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Sway Control
By A.W.Gerstel,
December 2001
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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.
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� 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
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� 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.
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� 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?
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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
!
!
&& &&& &&
&& &&&
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� 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)
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� 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.
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� 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)
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� 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
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� 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.
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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)
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� 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)
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� 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.
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� 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)
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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
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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.
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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
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� 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)
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� 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)
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� 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)
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� 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.
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N
t=tr -X t=tr t=X- -- ---
2*aN g=
N 2*N
x=-ax=0x=a
Fig. 5
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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
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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?