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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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Presentation on theme: "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."— Presentation transcript:

1 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 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 O4O4 A O2O2 ω2ω2 P V t AO 2 V t AO 4 V s AO 4 OVOV A

2 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderInversion 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. O4O4 A O2O2 ω2ω2 P

3 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVector loop: method 1 We define three position vectors to obtain a vector loop equation: R AO 2 = R O 4 O 2 + R AO 4 R AO 2 has a constant length but varying direction. Therefore its time derivative is a tangential velocity: V AO 2 = V t AO 2 R O 4 O 2 has constant length and direction. Its time derivative is zero: V O 4 O 2 = 0 R AO 4 has varying length and direction. Its time derivative consists of two components: a tangential velocity and a slip velocity: V AO 4 = V t AO 4 + V s AO 4 Then, the velocity equation is: V t AO 2 = V t AO 4 + V s AO 4 ► O4O4 RO4O2RO4O2 A R AO 2 O2O2 ω2ω2 R AO 4 Vector loop: method 1

4 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVelocity polygon: method 1 V t AO 2 = V t AO 4 + V s AO 4 We calculate V t AO 2 : V t AO 2 = ω 2 ∙ R AO 2 The direction is found by rotating R AO 2 90° in the direction of ω 2 The direction of V t AO 4 is perpendicular to R AO 4 The direction of V s AO 4 is parallel to R AO 4 Now we can draw the velocity polygon: V t AO 2 is added to the origin V t AO 4 starts at O V V s AO 4 ends at A We construct the polygon ► V t AO 2 V t AO 4 V s AO 4 O4O4 RO4O2RO4O2 A R AO 2 O2O2 ω2ω2 R AO 4 V t AO 2 OVOV A ► ► ► ► ► ►

5 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderAngular velocities: method 1 We can determine ω 4 : ω 4 = V t AO 4 / R AO 4 R AO 4 has to be rotated 90° counterclockwise to point in the same direction as V t AO 4. Therefore ω 4 is ccw ω 3 is equal to ω 4, since the sliding joint prohibits any relative rotation between link 3 and link 4. O4O4 ► RO4O2RO4O2 A R AO 2 O2O2 ω2ω2 R AO 4 V t AO 2 V t AO 4 V s AO 4 OVOV A ω4ω4

6 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVector 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. A 4 is a point on link 4 that has the same position as A As the crank rotates A will move away from A 4. We define four position vectors to obtain a vector loop equation: R AO 2 = R O 4 O 2 + R A 4 O 4 + R AA 4 R AA 4 has zero length We take the time derivative to perform a velocity analysis: V AO 2 = V O 4 O 2 + V A 4 O 4 + V AA 4 O4O4 RO4O2RO4O2 A R AO 2 O2O2 ω2ω2 RA4O4RA4O4 ►► ► ► Vector loop: method 2 ► R AA 4 A4A4

7 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVelocity equation: method 2 V AO 2 = V O 4 O 2 + V A 4 O 4 + V AA 4 R A 2 O 2 has constant length but varying direction. That means V AO 2 is a tangential velocity. R O 4 O 2 has constant length and direction; V O 4 O 2 equals 0. R A 4 O 4 has constant length but varying direction. Therefore V A 4 O 4 is a tangential velocity. R AA 4 has varying length and direction. That means V AA 4 has two components: V AA 4 = V t AA 4 + V s AA 4 The result is: V t AO 2 = V t A 4 O 4 + V t AA 4 + V s AA 4 V t AA 4 is proportional to the length of R AA 4 which is zero: V t AA 4 = 0 Therefore, the velocity equation becomes: V t AO 2 = V t AO 4 + V s AA 4 O4O4 RO4O2RO4O2 A RA2O2RA2O2 O2O2 ω2ω2 RA4O4RA4O4 R AA 4

8 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVelocity polygon: method 2 V t AO 2 = V t A 4 O 2 + V s AA 4 We calculate V t AO 2 : V t AO 2 = ω 2 ∙ R AO 2 The direction is found by rotating R AO 2 90° in the direction of ω 2 The direction of V t A 4 O 4 is perpendicular to R A 4 O 4 The direction of V s AA 4 is parallel to link 4 Now we can draw the velocity polygon: V t AO 2 is added to the origin V t A 4 O 4 starts at O V V s AA 4 ends at A We construct the polygon V t AO 2 VtA4O4 VtA4O4 V s AA 4 OVOV A4A4 A O4O4 RO4O2RO4O2 A R AO 2 O2O2 ω2ω2 RA4O4RA4O4 ► ► ► ► ► ► ► V t AO 2 R AA 4

9 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderAngular velocities: method 2 We can determine ω 4 : ω 4 = V t A 4 O 4 / R A 4 O 4 R A 4 O 4 has to be rotated 90° counterclockwise to point in the same direction as V t A 4 O 4. Therefore ω 4 is ccw ω 3 equals ω 4, since the sliding joint prohibits any relative rotation between link 3 and link 4. ► V t AO 2 OVOV A4A4 O4O4 RO4O2RO4O2 A R AO 2 O2O2 ω2ω2 RA4O4RA4O4 V t AO 2 ω4ω4 A VtA4O4 VtA4O4 V s AA 4 R AA 4

10 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVelocity 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. V t AO 2 OVOV A4A4 A VtA4O4 VtA4O4 V s AA 4 V t AO 2 V t AO 4 V s AO 4 OVOV A

11 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVelocity 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 R PO 4 R PO 4 has constant length, but varying direction. That means V PO 4 is a tangential velocity: V PO 4 = V t PO 4 We can calculate its magnitude as: V t PO 4 = ω 4 ∙ R PO 4 The direction is found by rotating R PO 4 90° in the direction of ω 4 O4O4 A O2O2 ω2ω2 ω4ω4 R PO 4 V PO 4 P ► ►

12 P. Nikravesh, AME, U of A Velocity Polygon for a Crank-SliderVelocity of point P: method (b) ω4ω4 P O4O4 A O2O2 ω2ω2 This process can be followed after we find the velocity polygon from either method 1 or method 2. We note that P, A 4 and O 4 lie on the same line on link 4. That means they also lie on the same line on the velocity polygon. The ratio PA 4 / A 4 O 4 on link 4 must be equal to PA 4 / A 4 O V on the velocity polygon. We can measure it on the mechanism; e.g., PA 4 / A 4 O 4 = PA 4 / A 4 O V = 0.32 Next we use this ratio to calculate the distance PA 4 on the velocity polygon as: PA 4 = 0.32 ∙ A 4 O V Now we draw V PO 4. It starts at the origin and ends at P ► ► V t AO 2 OVOV A4A4 A PA 4 A4O4A4O4 A4OVA4OV P V t PO 4 ►


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