Control of a Suspended Load Using Inertia Rotors

with Traveling Disturbance Yasuo Yoshida

Department of Mechanical Engineering Chubu University

1200 Matsumoto-cho, Kasugai-shi, Aichi 487-8501, Japan yyoshida@isc.chubu.ac.jp

Abstract

Rotational control and swing suppression of a crane suspended load model with traveling disturbance are

studied. A rotational free rigid body suspended by a

single rope is controlled using three inertia rotors. The

end-supporting-point of the single rope is forced to be traveled as disturbance. Control angles and angular

velocities are derived from measured data of fiber optic

gyros installed on the suspended load. Rotor angular velocities controlling the suspended load are obtained by

integrating the computed digital sliding-mode control

feedback accelerations based on the coupling system’s

dynamics. Experiments and simulations investigate simultaneous control of the load’s rotational orientation

and swing suppression for traveling disturbance.

1. Introduction A crane suspended load in a construction work is

frequently rotated and swung by wind pressure or inertia

force accompanied with movement of the crane. Most studies to control the suspended load have focused on

suppressing swing of the load as a lumped mass and

manipulating the crane. For example, Sakawa and

Nakazumi [1] treated a rotary crane and Lee [2] discussed the simultaneous traveling and traversing

trolley. The suspended load is actually a rigid body

different from above-mentioned a lumped mass model

and possible to be rotated. Kanki et al. [3] developed an active control device using gyroscopic moment to

control the rotation of a crane suspended load.

There are few studies to control both rotation and

swing of the suspended load as a rotational free rigid

body. Yoshida and Mori [4] studied simultaneous control

of rotation and swing of a pendulum having a rotational

free rigid body using inertia rotors and recently Yoshida and Yajima [5] studied for a rotational free rigid body

model suspended by a single rope with initial swing

disturbance.

This paper presents rotational control and swing suppression of a suspended load model with traveling

disturbance. The rotational free rigid body is suspended

by a single rope and controlled using three inertia rotors, where the end-supporting-point of the single rope is

forced to be traveled as disturbance. The suspended load

as a rigid body has non-actuating (passive) three-degree-

of-freedom and three inertia rotors have individually actuating (active) one-degree-of-freedom. The model

system has six-degree-of-freedom and passive degrees of

freedom are controlled by active degrees of freedom

based on the coupling system’s dynamics.

2. Dynamic Model and Controller

Figure 1 shows static state of the suspended load

model, where the coordinate x y z0 0 0 is nonmoving

base frame and xyz is moving frame fixed with the

load. The rope-end-point travels to x0 direction. Three

inertia rotors installed on the load. The purpose of this

study is to control the motion of the load using inertia

rotors. Fig.2 shows moving state, where the suspended

load is represented with the rope-end-point displacement

x , zxy Euler angles θ θ θ1 2 3, , and zxz Euler

angles θ θ θ1 2 3, , . Angular velocities of the load are

ω ω ωx y z, , with respect to ,x y and z axes.

Angles and angular velocities of inertia rotors are

Proceedings of the 2001 IEEE International Conference on Robotics & Automation

Seoul, Korea • May 21-26, 2001

0-7803-6475-9/01/$10.00© 2001 IEEE

θ θ θ4 5 6, , and ω θ ω θ ω θz x y+ + +, ,4 5 6 , and torques

are τ τ τ1 2 3, , with respect to ,z x and y axes.

Based on zxy Euler angles θ θ θ1 2 3, , of the

suspended load and θ θ θ4 5 6, , of inertia rotors, the

equation of motion can be written as

M M

M M

q

q

h

h

Xx

c c c f

f c f f

c

f

c

f

+

+

=

0

0

τ, (1)

where

M

m m m

m m

m mcc =

11 12 13

12 22

13 33

0

0

, M M

m m m

m m

mcf fc

T= =

14 15 16

24 25

36

0

0 0

,

M

I

I

Iff

R

R

R

=

3

1

2

0 0

0 0

0 0

,

[ ] [ ]q qc

T

f

T= =θ θ θ θ θ θ1 2 3 4 5 6, ,

[ ] [ ]h h h h h h h hc

T

f

T= =1 2 3 4 5 6, , [ ]X x x xT= 1 2 3 ,

[ ] [ ]0 0 0 0 1 2 3= =T T, .τ τ τ τ

qc shows controlled (non-actuating) angle state vector

and q f free (actuating) angle state vector. Mass matrix

M M Mcc cf fcT, ( )= are function of θ θ2 3, . h hc f, are

vector of Coriolis, centrifugal and gravity terms and

function of θ θ θ2 3 1 6), , ( ~i i = . X is vector of

traveling disturbance and function of θ θ θ1 2 3, , .

The system equation of controlled zxy Euler angle’s

state vector qc is linearized by using the control input

vector uc as

[ ]q u u u uc c

T= = 1 2 3. (2)

A digital sliding-mode controller by which the

controlled angles θ θ θ1 2 3, , of the suspended load

follow the desired values θ θ θ1 2 3d d d, , is designed from

equation (2). Digital state equation of control angle error

is obtained using sampling time T and time steps

k ( , , )=1 2 as

E pE q u ii k ik ik, ( ~ )+ = − =1 1 3 (3)

Ee

ep

Tq

T

Tik

ik

ik

=

=

=

, ,

.1

0 1

05 2

　

e e u uik idk ik ik idk ik ik ik idk= − = − = −θ θ θ θ θ, ,

Rope- end Traveling Direction

Rotor for yaw rotation

Rotor for roll swing Rotor for pitch swing

Suspended load

Rope

x

y

z

G

O

x0

y0z0

x

Fig.1 Suspended load model using inertia rotors with

traveling disturbance.

θ1

θ2

θ3

θ1

θ2

θ3

τ ω θ3 6, y

+

τ ω θ2 5, x

+

τ ω θ1 4, z

+ θ3

θ1

θ1

x

z

y

x

g

x0 y0

z0

( )z

( )x

( )y

Fig.2 Coordinates and angles of the suspended load

model.

Switching function σ ik is defined with scalar

vector ][S s s= 1 2 as follows

σ ik ikSE i= =( ~ )1 3 , (4)

where, σ ik = 0 gives switching line. Sliding-mode

control input uik can be represented as follows, where

u eqik is equivalent input on switching line and u nlik

is nonlinear input acting to reach switching line,

u u uik eqik nlik= + , (5)

u Sq S p I Eeqik ik= −−( ) ( )1 ,

u Sq fornlik ik= < <−η σ η( ) 1 0 2 .

From equation (2) to (5), sliding-mode controller is

obtained as

u K e K e for iik ik ik idk= + + =1 2 1 3( ~ )θ , (6)

KT

T s sK

T s s

T s s12 1

22 1

2 105

1

05=

+=

++

( / )

. ( / ),

( / )( / )

. ( / )

η η .

Estimated accelerations of inertia rotor’s angles q f

are taken as

[ ]q u u u uf f k k k

T= = 4 5 6. (7)

After putting equation (2), (7) into equation (1),

upper part of the equation of motion (1) becomes as

M u M u h Xxcc c cf f c+ + + = 0 . (8)

Above equation (8) shows the dynamic coupling

between controlled angles qc and free angles q f .

Then, estimated accelerations u f are given as

( )u M M u h Xxf cf cc c c= − + +−1 . (9)

By integrating equation (9) with the sampling time

T , manipulating velocities of inertia rotors q vf k=

[ ]= v v vk k k

T

4 5 6 are obtained as

v v u Tk k f= +−1. (10)

Fiber optic gyros installed on the suspended load

can measure the angles θ θ θ θ θ13 1 3 2 3( ), ,≡ + and the

angular velocities ω ω ωx y z, , .

zxy Euler angles θ θ θ1 2 3, , and zxz Euler

angles θ θ θ1 2 3, , are mutually convertible. zxz Euler

angles are obtained by measured angles as

θ θ θθ

θ

1 13 3

21

2 3

3 2 3 22

= −== −

−

,

cos ( ),

tan ( , ).

C C

A C S S

(11)

Under the condition that θ θ2 3, are directly

measured by fiber optic gyros, θ1 must be calculated

based on equation (11) as

θ1 1 3 1 2 3 1 3 1 2 32= + − +A C S S C C S S C C Ctan ( , ) , (12)

where Ci and Si

are defined as cosθi and

sinθi for i = 1 2 3, , , respectively.

Angular velocities , ,θ θ θ1 2 3 are obtained using

ω ω ωx y z, , and θ θ2 3, as

/ /θ

θθ

ωωω

1

2

3

3 2 3 2

3 3

2 3 2 3

0

0

1

=−

−

S C C C

C S

T S T C

x

y

z

, (13)

where T2 is tanθ2 . Fig.3 shows the control system.

θθθωωω

13

2

3

k

k

k

xk

yk

zk

θθθ

θ

θ

θ

1

2

3

1

2

3

k

k

k

k

k

k

θθθ

1

2

3

k

k

k

θ1dk

[ ]u K e K e

u u u u

ik i ik i ik idk

c k k k

T

= + +

=1 2

1 2 3

θ

[ ]( )u M M u h Xx

u u u

f cf cc c c

k k k

T

= − + +

=

−1

4 5 6

[ ]v v u T

v v v

k k f

k k k

T

= +

=

−1

4 5 6

i =1 3~

+

−

gyro

data

zxz

Euler

angle

zxy

Euler

angle

t etargdisturbance x

suspended load

Fig.3 Control system.

Simulation is performed for the sampling time

between t (=step k) and t+T (=step k+1). Following

simultaneous differential equations are solved using

Runge-Kutta numerical integration.

( )q

q

M M u h X xc

f

cc cf f c

=

− + +

0

, (14)

where u v v Tf k k= − −( ) /1.

Since the manipulating velocities are held and constant at

the step k, angular accelerations of free angle q f are

zeros in the time between sampling time T.

However, the manipulating velocities change at each

time step and therefore accelerations of free angle are

approximately taken as u f that is the change of the

manipulating velocity between k-1 and k steps. u f

assures the dynamic coupling in the simulation.

3. Experimental Results

Fig.4 shows the experimental suspended model

photograph. Length between support point and the load’s

center of gravity is l m= 1 22. , mass of the load including

inertia rotors m kg= 20 9. , moments of inertia of the load

I kgmx = 136 2. , I kgmy = 0 676 2. , I kgmz = 117 2. and inertia

rotors I I I kgmR R R1 2 320 0377= = = . . Maximum angular

velocities of servomotors driving inertia rotors are

θ4 173= rpm , θ5 184= rpm , θ6 168= rpm . Sliding-mode

digital gains are K K1 280 16= =, using sampling time

T = 0 05. sec and η = =05 012 1. , / .s s , selected from

simulation results using 2~0=η , 10~0/ 12 =ss . And,

digital poles of closed system obtained from equation (3)

Fig.4 Experimental suspended model’s photograph.

and (6) using these digital gains are 5.0=z and 6.0 ,

therefore the control system is stable. Simulation is used

by Runge-Kutta numerical integration in time by steps of

size 0 005. sec . Fig.5 shows the linear traveling of the

end-supporting-point of the single rope as disturbance

for the suspended load. Fig.5(a) is traveling displacement

moving 0 25. m within 15. sec . Fig.5(b) shows velocity

and acceleration, that is, accelerating to 0.5 sec, uniform

velocity to 1.3 sec and decelerating to 1.5 sec. Initial

angles of the suspended load are manually positioned

and it is difficult to give the accurate same initial angles

to the experiments of both without and with controls. But

the differences of initial zxy Euler angles are small

between θ1 25= , θ2 1= , θ3 0= in the case of

experimental values without control and θ1 27= ,

θ2 0= , θ3 0= with control. Zero desired values

θ θ θ1 2 3d d d, , are given for controlled angle θ θ θ1 2 3, , .

0 1 2 3 4 5-0.3

-0.2

-0.1

0.0

Linear Traveling

t (s)

x (m

)

(a) Traveling displacement.

0 1 2 3 4 5-1.0

-0.5

0.0

0.5

1.0

..

x .

x

...

x (m

/s2 )

x (m

/s),

Linear Traveling

t (s)

(b) Traveling velocity and acceleration.

Fig.5 Linear traveling disturbance.

3

Fig.6 shows the experimental time response of

controlled rotational angle θ1 within 30 sec. Thin line

indicates the case of without control and thick line with

control, and the following figures show in similar ways.

Thin line of without control moves from initial angle

25 to maximum − 66 at 12 sec and thereafter largely

fluctuates. This phenomenon is explained by Yoshida

and Mori [5] that both centrifugal force caused by swing

and the difference between moment of inertia Ix and

I y generate yaw rotation torque. Thick line of with

control moves rapidly from initial angle to zero desired

value within 2 sec and hereafter keep desired value.

Fig.7 shows the experimental swing angle’s

components. Fig.7(a) shows controlled angle θ2 . Thin

line of without control enlarges vibration amplitude to

maximum 6 between 5 sec to 15 sec and and thick line

with control keeps within 15. value from initial time to

30 sec. Fig.7(b) shows controlled angle θ3 . Both cases

of without and with control are same till the time 15 sec,

but after that time thin line of without control remains

4 constant vibration amplitude and thick line with

control damps to zero desired value.

Fig.8 shows the response of swing angle ( zxz Euler

angle) θ2 . The amplitudes in the case of with control

are suppressed under half of those of without control.

Fig.9 shows swing displacement x y, for the fixed

coordinate when the load is at rest. The vibration

displacement x is induced by travel disturbance.

0 5 10 15 20 25 30-80

-60

-40

-20

0

20

40Exp.

With Conrol

Without Conrol

t (s)

θ 1 (de

g)

Fig.6 Experimental time response of

rotational angle θ1 .

Thin line of x without control leaves vibration

amplitude at 30 sec, but thick line of x with control

damps to zero desired value.

0 5 10 15 20 25 30-8

-4

0

4

8

Exp.Without Conrol

With Conrol

t (s)

θ 2 (de

g)

(a) Time response of θ2 .

0 5 10 15 20 25 30-8

-4

0

4

8Exp.

With Conrol

Without Conrol

t (s)

θ 3 (de

g)

(b) Time response of θ3 .

Fig.7 Experimental swing angle’s components.

0 5 10 15 20 25 300

2

4

6

8

With Conrol

Without ConrolExp.

t (s)

- θ 2 (de

g)

Fig.8 Experimental swing angle θ2 .

0 5 10 15 20 25 30-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1 Exp.

yy

xx With ConrolWithout Conrol

t (s)

x(m

), y

(m)

Fig.9 Experimental swing displacements x y, .

0 5 10 15 20 25 30-40

-20

0

20

40

With Conrol

Cal.

Without Conrol

t (s)

θ 1 (de

g)

Fig.10 Simulation time response of

rotational angle θ1 .

0 5 10 15 20 25 30-0.5

-0.4

-0.3

-0.2

-0.1

0.0

0.1

y

xxy Without Conrol With Conrol

Cal.

t (s)

x(m

), y

(m)

Fig.11 Simulation swing displacements x y, .

Fig.10 shows simulation time response of rotational

angle θ1 corresponding to experimental result of Fig.6.

Torsional spring coefficient of the rope is considered as

01. /Nm rad in simulation. Fluctuation tendency of thin

line without control shows the same pattern of

experiment and thick line with control moves rapidly

from initial angle to zero value same as experiment.

Fig.11 shows simulation swing displacements x y,

corresponding to experimental result of Fig.9 and these

are similar as experimental ones.

4. Conclusions A rotational free rigid body suspended by a single rope model with traveling disturbance is controlled using

three inertia rotors. Velocity-command-type control

system is developed for inertia rotors by integrating the

computed feedback accelerations of digital sliding-mode control, based on the coupling system’s dynamics. The

experimental and simulation results show that inertia

rotors can control the rotation and swing of the

suspended load for traveling disturbance.

References [1] T. Sakawa and A.Nakazumi, “Modeling and Control

of a Rotary Crane”, ASME Journal of Dynamic Systems, Measurement, and Control, Vol.107, pp.200-205, 1985.

[2] H. H. Lee, “Modeling and Control of a Three-Dimensional Overhead Crane”, ASME Journal of Dynamic Systems, Measurement, and Control, Vol. 120, pp.471-476, 1998.

[3] H. Kanki and Y. Nekomoto et al., “Development of Suspender Controlled by CMG (in Japanese with English abstracts)”, Proceedings of JSME Dynamics and Design Conference ’95, No.95-8 (1), B, pp.34-37, 1995.

[4] Y. Yoshida and K. Mori, ”Simultaneous Control of Attitude and Swing of a Pendulum Having a Rotational Free Body (in Japanese with English abstracts)”, Trans. of JSME, Series C, Vol.64, No.628, pp.4660-4665, 1998.

[5]Y. Yoshida and M. Yajima, “Control of a Suspended Load Using Inertia Rotors”, Proceedings of the 1999 ASME Design Engineering Technical Conferences, Symposium on Motion and Vibration, DETC99/MOVIC-8418, pp.1-6, Las Vegas, 1999.