Segway - inverse pendulum

and what math tools you need to simulate it

January 9, 2019

Motivation

Segway

I A Segway PT - short for Segway Personal Transporter - is a two-wheeled,self-balancing personal transporter.

Source: http://de-de.segway.com/

I How does it balance itself?

Motivation

Goals

I We want to understand the function of the Segway by simulating the simplifiedmodel of an inverse pendulum on a moving platform:

I Throughout the presentation we will pay particular attention to the math toolsneeded for the derivation and simulation of the model.

Model

I The model, as depicted in Das Mathe-Praxis-Buch, J. Harterich, A. Rooch, 2014,Springer Verlag:

Source: Das Mathe-Praxis-Buch, p. 4, Abb. 1.1

I With M mass of the cart, m mass of the pendulum, l length of the pendulum, αangular displacement of the pendulum, g gravitional force, F force applied on thecart, x position of the cart, T disturbance (e.g. wind, pushing), Θ moment ofinertia.

Equations of motion

Derivation

I To obtain equations of motion for such a system we use Lagrangian mechanics.

I For our inverse pendulum we obtain:

x = −ml

M +mα2 sin(α) +

ml

M +mα cos(α) +

1

M +mF

α =3

4lx cos(α) +

3g

4lsin(α) +

3

4ml2T

I To reduce writing overhead we simplify:

x = c0α2 sin(α) + c1α cos(α) + c2F

α = c3x cos(α) + c4 sin(α) + c5T

I We obtain a system of coupled nonlinear ordinary differential equations ofsecond order.I ordinary differential equation - equation contains an entity u and its ordinary

derivatives e.g. u, ∂u∂x

I nonlinear - equation contains nonlinear expressions e.g. u2, sin(u)I second order - highest occuring derivativeI system - more than one equationI coupled - equation regarding u contains entities, that are not u or one of its derivatives

I Such an equation is in general not solveable by analytical means only.

Equations of motion

Partly decoupling

I By inserting equation 1 in equation 2 and vice versa we obain ”less” coupledequations:

x =(c0α

2 sin(α) + c1c4 sin(α) cos(α) + c2F + c1c5T)/(

1− c1c3 cos2(α))

α =(c0c3α

2 sin(α) cos(α) + c4 sin(α) + c2c3 cos(α)F + c5T)/(

1− c1c3 cos2(α))

I Now our system is more convenient but still highly nonlinear.

Linearization

I To simplify our model, we want to remove any nonlinear term. This procedure iscalled linearization.

I It relies heavily on approximation and therefor resulting systems only have alimited validity.

I We use Taylor’s theorem, more correct Taylor polynomials of first order, toapproximate the terms:

α2 ≈ 0

sin(α) ≈ αcos(α) ≈ 1

Equations of motion

Linearized

I After replacing nonlinear parts with linear approximations we obtain:

x =(c1c4α+ c2F + c1c5T

)/(1− c1c3

)α =

(c4α+ c2c3F + c5T

)/(1− c1c3

)I It is necessary to realize how small the margin is in which the approximation - and

therefor the linear model - holds.

I Now our system is linear but still of second order.I There are at least two reasons to reduce the order of the system of differential

equations:I We are interested in the position x of the cart as well as the angular displacement of

the pendulum α on it. So far the equations describe only the change of the velocity xof the cart and the change of the angular velocity α of the pendulum.

I We will later be able to apply iterative methods to solve the system numerically e.g. theclassical Runge-Kutta method.

System of differential equations

Reducing order

I The entities we want to observe are ~x =(x, x, α, α

)T. The time derivative is

~x =(x, x, α, α

)T.

I By appending the system with the equations

x = x, and α = α

we obtainxxαα

=

0 1 0 00 0 c1c4

1−c1c30

0 0 0 10 0 c4

1−c1c30

xxαα

+

0c2

1−c1c30

c2c31−c1c3

F +

0

c1c51−c1c3

0c5

1−c1c3

T

Keep in mind that F and T are scalars in this one-dimensional model problem.

I In matrix notation

~x = A~x+~b1F +~b2T

it is more visible that we now have a system of linear differential equations offirst order.

System of differential equations

Analytical solution

I When omitting F and T we restrict external influence T or a force F upon thecart. We then have the opportunity to solve the system

~x = A~x

by means of analytical methods.

I The solutions to the system will be a linear combination of

~xi(t) = eλit~vi,

where λi are the eigenvalues and ~vi are the eigenvectors of the systems matrixA.

I When A is diagonalizable, creating the solutions via linear combination is trivial.Unfortunately our A is not diagonalizable:I When the eigenvalues of A are complex we get real solutions via the real and the

imaginary part of the complex solutions.I When the eigenvalues of A are real, but their algebraic multiplicity differs from their

geometric multiplicity we have to find generalized eigenvectors to append oursolutions with.

Eigenvalues and Eigenvectors

Eigenvalues

I Eigenvalues are the roots of the characteristic polynomial of A, which iscomputed by the determinant:

det(A− λI) = P(λ).

I While the determinant of matrices up to the size 3× 3 can easily be obtained viathe rule of Sarrus, for higher-dimensional matrices we use Laplace’s formula.

I For (artificial) concrete values M = 0.5kg, m = 1.75kg, l = 0.5m andg = 10m/s2 we get λ1 = 0, λ2 = 0, λ3 = −6, λ4 = 6.

Eigenvectors

I Eigenvectors are the solutions to the homogeneous system of linear equations

(A− λiI)~vi = 0

I For λ3 and λ4 we find the related ~vi. For λ1,2 we additionally have to find ageneralized eigenvector ~w.

System of differential equations

Analytical solution

I The solution has the form:

~x(t) = d1~v1eλ1,2t + d2(t~v1 + ~w)eλ1,2t + d3~v3e

λ3t + d4~v4eλ4t

=

d1 − 7c4e−6t + 7d3e6t + d2(1 + t)

d2 + 42c4e−6t + 42d3e6t

−18d4e−6t + 18d3e6t

108d4e−6t + 108d3e6t

.

I Due to the natur of the e-function it is obvious that eλt will be unconstrained aslong as λ > 0 and tend to +∞, for growing t.

I It is necessary to find a way to control the eigenvalues of A.

Control

Control matrix

I We use a linear control approach F = −K~x with a matrix K, that is effectivelya vector - in our one-dimensional case - and get

~x = A~x+~b1F = A~x−~b1K~x

= (A−~b1K)~x = A′~x.

I By correct choice of K we can control the eigenvalues of A′. This procedure iscalled pole placement. In Matlab this method is provided in a function calledplace().

I If you are interested in how this method works have a look at Robust PoleAssignment in Linear State Feedback, J. Kautsky, N.K. Nichols, and P. VanDooren, International Journal of Control, 41 (1985), pp. 1129-1155.

Simulation

Iterative method

I There are many explicit and implicit iterative methods to solve systems of linearordinary differential equations of first order - and therefor single linear differentialequations of first order as well. The simplest is the Euler method.

I We use the classical Runge-Kutta method, due to its good properties in stabilityand stiffness:

~k1 = ~f(tn, ~yn),

~k2 = ~f(tn +1

2h, ~yn +

1

2h~k1),

~k3 = ~f(tn +1

2h, ~yn +

1

2h~k2),

~k4 = ~f(tn + h, ~yn + h~k3),

~yn+1 = ~yn +1

6h(~k1 + 2~k2 + 2~k3 + ~k4).

I Figuratively speaking, we obtain the new descent direction as a weighted averageof four different descent directions.

I The properties of such a method need to be proven mathematically.

Solution

Compare linear and nonlinear case

I We compare the numerical solutions of the linear (red) and the nonlinear (blue)case without control:

-0.02

0

0.02

0.04

x

Vergleich beider Modelle ohne Regulierung

0

50

100

150

200

-1

0

1

F

0.8 0.9 1 1.1 1.2 1.3 1.4

t

0

0.5

1

T

I We observe two characteristics:I Up until a value for α, there is barely a difference between the linear and nonlinear

cases - as we expected when we linearized the system.I The solution for the linear case tends to infinity - as we expected due to the positive

eigenvalue of A.

Solution

Compare linear and nonlinear case

I We compare the numerical solutions of the linear (red) and the nonlinear (blue)case with control:

-2

-1

0

x

Vergleich beider Modelle mit Regulierung

-20

0

20

-20

-10

0

10

F

0 1 2 3 4 5 6 7 8 9 10

t

0

0.5

1

T

I We observe only slight differences between the two cases that minimize over time,due to the control we applied.

Solution

Nonlinear case

I We review the numerical solution of the nonlinear (blue) case without control:

-0.05

0

0.05

0.1

x

Loesung mit nicht-linearem Modell ohne Regulierung

0

2000

4000

-1

0

1

F

0 1 2 3 4 5 6 7 8 9 10

t

0

0.5

1

T

I We observe that the momentum T we applied to the system reflects in the overallmovement of the cart over time.

Sources

I Das Mathe-Praxis-Buch, J. Harterich, A. Rooch, 2014, Springer Verlag

I http://de-de.segway.com/SegwayPortal/media/Segway/Productimages/i2-SE/360/i2 SE 03.png, 2019

I http://de-de.segway.com/SegwayPortal/media/Segway/Productimages/i2-SE/360/i2 SE 05.png, 2019

I Numerik gewohnlicher Differentialgleichungen, M. Eiermann, 2011, TUBAF

I The dynamics of a Mobile Inverted Pendulum (MIP), S. Ostovari, N. Morozovsky,T. Bewley, 2012, UCSD