1. Acceleration Polygon for a Four-bar Mechanism
Introduction
Acceleration Polygon for a Four-bar
Acceleration Polygon for a Four-bar Mechanism
This presentation shows how to construct the acceleration polygon for a four-bar mechanism. It is assumed that the dimensions for the links are known and the analysis is being performed at a given (known) configuration. Since the four-bar has one degree-of-freedom, the angular velocity and acceleration of one of the links must be given as well.
As an example, for the four-bar shown, we first construct the velocity polygon. Then we will learn:
How to construct the acceleration polygon shown on the right
How to extract acceleration information from the polygon OA
ABA AB AA A B A O4 B O2 ω2 α2

2. Example
Acceleration Polygon for a Four-bar A O4 B O2 ω2 α2
Example
This example shows us how to construct the acceleration polygon for a typical four-bar, such as the one shown on this slide.
It is assumed that: all the lengths are known and the four-bar is being analyzed at the configuration shown; the angular velocity and acceleration of the crank are given as well.

3. Vector loop, differentiation
Acceleration Polygon for a Four-bar B
RBA
Vector loop
We follow the same process as we did for the velocity polygon.
We first define four position vectors to obtain a vector loop equation:
RAO2 + RBA = RO4O2 + RBO4
The first time derivative provides the velocity loop equation:
VtA + VtBA = VtB
The second time derivative provides the acceleration loop equation:
AA + ABA = AB
We split each acceleration vector into a normal and a tangential component: A
RBO4
RAO2 ► O4
RO4O2 O2
AtA + AnA + AtBA + AnBA = AtB + AnB
We need the velocities to calculate some of the accelerations. Therefore we perform a velocity analysis first.

4. The direction is found by rotating RAO2 90° in the direction of ω2
Velocity analysis
Acceleration Polygon for a Four-bar
VtA B
RBA
Velocity polygon
We calculate VtA:
VtA = ω2 ∙ RAO2
The direction is found by rotating RAO2 90° in the direction of ω2
The direction of VtBA is perpendicular to RBA
The direction of VtB is perpendicular to RBO4
Now we construct the velocity polygon
Next we determine the angular velocities.
We will use these results to calculate the normal components of acceleration vectors. A
RBO4 ω2
RAO2 O4 ►
RO4O2 O2 ► A ►
VtA OV ►
VtBA
VtB B
VtA + VtBA = VtB

5. The absolute value of the angular velocities are computed as:
Acceleration Polygon for a Four-bar
VtA B ω3
RBA
Angular velocities
The absolute value of the angular velocities are computed as:
ω3 = VtBA ∕ RBA
ω4 = VtB ∕ RBO4
RBA has to be rotated 90° cw to head in the direction of VtBA. Therefore ω3 is cw
RBO4 has to be rotated 90° ccw to head in the direction of VtB. Therefore ω4 is ccw
We will use these results to calculate accelerations. A
RBO4 ω2
RAO2 ω4 O4
RO4O2 O2 ► A
VtA OV
VtBA ►
VtB B

6. AtA + AnA + AtBA + AnBA = AtB + AnB
Normal components
Acceleration Polygon for a Four-bar
AtA + AnA + AtBA + AnBA = AtB + AnB B ω3
RBA
Normal components
We first calculate the magnitude of all the normal components:
AnA = ω22 ∙ RAO2
AnBA = ω32 ∙ RBA
AnB = ω42 ∙ RBO4
The direction of each normal component of acceleration is opposite to the corresponding position vector A
RBO4 ω2
RAO2 ω4 O4
RO4O2 O2 B
AnBA ► A
AnB
AnA O4 O2

7. Tangential components
Acceleration Polygon for a Four-bar
AtA + AnA + AtBA + AnBA = AtB + AnB B
RBA
Tangential components
We first calculate the magnitude of AtA:
AtA = α2 ∙ RAO2
The direction is found by rotating RAO2 90° in the direction of α2
We also know that AtBA is on an axis perpendicular to RBA
Similarly we know that AtB is on an axis perpendicular to RBO4 A
RBO4
RAO2 α2 O4
RO4O2 O2 ► ►
AtA B
AnBA A ►
AnB
AnA O4 O2

8. AtA + AnA + AtBA + AnBA = AtB + AnB
Acceleration polygon
Acceleration Polygon for a Four-bar
AtA + AnA + AtBA + AnBA = AtB + AnB
Acceleration polygon
Now we are ready to draw the acceleration polygon.
First we select the origin and add AtA and AnA to obtain AA
AnBA is a added at A
We also know the axis of AtBA which would be added to AnBA
AnB is added to the origin
We also know the axis of AtB which would be added to AnB
The two lines intersect at B. We complete the acceleration polygon by drawing the missing accelerations
Next we determine the angular accelerations.
AtA B
AnBA A
AnB ►
AnA ► O4 O2
AtA ► ►
AnA OA AA
AnB ►
AnBA A AB
ABA ►
AtBA
AtB B

9. Angular accelerations
Acceleration Polygon for a Four-bar B α3
RBA
Angular accelerations
The absolute values of the angular accelerations are computed as:
α3 = AtBA ∕ RBA
α4 = AtB ∕ RBO4
RBA has to be rotated 90° cw to head in the direction of AtBA. Therefore α3 is cw
RBO4 has to be rotated 90° ccw to head in the direction of AtB. Therefore α4 is ccw
This completes the acceleration analysis of this four-bar. A
RBO4
RAO2 α4 α2 O4
RO4O2 O2
AtA
AnA OA ► AA
AnB
AnBA ► AB
ABA
AtBA
AtB