Velocity Polygon for a CrankSlider Introduction Velocity Polygon

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Velocity Polygon for a Crank-Slider Introduction Velocity Polygon for a Crank-Slider Mechanism This presentation

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 crankslider (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-offreedom, 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: 1. How to construct the polygon shown on the right 2. How to extract velocity information from the polygon A A Vt. BA ω2 O 2 P. Nikravesh, AME, U of A B Vt. AO 2 B Vs. BO 4 OV

Inversion 1 Velocity Polygon for a Crank-Slider Inversion 1 ω2 Whether the crank-slider is

Inversion 1 Velocity Polygon for a Crank-Slider Inversion 1 ω2 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. P. Nikravesh, AME, U of A A O 2 B A ω2 O 2 B

Vector loop Velocity Polygon for a Crank-Slider We define four position vectors to obtain

Vector loop Velocity Polygon for a Crank-Slider We define four position vectors to obtain a vector loop equation: ► RAO 2 + RBA = RO 4 O 2 + RBO 4 O 2 Time derivative: The lengths of RAO 2 and RBA are constants, therefore VAO 2 and VBA are tangential velocities. The axis of RBO 4 is fixed, but its length varies. Therefore VBO 4 consists only of a component parallel to RBO 4 which is a slip velocity. P. Nikravesh, AME, U of A RAO 2 RBA RO 4 O 2 VAO 2 + VBA = VO 4 O 2 + VBO 4 Since RO 4 O 2 is fixed to the ground, VO 4 O 2 = 0. A ω2 O 4 RBO 4 B The velocity equation can then be expressed as: Vt. AO 2 + Vt. BA = Vs. BO 4

Determine velocities Velocity Polygon for a Crank-Slider Vt. AO 2 + Vt. BA =

Determine velocities Velocity Polygon for a Crank-Slider Vt. AO 2 + Vt. BA = Vs. BO 4 We calculate Vt. AO 2 : Vt. AO 2 = ω2 ∙ RAO 2 The direction is found by rotating RAO 2 90° in the direction of ω2: ► The direction of Vt. BA is perpendicular to RBA ► RAO 2 Vt. BA starts at A Vs. BO 4 A Vt. BA ► starts at the origin loop equation shown above! P. Nikravesh, AME, U of A B RBO 4 ► The two lines intersect at► B. We add the missing velocities: This polygon represents►the velocity RBA RO 4 O 2 The direction of Vs. BO 4 is parallel to RBO 4 ► the velocity polygon: We draw Vt. AO 2 is added to the origin A ω2 Vt. AO 2 B Vs. BO 4 OV

Angular velocities Velocity Polygon for a Crank-Slider Vt. AO 2 We can determine ω3:

Angular velocities Velocity Polygon for a Crank-Slider Vt. AO 2 We can determine ω3: ω3 = Vt. BA / RBA has to be rotated 90° clockwise to point in the same direction as Vt. BA. Therefore ω3 is clockwise ► ω4 equals zero, since the sliding joint prohibits any rotation with respect to the ground. A ω2 O 2 ω3 RAO 2 RBA RO 4 O 2 O 4 B RBO 4 A Vt. BA Vt. AO 2 B Vs. BO 4 P. Nikravesh, AME, U of A OV