Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Introduction

Velocity Polygon for a Crank-Slider Mechanism

This presentation shows how to construct the velocity polygon for a crank-slider (inversion 2) mechanism.

As an example, for the crank-slider 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

3. How to determine the velocity of a point P on the output link

O4

A

O2

ω2

P

VtAO2

VtAO4Vs

AO4

OV

A

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Inversion 2

Inversion 2

This example shows the construction of the velocity polygon for the second inversion of a crank-slider.

In addition, this example shows how to find the velocity of a point P on the output link.

Two methods will be presented for constructing the velocity polygons and also two methods will be presented for determining the velocity of point P.

Like any other system, it is assumed that all the lengths are known and the system is being analyzed at a given configuration. Furthermore, it is assumed that the angular velocity of the crank is given.

O4

A

O2

ω2

P

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Vector loop: method 1

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

RAO2 = RO4O2

+ RAO4

RAO2 has a constant length but varying

direction. Therefore its time derivative is a tangential velocity:

VAO2 = Vt

AO2

RO4O2 has constant length and

direction. Its time derivative is zero:

VO4O2 = 0

RAO4 has varying length and direction.

Its time derivative consists of two components: a tangential velocity and a slip velocity:

VAO4 = Vt

AO4 + Vs

AO4

Then, the velocity equation is:

VtAO2

= VtAO4

+ VsAO4

►

O4

RO4O2

A

RAO2

O2

ω2

RAO4

Vector loop: method 1

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Velocity polygon: method 1

VtAO2

= VtAO4

+ VsAO4

We calculate VtAO2

:

VtAO2

= ω2 ∙ RAO2

The direction is found by rotating RAO2

90° in the direction of ω2

The direction of VtAO4

is

perpendicular to RAO4

The direction of VsAO4

is parallel to

RAO4

Now we can draw the velocity polygon:

VtAO2

is added to the origin

VtAO4

starts at OV

VsAO4

ends at A

We construct the polygon

►

VtAO2

VtAO4Vs

AO4

O4

RO4O2

A

RAO2

O2

ω2

RAO4

VtAO2

OV

A

►

►

►

►

►

►

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Angular velocities: method 1

We can determine ω4:

ω4 = VtAO4

/ RAO4

RAO4 has to be rotated 90°

counterclockwise to point in the same direction as Vt

AO4. Therefore

ω4 is ccw

ω3 is equal to ω4, since the sliding joint prohibits any relative rotation between link 3 and link 4.

O4►RO4O2

A

RAO2

O2

ω2

RAO4

VtAO2

VtAO2

VtAO4Vs

AO4

OV

A

ω4

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Vector loop: method 2

In this method we introduce an extra position vector in the vector loop equation.

We note that A is a point on link 2 and link 3. A4 is a point on link 4 that has the same position as A

As the crank rotates A will move away from A4.

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

RAO2 = RO4O2

+ RA4O4 + RAA4

RAA4 has zero length

We take the time derivative to perform a velocity analysis:

VAO2 = VO4O2

+ VA4O4 + VAA4

O4RO4O2

A

RAO2

O2

ω2

RA4O4

► ► ►

►

Vector loop: method 2

►

RAA4

A4

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Velocity equation: method 2

VAO2 = VO4O2

+ VA4O4 + VAA4

RA2O2 has constant length but

varying direction. That means VAO2

is a tangential velocity.

RO4O2 has constant length and

direction; VO4O2 equals 0.

RA4O4 has constant length but

varying direction. Therefore VA4O4 is

a tangential velocity.

RAA4 has varying length and

direction. That means VAA4 has two

components:

VAA4 = Vt

AA4 + Vs

AA4

The result is:

VtAO2

= VtA4O4

+ VtAA4

+ VsAA4

VtAA4

is proportional to the length of

RAA4 which is zero:

VtAA4

= 0

Therefore, the velocity equation becomes:

VtAO2

= VtAO4

+ VsAA4

O4RO4O2

A

RA2O2

O2

ω2

RA4O4

RAA4

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Velocity polygon: method 2

VtAO2

= VtA4O2

+ VsAA4

We calculate VtAO2 :

VtAO2

= ω2 ∙ RAO2

The direction is found by rotating RAO2

90° in the

direction of ω2

The direction of VtA4O4

is

perpendicular to RA4O4

The direction of VsAA4

is parallel

to link 4

Now we can draw the velocity polygon:

VtAO2

is added to the origin

VtA4O4

starts at OV

VsAA4

ends at A

We construct the polygon

VtAO2

VtA4O4Vs

AA4

OV

A4

A

O4RO4O2

A

RAO2

O2

ω2

RA4O4

►

►

►

►

►

►

►

VtAO2

RAA4

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Angular velocities: method 2

We can determine ω4:

ω4 = VtA4O4

/ RA4O4

RA4O4 has to be rotated 90°

counterclockwise to point in the same direction as Vt

A4O4.

Therefore ω4 is ccw

ω3 equals ω4, since the sliding joint prohibits any relative rotation between link 3 and link 4.

►

VtAO2 OV

A4

O4RO4O2

A

RAO2

O2

ω2

RA4O4

VtAO2

ω4

A

VtA4O4Vs

AA4

RAA4

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Velocity polygons

Note that the velocity polygons that are obtained from the two methods are identical. The difference is in how the vectors are labeled and viewed.

Viewing point A as two separate points (but coinciding), one on link 2 (or 3) and one on link 4, can be helpful in realizing the sliding velocity component in rotating bodies connected by a slider.

VtAO2 OV

A4

A

VtA4O4Vs

AA4

VtAO2

VtAO4Vs

AO4

OV

A

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Velocity of point P: method (a)

Velocity of point P: method (a)

After finding the angular velocity of link 4, whether from method 1 or method 2, we can determine the velocity of point P.

We define a position vector RPO4

RPO4 has constant length, but varying

direction. That means VPO4 is a

tangential velocity:

VPO4 = Vt

PO4

We can calculate its magnitude as:

VtPO4

= ω4 ∙ RPO4

The direction is found by rotating RPO4

90° in the direction of ω4

O4

A

O2

ω2

ω4

RPO4

VPO4

P

►

►

Velocity Polygon for a Crank-Slider

P. Nikravesh, AME, U of A

Velocity of point P: method (b)

ω4

P

O4

A

O2

ω2

Velocity of point P: method (b)

This process can be followed after we find the velocity polygon from either method 1 or method 2.

We note that P, A4 and O4 lie on the same line on link 4. That means they also lie on the same line on the velocity polygon.

The ratio PA4 / A4O4 on link 4 must be equal to PA4 / A4OV on the velocity polygon. We can measure it on the mechanism; e.g.,

PA4 / A4O4 = PA4 / A4OV = 0.32

Next we use this ratio to calculate the distance PA4 on the velocity polygon as:

PA4 = 0.32 ∙ A4OV

Now we draw VPO4. It starts at the origin

and ends at P

►

►

VtAO2 OV

A4

A

PA4

A4O4

A4OV

PA4P Vt

PO4

►