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 1) mechanism. It is assumed that the dimensions for the links are known and the analysis is being performed at a given (known) configuration of the mechanism. Since the crank-slider has one degree-of-freedom, the angular velocity of one of the links must be given as well.
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
VtAO2 OV B
VtBA
VsBO4 A B O2 ω2 A

2. Inversion 1
Velocity Polygon for a Crank-Slider
Inversion 1
Whether the crank-slider is offset or not, the process of constructing a velocity polygon remains the same. Therefore, in the first example we consider the more general case; I.e., an offset crank-slider.
As for 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. B O2 ω2 A B O2 ω2 A

3. We define four position vectors to obtain a vector loop equation:
Velocity Polygon for a Crank-Slider
We define four position vectors to obtain a vector loop equation:
RAO2 + RBA = RO4O2 + RBO4
Time derivative:
VAO2 + VBA = VO4O2 + VBO4
Since RO4O2 is fixed to the ground, VO4O2 = 0.
The lengths of RAO2 and RBA are constants, therefore VAO2 and VBA are tangential velocities.
The axis of RBO4 is fixed, but its length varies. Therefore VBO4 consists only of a component parallel to RBO4 which is a slip velocity. ω2 A ► O2
RAO2
RBA
RO4O2 B O4
RBO4
The velocity equation can then be expressed as:
VtAO2 + VtBA = VsBO4

4. The direction is found by rotating RAO2 90° in the direction of ω2:
Determine velocities
Velocity Polygon for a Crank-Slider
VtAO2 + VtBA = VsBO4
We calculate VtAO2 :
VtAO2 = 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 VsBO4 is parallel to RBO4
We draw the velocity polygon:
VtAO2 is added to the origin
VtBA starts at A
VsBO4 starts at the origin
The two lines intersect at B. We add the missing velocities:
This polygon represents the velocity loop equation shown above!
VtAO2 ω2 A O2
RAO2
RBA ►
RO4O2 B ► O4
RBO4 ► A ► ►
VtBA
VtAO2 ► B
VsBO4 OV ►

5. VtAO2 We can determine ω3: ω3 = VtBA / RBA
Angular velocities
Velocity Polygon for a Crank-Slider
VtAO2
We can determine ω3:
ω3 = VtBA / RBA
RBA has to be rotated 90° clockwise to point in the same direction as VtBA. Therefore ω3 is clockwise
ω4 equals zero, since the sliding joint prohibits any rotation with respect to the ground. ω2 A O2
RAO2 ω3
RBA
RO4O2 ► B O4
RBO4 A
VtBA
VtAO2 B
VsBO4 OV