1. Velocity Polygon for a Crank-Slider Mechanism
Introduction
Velocity Polygon for a Crank-Slider
Velocity Polygon for a Crank-Slider Mechanism
This presentation shows how to construct the velocity polygon for a crank-slider (inversion 3) 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 O4 A ω2 O2 O3 P OV A
VtAO2
VsO3O4
VtAO3 O3

2. Inversion 3
Velocity Polygon for a Crank-Slider
Inversion 3
This example shows the construction of the velocity polygon for the third 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 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 ω2 O2 O3 P

3. Vector loop
Velocity Polygon for a Crank-Slider
Vector loop
We note that O4 is a point on link 4 that is fixed to the ground. We define another point O3 at the same position, but fixed to link 3
We define four position vectors to obtain a vector loop equation:
RAO2 = RO4O2 + RO3O4 + RAO3
where RO3O4 has zero length.
We take the time derivative to perform a velocity analysis:
VAO2 = VO4O2 + VO3O4 + VAO3 A ω2
RAO3
RAO2
RO3O4 O3 ► O2
RO4O2 O4 ► ►

4. Reduce velocity equation
Velocity Polygon for a Crank-Slider
VAO2 = VO4O2 + VO3O4 + VAO3
RA2O2 has constant length, but varying direction. That means VA2O2 is a tangential velocity.
RO4O2 has constant length and direction; VO4O2 equals 0.
RO3O4 has varying length and direction. That means VO3O4 has two components:
VO3O4 = VtO3O4 + VsO3O4
RAO3 has constant length, but varying direction. That means VAO3 is a tangential velocity.
The result is:
VtAO2 = VtO3O4 + VsO3O4 + VtAO3 A ω2
RAO3
RAO2
RO3O4 O2
RO4O2 O4 O3
VtO3O4 is proportional to the length of RO3O4, which is zero:
VtO3O4 = 0
Therefore the velocity equation becomes:
VtAO2 = VsO3O4 + VtAO3

5. The direction is found by rotating RAO2 90° in the direction of ω2
Velocity polygon
Velocity Polygon for a Crank-Slider
VtAO2
VtAO2 = VsO3O4 + VtAO3
We can calculate VtAO2 :
VtAO2 = ω2 ∙ RAO2
The direction is found by rotating RAO2 90° in the direction of ω2
The direction of VsO3O4 is parallel to link 3
The direction of VtAO3 is perpendicular to RAO3
Now we can draw the velocity polygon:
VtAO2 is added to the origin
VsO3O4 starts at OV
VtAO3 ends at A
We construct the polygon A ω2
RAO3
RAO2
RO3O4 ► O2
RO4O2 O4 O3 ► A ►
VtAO3
VtAO2 O3 ►
VsO3O4 ► OV ► ►

6. VtAO2 We can determine ω3: ω3 = VtAO3 / RAO3
Angular velocities
Velocity Polygon for a Crank-Slider
VtAO2
We can determine ω3:
ω3 = VtAO3 / RAO3
RAO3 has to be rotated 90° clockwise to point in the same direction as VtAO3. Therefore ω3 is cw.
ω4 equals ω3, since the sliding joint prohibits any relative rotation between link 3 and link 4. A
RAO3 ω2
RAO2 ω3
RO3O4 ► O2
RO4O2 O4 O3 A
VtAO3
VtAO2 O3
VsO3O4 OV

7. Velocity of point P: method (a)
Velocity Polygon for a Crank-Slider
Velocity of point P: method (a)
We define a position vector RPA.
We can position point P as
RPO2 = RAO2 + RPA
Which yields:
VPO2 = VtAO2 + VPA
RPA has constant length, but varying direction. That means VPA is a tangential velocity: VPA = VtPA
We can calculate its magnitude:
VtPA = ω3 ∙ RPA
The direction is found by rotating RPA 90° in the direction of ω3
VtPA is added to VtAO2 to find point P in the velocity polygon
VPO2 is the vector that starts at OV and ends at P
VtAO2 ► A
RPA ω2
RAO2 ω3 O3 O2
RO4O2 O4 P
VtPA A
VtPA
VtAO2 P ►
VPO2 OV ► ►

8. Velocity of point P: method (b)
Velocity Polygon for a Crank-Slider
VtAO2
Velocity of point P: method (b)
An easier way to determine the velocity of P is to use the velocity polygon directly.
P, A4 and O4 lie on the same line on link 4. That means they also lie on the same line in the velocity polygon.
Also the ratio AO3 / O3P on link 3 equals AO3 / O3P in the velocity polygon.
That means the line in the polygon is a scaled version of the line on the link.
VPO4 starts at the origin and ends at P. A
AO3
RPA ω2
RAO2 ω3
O3P O3 O2
RO4O2 O4 P ► A
VtAO3
VtAO2 O3 P
VPO4 ► OV ►