Velocity Polygon for a Fourbar Introduction Velocity Polygon

Velocity Polygon for a Four-bar Introduction Velocity Polygon for a Four-bar Mechanism This presentation shows how to construct the velocity polygon for a given fourbar 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 B A A V t. A OV V t. BA ω2 O 2 P. Nikravesh, AME, U of A V t. B O 4 B

Example 1 Velocity Polygon for a Four-bar 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. B A Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction). ω2 O 2 P. Nikravesh, AME, U of A O 4

Vector loop Velocity Polygon for a Four-bar Velocity analysis We define four position vectors to obtain a vector loop equation: ► A VAO 2 + VBA = VO 4 O 2 + VBO 4 Since RO 4 O 2 is fixed to the ground, VO 4 O 2 = 0. Since any velocity vector with respect to a fixed point only needs one subscript, we can further simplify the velocity loop equation: P. Nikravesh, AME, U of A B RAO 2 + RBA = RO 4 O 2 + RBO 4 The time derivative of this equation yields the velocity loop equation: RBA RBO 4 ω2 O 2 RO 4 O 2 VA + VBA = VB For clarification purposes we assign superscript “t” to these vectors indicating they are tangential : V t. A + V t. BA = V t. B O 4

VA and lines of action Velocity Polygon for a Four-bar V t. A + V t. BA = V t. B V t. A A RBA RAO 2 B RBO 4 ω2 RO 4 O 2 We can calculate V t. A: V t. A = ω2 ∙ RAO 2 The direction is found by rotating RAO 2 90° in the direction of ω2: ► P. Nikravesh, AME, U of A O 4 The direction of V t. BA is perpendicular to RBA: ► The direction of V t. B is perpendicular to RBO 4: ►

Velocity polygon Velocity Polygon for a Four-bar V t. A + V t. BA = V t. B V t. A A RBA B A OV V t. BA RAO 2 RBO 4 V t. B ω2 O 2 V t. A B RO 4 O 2 O 4 To construct the velocity polygon we select an origin and draw V t. A: ► Now we can complete the velocity polygon: ► V t. B starts at the origin. We know the line of action: ► Note that this polygon represents the velocity loop equation shown above! V t. BA starts at A. We also know the line of action: ► P. Nikravesh, AME, U of A

Angular velocities Velocity Polygon for a Four-bar ω3 V t. A A RBO 4 RAO 2 We can determine ω3: ω3 = V t. BA / RBA has to be rotated 90° clockwise to point in the same direction as V t. BA. Therefore ω3 is clockwise: ► P. Nikravesh, AME, U of A OV V t. B B RO 4 O 2 V t. A V t. BA ω4 ω2 O 2 B O 4 To determine ω4: ω4 = V t. B / RBO 4 has to be rotated 90° counter-clockwise to point in the same direction as V t. B. Therefore ω4 is ccw: ►

Example 2 Velocity Polygon for a Four-bar 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. O 2 ω2 B A P Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction). O 4 P. Nikravesh, AME, U of A

Vector loop Velocity Polygon for a Four-bar Velocity analysis We define four position vectors to obtain a vector loop equation: ► RAO 2 + RBA = RO 4 O 2 + RBO 4 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: O 2 ω2 RAO 2 RBA A P RO 4 O 2 RBO 4 V t. AO 2 + V t. BA = V t. O 4 O 2 + V t. BO 4 Discard zero vectors and subscripts referring to non-moving points: V t. A + V t. BA = V t. B P. Nikravesh, AME, U of A B O 4

VA and lines of action Velocity Polygon for a Four-bar ω2 V t. A + V t. BA = V t. B O 2 RAO 2 V t. A B RBA A P RO 4 O 2 RBO 4 We can calculate V t. A: V t. A = ω2 ∙ RAO 2 The direction is found by rotating RAO 2 90° in the direction of ω2: ► P. Nikravesh, AME, U of A The direction of V t. BA is perpendicular to RBA: ► The direction of V t. B is perpendicular to RBO 4: ►

Velocity polygon Velocity Polygon for a Four-bar V t. A + V t. BA = V t. B ω2 O 2 RAO 2 V t. A B RBA A A OV P RO 4 O 2 RBO 4 V t. BA V t. B B O 4 To construct the velocity polygon we select the origin and draw V t. A: ► Now we can complete the velocity polygon: ► V t. B starts at the origin. We know the line of action: ► Note that this polygon represents the velocity loop equation shown above! V t. BA starts at A. We also know the line of action: ► P. Nikravesh, AME, U of A

Velocity of coupler point P Velocity Polygon for a Four-bar ω2 O 2 RAO 2 V t. A B RBA A V t. A A P P RO 4 O 2 RBO 4 OV VP V t. B O 4 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: ► P. Nikravesh, AME, U of A The point of intersection is point P in the velocity polygon. Now we can draw VP: ► V t. BA B

Angular velocities Velocity Polygon for a Four-bar ω2 O 2 RAO 2 V t. A ω3 RBA B A V t. A A P P OV RBO 4 RO 4 O 2 V t. P V t. B ω4 O 4 We can determine ω3: To determine ω4: ω3 = V t. BA / RBA ω4 = V t. B / RBO 4 RBA has to be rotated 90° clockwise to point in the same direction as V t. BA. Therefore ω3 is clockwise: ► RBO 4 has to be rotated 90° clockwise to point in the same direction as V t. B. Therefore ω4 is clockwise: ► P. Nikravesh, AME, U of A V t. BA B