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Velocity Polygon for a Crank-Slider

P

A

Ï‰2

O4

VtAO2

O2

OV

A

VtAO4

VsAO4

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:

How to construct the polygon shown on the right

How to extract velocity information from the polygon

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

Velocity Polygon for a Crank-Slider

P

A

Ï‰2

O4

O2

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.

Velocity Polygon for a Crank-Slider

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 = VtAO2

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 = VtAO4 + VsAO4

â–º

A

RAO4

RAO2

Ï‰2

O4

RO4O2

O2

Then, the velocity equation is:

VtAO2 = VtAO4 + VsAO4

Velocity Polygon for a Crank-Slider

VtAO2 = VtAO4 + VsAO4

We calculate VtAO2 :

VtAO2 = Ï‰2âˆ™ RAO2

The direction is found by rotating RAO2 90Â° in the direction of Ï‰2

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

A

RAO4

RAO2

Ï‰2

â–º

O4

RO4O2

O2

â–º

â–º

VtAO2

OV

A

â–º

VtAO4

VsAO4

â–º

â–º

â–º

Velocity Polygon for a Crank-Slider

We can determine Ï‰4:

Ï‰4 = VtAO4 / RAO4

RAO4 has to be rotated 90Â° counterclockwise to point in the same direction as VtAO4. Therefore Ï‰4 is ccw

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

VtAO2

A

RAO4

Ï‰4

RAO2

Ï‰2

O4

â–º

RO4O2

O2

VtAO2

OV

A

VtAO4

VsAO4

Velocity Polygon for a Crank-Slider

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

RAA4

A4

A

RA4O4

RAO2

â–º

â–º

â–º

Ï‰2

O4

RO4O2

O2

â–º

â–º

Velocity Polygon for a Crank-Slider

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 = VtAA4 + VsAA4

The result is:

VtAO2 = VtA4O4 + VtAA4 + VsAA4

RAA4

A

RA4O4

RA2O2

Ï‰2

O4

RO4O2

O2

VtAA4 is proportional to the length of RAA4 which is zero:

VtAA4 = 0

Therefore, the velocity equation becomes:

VtAO2 = VtAO4 + VsAA4

Velocity Polygon for a Crank-Slider

VtAO2 = VtA4O2 + VsAA4

We calculate VtAO2:

VtAO2 = Ï‰2âˆ™ RAO2

The direction is found by rotating RAO2 90Â° in the direction of Ï‰2

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

RAA4

VtAO2

A

RA4O4

RAO2

Ï‰2

â–º

O4

RO4O2

O2

â–º

â–º

VtAO2

OV

A

VtA4O4

VsAA4

â–º

A4

â–º

â–º

â–º

Velocity Polygon for a Crank-Slider

We can determine Ï‰4:

Ï‰4 = VtA4O4 / RA4O4

RA4O4 has to be rotated 90Â° counterclockwise to point in the same direction as VtA4O4. Therefore Ï‰4 is ccw

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

RAA4

VtAO2

A

RA4O4

RAO2

Ï‰4

Ï‰2

â–º

O4

RO4O2

O2

VtAO2

OV

A

VtA4O4

VsAA4

A4

Velocity polygons

Velocity Polygon for a Crank-Slider

VtAO2

OV

VtAO2

OV

A

A

VtA4O4

VsAA4

VtAO4

VsAO4

A4

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.

Velocity Polygon for a Crank-Slider

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 = VtPO4

We can calculate its magnitude as:

VtPO4 = Ï‰4âˆ™ RPO4

The direction is found by rotating RPO4 90Â° in the direction of Ï‰4

P

A

RPO4

VPO4

Ï‰4

â–º

Ï‰2

O4

O2

â–º

Velocity Polygon for a Crank-Slider

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 âˆ™ A4OV

Now we draw VPO4. It starts at the origin and ends at P

PA4

P

A

A4O4

Ï‰4

Ï‰2

O4

O2

â–º

VtAO2

OV

A

A4OV

A4

VtPO4

PA4

P

â–º

â–º