Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

Introduction

Velocity Polygon for a Four-bar Mechanism

This presentation shows how to construct the velocity polygon for a given 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 of one of the links must be given as well.

As an example, for the four-bar shown on the left we will learn:

1. How to construct the polygon shown on the right

2. How to extract velocity information from the polygon

A

O4

B

O2

ω2

V tA

V tB

V tBA

A

B

OV

Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

Example 1

Example 1

The first example shows us how to construct the velocity 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; i.e., all the angles are known.

Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction).

A

O4

B

O2

ω2

Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

Velocity analysis

We define four position vectors to obtain a vector loop equation:

RAO2 + RBA = RO4O2

+ RBO4

The time derivative of this equation yields the velocity loop equation:

VAO2 + VBA = VO4O2

+ VBO4

Since RO4O2 is fixed to the

ground, VO4O2 = 0.

Since any velocity vector with respect to a fixed point only needs one subscript, we can further simplify the velocity loop equation:

A

O4

B

RBO4

RO4O2

RBA

RAO2

Vector loop

O2

ω2

VA + VBA = VB

For clarification purposes we assign superscript “t” to these vectors indicating they are tangential :

V tA + V t

BA = V tB

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Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

We can calculate V tA:

V tA = ω2 ∙ RAO2

The direction is found by rotating RAO2

90° in the

direction of ω2:

The direction of V tBA is

perpendicular to RBA:

The direction of V tB is

perpendicular to RBO4:

VA and lines of action

A

O4

B

RBO4

RO4O2

RBA

V tA

O2

V tA + V t

BA = V tB

RAO2

ω2

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Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

To construct the velocity polygon we select an origin and draw V t

A:

V tB starts at the origin. We know the

line of action:

V tBA starts at A. We also know the

line of action:

Now we can complete the velocity polygon:

Note that this polygon represents the velocity loop equation shown above!

Velocity polygon

A

O4

B

RBO4

RO4O2

V tA

V tB

V tBA

V tA

O2

V tA + V t

BA = V tB

A

B

OV

RAO2

ω2

RBA

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Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

We can determine ω3:

ω3 = V tBA / RBA

RBA has to be rotated 90° clockwise to point in the same direction as V t

BA. Therefore ω3 is clockwise:

To determine ω4:

ω4 = V tB / RBO4

RBO4 has to be rotated 90°

counter-clockwise to point in the same direction as V t

B. Therefore ω4 is ccw:

ω4

ω3

Angular velocities

A

O4

B

RBO4

RO4O2

RBA

O2

V tA

V tB

V tBA

V tA

RAO2

ω2

A

B

OV

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Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

Example 2

Example 2

The second example shows how to construct the velocity polygon for another typical four-bar. In addition, we learn how to determine the velocity of a coupler point from the polygon.

Similar to the first example, it is assumed that all the lengths are known and the four-bar is being analyzed at the configuration shown; i.e., all the angles are known.

Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction).

P

A

O4

B

O2

ω2

Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

P

Velocity analysis

We define four position vectors to obtain a vector loop equation:

RAO2 + RBA = RO4O2

+ RBO4

Note that we first construct the vector loop containing the primary links and points. Point P is not a primary point--it will be consider later for analysis.

Since the vectors have constant lengths the time derivatives are tangential velocities:

V tAO2

+ V tBA = V t

O4O2 + V t

BO4

Discard zero vectors and subscripts referring to non-moving points:

V tA + V t

BA = V tB

A

O4

B

RBO4RO4O2

RBA

RAO2

O2

ω2

Vector loop

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Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

A

O4

RBO4RO4O2

RAO2 V tA

O2

ω2

VA and lines of action

We can calculate V tA:

V tA = ω2 ∙ RAO2

The direction is found by rotating RAO2

90° in the

direction of ω2:

The direction of V tBA is

perpendicular to RBA:

The direction of V tB is

perpendicular to RBO4:

P

BRBA

►

►

►

V tA + V t

BA = V tB

Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

V tA

V tBA

V tB

OV

A

B

Velocity polygon

A

O4

RBO4RO4O2

RAO2 V tA

O2

ω2

To construct the velocity polygon we select the origin and draw V t

A:

V tB starts at the origin. We know the

line of action:

V tBA starts at A. We also know the

line of action:

Now we can complete the velocity polygon:

Note that this polygon represents the velocity loop equation shown above!

P

BRBA

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V tA + V t

BA = V tB

Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

V tA

V tBA

V tB

OV

A

B

Velocity of coupler point P

A

O4

B

RBO4RO4O2

RAO2 V tA

O2

ω2

P

In order to determine the Velocity of point P we rotate the line AP 90° and move it to point A in the velocity polygon:

Next we rotate the line BP 90° and move it to point B in the velocity polygon:

The point of intersection is point P in the velocity polygon. Now we can draw VP:

PVP

RBA

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Velocity Polygon for a Four-bar

P. Nikravesh, AME, U of A

V tA

V tBA

V tB

OV

A

B

Angular velocities

A

O4

B

RBO4RO4O2

RBA

RAO2 V tA

O2

ω2

P

PV t

P

We can determine ω3:

ω3 = V tBA / RBA

RBA has to be rotated 90° clockwise to point in the same direction as V t

BA. Therefore ω3 is clockwise:

To determine ω4:

ω4 = V tB / RBO4

RBO4 has to be rotated 90°

clockwise to point in the same direction as V t

B. Therefore ω4 is clockwise:

ω3

ω4

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