evolution of u depend only on q4, q5, q16, q17, u1 and u2

There are only six variables that do anything interesting

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So we can define a reduced state

€

x =

q4

q5

q16

q17

u1

u2

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

and write the state equations in the usual linear form

€

˙ x = A

q4

q5

q16

q17

u1

u2

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

+ Bτ 2

τ 4

⎧ ⎨ ⎩

⎫ ⎬ ⎭

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We’ll need to look at the generalized forces to fill in what is happening

Let me defer that a bit to see how stability analysis works for this problem

We drop the generalized forces and look at the reduced matrix

€

A =

−0.2753 0 0.2753 0 1.3269 −0.09420 0 0 0 1 0

−0.2753 0 0.2753 0 4.07435 0.84550 0 0 0 0 0.3420

0.0754ω02 0.6444 0.0754ω0

2 0.1631 −0.1950ω0 0.0964ω0

1.9928ω02 2.2204 1.9928ω0

2 2.8384 −24.254ω0 −4.1868ω0

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

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This matrix is a 6 x 6. It has two zero eigenvalues.

Its nonzero eigenvalues are the same as the nonzero eigenvalues of the 25 x 25 matrix

The other 19 zero eigenvalues from the 25 x 25 problem are irrelevantfor the physics of the problem, which is a good thing

It’s worth noting that this simplification does not appear to extend to the full problem

I suspect that that is because I have not given it enough thoughtthere must be a comparable reduction

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We now should be confident that we can restrict our linear analysisincluding possible control to the reduce 6 x 6 system

The system is unstable — we’d like a control to stabilize it

This means that the time has come to work out the generalized forces

There are two torques, one applied to the fork and one applied to the rear wheel

Each has a reaction torque

25

I can write the rate of work in terms of the vector torques and rotations

€

˙ W = τ 4K 4 ⋅ ω4 − ω1( ) + τ 2K 2 ⋅ ω2 − ω1( )

I take the positive sense to be action on the receiving link

I can calculate the generalized force in the usual way

€

Qi = ∂ ˙ W ∂˙ q i

These need to be reduced, because they contribute to the reduced Hamilton’s equations

€

˜ Q i = ∂ ˙ W ∂˙ q j

Sij

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This is quite monstrous expression in the general nonlinear situationbut it pretty simple in the linear situation, which is what I need for control

€

˜ Q i = ∂ ˙ W ∂˙ q j

S0ij = cosθ2 cot γ + sin φ1 − φ2( )sinθ2( )τ 2 τ 2 τ 4{ }

These go into the reduced Hamilton equations, but the equations we are dealing with have been solved for

€

˙ u j

So we need to do a few more things before we can find B and look at controllability

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Substitute our equilibrium (we can do that at any time)

Multiply by the inverse of the matrix multiplying

€

˙ u j

Drop the last term, because u3 is not in the reduced state

After this we find that the forcing depends only on t2

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We have

€

B =

0000

0.035980.42842

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

€

A =

−0.2753 0 0.2753 0 1.3269 −0.09420 0 0 0 1 0

−0.2753 0 0.2753 0 4.07435 0.84550 0 0 0 0 0.3420

0.0754ω02 0.6444 0.0754ω0

2 0.1631 −0.1950ω0 0.0964ω0

1.9928ω02 2.2204 1.9928ω0

2 2.8384 −24.254ω0 −4.1868ω0

⎧

⎨

⎪ ⎪ ⎪

⎩

⎪ ⎪ ⎪

⎫

⎬

⎪ ⎪ ⎪

⎭

⎪ ⎪ ⎪

29

The linear problem turns out not to be controllablethe rank of W is five, not six

We can explore the consequences of this better in Mathematica, so . . .

lecture 23 the spherical bicycle ii

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