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Introduction

Introduction. A. V t AO 3. V t AO 2. O 3. A. ω 2. V s O 3 O 4. O 4. O V. P. O 2. O 3. Velocity Polygon for a Crank-Slider Mechanism This presentation shows how to construct the velocity polygon for a crank-slider (inversion 3) mechanism.

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Introduction

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  1. Introduction Velocity Polygon for a Crank-Slider A VtAO3 VtAO2 O3 A ω2 VsO3O4 O4 OV P O2 O3 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

  2. Inversion 3 Velocity Polygon for a Crank-Slider A ω2 O4 P O2 O3 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.

  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 RAO3 ω2 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 RAO3 ω2 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. 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 RAO3 ω2 RAO2 RO3O4 ► O2 RO4O2 O4 O3 ► A ► VtAO3 VtAO2 O3 VsO3O4 ► ► OV ► ►

  6. 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 ω3 RAO2 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 canposition 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 ω3 RAO2 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 ►

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