Sway Control

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Sway Control

By A.W.Gerstel,

December 2001



Sway of the load during traveling of the

trolley

Introduction

� The load will sway during traveling of the trolley

in an over head crane.

� The question is how to choose the traveling

function xk =xk (t) of the trolley such that all

swaying has been eliminated when the trolley

has arrived at its destination point.

� In the following calculation xk (t) is assumed to be

defined as shown in Fig. 1.





� It is:

± Stage I : constant acceleration a of the trolley during X

± Stage II : constant speed vr of the trolley

± Stage III: constant deceleration -a of the trolley during X

� Fig. 2 shows the motion model of the trolley and the load.It is also called ´definition diagram´.

� With respect to the diagram:

± Mass of the trolley: mk [kg]

± Mass of the load:

ml [kg]

± Gravity force: F=ml *g[N] g=9.81[ m/s2]

± Force in cable: S=S(t) S is a function of time



� It is further assumed:

± All mass of the load is concentrated in one point, so the

dimensions of the grab or container are neglected± The load is suspended by only one cable

± The cable is not elastic

± There is no hoisting during trolley traveling, so the length l

of the cable is constant

� A more detailed motion model of the load, the cables

and the trolley representing reality more accurate should

be analyzed with Matlab or Adams.

� A definition diagram as presented in Fig. 2 must contain

and define all variables being relevant in the

corresponding equations of motion, including the positive

directions of the variables.





� The arrows of xk , N , F , etc. in Fig. 2 define the positivedirections of these variables.

� This is also applicable to the axes x and y implying thatthe force F - due to its chosen direction opposite to thepositive direction of the y-axis - has to be put into thedifferential equations with a negative sign!

Questions to be answered

± How is the time function of the sway angle N =N (t) of the

load related to the movement function xk =xk (t) of thetrolley?

± Under which conditions is the sway of the load increased

or just eliminated at th

e end of th

e trolley travelingmovement?

± Which are the optimal values of the variables a, vr , X, t r , l

and mk given a trolley traveling distance sk and a mass ml

of the load?



Differential equations� The coordinates x and y of the load can be related to xk , l

and N .� As indicated before, the positive directions of the

variables has to be taken strictly into account. That is:

� The second order time derivatives become:

sin

cos

k x x l

y l

N

N

!

! (1)

(2)

2

2

cos sin

sin cos

k x x l l

y l l

N N N N

N N N N

!

!

&& &&& &&

&& &&&



� Using these expressions the accelerations ar and at

along and perpendicular to the direction of motion of the

load can be determined:

2

cos sin cos

sin cos sin

t k

r k

a x y x l

a x y x l

N N N N

N N N N

! !

! !

&&&& && &&

&&& && &&

(3)

� On the basis of Newton's law, the equations of motion of

the load are formulated as:

sin

cos

where

l t

l r

l

F a

S F a

F g

N N

! !

!

(4)



� Accelerations at and ar in these equations are eliminated

with formula (3):

_ a _ a2

sin cos

cos sin

l l k

l l k

m g m x l

S m g m x l

N N N

N N N

!

!

&&&&

&&&

(5)

� Having chosen a function xk =xk (t), in this case according

to Fig. 1, these differential equations can be solved

resulting in the time functions N =N (t) and S=S(t).

� The results maybe obtained analytically in any case in anumerical way, for example with the Matlab software.



� Because N is small it is acceptable to assume:

3 5

sin ...3! 5!

N N

N N N ! }

2 4

cos 1 ... 12! 4!

N N N ! }

(6)

(7)

� Matlab can be used to verify the consequences of these

assumptions.

� Substitution of (6) and (7) into (5) results in:..

.. k g xl l

N N !

2.. .k l l S m g m x l N N

® ¾! ¯ ¿

° À

(8)

(9)



� Further attention will only be paid to equation (8). Once

N =N (t) has been obtained, one can easily substitute this

result in (9) to arrive at S=S(t).� Applying mathematics, the solution of (8) appears to be:

cos sinh

A t B t N [ [ !

2, where [1/ ]

..

p

k x g s

g l T

T

N [ ! ! !

(10)

(11)

Homogenous Part

Particular Part

Where:A, [rad ]: integration constants

g[ m/s2]: gravitational acceleration ,g=9.81[m/s2]

l [ m]: length of the hoisting cable between the trolley and

the load

T[ s]: swaying time of the cable and the load



� For all three stages of motion I, II and III of the trolleyfrom starting to destination point, owing to (10) and (11):

cos sin..

k xA t t

g N [ [!

sin cos

.

A t B t N [ [ [ [ !

(12)

(13)

� Now a best and a worst trolley motion procedure are

considered with respect to the occurrence of sway of the

load.



Best procedure to move the trolley

± With "best" is meant such a traveling of the trolley that all

sway of the load

has disappeared just at arrival of t

he trolleyat its destination point.

± In that case the load can immediately be released, e.g. a

container onto an Automatic Guided Vehicle as used for

internal transport on container terminals.

� To determine the constants of integration, at the departurepoint of the trolley no sway of the load is assumed:

0 0 and 0

...

k xt A B

g N N ! ! ! ! !

_ acos 1

..k x t

g N [ !

sin

...

k xt

g N [ !

(14)

(15)

(16)



� In stage I of the trolley traveling, it is accelerated

according to:

_ acos 1a

t g

N [ !

sin. a

t

g

N [!

(17)

(18)

� On t=X at the end of stage I it is:

_ acos 1a

g N [X!

sin. a

g N [ X!

(19)

(20)



� Now it is chosen the acceleration time X of the trolley

equal to a multiple of T :

2 1, 2,3,...k T k T k k X [ X [ T ! p ! ! !

� Then according to (19) and (20):

( ) ( ) 0.N X N X! ! (21)

� This means no sway of the load just at the beginning of

stage II.

� Consequently the load will remain perpendicular under

the trolley in stage II, as long as the trolley is traveling

with a constant velocity.

� Note: Wind and air resistance forces are neglected.



� When the deceleration in stage III is chosen equal to theacceleration a in stage I and the deceleration time X

equal to th

e acceleration time, so X=k *

T , the motion of the trolley and the load in stage III is mirrored relative to

stage I. This results in a full elimination of the sway of the load just at the time the trolley has arrived at its pointof destination.

� Fig. 3, next slide, shows the motion functions of thetrolley and the load in the 3 stages.

� The trolley traveling time tr and the acceleration time Xcan be expressed in sk , a and vr as defined in Fig. 1.

2

22 2 22

2

k r r r r

k k r r r r r r r r

r r

as t v a t v

s s vv t v v t v t

v v a

XX X X

X X X X

¨ ¸! ! © ¹© ¹ª º

! ! p ! !

(22)



22 6

r r r v v vl g g k T k a

a g k l k l X T

T ! ! ! p !

(23)

y

t[s]0 X tr -X tr

-

t[s]X tr -X0

-- ---

tr

xk[m]

Sk=xk

a*X2/2

a*X2/2

(tr -2*X)*vr vr =a*X

y

y

!!y

0

yy

y

Fig. 3



Wor st procedure to move the trolley

� With "worst" is meant such a traveling of the trolley that

the sway of the load is at maximum when the trolley hasarrived at its destination point.

� Now it is chosen with respect to the acceleration time Xof the trolley:

1/ 2 1 / 2 1/ 2 2 1,2,3,...k T k T k k X [ X [ T ! ! ! !

2

( ) ( ) 0

.a

g N X N X

! !

� Substitution of (24) in (19) and (20) results in:

(24)

(25)

� Fig. 4, next slide, shows the position of the load after

X=k *T [ s] and X=(k+1/2)*T [ s] respectively.



t=X=k*T k=1,2,3,«

x-=a*X2/2

t=0

t=X=(k+1/2)*T

x-=a*X2/2

t=0

p:q=g:(2*a)

q

N =-(2*a)/:gN

p

No sway after X

Max sway after X

Fig. 4



� As a result of (25) the load will remain swaying with an

amplitude 2*a/g in stage II, during traveling of the trolley

with a constant velocity.� Assume with respect to the traveling time t II in this stage:

2 1, 2,3,...II r t T so t T N X! ! !

� Then the load is in the same position relative to the

trolley as at the beginning of stage II, so the same data

as presented in (25) apply at the beginning of stage III.

� In this stage the trolley is decelerated according to:

..k x a!

(26)

(27)



� Introducing a new time variable t'=t-(t r -X), similar equations as

(12) and (13) describe the movement of the load in stage III.

�Substituting (27) in t

hese equations, it follows:

_ a

2

3

3

' cos ' sin ' (28)

' sin ' cos ' (29)

On the basis of (25) it is:

( ' 0) ( ' 0) 0 (30)

Combining (28), (29) and (30) results in:

' cos ' 1 (31)

' sin

.

.

.

at A t B t

g

t A t B t

at t

g

at t

g

at t

g

N [ [

N [ [ [ [

N N

N [

[N [

!

!

! ! ! !

!

�! ' (32)



� With reference to (24) it is on time t'=X at the end of the

deceleration stage:

1/ 2' ' 2 1,2,3,...t t k T k k X [ [ X [ T T ! ! ! ! !

� Substitution of (33) in (31) and (32) and making use that

t'=X corresponds to t=tr :

2

4

4 0.75 36017.5[degree]

9.81 2

(34)

0 (35

)In practice a 0.75[ / ] which means:

.

r

r

at t

g

t t m s

Sway ang le

N

N

T

! !

! !}

! !

g

(33)



� As a consequence of (34) the load will sway between+17.5 and -17.5 degree after the trolley has arrived itsdestination point.

� This is definitely unacceptable with respect to thelowering of the container onto an Automatic GuidedVehicle as used on container terminals or into the cellguides in the hold of a container ship !!!

� The positions of the load during trolley traveling and atthe end of the traveling are shown in Fig. 5, next slide.



N

t=tr -X t=tr t=X- -- ---

2*aN g=

N 2*N

x=-ax=0x=a

Fig. 5



Operation in practice

� In reality trolley traveling and hoisting are done

simultaneously.� The motion control of the trolley and the load then

depends on:

± the experience of the crane operator

± if installed, the automatic control of trolley traveling andhoisting by means of feedback algorithms implemented in

the drive of these motions

± the mostly installed possibility to apply a low trolley velocity

when approaching its destination point



Some more questions

� Wi th referenc e to (9) whic h i s the mathemat ic al

expressi on of the func t i on S=S(t) of the for c e in thehoi st ing c able?

� I s the travel ing trolley func t i on xk =xk (t) acc ord ing to F ig. 1

the best one to apply in prac t ic e, assumed suc h a c hoic e

of a, vr

and X that no sway occ urs after the trolley has

arr i ved at i ts dest inat i on point?

� I f not whic h k ind of func t i on would be better from v i ew

point of dy namic s of the trolley dr i ve and the frame of the

c rane?

� I s i t then st i ll possi ble to shape that func t i onmathemat ic ally suc h that no sway occ urs after the trolley

has arr i ved at i ts dest inat i on point?

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