p. nikravesh, ame, u of a velocity polygon for a four-bar introduction velocity polygon for a...
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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
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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 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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