Modelling and Simulation of an Inverted Double Pendulum

1.0 Description

The inverted Pendulum, is a pendulum which has its mass above its pivot point, it is

implemented with pivot point mounted on a cart that can move horizontally and called a

CART and POLE (as shown in the diagram below).

An inverted Pendulum is inherently unstable and must be actively balanced in order to remain

upright by either applying a torque at the pivot or by moving the pivot point horizontally as

part of a feedback system.

Generally the inverted pendulums are widely used as a benchmark for testing control

algorithm; the inverted pendulum is also applied in many industrial and engineering products

such as high-precision control of robot arm, stability control of launching rocket, and attitude

control of satellite.

Fig 1. The Simple pendulum system

1.1 The Article Description:

The paper main aim is to show the relationship between controllability and parameters of the

system. The paper is based on the modelling of the double pendulum on a cart using Newton’s

Laws of motion and linearised on an approximately small angle.

However we modelled the using the Lagrange’s Equation hence the difference in the

differential equations obtained.

Fig 2. The Simplified Double Pendulum system

1.2 Modelling and Parameters definition;

M is mass of trolley; 𝒙𝒙 is its displacement; F is the force exerted on trolley; 𝒎𝒎𝟏𝟏 and 𝒎𝒎𝟐𝟐 are

the mass of left and right rocker; 2𝑳𝑳𝟏𝟏 and 2𝑳𝑳𝟐𝟐 are the length of rockers; 𝐽𝐽1 and 𝐽𝐽2 are the

inertia moment of rockers; 𝜽𝜽𝟏𝟏 and 𝜽𝜽𝟐𝟐 are the angle between rockers and the vertical direction.

In the following we derive the differential equations which describe the dynamics of the inverted pendulum system use Lagrange’s equation. Two rockers have similar situation. The kinetic energy of cart is:

𝑇𝑇1 =12𝑀𝑀�̇�𝑥2

The kinetic energy of the left bob is

𝑇𝑇2 =12𝑚𝑚1(𝑥𝑥1

2̇ + �̇�𝑧12)

where

𝑥𝑥1 = 𝑥𝑥 + 2𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1, 𝑧𝑧1 = 2𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1

So that

𝑥𝑥1̇ = �̇�𝑥 + 2𝐿𝐿1�̇�𝜃1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1, �̇�𝑧1 = −2𝐿𝐿1𝜃𝜃1̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1

The kinetic energy of the right bob is

𝑇𝑇3 =12𝑚𝑚2(𝑥𝑥2

2̇ + �̇�𝑧22)

𝑥𝑥2 = 𝑥𝑥 + 2𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2, 𝑧𝑧2 = 2𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2

so that

𝑥𝑥2̇ = �̇�𝑥 + 2𝐿𝐿2�̇�𝜃2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2, �̇�𝑧2 = −2𝐿𝐿2𝜃𝜃2̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2

Therefore, the total kinetic energy is

𝑇𝑇 = 𝑇𝑇1 + 𝑇𝑇2 + 𝑇𝑇3

=12𝑀𝑀𝑥𝑥2̇ +

12𝑚𝑚1��̇�𝑥2 + 4�̇�𝑥�̇�𝜃1𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + (2𝐿𝐿1)2�̇�𝜃1

2� +12𝑚𝑚2(�̇�𝑥2 + 2�̇�𝑥�̇�𝜃22𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2+(2𝐿𝐿2)2�̇�𝜃2

2)

The potential energy is equivalent to:

𝑈𝑈 = 𝑈𝑈1 + 𝑈𝑈2 = 𝑚𝑚1𝑔𝑔2𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 𝑚𝑚2𝑔𝑔2𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2

The Lagrange equation

𝐿𝐿 = 𝑇𝑇 − 𝑈𝑈 =

12𝑀𝑀𝑥𝑥2̇ + 1

2𝑚𝑚1��̇�𝑥2 + 2�̇�𝑥�̇�𝜃12𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + (2𝐿𝐿1)2�̇�𝜃1

2� + 12𝑚𝑚2��̇�𝑥2 + 2�̇�𝑥�̇�𝜃22𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + (2𝐿𝐿2)2�̇�𝜃2

2� −

𝑚𝑚1𝑔𝑔2𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 −𝑚𝑚2𝑔𝑔2𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2;

The generalized coordinates are selected as 𝑞𝑞 = [𝑥𝑥 𝜃𝜃1 𝜃𝜃2]𝑇𝑇 so that Lagrange’s equations are

𝑑𝑑𝑑𝑑𝑑𝑑�𝜕𝜕𝐿𝐿𝜕𝜕�̇�𝑥� −

𝜕𝜕𝐿𝐿𝜕𝜕𝑥𝑥

= 𝐹𝐹

𝑑𝑑𝑑𝑑𝑑𝑑 �

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1̇

� −𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1

= 0

𝑑𝑑𝑑𝑑𝑑𝑑 �

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2̇

� −𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2

= 0

1) 𝑞𝑞 = 𝑥𝑥 𝜕𝜕𝐿𝐿𝜕𝜕𝑥𝑥

= 0;

𝜕𝜕𝐿𝐿𝜕𝜕�̇�𝑥

= 𝑚𝑚1�̇�𝑥 + 2𝑚𝑚1𝜃𝜃1̇𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 +𝑚𝑚2�̇�𝑥 + 2𝑚𝑚2𝜃𝜃2̇𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 +𝑀𝑀�̇�𝑥

𝑑𝑑𝑑𝑑𝑑𝑑�𝜕𝜕𝐿𝐿𝜕𝜕�̇�𝑥� =

𝑚𝑚1�̈�𝑥 − 𝑚𝑚1𝜃𝜃12̇2𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 2𝑚𝑚1𝐿𝐿1𝜃𝜃1̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 𝑚𝑚2�̈�𝑥 − 𝑚𝑚2𝜃𝜃2

2̇2𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 2𝑚𝑚2𝐿𝐿2𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 𝑀𝑀�̈�𝑥

𝑑𝑑𝑑𝑑𝑑𝑑�𝜕𝜕𝐿𝐿𝜕𝜕�̇�𝑥� −

𝜕𝜕𝐿𝐿𝜕𝜕𝑥𝑥

= 𝐹𝐹

𝑚𝑚1�̈�𝑥 − 𝑚𝑚1𝜃𝜃12̇2𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 2𝑚𝑚1𝐿𝐿1𝜃𝜃1̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 𝑚𝑚2�̈�𝑥 − 𝑚𝑚2𝜃𝜃2

2̇2𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 2𝑚𝑚2𝐿𝐿2𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 𝑀𝑀�̈�𝑥 = 𝐹𝐹

2) 𝑞𝑞 = 𝜃𝜃1

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1

= 2𝑚𝑚1𝑔𝑔𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 − 2𝑚𝑚1�̇�𝑥𝜃𝜃1̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1̇

= 2𝑚𝑚1�̇�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̇𝑚𝑚1

𝑑𝑑𝑑𝑑𝑑𝑑 �

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1̇

� = 2𝑚𝑚1�̈�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 − 2𝑚𝑚1�̇�𝑥𝜃𝜃1̇𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̈𝑚𝑚1

𝑑𝑑𝑑𝑑𝑑𝑑 �

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1̇

� −𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃1̇

= 0

2𝑚𝑚1�̈�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 − 2𝑚𝑚1�̇�𝑥𝜃𝜃1̇𝐿𝐿𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̈𝑚𝑚1 − 2𝑚𝑚1𝑔𝑔𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 2𝑚𝑚1�̇�𝑥𝜃𝜃1̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 = 0

2𝑚𝑚1�̈�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̈𝑚𝑚1 − 2𝑚𝑚1𝑔𝑔𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 = 0

2) 𝑞𝑞 = 𝜃𝜃2 𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2

= 2𝑚𝑚2𝑔𝑔𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 − 2𝑚𝑚2�̇�𝑥𝜃𝜃2̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2̇

= 2𝑚𝑚2�̇�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̇𝑚𝑚2

𝑑𝑑𝑑𝑑𝑑𝑑 �

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2̇

� = 2𝑚𝑚2�̈�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 − 2𝑚𝑚2�̇�𝑥𝜃𝜃2̇𝐿𝐿𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̈𝑚𝑚2

𝑑𝑑𝑑𝑑𝑑𝑑 �

𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2̇

� −𝜕𝜕𝐿𝐿𝜕𝜕𝜃𝜃2̇

= 0

2𝑚𝑚2�̈�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 − 2𝑚𝑚2�̇�𝑥𝜃𝜃2̇𝐿𝐿𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̈𝑚𝑚2 − 2𝑚𝑚2𝑔𝑔𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 2𝑚𝑚2�̇�𝑥𝜃𝜃2̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 = 0

2𝑚𝑚2�̈�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̈𝑚𝑚2 − 2𝑚𝑚2𝑔𝑔𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 = 0

We derive the nonlinear equation

𝑚𝑚1�̈�𝑥 − 𝑚𝑚1𝜃𝜃12̇2𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 2𝑚𝑚1𝐿𝐿1𝜃𝜃1̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 𝑚𝑚2�̈�𝑥 − 𝑚𝑚2𝜃𝜃2

2̇2𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 2𝑚𝑚2𝐿𝐿2𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 𝑀𝑀�̈�𝑥 = 𝐹𝐹

2𝑚𝑚1�̈�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̈𝑚𝑚1 − 2𝑚𝑚1𝑔𝑔𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 = 0

2𝑚𝑚2�̈�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̈𝑚𝑚2 − 2𝑚𝑚2𝑔𝑔𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 = 0

With the parameters defined from the article as below;

𝑀𝑀 = 1.5 ,𝑚𝑚1 = 0.5,𝑚𝑚2 = 0.5,𝑔𝑔 = 9.8, 𝐿𝐿1 = 0.8,𝐿𝐿2 = 0.8 substituting these parameters into the

above 3 equations.

𝑚𝑚1�̈�𝑥 − 𝑚𝑚1𝜃𝜃12̇2𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 2𝑚𝑚1𝐿𝐿1𝜃𝜃1̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 𝑚𝑚2�̈�𝑥 − 𝑚𝑚2𝜃𝜃2

2̇2𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 2𝑚𝑚2𝐿𝐿2𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 𝑀𝑀�̈�𝑥 = 𝐹𝐹

2𝑚𝑚1�̈�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̈𝑚𝑚1 − 2𝑚𝑚1𝑔𝑔𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 = 0

2𝑚𝑚2�̈�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̈𝑚𝑚2 − 2𝑚𝑚2𝑔𝑔𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 = 0

Simplifying the above equations,

(𝑀𝑀 + 𝑚𝑚1 + 𝑚𝑚2)�̈�𝑥 − 𝑚𝑚1𝜃𝜃12̇2𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 2𝑚𝑚1𝐿𝐿1𝜃𝜃1̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 −𝑚𝑚2𝜃𝜃2

2̇2𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 2𝑚𝑚2𝐿𝐿2𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 = 𝐹𝐹 ..eqn1

2𝑚𝑚1�̈�𝑥𝐿𝐿1𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 4𝐿𝐿12𝜃𝜃1̈𝑚𝑚1 − 2𝑚𝑚1𝑔𝑔𝐿𝐿1𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 = 0 ..eqn2

2𝑚𝑚2�̈�𝑥𝐿𝐿2𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 4𝐿𝐿22𝜃𝜃2̈𝑚𝑚2 − 2𝑚𝑚2𝑔𝑔𝐿𝐿2𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 = 0 ..eqn3

𝑀𝑀 = 1.5 ,𝑚𝑚1 = 0.5,𝑚𝑚2 = 0.5,𝑔𝑔 = 9.8, 𝐿𝐿1 = 0.8,𝐿𝐿2 = 0.8

2.5 �̈�𝑥 − 0.8𝜃𝜃12̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 + 0.8𝜃𝜃1 ̈ 𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 − 0.8𝜃𝜃2

2̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 + 0.8𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 = 𝐹𝐹 ..eqn4

0.8�̈�𝑥𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 1.28𝜃𝜃1̈ − 7.84𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 = 0 ..eqn5

0.8�̈�𝑥𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 1.28𝜃𝜃2̈ − 7.84𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 = 0 ..eqn6

�̈�𝑥 = 0.32𝜃𝜃12̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 − 0.32𝜃𝜃1 ̈ 𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 0.32𝜃𝜃2

2̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 − 0.32𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 0.4𝐹𝐹(𝑑𝑑) .eqn7

𝜃𝜃1̈ = −0.625�̈�𝑥𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 6.125𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 .eqn8

𝜃𝜃2̈ = −0.625�̈�𝑥𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 6.125𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 .eqn9

Using a unit step input as the force F(t).

2.0 SIMULATION RESULTS AND GRAPHS

Fig 3. The Step response of the model (Non-linear model) with different scopes

Fig 4. The Step response of the model displacement (Non-linear model)

Fig 5. The Step response of the model rocker 1 pivot angle (Non-linear model)

Fig 6. The Step response of the model rocker 2 pivot angle (Non-linear model)

Fig 7. The simulation block for model (Non-linear) one scope

Fig 8. The Step response of the model (Non-linear model)

lengend

Step response curve of trolley displacement

Step response curve of the two rocker pivot angle

3.0 Linearization of the Non-linear model

Linearization of the non linear model

�̈�𝑥 = 0.32𝜃𝜃12̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1 − 0.32𝜃𝜃1 ̈ 𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 0.32𝜃𝜃2

2̇𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2 − 0.32𝜃𝜃2̈𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 0.4𝐹𝐹

𝜃𝜃1̈ = −0.625�̈�𝑥𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃1 + 6.125𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃1

𝜃𝜃2̈ = −0.625�̈�𝑥𝑐𝑐𝑐𝑐𝑠𝑠𝜃𝜃2 + 6.125𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃2

Linearizing the above model we obtain linear model to be

Sin 𝜃𝜃1 ≈ 𝜃𝜃1, sin𝜃𝜃2 ≈ 𝜃𝜃2, cos𝜃𝜃1 = 1, cos𝜃𝜃2 = 1,

)(4.032.032.0 21 tFx +∆−∆−=∆ θθ

11 125.6625.0 θθ ∆+∆−=∆ x

22 125.6625.0 θθ ∆+∆−=∆ x

Assuming that 𝜃𝜃1 𝑎𝑎𝑠𝑠𝑑𝑑 𝜃𝜃2 is very small, then the linearised model is as below,

)(4.032.032.0 21 tFx +−−= θθ

11 125.6625.0 θθ +−= x

22 125.6625.0 θθ +−= x

The simulation model of the linear model using simulink

Fig 9. The Simulink of the model (Linear model)

Fig 10. The Step response of the model (Linear model)

lengend

Step response curve of trolley displacement

Step response curve of the two rocker pivot angle

4.0 Discussion on the accuracy of the results

From the results obtained in the simulation of the non-linear and linear model obtained from

our modelling and that of the results of the article, its shows that our model are accurate

representation of the model as the modelled in the article.

As side the difference in the techniques used in the modelling (the use of the Lagrange as

against Newton’s Equation) we obtain an accurate model which can be used in the

verification of the objective of the article.

However if the linearization tent is extended beyond the very small angle for pivot the system

becomes very unstable and hence the pendulum will fall from the inverted position.