6 Acceleration Polygon

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: A A , α and ω

Determine: AB

A B

AA

P.2 Known: A A and AB Determine: α

A B AA

BA

P.3

Known: A A , α , and ω

Determine: AB , AC and ABC

A

B

C

AA

P.4 Known: A A and AB

Determine: AC

A

B

C AA

BA

P.5

Known: A A , α , and ω

Determine: AB , AC and ABC

A B C AA

P.6 Known: VA , VB , A A and AB

Determine: AC

A B C VA

BV

BA

AA

P.7

Known: ω , VBAs A A , α , ABA

s ,

VBAs = 1 unit/sec positive, and

P.8 Known: ω , VBA

s A A , α , ABAs ,

VBAs = 1 unit/sec positive, and


P.E. Nikravesh 6-2

ABAs = 1 unit/sec2 positive

Determine: AB and AC

A B C

AA

ABAs = 1 unit/sec2 negative


A B C

AA

P.9

Known: ω , VBAs , A A , AB ,

VBAs = 1 unit/sec negative

Determine: α and AC

A B C

AABA

P.10

Known: ω i = 1 rad/sec CCW, ω j = 1

rad/sec CW, α i = 1 rad/sec2 CCW,

α j = 1 rad/sec2 CW, and A A


A B

C

(i)

(j) AA

P.11

Known: ω i , ω j , A A , and AC

Determine: AB , α i 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.

ABRBA

O2O4

RAO2

RBO4

RO4O2

O V

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:

AAn +AA

t +ABAn +ABA

t − ABn − AB

t = 0

Or,

−ω2

2RAO2+α2RAO2

−ω 32RBA +α 3RBA − (−ω 4

2RBO4)−α 4RBO4

= 0

We note that since ω2 , ω 3 , ω 4 , and α2 are known, AAn , AA

t , ABAn , and AB

n can completely be


P.E. Nikravesh 6-3

constructed. The remaining components, ABAt 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

2RAO2+α2RAO2

−ω 32RBA − (−ω 4

2RBO4)+α 3RBA −α 4RBO4

= 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 AAn as RAO2

ω22 . From

OV construct vector AAn in the opposite direction

of RAO2.

3. Compute the magnitude of AAt as RAO2

α2 . The

direction of AAt is determined by rotating RAO2

90o in the direction of α2 . Add this vector to AAn .

Note that the sum of AAn and AA

t is AA .

3. Compute the magnitude of ABAn as RBAω 3

2 . Add

this vector in the opposite direction of RBA to the

other two vectors.

4. Compute the magnitude of ABn as RBω 4

2 . Note that

ABn is in the opposite direction of RBO4

. Since

ABn 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 ABAt must be perpendicular to RBA , draw a

line perpendicular to RBA in anticipation of

adding ABAt to the diagram.

6. Since ABt must be perpendicular to RBO4

, draw a

line perpendicular to RBO4 closing (completing)

the polygon.

7. Construct vectors ABAt and AB

t on the polygon.

8. Determine the magnitude of ABAt from the

polygon. Compute α3 as α 3 = ABAt / RBA (in this

diagram it is CW). 9. Determine the magnitude of AB

t from the

polygon. Compute α4 as α 4 = ABt / RBO4

(in this

diagram it is CCW).

A

O2

RAO2

O A

AAn

AAt

RBAAB

O A

ABAn

AAn

AAt AB

n

RBO4

RBAAB

O A

ABAn

AAn

AAt

RBO4

ABn

ABAt

ABt

ABAn

AAn

AAt

ABn


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 R PO2

= R AO2+ R PA , the

acceleration expression becomes

AP = A A + APA = A An + A A

t + APAn + APA

t

= −ω22R AO2

+α2R AO2−ω3

2R PA +α3R PA

All four vectors can be constructed graphically. The vector sum is the acceleration of P.

O A

AAPA

n

APAt

AP

AAn

AAt

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

A Bt ABA

n

AAn

AAt

ABn

O

AAn

AAt

AP

APAn 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

VBO V

VAA

B

RBA

O2

RAO2

RBO2


P.E. Nikravesh 6-5

Or, in terms of the components of the acceleration vectors, we have

AAn +AA

t +ABAn +ABA

t − ABs = 0

Or,

−ω2

2RAO2+α2RAO2

−ω 32RBA +α 3RBA − AB

s = 0

The first three components are completely and the last two components are partially known. Acceleration polygon 1. Select a point in a convenient position as the reference

for zero acceleration, OA .

2. Compute AAn = RAO2

ω22 . From OV construct AA

n in the

opposite direction of RAO2.

3. Compute AAt = RAO2

α2 . The direction of AAt is

determined by rotating RAO2 90o in the direction of

α2 . Add this vector to the diagram.

A

O2

RAO2

O A

AAn

AAt

4. Compute ABAn = RBAω 3

2 . Construct ABAn in the opposite

direction of RBA .

5. ABAt must be perpendicular to RBA . Draw a line

perpendicular to RBA in anticipation of adding ABAt to

ABAn .

6. From OA draw a line parallel to the sliding axis. AB

must reside on this line.

RBA

B

A

O2

O A

AAn

AAt

ABAn

7. Construct vectors ABAt and AB .

8. Determine the magnitude of ABAt . Compute α3 as

α 3 = ABAt / 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.

ABAt

ABs

O A

AAn

AAt

ABAn

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

AAn

AAt

ABAn

ABAt

AB


P.E. Nikravesh 6-6

Slider-crank (inversion 2)

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

O V VAO2

A

O2

RAO2

O4

RAO4

RO4O2

The velocity polygon for this mechanism has already been obtained; i.e.,ω 4 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

2RAO2+α2RAO2

− (−ω 42RAO4

)−α 4RAO4− AAO4

s − 2ω 4VAO4

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:

−ω22RAO2

+α2RAO2− (−ω 4

2RAO4)− 2ω 4VAO4

s

− AAO4

s −α 4RAO4= 0

Acceleration polygon

1. Select the origin of accelerations, OA , in a

convenient position.

2. Compute AAn = RAO2

ω22 . From OV construct vector

AAn in the opposite direction of RAO2

.


α2 . Determine the direction

of AAt . Add this vector to AA

n .

4. Compute AAO4

n = RAO4ω 4

2 . Construct vector AAO4

n

in the opposite direction of RAO4 and add it to the

polygon.

5. Determine the Coriolis acceleration AAO4

c . The

magnitude of this vector is 2VAO4

s ω 4 , and its

direction is found by rotating VAO4

s 90o in the

direction of ω4 . Add this vector to the polygon.

O A

AAO2

n

AAO2

t

O2

RAO2

O4

RAO4

RO4O2

AAO4

n

AAO4

cO A

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

cO A

AAO2

n

AAO2

t

O4

RAO4

9. Determine the magnitude of AA4O4

t from the

polygon. Compute α4 as α 4 = 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

tAAO4

s

AAO4

n

AAO4

cO A

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= −ω 4

2RPO4+α 4RPO4

Since the angular velocity and acceleration of link 4 are both known, the two components of the acceleration vector can be constructed.

RPO4

O2RO4O2 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

Slider-crank (inversion 3) 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 ω 3 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 + VO4At + VO4A

s = 0

The velocity polygon for this mechanism has already been obtained. From this polygon, ω 3 and the velocity of the slider-block have been determined.

O2 O4

(3)

(4)

A

RAO2

RO4O2

RO4A

VAO2

VO4At

VO4As

The acceleration equation is obtained from the time

derivative of the velocity equation: AAO2

n +AAO2

t + AO4An + AO4A

t +AO4As + AO4A

c = 0

or,

−ω2

2RAO2+α2RAO2

−ω 32RO4A + 2ω 3VO4A

s +AO4As +α 3RO4A = 0

Acceleration polygon 1. Select the origin of accelerations, OA , in a

convenient position. 2. Compute AA

n = RAO2ω2

2 . From OV construct vector

AAn in the opposite direction of RAO2

.


α2 . Determine the direction of

AAt based on the direction of α2 . Add this vector to

AAn .

4. Compute AO4A

n =ω 32RO4A . The direction of AO4A

n is

opposite of RO4A . Add this vector to the polygon.

5. Determine the Coriolis acceleration AO4Ac . The

magnitude of this vector is 2ω 3VO4As , and its direction

is found by rotating VO4As 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 AO4At perpendicular to RAO4

.

O2

A

RAO2

O A

AAO2

n

AAO2

t

RO4 A

O4

(3)

(4)

A

AO4 An

AO4 Ac

RO4 A

O4

(3)

(4)

A

AO4 An

AO4 Ac


P.E. Nikravesh 6-9

8. Construct vectors AAO3

t and AO3O4

s .

9. Determine the magnitude of AO4At from the polygon.

Compute α3 as α 3 = AO4At / RO4A .

10. Determine the direction of α3 . In this example it is

CW since RO4A must rotate 90o CW to line up with

AO4At .

AO4 At AO4 A

s

AO4 An

AAO2

t

AAO2

n

AO4 Ac

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 At

AO4 As

AO4 An

AAO2

t

AAO2

n

AO4 Ac

O

APAt

APAn

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)

6 Acceleration Polygon

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