Kinematic chain notations
Kinematic scheme Structural scheme
GRAF
Matrix of the structure
Contours
1
4
0
2
5
3B
C
E FG
AD
C
FE I
I
2
DA
B
I
I
13
5
I
I
G
4
0
I0
1
2
3
4
5
-
A
-
D
-
G
A
-
B
-
-
-
-
B
-
C
E
-
D
-
C
-
-
-
0 1 2 3 4 5
-
-
E
-
-
F
G
-
-
-
F
-
5
3F
D
G
C4
2
A
B
E1
0
K1= 0 –A – 1 – B – 2 – C – 3 – D – 0
K2= 0 – D – 3 – C – 2 – E – 4 – F – 5 – G - 0
K3(outer)= 0 –A – 1 – B – 2 – E – 4 – F – 5 – G - 0
Using Graf
Find direction of VK
3
2
20
B
20
K
BS
v
KS
v
AB 1Bv
Determining the motion of the mechanism:
2
1
43
S14
S13
F1F2F3
VELOCITY AND ACCELERATION
VECTOR EQUATIONS – v and a polygons
(2D) – M & N – points of one link
0t
v and a vector equations (2D complex motion)
NMMN vvv
MNkNM rωv
t
NM
n
NMMNMMN aaaaaa
MNkMNkk
n
NM rrωωa2
MNk
t
NM rεa
velocity:
acceleration:
B
AC
2
pv
vB
d.vCB
vC
vCB
2
14
3
d.vBA
vB
d.vCB
d.vC
d.vC
vC
Data: 2 = const, e2 = 0
Find: vB, vC, 3 , aB, aC, e3
vB = vA + vBA
vA = 0
vBA = 2 BA
vB = vBA= 2 BA
vC = vB + vCB
3 vCB /CB
aB = aA + aBAn + aBA
t
aA = 0
aBAn = 2
2 BA
aBAt = e2 BA =0
aC = aB + aCBn + aCB
t
aCBn = 3
2 CB
aCBt = e3 CB
e3 aCBt /CB
3
pa
d.aCBt
d.aC
d.aCBt
aC
aCBn
aCBn
aB= aBAn
d.aC
aCBt
aC
aCBte3
aBAn = aB
Example v and a (crank-slider)
A
B
C
K
D0
1
2
31 const
ABωv 1B1B ABωv
A
B
C
K
D0
1
2
31 constdvC
dvCB
CBBC vvv dvCBdvC
vB vCB
vCpv
c
b
A
B
C
K
D0
1
2
31 const
dvKB
dvKC
KCCKBB
KCCK
KBBK
vvvv
vvv
vvv
vB
vCB
vCpv
c
b
kdvKBvKB
vK
dvKC
A
B
C
K
D0
1
2
31 const
kvKB
kvBC
kvKC
dvCB
vB
vCB
vCpv
c
b
k
dvKC
dvKBvKB
vKkc
KC
bk
BK
bc
BC
ΔbckΔBCK
~
Triangles BCK and bck
are similar
A
B
C
K
D0
1
2
31 const
t
B
n
BB aaa
AB
vABωa)(
2
B2
1
n
B11
n
B ABωωa
0AB0ABεa 1
t
B1
t
B ABεa
A
B
C
K
D0
1
2
31 const
Ca
t
CBa
c
b
Cda
Ban
CBa
CBa
t
CBda
n
CBda
pat
CB
n
CB
t
B
n
BC
CBBC
aaaaa
aaa
CB
va
0since0a,AB
va
2
CBn
CB
1
t
B
2
2
1
n
BB
e AB
A
B
C
K
D0
1
2
31 const
KCCKBB
KCCK
KBBK
aaaa
aaa
aaa
t
KC
n
KCC
t
KB
n
KBB
a
a
aa
aa
KC
va,
KB
va
2
KCn
KC
2
KBn
KB t
KBa
Ca
n
KBa
pac
bk
Ba
n
KCa
t
KCa
CBa
t
KBdat
KCda
ΔbckΔBCK
polygon) accel its and(link similarity
~
A
B
C
K
D0
1
2
31 const
Ca pac
bk
Ba
VELOCITY AND ACCELARATION – extension
In last 2D example: M and N – points of one link
NMMN vvv
t
NM
n
NMMNMMN aaaaaa
(2D) – points J and K belong to j and k
KJJK vvv
ρωv k
j
KJ
C
KJ
t
KJ
n
KJJKJJK aaaaaaa
ρρωωa2
k
j
k
j
k
jn
KJ
ρεa k
jt
KJ
KJj
C
KJ vωa 2
- radius of curvature
dir
C
KJ
t
KJJKJJK aaaaaa
02 ρρωωa k
j
k
j
k
jn
KJ
Since
we have
dir