Notes_03_04
page 1 of 9
POLYCENTRIC HINGE - SKELETAL DIAGRAM
D
3
4
F
C
8
G
B
5
E
6
A
H
7
2
1
1
POLYCENTRIC HINGE - VECTORS
R10
D
R4
C
R3
R9
R6
F
G R7
R8
R2
B
A
Vector
Position of
1
2
3
4
5
6
7
8
9
10
H wrt A
B wrt A
C wrt B
D wrt E
E wrt F
F wrt H
G wrt H
G wrt A
C wrt H
D wrt C
R5
Length
(mm)
21.67
26.69
26.00
14.00
17.83
22.47
22.03
3.23
31.33
24.62
R1
constant
constant
constant
constant
constant
constant
var
constant
constant
constant
H
Angle
(deg)
0.00
181.61
36.95
99.38
-15.87
146.31
171.56
92.21
151.64
15.36
R 2  R 3  R 9  R1  0
R 2  R 3  R 10  R 4  R 5  R 6  R 1  0
R 8  R 7  R1  0
E
constant
var
var
driver
var
9 - 5.33
var
2 - 89.40
var
3 - 21.59
Notes_03_04
page 2 of 9
Sewing Machine
Determine angles ,  and  as well as distance s for this sewing machine linkage at the position
shown below.

D
AD2 = AG2 + DG2

AD = 4.066 cm
D
C
tan GDA = AG / DG
GDA = 20.44
4

E

G
3
20.44
D
5
B
2

45
2
G
A
A
AB
BC
DC
CE
DE
EF
AG
DG
= 1.60 cm
= 3.57 cm
= 2.24 cm
= 1.60 cm
= 2.74 cm
= 3.81 cm
= 1.42 cm
= 3.81 cm
s


20.44
6

e
e
B
B
G
F
C
D
DCE = 90
CDE = 35.7
A
24.56
D
e2 = AB2 + AD2 - 2 (AB) (AD) cos 24.56
sin  / AB = sin 24.56 / e
 = 14.29

E
G
F
e2 = BC2 + CD2 - 2 (BC) (CD) cos 
 = 48.94
sin / BC = sin / e
 = 87.77 or 92.23 ??? check which one
BC2 = e2 + CD2 + - 2 e (CD) cos 
 = 92.23
GDA -  +  = CDG =  + CDE
 = 62.65
 + CDE +  +  = 180
 = 32.68
A
s
e = 2.694 cm

DE sin  = AG + EF sin 
 = 15.43
s + DG = DE cos  + EF cos 
s = 1.12 cm
Notes_03_04
page 3 of 9
Notes_03_04
page 4 of 9
Determine the angular velocity of links 2, 3, 4 and 5 as well as the velocity of needle 6 for the
sewing machine linkage as shown below when sewing at 4 stitches per second constant speed.
D
2 = 4 stitches/sec = +8 rad/sec
VB = AB 2 = 40.21 cps
C
62.7
 VB
VC
4
VC / B
AB 2
AB
?
CD
98.4

?
BC
E
32.7
3
VB / sin 48.9 = VC / sin 77.7 = VC/B / sin 53.4
VC = 52.31 cps
4 = VC / CD = 23.27 rad/sec CW
VC/B = 42.84 cps
3 = VC/B / BC = 12.00 rad/sec CW
5
constant
2
B
45 15.4
2
G
A
F
AB
BC
DC
CE
DE
EF
AG
DG
= 1.60 cm
= 3.57 cm
= 2.24 cm
= 1.60 cm
= 2.74 cm
= 3.81 cm
= 1.42 cm
= 3.81 cm
45
45
27.3
VB
53.4
VB
45
VC
77.7
32.7
VC
48.9
VC/B
DCE = 90
CDE = 35.7
6
8.4
VC/B
VE = DE 4 = 63.77 cps
62.7
VF
VE
VE
VF
15.4
VF
VF/E
78.1
VF/E
74.6
 VE
?
vertical

DE 4
DE
VF / E
?
EF
VE / sin 74.6 = VF / sin 78.1 = VF/E / sin 27.3
VF = 64.72 cps
VF/E = 30.34 cps
5 = VF/E / EF = 7.96 rad/sec CCW
Notes_03_04
page 5 of 9
Determine the angular velocity of links 2, 3, 4 and 5 as well as the velocity of needle 6 for the
sewing machine linkage as shown below when sewing at 4 stitches per second constant speed.
D
C
R4
D
C
R3
R1
62.7
4
E
32.7
3
B
R2
5
constant
2
B
45 15.4
2
G
A
F
6
A
AB
BC
DC
CE
DE
EF
AG
DG
= 1.60 cm
= 3.57 cm
= 2.24 cm
= 1.60 cm
= 2.74 cm
= 3.81 cm
= 1.42 cm
= 3.81 cm
DCE = 90
CDE = 35.7
2 = +8 rad/sec
r [cm]
1
4.07
2
1.60
3
3.57
4
2.24
 [deg]
110.4
135.0
57.3
8.4
3 = -r2 2 sin(2-4) / r3 sin(3-4) = -12.00 rad/s
4 = -r2 2 sin(2-3) / r4 sin(3-4) = -23.27 rad/s
2* = 4 = -23.27 rad/sec
r* [cm]
* [deg]
1
4.93
270
2
2.74
332.7
3
3.81
254.6
4
1.42
0
D
R2 *
E
3* = - r2* 2*cos(2*-1*) / r3* cos(3*-1*) = + 7.96 rad/s
r1 * = - r2* 2* sin(2*-3*) / cos(3*-1*) = + 64.71cps
5 = 3* = 7.96 rad/sec CCW
VF = 64.71 cps down
R1 *
F
G
R4 *
R3 *
Notes_03_04
page 6 of 9
Determine the angular velocity of links 2, 3, 4 and 5 as well as the velocity of needle 6 for the
sewing machine linkage as shown below when sewing at 4 stitches per second constant speed.
R4
D
C
C
D
R3
R7
R8
62.7
4
E
E
32.7
B
3
R2
5
A
constant
2
AB
BC
DC
CE
DE
EF
AG
DG
B
45 15.4
2
G
A
F
= 1.60 cm
= 3.57 cm
= 2.24 cm
= 1.60 cm
= 2.74 cm
= 3.81 cm
= 1.42 cm
= 3.81 cm
DCE = 90
CDE = 35.7
6
R 2  R 3  R 4  R 7  R1  0
r2 e
j 2
 r3 e
j 3
 r4 e
j 4
 r7 e
j 7
 r1e
j1
G
R1
R5
R6
F
1
2
3
4
5
6
7
8
r [cm]
1.42 constant
1.60 constant
3.57 constant
2.24 constant
3.81 constant
1.12 variable
3.81 constant
1.60 constant
 [deg]
180 constant
135 driver
57.3 variable
8.4 variable
254.6 variable
270 constant
90
constant
4 - 90 constraint
0
j
j
j
j r2  2 e 2  j r3  3 e 3  j r4  4 e 4  0
REAL:  r2  2 sin  2  r3  3 sin  3  r4  4 sin  4  0
IMAG: j r2  2 cos  2  j r3  3 cos  3  j r4  4 cos  4  0
R2  R3  R8  R5  R6  0
r2 e
j 2
 r3 e
j 3
 r8 e
j 8
 r5 e
j 5
 r6 e
j 6
0
j
j
j
j
j
j r2  2 e 2  j r3  3 e 3  j r8  8 e 8  j r5  5 e 5  j r6  6 e 6  0
FOR
REAL:  r2  2 sin  2  r3  3 sin  3  r8  4 sin 8  r5  5 sin  5  r6 cos  6  0
IMAG: j r2  2 cos  2  j r3  3 cos  3  j r8  4 cos  8  j r5  5 cos  5  j r6 cos  6  0
 8   4
Notes_03_04
 r3 sin  3
 r cos 
3
 3
 r3 sin  3

 r3 cos  3
r4 sin  4
0
 r4 cos  4
0
 r8 sin  8
 r5 sin  5
r8 cos  8
r5 cos  5
0
 3.004 0.327
 1.929  2.216
0

 3.004 1.583
3.673

0.234  1.012
 1.929
using MATLAB
 3    12.00 rad / sec 
  

 4   23.27 rad / sec 
   

 5    7.96 rad / sec 
 r6    65.07 cm / sec 
0  3   r2  2 sin  2 
  

0  4   r2  2 cos  2 

  

0  5   r2  2 sin  2 

1  r6   r2  2 cos  2 
0  3  28.43
 
0  4  28.43
 

0  5  28.43

1  r6  28.43
page 7 of 9
Notes_03_04
page 8 of 9
Determine the angular acceleration of links 2, 3, 4 and 5 as well as the acceleration of needle 6
for the sewing machine linkage as shown below when sewing at 4 stitches per second constant
speed.
D
C
62.7
4
E
3
32.7
5
constant
2
B
45 15.4
2
G
A
F
6
2 = +8 rad/sec
r [cm]
1
2
3
4
4.07
1.60
3.57
2.24
r3 sin 3
 r cos 
3
 3
R4
D
2 = 0
 [deg]
110.4
135.0
57.3
8.4
C
R3
R1
AB
BC
DC
CE
DE
EF
AG
DG
= 1.60 cm
= 3.57 cm
= 2.24 cm
= 1.60 cm
= 2.74 cm
= 3.81 cm
= 1.42 cm
= 3.81 cm
B
R2
A
DCE = 90
CDE = 35.7
angular velocities from velocity solution
 [rad/sec] r 2 [cpss] 
r [cpss]
[rad/s/s]
+25.13
-12.00
-23.27
+1010.6
+514.1
+1212.9
0
?
?
0
r4 sin  4  3   r2 2 sin  2  r2  22 cos  2  r3  32 cos 3  r4  24 cos  4 
 

 r4 cos  4   4   r2 2 cos  2  r2  22 sin  2  r3  32 sin 3  r4  24 sin  4 
3.004 0.327  3   1636.8 cpss 
 1.929  2.216     970.0 cpss 

  4  

3   0.3677  0.0543  1636.8 549.3 rad / s / s
   
  970.0    40.4 rad / s / s 

0
.
3201

0
.
4985



 

4
 
closed form
 = ( - r2  sin(2-4) - r2  2 cos(2-4) - r3  2 cos(3-4) + r4  2 ) / r3 sin(3-4) = 549.3 rad/s/s
3
3
2
2
4
 = ( - r2  sin(2-3) - r2  2 cos(2-3) - r3  2 + r4  2 cos(4-3)) / r4 sin(3-4) = 40.4 rad/s/s
3
4
2
2
4
Notes_03_04
page 9 of 9
D
R2 *
E
R1 *
F
R3 *
G
R4 *
2* = 4 = -23.27 rad/sec
r* [cm]
* [deg]
1
4.93
270
2
2.74
332.7
3
3.81
254.6
4
1.42
0
r3 sin 3
 r cos 
3
 3
2* = 4 = +40.4 rad/s/s
3* = 5 = -23.27 rad/sec
r 2 * [cpss]  * [rad/s/s]
 * [rad/sec]
r * [cpss]
-23.27
+7.96
+1483.7
+241.4
+40.4
?
+110.7
 cos 1  3   r2  2 sin  2  r2  22 cos  2  r3  32 cos 3 
 

 sin 1   r1   r2  2 cos  2  r2  22 sin  2  r3  32 sin 3 
 3.673 0  3   1203.6 cpss 
  1.012 1      1011.6 cpss 

  r1  

 3  0.2723 0  1203.6 cpss   327.7 rad/s/s 
 
  1011.6 cpss    - 680.0 cm/s/s 


0
.
2755
1
r


 

 1