Example –mechanism with point to
curve constrain
A
B
C
0 D 1
2
3
w1= w2 = const.
e3 w1
w3
dvCB
dvC vB
vCB
vC
pv
c
b
IIAB
AB CD
d
ω AB v
B 11 B
=
= ω AB v
CB B
C
v v
v =
CD
C 3
= v
w
A
B
C
0 D 1
2
3
w1= w2 = const.
e3 w1
w3
w1
c
a
CBv
CBC CB t
CB n
CB t
B n
B t
C n
C
a a a a a a
a =
=
=
=
=
0 , since a v
CD , a v
2 n CB
CB 2
n C C
ε 0 since 0,
ε AB a
AB ,
a v
tB 1 12 n B
B
= = = =
CB 1
C CB CB
1 C
CB
= 2 ω v a = 2 ω v a
CB B
C
a a
a =
A
B
C
0 D 1
2
3
w1= w2 = const.
e3 w1
w3 t
CB C
CB n
B t
C n
C
a a a a
a =
b
B c
a
c
a
CBt
d. a
CBt
.
Cd a
pa
n
a
CA
B
C
0 D 1
2
3
w1= w2 = const.
e3 w1
w3
a
Ct
a
Cb
a
Bc
a
CBpa
t
a
CB na
Ct CB C
CB n
B t
C n
C
a a a a
a =
c
CD
t C 3
= a
e
w
3w
4v
CBp
vv
Bv
Bv
C=v
CDv
B= v
A+ v
BAv
A= 0
v
B= v
BA=
v
ww
3= v
CB/CB
w
4= v
CD/CD v
C= v
B+ v
CBv
C= v
D+ v
CDv
D= 0
v
C=v
CDd.v
CDd.v
CBv
CBExample 2
Example 2
w
3w
4a
B= a
BAca
BAca
CBna
CBna
CDna
CDna
Ca
CDta
CBtp
ae
4d.a
CBtd.a
CDtv
BAa
B= a
A+ a
BAn+ a
BAt+ a
BACa
A= 0
a
BAt= dv
BA/dt=0 , bo v
BA=v
w=const.
a
BAn= v
BA2/ = 0, bo →∞
a
BAC= 2w
3x v
BAa
B= a
BACa
C= a
B+ a
CBn+ a
CBta
CBn= w
32CB
a
C= a
D+ a
CDn+ a
CDta
CDn= w
42CD
a
D= 0 e
4= a
CDt/CD
d.a
CBtd.a
CDte
3= a
CBt/CB
e
3Complex planar mechanism
A
C B
E
D
F
v6
1 2
4
3
5
6
Known: v
6= v
FEF F
E
v v
v =
We have equation:
Direction of v
E???
Instant center ? Trajectory ?
P
A
C B
E
D
F
v6
1 2
4
3
5
6 d.vBAII d.vPB
d.vPA
d.vEF II d.vPE d.vPF
PE E
P
EF F
E
v v
v
v v
v
=
=
) (
EF PEF PE
EF F
P
v v v v v v
v = =
PF F
P
v v
v = Common direction
P
A
C B
E
D
F
v6
1 2
4
3
5
6 d.vBAII d.vPB
d.vPA
d.vEF II d.vPE d.vPF
PB B
P
BA A
B
v v
v
v v
v
=
=
) (
BA PBA PB
BA A
P
v v v v v v
v = =
PA A
P
v v
v = Common direction
=
=
=
PF F
P
PA PA
A P
v v
v
v v
v v
PF F
PA
v v
v =
P
A
C B
E
D
F
v6
1 2
4
3
5
6 d.vBA d.vPB
d.vPA
d.vEF d.vPE d.vPF
d.vBAII d.vPB d.vPA
d.vEFII d.vPE d.vPF
vF
vPF vPA=vP
pv
p f
P
A
C B
E
D
F
d.vC
v6
1 2
4
3
5
6 d.vBAII d.vPB
d.vPA
d.vEF II d.vPE d.vPF
d.vCP
=
=
=
CD CD
D C
CP P
C
v v
v v
v v
v
CD CP
P
v v
v =
d.vEF II d.vPE d.vPF
vF f
d.vC
vCP vC
vPF d.vCP
pv c
p
P
A
C B
E
D
F
d.vC
v6
1 2
4
3
5
6 d.vBAII d.vPB
d.vPA
d.vEF II d.vPE d.vPF
d.vCP
d.vEF II d.vPE d.vBAII d.vPB
d.vP
A
d.vPF vF f
d.vC
vCP vC
vPF d.vCP
pv b
c
p
Δbcp
ΔBCP~
P
A
C B
E
D
F
v6
1 2
4
3
5
6
t PE t
EF d da II a
t
daPF t
PB t
BA d da II a
t
daPA
n
daBA n
daPB
n
daPE n
daEF
t PF n
PE n
EF F
P
a a a a
a =
) (
tEF tPEn PE n
EF F
P
t PE n
PE t
EF n
EF F
PE E
P
t EF n
EF F
EF F
E
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
=
=
=
=
=
Common direction
t PA n
PB n
BA A
P
a a a a
a =
) (
tBA PBtn PB n
BA A
P
t PB n
PB t
BA n
BA A
PB B
P
t BA n
BA A
BA A
B
a a
a a
a a
a a
a a
a a
a a
a a
a a
a a
=
=
=
=
=
Common direction
P
A
C B
E
D
F
v6
1 2
4
3
5
6
t PE t
EF d da II a
t
daPF t
PB t
BA d da II a
t
daPA
n
daBA n
daPB
n
daPE n
daEF
n
aPE n
aEF
t
kaPF n
aBA n
aPB
t
kaPA
p pa
; a
; a
; a
EF ; a v
;
0
nPE nBA nPB2 n EF
EF A
F
=a = = = = =
a
t PA n
PB n
BA A
P
a a a a
a =
t PF n
PE n
EF F
P
a a a a
a =
P
A
C B
E
D
F
v6
1 2
4
3
5
6
t PE t
EF d da II a
t
daPF t
PB t
BA d da II a
t
daPA
n
daBA n
daPB
n
daPE n
daEF
j
k J
K
O
kO
jv
JKAcceleration – equivalent mechanism Cam pair (joint)
Velocity: v
J= v
K+ v
JKCam mechanism - velocity
1
2
A 0
B C
D d.vCB
w1
1
2
A 0
B C
D d.vCB
w1
vB d.vC
CB B
C
v v
v =
vB
vC vCB
p
vw2
CD
C 2
= v
w
Cam mechanism – equivalent mechanism (4 bar linkage)
1
2
0 A
O1
D O2
w1
w2
