Lecture 23 The Spherical Bicycle II
(This is something of a guide to the Mathematica we will look at at the end
whichsopicks
The story
far: up where we left off last time.)
We’ve put together a model
applied simple holonomic constraints
and made a Lagrangian
We found the constraint matrix and the null space matrix
We “cheated” a little bit in assessing the rank of the constraint matrix
but I have some confidence that it is actually full rank as found
1
Here’s the picture to remind us
2
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
3
Next we need equations of motion — velocity and momentum
The velocity equations are the usual qÝi  S ij 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
4
The components of q are
q i  x1

y1 z1 1 1 1
x4
y4
z4  4 
T
The components of u are
Ý

1


Ý2 
u j  


Ý
 4 

5
The momentum equations are the usual reduced Hamilton equations
which we have from last time — I’ll spare the details, which are posted
 M ij j 1 M
 m
Srj n  i r s V i
n
i
mn
M S S uÝ   m Ss 
S
S

M
S
S
u
u

S

Q
S


s
r
ij
s
p
p
i
p
2 q i
q n 
q i

 q
j
ij k
i
p
k
which is of the form
Apk uÝk  Z prsur us  V˜p  Q˜ p

where
V
V˜p  i S pi
q


6
We have a total of 25 quite complicated equations and several tasks
Questions:
Is this system
system (infinitesimally,
(infinitesimally,linearly)
linearly)stable?
stable?
Can it be controlled?
Build a simulation:
Generalized forces
Numerical issues
7
Is this system (infinitesimally, linearly) stable?
The stability question requires:
an equilibrium position
linearized equations with no generalized forces
I seek an equilibrium for which the bicycle is erect
and moving in a straight line at a constant speed
 0 
 
j
u0   0 
 
 0 
8
What does this means for q0?



1  10, 1   , 2  10  , 2  , 3  10, 3 
2

2
2
 0 
 
j
suffices to satisfy the reduced Hamilton’s equations for u0   0 
 
 0 
We can deduce the rest of the equilibrium variables from the velocity equations

9
x i  rW  0 t cos 1, y i  rW  0 t sin 1, i 1 4
3   0 t, 4   0t
There are five variables that the differential equations don’t care about

zi  rW0t cos 1, i 1 4, 1
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
10
equilibrium and stability condensed from last time
We have an infinite set of equilibria

1  10, 1   , 1 fixed
2
x i  rW  0 t cos 10, y i  rW  0 t sin 10, zi fixed
 2  10 
 3  10,  3 

2
, 2   ,  2  

,  3   0 t,  4   0 t
2
u1  0  u 2 , u 3   0
When we get to numbers we’ll be looking at 10 



4
11
equilibrium and stability
Stability
stability here is a little tricky,
but we can follow the idea of stability by expanding around the equilibrium
q i  q0i  qi , u j  u0j  uj
The velocity equations

S3i k
qÝ   0 k q  S0i j uj
q
i

12
equilibrium and stability
For the u’ equations we can start with the symbolic homogeneous version
Apk uÝk  Z prsur us  V˜p
After considerable manipulation (see last lecture) we arrive at

 Zp 33 V˜p 
k
r
A0 pj uÝ   02

q

2

Z
u



0 0 pr3
q k
q k 

j

13
equilibrium and stability
We have a pair of linear vector equations that can be manipulated
S3i k
qÝ   0 k q  S0i j uj
q
i



 Zp 33 V˜p 
k
r
A0 pj uÝ   02

q

2

Z
u



0 0 pr3
q k
q k 

j
 2 Z p 33 V˜p  k
uÝ  A0 pj  0
 k q  2 0 A0 pj Z 0 pr3 ur
k
q 
 q
j
14
Now we can think about a state space picture of this
S3i k
qÝ   0 k q  S0i j uj
q
i
 2 Z p 33 V˜p  k
uÝ  A0 pj  0
 k q  2 0 A0 pj Z 0 pr3 ur
k
q 
 q

j




S3i
i
0 k
S0 j

qÝi  


q
xÝ  j   
x
˜



Z

V
uÝ  A  2 p 33  p  2 A Z 
0 pj
0
0 0 pj 0 pr3
k


q k 
  q

15
The block matrix elements have dimensions
22  22 22  3


 3  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  a2s2  a3s  a4  0
There are 21 zero roots, and four nonzero roots

16
The nonzero roots depend on the value of 0, but not 10
For the parameters of the present model there is:
a pair of complex conjugate roots with negative real parts
which grow rapidly with 0
one real negative root
one real positive root
the system is unstable!
17
The latter two roots vs. 0
unreal numerical glitch
18
Now we have an interesting question that we need to think about
What do the 21 zero roots represent?
This immediately leads us to another question
We have 22 generalized coordinates and nineteen constraints — a three DOF system
How many exponents do we expect for a three DOF system?
SIX
How can we reconcile this? Will that help with the first question?
19
If we look at the perturbation equations for q, we find that
q 3  z1, q 6  1, q 9  z2 , q11   2, q15  z3 , q 21  z4
do not vary in the linear limit

If we look at the linear equations for the evolution of q
we 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
20
q 4 
 5 
q 
q16 
x   17 
q 
u1 
 2 
u 
So we can define a reduced state
and write the state equations in the usual linear form

q 4 
 5 
q 
q16   2 
xÝ A  17  B 
q   4 
u1 
 2 
u 
21

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
 0.2753
0
0.2753
0
1.3269
0.0942 


0
0
0
0
1
0


 0.2753
0
0.2753
0
4.07435
0.8455 
A  

0
0
0
0
0.3420 
 0
0.0754 02 0.6444 0.0754 02 0.1631 0.1950 0 0.0964 0 


2
2
1.9928

2.2204
1.9928

2.8384
24.254

4.1868


0
0
0
0 
22
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 irrelevant
for 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 thought
there must be a comparable reduction
23
We now should be confident that we can restrict our linear analysis
including 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
24
I can write the rate of work in terms of the vector torques and rotations
I take the positive sense to be action on the receiving link
WÝ   4K 4   4  1   2K 2   2  1
I can calculate the generalized force in the usual way

WÝ
Qi  i
qÝ
These need to be reduced, because they contribute to the reduced Hamilton’s equations


WÝ j
˜
Qi  j Si
qÝ
25
This is quite monstrous expression in the general nonlinear situation
but it pretty simple in the linear situation, which is what I need for control
WÝ j
˜
Qi  j S0i  cos2 cot   sin 1  2 sin 2  2  2  4 
qÝ
These go into the reduced Hamilton equations,

but the equations we are dealing with have been solved for uÝj
So we need to do a few more things before we can findB and look at controllability
26
Substitute our equilibrium (we can do that at any time)
Multiply by the inverse of the matrix multiplying uÝj
Drop the last term, because u3 is not inthe reduced state
After this we find that the forcing depends only on 2
27

We have
 0.2753
0
0.2753

0
0
 0
 0.2753
0
0.2753
A  
0
0
 0
0.0754 02 0.6444 0.0754 02

2
2
1.9928 0 2.2204 1.9928 0
0
0
0
1.3269
1
4.07435
0
0
0.1631 0.1950 0
2.8384 24.254 0





0.3420 
0.0964 0 

4.1868 0 
0.0942
0
0.8455
 0 


0


 0 
B  

 0 
0.03598


0.42842
28
The linear problem turns out not to be controllable
the rank of W is five, not six
We can explore the consequences of this better in Mathematica, so . . .
29