P. Nikravesh, AME, U of A Velocity Polygon for a Crank-Slider Introduction Velocity Polygon for a Crank-Slider Mechanism This presentation shows how to.
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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
►
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