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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

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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

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WHAT 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?

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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

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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 todaybut it’s worth taking a look at them in a (partially) abstract sense

What can you tell me about this?

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For orientation I need body axes

Simple conventions:

Align the two wheel Ks with the axlestheir 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”

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Orientation constraints

€

K 2 = −sinγI1 − cosγJ1

The fork K axis points up and is tilted back an angle g

€

K 3 = J2

The front wheel K axis is aligned with the fork J axis

€

K 4 =K1

The rear wheel K axis is aligned with the frame K axis

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We learned how to align K axes with arbitrary directions (chapter 2 pp. 19-21)

Our target vectors are unit vectors and can be written directly as

€

V =

v1

v2

v3

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪=

cosα sinβ

sinα sinβ

cosβ

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

We see that cosb = v3 and tana = v2/v1.

a lies on (0, 2π) and b lies on (0, π)

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The K4 condition is trivial: f4 = f1 and q4 = q1

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

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Connectivity constraintswhich are best written in terms of vectors in the body frames

and we know all the necessary body axis vectors at this point

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Connect 1 to 2 and 1 to 4in the frame frame

Connect 2 to 3 in the fork frame

(The 1 and 2 CMs are nontrivial)

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Simple block model containing most of the dynamics

Frame is green, the fork is blue and the two wheels are red

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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

€

L = T −V

T denotes the kinetic energy and V the potential energy

€

T =1

2m ˙ x 2 + ˙ y 2 + ˙ z 2( ) +

1

2AΩX

2 +1

2BΩY

2 +1

2CΩZ

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Each link has six degrees of freedom before it is constrainedand its kinetic energy is

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Mechanisms operating in a gravitational field have a potential energy for each link

€

V = mg ⋅r

€

mgz, generally

Springs connect links, and so their potential energy involves more than one link

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

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The kinetic energy is the sum of the individual kinetic energiesand the potential energy is the sum of the gravitational energies

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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)

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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

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The rotations get defined at the top

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At this point I have built a candidate Lagrangiancontaining 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

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Holonomic constraints can also be divided into orientation and connectivity constraints

Orientation constraints are generally simpleconnectivity 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 coordinateswhether or not to apply them immediately

or convert them to pseudononholonomic constraints is a matter of judgment

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After you have applied whatever holonomic constraints you wishyou 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

€

L =1

2M ij qk

( ) ˙ q i ˙ q j −V qk( )

If there are springs or effective springs, their contributions are supposed to have been included

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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 equationsare 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

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Recall the rolling constraint

€

v =ω × r

In our most usual convention, where the wheel axle is the body axis K

€

˙ x

˙ y

˙ z

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪=

ωyrz −ωzry

ωxrz −ωxrz

ωxry −ωyrx

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

and

€

r = rW

−cosθ sinφ

cosθ cosφ

sinθ

⎧

⎨ ⎪

⎩ ⎪

⎫

⎬ ⎪

⎭ ⎪

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The z component of this is actually integrable

€

˙ z = rW cosθ ˙ θ ⇒ z = rW sinθ

although we generally carry it along

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We learned how to handle nonholonomic constraints using Lagrange multipliersbut 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

€

C ji qk( ) ˙ q j = 0

€

˙ q jThis 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

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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)

€

˙ q j = Sij qk( )ui

Once you have the appropriate null space vectors, put them in a matrix and write

where u is a new vector

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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

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Let’s move on and see where this can take us.

€

M ijSkj is a function only of the qs

€

˙ p i = M ijSkj ˙ u k +

d M ijSkj

( )

dtuk = M ijSk

j ˙ u k +∂ M ijSk

j( )

∂qm˙ q muk

€

˙ p i = M ijSkj ˙ u k +

∂ M ijSkj

( )

∂qmSn

munuk

€

˙ p i =∂L

∂qi+ λ jCi

j + Qithe momentum equations

€

pi = M ij ˙ q j = M ijSkjukthe momentum

combine in two steps

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€

˙ p i =∂L

∂qi+ λ jCi

j + Qi

€

M ijSkj ˙ u kSp

i = −∂ M ijSk

j( )

∂qmSn

munukSpi +

∂L

∂qiSp

i + λ jCijSp

i + QiSpi

And finally we can get rid of the Lagrange multipliers

0!

combine the momentum equation

€

˙ p i = M ijSkj ˙ u k +

∂ M ijSkj

( )

∂qmSn

munuk

with what we just did

€

M ijSkj ˙ u k = −

∂ M ijSkj

( )

∂qmSn

munuk +∂L

∂qi+ λ jCi

j + Qi

momentum equationin terms of u

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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€

M ijSkj ˙ u kSp

i = −∂ M ijSk

j( )

∂qmSn

munukSpi +

∂L

∂qiSp

i + QiSpi

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