Introduction

1. Introduction Velocity Polygon for a Four-bar A V tA OV V tBA V tB B B A ω2 O2 O4 Velocity Polygon for a Four-bar Mechanism This presentation shows how to construct the velocity polygon for a given four-bar 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: How to construct the polygon shown on the right How to extract velocity information from the polygon

2. Introduction Example 1 Velocity Polygon for a Four-bar B A ω2 O2 O4 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. Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction).

3. Vector loop Velocity Polygon for a Four-bar Velocity analysis We define four position vectors to obtain a vector loop equation: RAO2 + RBA = RO4O2 + RBO4 The time derivative of this equation yields the velocity loop equation: VAO2 + VBA = VO4O2 + VBO4 Since RO4O2 is fixed to the ground, VO4O2 = 0. Since any velocity vector with respect to a fixed point only needs one subscript, we can further simplify the velocity loop equation: B A RBA ► RAO2 RBO4 ω2 O2 RO4O2 O4 VA + VBA = VB For clarification purposes we assign superscript “t” to these vectors indicating they are tangential : V tA + V tBA = V tB

4. VA and lines of action Velocity Polygon for a Four-bar V tA + V tBA = V tB V tA B A RBA RAO2 RBO4 ω2 RO4O2 O4 O2 We can calculate V tA: V tA = ω2∙ RAO2 The direction is found by rotating RAO2 90° in the direction of ω2: The direction of V tBA is perpendicular to RBA: The direction of V tB is perpendicular to RBO4: ► ► ►

5. Velocity polygon Velocity Polygon for a Four-bar V tA V tA V tA + V tBA = V tB B A A RBA OV V tBA RAO2 RBO4 V tB ω2 B RO4O2 O4 O2 To construct the velocity polygon we select an origin and draw V tA: V tB starts at the origin. We know the line of action: V tBA starts at A. We also know the line of action: Now we can complete the velocity polygon: Note that this polygon represents the velocity loop equation shown above! ► ► ► ►

6. Angular velocities Velocity Polygon for a Four-bar V tA ω3 V tA B A A RBA OV RBO4 V tBA RAO2 V tB ω4 ω2 B O4 RO4O2 O2 We can determine ω3: ω3 = V tBA / RBA RBA has to be rotated 90° clockwise to point in the same direction as V tBA. Therefore ω3 is clockwise: To determine ω4: ω4 = V tB / RBO4 RBO4 has to be rotated 90° counter-clockwise to point in the same direction as V tB. Therefore ω4 is ccw: ► ►

7. Example 2 Velocity Polygon for a Four-bar O2 ω2 B A P O4 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. Furthermore, it is assumed that the angular velocity of the crank (link 2) is given (known magnitude and direction).

8. Vector loop Velocity Polygon for a Four-bar Velocity analysis We define four position vectors to obtain a vector loop equation: RAO2 + RBA = RO4O2 + RBO4 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: V tAO2 + V tBA = V tO4O2 + V tBO4 Discard zero vectors and subscripts referring to non-moving points: V tA + V tBA = V tB ► O2 ω2 RAO2 RBA B A P RO4O2 RBO4 O4

9. VA and lines of action Velocity Polygon for a Four-bar ω2 V tA + V tBA = V tB O2 RAO2 V tA B RBA A P RO4O2 RBO4 O4 We can calculate V tA: V tA = ω2∙ RAO2 The direction is found by rotating RAO2 90° in the direction of ω2: The direction of V tBA is perpendicular to RBA: The direction of V tB is perpendicular to RBO4: ► ► ►

10. Velocity polygon Velocity Polygon for a Four-bar V tA + V tBA = V tB ω2 O2 RAO2 V tA A V tA B RBA A OV V tBA P V tB RO4O2 RBO4 B O4 To construct the velocity polygon we select the origin and draw V tA: V tB starts at the origin. We know the line of action: V tBA starts at A. We also know the line of action: Now we can complete the velocity polygon: Note that this polygon represents the velocity loop equation shown above! ► ► ► ►

11. Velocity of coupler point P Velocity Polygon for a Four-bar ω2 O2 RAO2 V tA A B RBA V tA A P VP P OV V tBA RO4O2 RBO4 V tB B O4 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: The point of intersection is point P in the velocity polygon. Now we can draw VP: ► ► ►

12. Angular velocities Velocity Polygon for a Four-bar ω2 O2 RAO2 V tA A B RBA ω3 V tA A P V tP P OV V tBA RBO4 RO4O2 V tB B ω4 O4 We can determine ω3: ω3 = V tBA / RBA RBA has to be rotated 90° clockwise to point in the same direction as V tBA. Therefore ω3 is clockwise: To determine ω4: ω4 = V tB / RBO4 RBO4 has to be rotated 90° clockwise to point in the same direction as V tB. Therefore ω4 is clockwise: ► ►