Solution of breaking problem of moving mechanical object is discussed. Five boundary values are used for the problem. Mathematical model of the process has been obtained. Also all dynamic functions and characteristics are represented. They shoe the accuracy of terminal positioning of deceleration.Terminal control, moving objects, boundary problems, adaptive control, reduction, acceleration, transient process, deceleration, positioning of deceleration

Cabir ErgÃ¼ven

English

Research Papers in Economics

Modeling of Deceleration Process of Spatial Movements
Cabir ERGÜVEN
Abstract: Solution of breaking problem of moving mechanical object is 
discussed. Five boundary values are used for the problem. Mathematical model of 
the process has been obtained. Also all dynamic functions and characteristics are 
represented. They shoe the accuracy of terminal positioning of deceleration.
Keywords: Terminal control, moving objects, boundary problems, adaptive 
control, reduction, acceleration, transient process, deceleration, positioning of 
deceleration
Introduction
The spinor model of the kinematics of spatial rotations developed 
on  the  basis  of  spinor  representation  of  generalized  spatial  rotations 
[Erguven (2007), Milnikov A.A., Prangishvili A.I., Rodonaia 2005] and the 
methods of the control theory of terminal states of motion of mechanical 
objects [Erguven (2007), Batenko A.P. (1977)] made it possible to create 
simple methods of controlling terminal states of spatial rotations of robot-
manipulators. The result has enabled us to reduce the three-dimensional 
problem of spatial motion control to the one-dimensional problem.
Basics of Model
Let us consider the following technologic task. It is required to 
bring by means of rotation a mechanical object of control (for instance, a 
1 2 3 gripping device or a spherical link) with coordinates x (x , x , x ) to the point 
1 2 3 of a three-dimensional space with coordinates y (y , y , y ). An intermediate 
rotating vector                      performs rotation by an angle defined by the 
terminal and initial points of rotation - 
) ; ; (
3 2 1 x x x x
)
) , (
cos( )
*
) , (
cos( 2
x
y x
ar
y x
y x
ar f = = g
Cabir ERGÜVEN is an associate professor in Faculty of Computer Technologies and Engineering at 
International Black Sea University
145
IBSU Scientific Journal      2 (1), 2008We have obtained the kinematics expressions for the rotating vectors. 
[Erguven (2007)]
(1)
where        ,
       ;
       .
We have used the rotation angle function satisfying  the 
following quite obvious condition:
        .
Let us divide the interval      into three segments:               
                          , where a , a  < 1 u a  = a . It is clear that the first  2 1 2 1
segment corresponds  to the beginning of the motion process, the second 
segment  to  uniform  motion  and,  finally,  the  third  subinterval  to 
deceleration. If we give a  = a , then there will be  no uniform motion (the  2  1 
length of the second segment is equal to zero), i.e. the initial stage of motion 
is immediately followed by the deceleration stage. Later we are discussing 
only the deceleration stage.
For the deceleration process ending in a complete stop we need to 
use the problem with five conditions since it is clear that at the end of the 
rotation process the acceleration must be equal to zero. Therefore the 
boundary conditions take the following form:  [Erguven Cabir, (2006)]
t=0;                 ;          ,      t=T;     ;    ; . (2)
Substituting
) ) ( cos )) ( (cos( ) (
1 1
2
1
y x f r t r t
r
x
t g g g x - - =
2
) ) ( cos )) ( (cos( ) (
2 2
2
2
y x f r t r t
r
x
t g g g x - - =
2
) ) ( cos )) ( (cos( ) (
3 3
2
3
y x f r t r t
r
x
t g g g x - - =
2
) ; ; (
1 2 2 1 3 1 1 3 2 3 3 2 y x y x y x y x y x y x r - - - =
) ; ; (
1 2 2 1 3 1 1 3 2 3 3 2 x r x r x r x r x r x r rx - - - =
) ; ; (
1 2 2 1 3 1 1 3 2 3 3 2 y r y r y r y r y r y r ry - - - =
) (t g
)
) , (
cos( ) ( 0 2
x
y x
ar t f = £ £ g g
] ; 0 [ f g ] ; 0 [ 1 f g a
] ; [ 2 1 f f g a g a ] ; [ 2 f f g g a
904 . 0 = g 1 = g& 355 . 1 = g 0 = g& 0 = g& &
146
IBSU Scientific Journal      2 (1), 2008into (1) we obtain the following expressions for the rotating vector 
coordinates as functions of time:
(3)
where  K , a (i = 0, 1, 2, 3, 4) and β are defined in [Erguven Cabir,  ω i
(2006)]. It is not difficult either to calculate the derivatives for (3), but we 
omit these calculations here because they are too long and tedious.
The values of                    are equal to the initial deviations from the 
synthesized control trajectories and define the presence of a transient 
 
å
=
-
+
-
- + - =
4
0
0 10
1 10 0 10
2 ) sin )
2
) (
( cos ) (( ) (
i
i
i
t K
t a t
a
K a t a e t b
g
g b g g w
w
&
); ) ) sin )
2
) (
(
cos ) (( cos( ) ) sin )
2
) (
(
cos ) (( (cos( ) (
1
4
0
0 10
1 10
0 10
2 1
4
0
0 10
1 10
0 10
2
2
1
y
i
i
i
t K
x
i
i
i
t K
f
r t a t
a
K a
t a e r t a t
a
K a
t a e
r
x
t
å
å
=
-
=
-
2
+
-
- +
+ - - +
-
- +
+ - - =
b
g
g
b g b
g
g
b g g x
w
w
w
w
&
&
); ) ) sin )
2
) (
(
cos ) (( cos( ) ) sin )
2
) (
(
cos ) (( (cos( ) (
2
4
0
0 10
1 10
0 10
2 2
4
0
0 10
1 10
0 10
2
2
2
y
i
i
i
t K
x
i
i
i
t K
f
r t a t
a
K a
t a e r t a t
a
K a
t a e
r
x
t
å
å
=
-
=
-
2
+
-
- +
+ - - +
-
- +
+ - - =
b
g
g
b g b
g
g
b g g x
w
w
w
w
&
&
), ) ) sin )
2
) (
(
cos ) (( cos( ) ) sin )
2
) (
(
cos ) (( (cos( ) (
3
4
0
0 10
1 10
0 10
2 3
4
0
0 10
1 10
0 10
2
2
3
y
i
i
i
t K
x
i
i
i
t K
f
r t a t
a
K a
t a e r t a t
a
K a
t a e
r
x
t
å
å
=
-
=
-
2
+
-
- +
+ - - +
-
- +
+ - - =
b
g
g
b g b
g
g
b g g x
w
w
w
w
&
&
0 0 g g & and
147
IBSU Scientific Journal      2 (1), 2008process. When they are equal to the boundary conditions (2) for t = 0, this 
means that there is no transient process at all. [Erguven Cabir, (2006)]
3. Results of Realization of the Model Deceleration Process
Figures  1?4  show  the  dynamic  characteristics  of  the  control 
process on the deceleration segment when
i.e. when there is no transient process – this is clearly seen from Figures. 1.b 
and 2.b Therefore the curves in Figure. 1.a and Figure. 2.a coincide, since 
the transient component is absent. Again we see that the control fully 
satisfies the boundary conditions and in this case the acceleration and the 
velocity become equal to zero at the end of the given time interval (Figure. 
2.c and Figure. 4) , which results in a complete stop.
Figure 1: The deceleration segment (there is no transient process): 
              the rotation  angle value as a function of time: 
    a) the forced component       
    b) the transient  component
    c) the complete solution: the sum of  the forced and transient components
 
1 905 . 0 0 10 0 10 = = = = g g g g & & and
b)
a) 
c) 
 
148
IBSU Scientific Journal      2 (1), 2008Figure 2: The deceleration segment (there is no transient process): 
              the rotation angle value as a function of time: 
   а) the forced component
 b) the transient component
 c) the complete solution: the sum of the forced and transient components
Figure 3: The deceleration segment (there is no transient process): the phase trajectory 
149
IBSU Scientific Journal      2 (1), 2008Figure 4:  The deceleration segment (there is no transient process):  
                 Acceleration as a function time 
Figure 5: The deceleration segment (there is a transient process): 
        The rotation angle value as a function of time: 
a) the forced component
b) the transient  component
c) the complete solution: the sum of the forced and transient components
150
IBSU Scientific Journal      2 (1), 2008Figure 6: The deceleration segment (there is a transient process): 
The angular velocity as a function of time 
a) the forced component 
b) the transient component;   
c) the sum of the forced and transient components
Figure 7: The deceleration segment (there is a transient process):   The phase trajectory:
151
IBSU Scientific Journal      2 (1), 2008Figure 8: The deceleration segment (there is a transient process):
Acceleration as a function of time
Figures  5?8 show the dynamic characteristics when 
In this case, as seen from Figures. 5.b and 6.b there 
exists a transient process. As different from the preceding motion stages, in this case the 
intensity of transient processes is quite comparable with stationary functions though these 
transient processes damp down soon. It is obvious that the intensity of transient processes 
explains an essential difference between the stationary and complete functions of angular 
motion (Figure. 5.a and 5.b) and its velocity (Figure. 6.a and 6.b). Nevertheless the control 
again satisfies the boundary conditions which gives for the deceleration stage the values
                           and            .
References
1. Erguven Cabir (2007),  Analysis and Synthesis of Terminal Control Reduction 
Problem, Tbilisi,  IBSU Journal    
2.  Erguven  Cabir  (2007),  Control  of  Terminal Acceleration,  Tbilisi,  IBSU 
Journal.
3.  Erguven  Cabir,  (2006),    Algorithms  of  Terminal  Control  of  Spatial 
Movements of Robots, PhD thesis, Tbilisi, Institute of Mechanics of Machines named 
after R.Dvali
4. Batenko A.P. (1977) Control of termınal statae of movıng objects, Moscow, 
Sovetskoe  Radıo  (Батенко  А.П.  Управление  конечным  состоянием  движущихся 
объектов. М.: Советское Радио, 1977, 256 с.
5.  Milnikov A.A.,  Prangishvili A.I.,  Rodonaia  I.D. A  Spinor  Model  of  3-
Dimensional Generalized Rotations//  // Automation and Remote Control, Moscow, 2005, 
№ 6, p. 30-37
. 0 0 0 10 0 10 = = = = g g g g & & and
355 . 1 ) ( 1 = = f T g g 55 ) ( ) 0 ( 1 = = = T x x x
152
IBSU Scientific Journal      2 (1), 2008

A Spinor Model of 3-Dimensional Generalized Rotations// // Automation and Remote Control,

Algorithms of Terminal Control of Spatial Movements of Robots,

Analysis and Synthesis of Terminal Control Reduction Problem, Tbilisi,

Control of termınal statae of movıng objects, Moscow, Sovetskoe Radıo (Батенко А.П. Управление конечным состоянием движущихся объектов. М.: Советское Радио,

Modelling of Deceleration Process of Spatial Movements

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Modelling of Deceleration Process of Spatial Movements

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