112

AME 352 GRAPHICAL ACCELERATION ANALYSIS
P.E. Nikravesh 6-1
6. GRAPHICAL ACCELERATION ANALYSIS
In kinematic analysis of mechanisms, acceleration analysis is usually performed following a
velocity analysis; i.e., the positions and orientations, and the velocities of all the links in a
mechanism are assumed known. In this chapter we concentrate on one graphical method for
acceleration analysis of planar mechanisms.
We start this chapter with some exercises to ensure that the fundamentals of acceleration
analysis are well understood. You may review these fundamentals in Chapter 2 of these notes.
Exercises
In these exercises take direct measurements from the figures for link lengths and the
magnitudes of velocity and acceleration vectors. If it is stated that the angular velocity and
acceleration are known, assume ω = 1 rad/sec CCW and α = 1 rad/sec2
CW unless it is stated
otherwise. Write the position, velocity, and acceleration vector equations, and then graphically
determine the unknown acceleration(s).
P.1
Known: AA
, α and ω
Determine: AB
A B
A
A
P.2
Known: AA
and AB
Determine: α
A B
A
A
B
A
P.3
Known: AA
, α , and ω
Determine: AB
, AC
and ABC
A
B
C
AA
P.4
Known: AA
and AB
Determine: AC
A
B
C AA
B
A
P.5
Known: AA
, α , and ω
Determine: AB
, AC
and ABC
A B C
A
A
P.6
Known: VA
, VB
, AA
and AB
Determine: AC
A B CVA
BV
BA
AA
P.7
Known: ω , VBA
s
AA
, α , ABA
s
,
VBA
s
= 1 unit/sec positive, and
P.8
Known: ω , VBA
s
AA
, α , ABA
s
,
VBA
s
= 1 unit/sec positive, and

P.E. Nikravesh 6-2
ABA
s
= 1 unit/sec2
positive
Determine: AB
and AC
A B C
A
A
ABA
s
= 1 unit/sec2
negative
Determine: AB
and AC
A B C
A
A
P.9
Known: ω , VBA
s
, AA
, AB
,
VBA
s
= 1 unit/sec negative
Determine: α and AC
A B C
A
A
BA
P.10
Known: ωi
= 1 rad/sec CCW, ω j
= 1
rad/sec CW, αi
= 1 rad/sec2
CCW,
α j
= 1 rad/sec2
CW, and AA
Determine: AB
and AC
A B
C
(i)
(j)A
A
P.11
Known: ωi
, ω j
, AA
, and AC
Determine: AB
, αi
and α j
A B
C
(i)
(j)
AA
CA
Polygon Method
Four-bar Mechanism
For a known four-bar mechanism, in a given
configuration and known velocities, and a given angular
acceleration of the crank, α2
(say CCW), construct the
acceleration polygon. Determine α3
and α4
.
The position and velocity vector loop equations are:
RAO2
+RBA − RBO4
− RO4O2
= 0
VA +VBA − VB = 0
It is assumed that for this given configuration a velocity
analysis has already been performed (e.g., velocity
polygon) and all of the unknown velocities have been
determined.
A
BRBA
O2
O4
RAO2
RBO4
RO4O2
OV
VA
A
B
VBA
VB
The acceleration equation is obtained from the time derivative of the velocity equation as
AA +ABA = AB . Since RAO2
, RBA , and RBO2
are moving vectors with constant lengths, their
acceleration vectors have normal and tangential components:
AA
n
+AA
t
+ABA
n
+ABA
t
− AB
n
− AB
t
= 0
Or,
−ω2
2
RAO2
+α2RAO2
−ω3
2
RBA +α3RBA − (−ω4
2
RBO4
)−α4RBO4
= 0
We note that since ω2 , ω3 , ω4 , and α2
are known, AA
n
, AA
t
, ABA
n
, and AB
n
can completely be

P.E. Nikravesh 6-3
constructed. The remaining components, ABA
t
and AB
t
, have known axes but unknown
magnitudes. We rearrange the terms such that these unknown terms appear as the last
component in the equation:
−ω2
2
RAO2
+α2RAO2
−ω3
2
RBA − (−ω4
2
RBO4
)+α3RBA −α4RBO4
= 0
Acceleration polygon
1. Select a point in a convenient position as the
reference for zero acceleration. Name this point
OA
(origin of accelerations).
2. Compute the magnitude of AA
n
as RAO2
ω2
2
. From
OV
construct vector AA
n
in the opposite direction
of RAO2
.
3. Compute the magnitude of AA
t
as RAO2
α2 . The
direction of AA
t
is determined by rotating RAO2
90o
in the direction of α2 . Add this vector to AA
n
.
Note that the sum of AA
n
and AA
t
is AA .
3. Compute the magnitude of ABA
n
as RBAω3
2
. Add
this vector in the opposite direction of RBA to the
other two vectors.
4. Compute the magnitude of AB
n
as RBω4
2
. Note that
AB
n
is in the opposite direction of RBO4
. Since
AB
n
itself appears with a negative sign in the
acceleration equation, it should be added to the
other vectors in the diagram as shown; i.e., head-
to-tail.
5. Since ABA
t
must be perpendicular to RBA , draw a
line perpendicular to RBA in anticipation of
adding ABA
t
to the diagram.
6. Since AB
t
must be perpendicular to RBO4
, draw a
line perpendicular to RBO4
closing (completing)
the polygon.
7. Construct vectors ABA
t
and AB
t
on the polygon.
8. Determine the magnitude of ABA
t
from the
polygon. Compute α3
as α3 = ABA
t
/ RBA (in this
diagram it is CW).
9. Determine the magnitude of AB
t
from the
as α4 = AB
t
/ RBO4
(in this
diagram it is CCW).
A
O2
RAO2
OA
AA
n
AA
t
RBA
A
B
OA
ABA
n
AA
n
AA
t
AB
n
RBO4
RBA
A
B
OA
ABA
n
AA
n
AA
t
RBO4
AB
n
ABA
t
AB
t
ABA
n
AA
n
AA
t
AB
n

P.E. Nikravesh 6-4
Secondary equation(s)
We can use the polygon method to determine the
acceleration of a coupler point, such as P. It is
assumed that all the angular velocities and
accelerations have already been determined.
For the position vector RPO2
= RAO2
+ RPA
, the
acceleration expression becomes
AP
= AA
+ APA
= AA
n
+ AA
t
+ APA
n
+ APA
t
= −ω2
2
RAO2
+α2
RAO2
−ω3
2
RPA
+α3
RPA
All four vectors can be constructed graphically. The
vector sum is the acceleration of P.
OA
A
APA
n
APA
t
AP
AA
n
AA
t
A
B
RPA
P
O2
RAO2 RPO2
x
y
Example FB-AP-1
This is a continuation of
Example FB-VP-1. Assume
an angular acceleration of
α2
= 1 rad/sec2
CW for the
crank.
Acceleration polygons are
constructed and the following
accelerations are obtained
from the polygons: α3
= 0.14
CW, α4
= 0.46 CW, AP
= 1.7
in the direction shown on the
polygon.
A
P
O2
B
O4
O
ABA
t
AB
t ABA
n
AA
n
AA
t
AB
n
O
AA
n
AA
t
AP
APA
n
APA
t
Slider-crank (inversion 1)
For a known slider-crank mechanism (inversion 1) in a given configuration and for known
velocities, the acceleration of the crank, α2
, is given. Construct the acceleration polygons, then
determine α3
and the acceleration of the slider block. Assume α2
is given to be CCW.
The position and velocity vector loop equations are:
RAO2
+RBA − RBO2
= 0
VA +VBA − VB = 0
Assume that all the velocities have already been
obtained.
The acceleration equation is obtained from
the time derivative of the velocity equation:
AA +ABA − AB = 0
VBA
VB
OV
VA
A
B
RBA
O2
RAO2
RBO2

P.E. Nikravesh 6-5
Or, in terms of the components of the acceleration vectors, we have
AA
n
+AA
t
+ABA
n
+ABA
t
− AB
s
= 0
Or,
−ω2
2
RAO2
+α2RAO2
−ω3
2
RBA +α3RBA − AB
s
= 0
The first three components are completely and the last two components are partially known.
1. Select a point in a convenient position as the reference
for zero acceleration, OA
.
2. Compute AA
n
= RAO2
ω2
2
. From OV
construct AA
n
in the
opposite direction of RAO2
.
3. Compute AA
t
= RAO2
α2 . The direction of AA
t
is
determined by rotating RAO2
90o
in the direction of
α2 . Add this vector to the diagram.
A
O2
RAO2
OA
AA
n
AA
t
4. Compute ABA
n
= RBAω3
2
. Construct ABA
n
in the opposite
direction of RBA .
5. ABA
t
must be perpendicular to RBA . Draw a line
perpendicular to RBA in anticipation of adding ABA
t
to
ABA
n
.
6. From OA
draw a line parallel to the sliding axis. AB
must reside on this line.
RBA
B
A
O2
OA
AA
n
AA
t
ABA
n
7. Construct vectors ABA
t
and AB .
8. Determine the magnitude of ABA
t
. Compute α3
as
α3 = ABA
t
/ RBA . Determine the direction of α3
(in this
example it is CCW).
9. Determine the magnitude of AB from the polygon.
The direction in this example is to the left.
ABA
t
AB
s
O
A
AA
n
AA
t
ABA
n
Example SC_AP-1
Continue with Example SC-VP-1
from the velocity polygon section.
Assume an angular acceleration of
α2
= 1 rad/sec2
CCW for the crank.
Using the results from the
velocity analysis, the acceleration
polygon is constructed. The results
are: α3
= 1.0 rad/sec2
CCW;
AB
= 0.76 to the left.
A
O2
B
O
AA
n
AA
t
ABA
n
ABA
t
AB

P.E. Nikravesh 6-6
For a known slider-crank mechanism
(inversion 2), in a given configuration and for
known velocities, the acceleration of the crank, α2
,
is given (say CW). Construct the acceleration
polygons and determine α3
.
We draw the slider-crank in the given
configuration and define position vectors to form a
vector loop equation:
RAO2
− RAO4
− RO4O2
= 0
The velocity equation is:
VAO2
t
− VAO4
t
− VAO4
s
= 0
VAO4
s
VAO4
t
OV
VAO2
A
O2
RAO2
O4
RAO4
RO4O2
The velocity polygon for this mechanism has already been obtained; i.e.,ω4 and VAO4
s
are
assumed known.
The acceleration equation is obtained from the time derivative of the velocity equation:
AAO2
n
+ AAO2
t
− AAO4
n
− AAO4
t
− AAO4
s
− AAO4
c
= 0
Or,
−ω2
2
RAO2
+α2RAO2
− (−ω4
2
RAO4
)−α4RAO4
− AAO4
s
− 2ω4VAO4
s
= 0
All the terms are fully known except for AAO4
s
and AAO4
t
. Re-arranging the terms in order to have
the partially known terms as the last two terms:
−ω2
2
RAO2
+α2RAO2
− (−ω4
2
RAO4
)− 2ω4VAO4
s
− AAO4
s
−α4RAO4
= 0
1. Select the origin of accelerations, OA
, in a
convenient position.
2. Compute AA
n
= RAO2
ω2
2
. From OV
construct vector
AA
n
in the opposite direction of RAO2
.
3. Compute AA
t
= RAO2
α2 . Determine the direction
of AA
t
. Add this vector to AA
n
.
4. Compute AAO4
n
= RAO4
ω4
2
. Construct vector AAO4
n
and add it to the
polygon.
5. Determine the Coriolis acceleration AAO4
c
. The
magnitude of this vector is 2VAO4
s
ω4 , and its
direction is found by rotating VAO4
s
90o
in the
direction of ω4
. Add this vector to the polygon.
OA
AAO2
n
AAO2
t
O2
RAO2
O4
RAO4
RO4O2
AAO4
n
AAO4
c
OA
AAO2
n
AAO2
t
O4
RAO4

P.E. Nikravesh 6-7
6. Draw an axis for AAO4
s
parallel to RAO4
.
7. Draw another axis perpendicular to RAO4
. AAO4
t
will be on this axis.
8. Construct vectors AAO4
s
and AAO4
t
to complete the
polygon.
AAO4
n
AAO4
c
OA
AAO2
n
AAO2
t
O4
RAO4
9. Determine the magnitude of AA4O4
t
from the
as α4 = AA4O4
t
/ RA4O4
.
Determine the direction of α4
(in this example it
is CW).
10. Determine the magnitude of AAO4
s
from the
polygon.
AAO4
t
AAO4
s
AAO4
n
AAO4
c
OA
AAO2
n
AAO2
t
Secondary point
In order to determine the acceleration of point P
on link 4, we express its acceleration as
RP = RO4O2
+ RPO4
AP = APO4
= −ω4
2
RPO4
+α4RPO4
Since the angular velocity and acceleration of link 4
are both known, the two components of the
acceleration vector can be constructed.
RPO4
O2
RO4O2 O4
P
y
x
AP
APO4
n
APO4
t
Example SC-AP-2
This is a continuation of Example SC-VP-2.
Assume an angular acceleration of 0.5 rad/sec2
,
CCW, for the crank.
The acceleration polygon is constructed and the
following accelerations are determined from the
polygon: α3
= α4
= 0.24 rad/sec2
, CCW; AAO4
s
= 3.9
in the direction shown.
The acceleration of point P is determined from a
second polygon. This acceleration has a magnitude
of AP
= 1.0 in the direction shown.
O2 O4
P
A
AAO4
n
O
AAO2
n
AAO2
tAAO4
c
AAO4
s
AAO4
t
O
AP

P.E. Nikravesh 6-8
For a known slider-crank mechanism (inversion 3), in a given configuration and for known
angular velocity and acceleration of the crank, ω2 and α2
(assume both CW), construct the
velocity and acceleration polygons, and then determine ω3 and α3
.
Draw the slider-crank in the given configuration and
define position vectors to form a vector loop equation:
RAO2
+ RO4A − RO4O2
= 0
The velocity equation is
VAO2
t
+ VO4A
t
+ VO4A
s
= 0
The velocity polygon for this mechanism has already been
obtained. From this polygon, ω3 and the velocity of the
slider-block have been determined.
O2 O4
(3)
(4)
A
RAO2
RO4O2
RO4 A
VAO2
VO4A
t
VO4A
s
The acceleration equation is obtained from the time
derivative of the velocity equation:
AAO2
n
+AAO2
t
+ AO4A
n
+ AO4A
t
+AO4A
s
+ AO4A
c
= 0
or,
−ω2
2
RAO2
+α2RAO2
−ω3
2
RO4A + 2ω3VO4A
s
+AO4A
s
+α3RO4A = 0
1. Select the origin of accelerations, OA
, in a
convenient position.
2. Compute AA
n
= RAO2
ω2
2
. From OV
construct vector
AA
n
.
3. Compute AA
t
= RAO2
α2 . Determine the direction of
AA
t
based on the direction of α2 . Add this vector to
AA
n
.
4. Compute AO4A
n
= ω3
2
RO4A . The direction of AO4A
n
is
opposite of RO4A . Add this vector to the polygon.
5. Determine the Coriolis acceleration AO4A
c
. The
magnitude of this vector is 2ω3VO4A
s
, and its direction
is found by rotating VO4A
s
90o
in the direction of ω3
.
Add this vector to the polygon.
6. Draw an axis for AO4A
s
parallel to RAO4
.
7. Draw an axis for AO4A
t
perpendicular to RAO4
.
O2
A
RAO2
OA
AAO2
n
AAO2
t
RO4 A
O4
(3)
(4)
A
AO4 A
n
AO4 A
c
RO4 A
O4
(3)
(4)
A
AO4 A
n
AO4 A
c

P.E. Nikravesh 6-9
8. Construct vectors AAO3
t
and AO3O4
s
.
9. Determine the magnitude of AO4A
t
from the polygon.
Compute α3
as α3 = AO4A
t
/ RO4A .
10. Determine the direction of α3
. In this example it is
CW since RO4A must rotate 90o
CW to line up with
AO4A
t
.
AO4 A
t AO4 A
s
AO4 A
n
AAO2
t
AAO2
n
AO4 A
c
Example SC-AP-3
This is a continuation of Example SC-
VP-3. Assume an angular acceleration of
1 rad/sec2
CW for the crank.
Acceleration polygon (right) is
constructed and the following accelerations
are determined from the polygon:
α3
= α4
= 2.67 rad/sec2
CW; AO4 A
s
= 3.8,
and AP
= 0.4 in the direction shown.
A second polygon (left) is constructed
to determine the acceleration of point P as:
AP = 2.7 in the direction shown.
O
AO4 A
t
AO4 A
s
AO4 A
n
AAO2
t
AAO2
n
AO4 A
c
O
APA
t
APA
n
AAO2
t
AAO2
n
AP
Exercises
In these exercises take direct measurements from the figures for link lengths and the
magnitudes of velocity and acceleration vectors.
Exercises P.1 – P.4 are examples of four-bar mechanism. Assume ω2
and α2
are given.
Determine α3
, α4
, and AP
.
P.1
(2)
(3)
(4)
P
P.2
(2)
(3)
(4)
P

P.E. Nikravesh 6-10
P.3
(2)
(3)
(4)
P
P.4
(2)
(3)
(4)
P
Exercises P.5 – P.8 are examples of slider-crank mechanism. Assume ω2
and α2
are given:
For P.5 and P.6 determine α3
, α4
, and the acceleration of the slider block; For P.7 and P.8
determine α3
, α4
, and AP
.
P.5
(2)
(3)
(4)
P.6
(2)
(3)
(4)
P.7
(2)(3)
(4)
P
P.8
(2)
(3)
(4)
P
P.9
For this six-bar mechanism ω2
and α2
are given. Determine α5
,
acceleration of P, and the
acceleration of the slider block 6. (6)
(2)
(3)
(4)
P
(5)
P
P.10
For this six-bar mechanism ω2
and α2
are given. Determine α5
and
the acceleration of the slider block 6.
(2)
(3)
(4)
(5)
(6)

112

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