Acceleration Analysis

Acceleration Analysis
Chapter 7

Definition
• Rate of change of velocity with respect to
time
– Angular
– Linear
– Position Vector
A
d
dt
dV
dt
R
PA
pe
j
– Velocity
V
pa
R
dt
PA
pje
j
d
dt
pe
j

Definition
dt
je
p
d
dt
d
j
PA
PA
A
A
sin
cos
cos
sin
2
j
p
j
p
PA
A
"Absolute"
PA A P
A
dt
d
je
dt
d
e
jp
j
j
PA
A
j
j
PA
e
p
je
p
2
A
PA
n
PA
t
PA
A
A
A

Definition
– Acceleration (difference)
A A A
P
A
PA
t n
t n
t n
A A A A A A
P
P
A
A
PA
PA
– Relative Acceleration
A
PA
A
P
A
A
"on thesame body "

Graphical Analysis
• Graphical Acceleration Analysis
A
A
A
P
t
n
A
A t
A n
A
– Solve for
A
r
PA
2
r
angular acceleration ; 3
, 4
linear acceleration; A , B,
C

Graphical Analysis
• Example 7-1
– Given θ 2 , θ 3 , θ 4 , ω 2 ,ω 3 , ω 4 ,α 2
find α 3 , α 4 , A A , A B and A C
– Velocity analysis already
performed
– 1. Start at the end of the
linkage about which you have
the most information.
Calculate the magnitude of
the acceleration of point A,
A n AO 2 A 2
2
A t A
AO 2
2

Graphical Analysis
• Example 7-1
– 2. Draw the acceleration A A
– 3. Move next to a point which
you have some information,
point B. Draw the
construction line pp through
B perpendicular to BO 4
– 4. Write the acceleration
difference equation for point
B vs. A
A
B
A
A
A
BA
t n
t n
t n
A A A A A A
B
2 B
4
B
A
A
BA
BA
A
n
A n B
BO 4

Graphical Analysis
– 5. Draw construction line qq
through point B and
perpendicular to BA to
represent the direction of A BA
A n BA
BA
2 3
– 6. The vector equation can be
solve graphically by drawing
the following vector diagram
A
A
t n t n t n
A
A
A
A
A
A
t
A
B
B
n
A
A
B
A
A
A
n
BA
BA
A
n
A
t
A A
A &
t BA
AB
?
B
BA
BA

Graphical Analysis
– 7. The angular velocities of link
3 and 4 can be calculated,
4
A t B
BO
4
3
A t BA
BA
– 8. Solve for V C
A A A
C
A
c 3
A t CA 2
A n CA c 3
CA

Graphical Analysis

Analytical Analysis

Analytical Solution
• Fourbar Pin-Joint Linkage
– Vector Loop
R
2
R3
R4
R1
0
– Position
ae
j
2 j3
j4
j1
be
ce
de
0
– Velocity
– Acceleration
ja
j2
j3
j4
2e
jb3e
jc4e
0
j
2 2 j2
j3
2 j3
j4
2 j4
a
je a
e b
je b
e c
je c
e 0
2 2
3
3
4
4

Analytical Solution
• Fourbar Pin-Joint Linkage
A
A
A
BA
A
B
0
A
A
A
A
BA
B
t n
j
2 2 j2
A
A
A
A
a
2
je a2e
t n
j
3 2 j3
ABA
ABA
b3
je b3
e
t n
j
4 2 j4
A A c
je c
e 0
B
B
4
Euler identity
real part
imaginary part
4

• Fourbar Pin-Joint Linkage
Analytical Solution
4
2
4
3
2
3
2
2
2
2
2
3
4
cos
cos
cos
sin
sin
sin
c
b
a
a
C
b
B
c
A
BD
AE
AF
CD
3
BD
AE
BF
CE
4
4
2
4
3
2
3
2
2
2
2
2
3
4
sin
sin
sin
cos
cos
cos
c
b
a
a
F
b
E
c
D

Analytical Solution
• Fourbar Pin-Joint Linkage
A
A
A
BA
A
B
0
A
A
A
A
BA
B
a
c
2
b
4
2
sin
2
j cos
2
a2
cos
2
j sin
2
2
3
sin3
j cos
3
b3
cos
3
j sin3
2
sin
j cos
c
cos
j sin
4
4
4
4
4

Analytical Solution
R
– Slider-Crank
2
R3
R4
R1
0
ae
j
2 j3
j4
j1
be
ce
de
0
ja
j2
j3
2e
jb3e
d
0
j
2 2 j2
j3
2 j3
a
je a
e b
je b
e d
0
2 2
3
3

Analytical Solution
A
A
A
A
B
– Slider-Crank
BA
A
A
t n
j
2 2 j2
A
A
A
A
a
2
je a2e
t n
j
j
A A b je b e
3 2 3
A
t
B
A
A
A
B
BA
d
BA
A
BA
B
BA
A BA
A
A
A
0
BA
3
3

Analytical Solution
– Slider-Crank
3
2
3
3
3
2
2
2
2
2
3
3
2
3
2
2
2
2
2
3
cos
sin
cos
sin
cos
sin
sin
cos
b
b
a
a
d
b
b
a
a
Euler identity
part
real
imaginary part

Analytical Solution
• Coriolis Acceleration
– Is an additional component
of acceleration present when
a sliding joint is on a rotating
link
– Thy position vector is both
rotating and changing length
as the system moves
R
P
pe
j 2

Analytical Solution
• Coriolis Acceleration
– The velocity expression is
V
P
V
P
p
e
V
– The acceleration,
2
j
2
V
p
e
Ptrans Pslip
j
2
A
P
p
j2
2 2 j2
2
je p2
j e 2
j2
p
je
2
p
e
j
2
A
P
A
Ptangetial
A
Pnormal
A
Pcoriolis
A
Pslip

Analytical Solution
• Coriolis Acceleration
– The acceleration,
A
P
p
j2
2 2 j2
2
je p2
j e 2
j2
p
je
2
p
e
j
2
A
P
A
Ptangetial
A
Pnormal
A
Pcoriolis
A
Pslip

Acceleration of Any Point
• Once the angular acceleration of all the links are
found it is easy to define and calculate the
acceleration of any point on any link for any input
position of the linkage

Acceleration of Any Point
• To find the acceleration of points S
2 2
scos
j
R S 0
R S
sin
A S
se j
2 2 2
2
2
V S
2
jse j
s
sin
s
2
2
2
sin
2
s
j
cos
2 2 2 2
cos
2
2
j cos
2
2
j sin
2
2
2
2
2
2

Acceleration of Any Point
R U
• To find the acceleration of points U
4 4
ucos
j
ue j
4 4 4
4
4
0
sin
V U
jse j
4
sin
4
u
j
4 4 4 4
cos
4
4
A U
u
4
u
sin
2
4
cos
4
4
j cos
4
4
j sin
4
4
4
4

Acceleration of Any Point
R
R
• To find the acceleration of point P
PA
P
A
A
V
V
P
PA
pe
R
PA
P
A
j
3
pcos
j
3
R
jpe
A
V
A
A
p
3
p
PA
j
3 3
sin
3
p
sin
j
3
V
A
2
3
PA
PA
sin
cos
3 3
3 3
cos
3
3
j cos
3
3
j sin
3
3
3
3
3
3
3
3

Important topics to review
• Jerk –page 367-369
• Human Tolerance to Acceleration – page 364-366
• Fourbar Inverted Slider-Crank – page 354-360
• Acceleration Analysis of the Geared Fivebar- page
361 -362