Geometry of five link mechanism
with two degrees of freedom
David Tavkhelidze
Internal combustion engine
E
A-Crankshaft;
B-Connecting rod;
C-Slider (piston);
D-Frame;
E-Valve mechanism
Kinematic pairs
Degree of freedom
Degree of freedom for spatial mechanism
W=6n-P1-2P2-3P3-4P4-5P5
Degree of freedom of planar mechanism
n
2
3
P5
W=3n-2P5
3n=2P5
n 
2
3
P5
Kinematic chains
Four link mechanism
w  3 3  2  4  0  1
Five link mechanism
W  34  2 5  2
Mechanisms used in technological machines
Four link mechanism with
rotating kinematic pairs
Six link gear mechanism
Four link slider - crank
mechanism
Five link mechanism with gear pair, reducing
number degrees of freedom of the mechanical
system
Mechanisms with two degrees of freedom
Kinematic scheme of five link mechanism with two
degrees of freedom
Various scheme of mechanisms with two degrees of freedom
.
Straight geometrical task
The design diagram of five link mechanism
For the closed kinematic chain it is necessary that the product of matrices of
transformation between coupled coordination systems connected with all
incoming links has to be equal to unit matrix
T
(1 , 0 )
T
( 2 ,1 )
T
(3, 2 )
T
( 4 ,3)
T
(0,4)
(1)
 E
For simplification of calculation it would be written
T
( 4 ,3)
T
(0,4)
 T
( 2 ,3)
T
(1 , 2 )
T
( 0 ,1 )
(2)
Straight geometrical task
The transformation matrix between of two sequential i-1 and i plain coordinate
systems has the following general form:
(3)
Taking in the account the previous equations will be obtained
(4)
1
0
l 3  l 4 cos  43
cos(  43   40 )
l 4 sin  43
sin(  43   40 )
1
  l 2 cos  23  l 1 cos(  12   23 )  l 0 cos(  01   12   23 )
l 2 sin  23  l 1 sin(  12   23 )  l 0 sin  01   12   23 
0
 sin(  43   40 ) 
cos(  43   40 )
0
cos(  01   12   23 )
 sin(  01   12   23 )
0
sin(  01   12   23 ) .
cos(  01   12   23 )
.
.
,
Straight geometrical task
The 4th matrix equation of five-link mechanism blockage contains full
information about parameters of link motion characteristics. In order to
determine relative and absolute displacement of links the respective elements
of left and right parts of equation should be equated and receive system of
algebraic equations the solution of which will enable to determine
displacements of mechanism links.
l 3  l 4 cos  43   l1 cos(  12   23 )  l 2 cos  23  l 0 cos(  01   12   23 )
l 4 sin  43  l 2 sin  23  l1 sin  12   23   l 0 sin  01   12   23 
sin  43   40    sin  01   12   23 
(5)
cos  43   40   cos  01   12   23 
Besides these equations, in order to solve the problem the subsidiary condition should be
added according to which the sum of internal angles of any five link is equal to 3π.
 01   12      23      34      40  3
(6)
.
,
Straight geometrical task
.
After transformations we get the following quadratic equation
C
2
sin 3   01   E
Here:
2
 cos
2
 34  2 AE cos  34  A  C sin 3   01   0
2
2
(7)
A  l 0  l1  l 2  l 3  l 4  2 l 2 l 3  2 l 0 l 3  2 l 0 l 2 cos 3   01 
2
2
2
2
2
B  2l3 l 4  2l 2 l 4
C  2l 0 l 4
E  B  C cos 3   01 
From the equation (7) we will obtain meanings of angelsϕ34 and ϕ23 that
determines position of the point C of the mechanism.
 34  arccos
2 AE 
4 A E  4  C sin  3   01   E
2
2
2
2  C sin  3   01
2
 23  arcsin
 C
 E 
2
2
sin  3   01   A
2
l 4 sin  34  l 0 sin 6    01   34   40
l1

2

(8)
(9)
And hence, in case of differentiating on time the received values of obtaining
equations, we shall receive values of speeds and acceleration of the links of
the mechanism.
The inverse geometrical task
In spite of the straight geometrical problem, here on the basis of the given
angels of rotation of the actuators mounted on the frame of the mechanism the
trajectory of the output link of the considered mechanical system is defined.
The formulation of inverse task of kinematics of five link mechanism is done
in the following way: the location of C point of mechanism i.e. its coordinates
in coordinate system connected with base, is given and it’s necessary to find
generalized coordinates of the mechanism which provide the location of C
point.
The design diagram of five link mechanism for inverse task
The inverse geometrical task
For this we take C point radius vector from the origin of coordinates and
represent it as the sum of two vectors:



R C  l1  l 2
(10)
Projections of these vectors in immovable coordinate system are expressed as:

l 1 cos  01
l1 
l 1 sin  01
l2 
l 2 cos   01   12 
l 2 sin   01   12 
(11)
In projections formula (10) will have the following form:
X C  l1 cos  01  l 2 cos  01   12
Y C  l1 sin  01  l 2 sin  01   12


(12)
The inverse geometrical task
In order to find two generalized and coordinates determining the location of
BC kinematic chain the expressions (12) should be squared and summed up:
X C  l1 cos  01  2 l1 l 2 cos  01 cos  01   12   l 2 cos
2
 01  2 l1 l 2 sin  01 sin  01   12   l sin
2
Y
2
2
2
2
C
 l sin
2
2
1
2
2
2
X
2
C
 01
  12 
 01   12 
(13)
 Y C  l1  2 l1 l 2 cos  12  l 2
2
2
2
The obtained expressions (13) allows to calculate values of angels  01 and  12
Hence, we will have:
co s  0 1 
l 2 YC sin  1 2  X C  l 2 co s  1 2  l1 
co s  1 2 
 l 2 co s  1 2
X
2
C
 l1   l sin  1 2
2
2
2
 YC  l1  l 2
2
2 l1l 2
2
(14)
2
2
(15)
The inverse geometrical task
We behave similarly when we determine the location of CDO kinematic chain.
We present radius vector of C point in the form of the following vectors sum:




R C  l0  l4  l3
(16)
And hence we can obtain:
cos  34 
cos  40 
X C
 l 0   YC  l3  l 4
2
2
2
2
(17)
2l3l 4
l 0  l 3 sin  40 sin  34  X C
(18)
l 3 cos  34  l 4
Based on derivations of the given formulas the values of velocities and
accelerations of the links of the investigated mechanism have obtained .
The inverse geometrical task
Based on usage of MATLAB software here are given curves of alternations of
phase angles of the five bar mechanism, when the two link junction point C is
performing movement along the circle.
The inverse geometrical task
Curves of alternation of angles
kuT xeebi
3
fi01
2.5
2
1.5
0
2
4
6
8
10
t
12
14
16
18
20
0
2
4
6
8
10
t
12
14
16
18
20
-1
fi12
-1.5
-2
-2.5
The inverse geometrical task
2.5
fi34
2
1.5
1
0
2
4
6
8
10
t
12
14
16
18
20
0
2
4
6
8
10
t
12
14
16
18
20
3
fi40
2.5
2
1.5
The inverse geometrical task
Values of angular velocities
kuT xur i
si Cqar eebi
1
ffi01
0
-1
-2
0
2
4
6
8
10
t
12
14
16
18
20
0
2
4
6
8
10
t
12
14
16
18
20
ffi12
0.5
0
-0.5
The inverse geometrical task
1
ffi34
0.5
0
-0.5
0
2
4
6
8
10
t
12
14
16
18
20
0
2
4
6
8
10
t
12
14
16
18
20
0.4
ffi40
0.2
0
-0.2
-0.4
The inverse geometrical task
Values of angular accelerations
kuT xur i aCqar ebebi
30
fffi01
20
10
0
-10
0
2
4
6
8
10
t
12
14
16
18
20
0
2
4
6
8
10
t
12
14
16
18
20
10
fffi12
0
-10
-20
-30
The inverse geometrical task
0.4
fffi34
0.2
0
-0.2
-0.4
0
2
4
6
8
10
t
12
14
16
18
20
0
2
4
6
8
10
t
12
14
16
18
20
0.5
fffi40
0
-0.5
-1
Kinetostatics of five bar planar
mechanisms
On the links of mechanical system are acting two
type of force factors - External forces and
Internal forces.
The internal forces – forces of weight, reduction
forces of inertia and moments of inertia of
force couples


Forces of inertia- F us   m l 2W s
Moments of inertia of force couples- M us   I s  z
Kinetostatics of five bar planar
mechanism
Reduction forces and moments of inertia acting on
the links of five bar mechanism
Determination of forces and
torques
Lagrange equation relatively to generalized
coordinate  4
d  T

dt    1

T

 M

1

D1
Lagrange equation relatively to generalized
coordinate  4
d  T


dt    4
 T

   M
4

D4
Determination of forces and
torques
Equitant for determination of torque acting on A kinematic pair.
I 11
d1
 I 14
d 4

1  I 11
dt
2 1
  I 14
1  I 44
 

 
2 1
4


  42  M


dt

2
1

 I 11

 1 4
4
D1
Equitant for determination of torque acting on O kinematic pair.
I 41
d 4
dt
 I 14
d1
dt
  I 14
1  I 11
 

 
2  4
1


1  I 44
2 

  12  M


 42 
4
D 4
 I 44
1
 1 4
Determination of force factors
Determination of torque acting on A kinematic pair.
90
80
70
60
M1p
50
40
30
20
10
0
-10
0
2
4
6
8
10
t
12
14
16
18
20
Determination of force factors
Determination of torque acting on O kinematic pair.
3
2
M4p
1
0
-1
-2
-3
0
2
4
6
8
10
t
12
14
16
18
20
Thank you