1. P. Nikravesh, AME, U of A 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 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: 1. How to construct the polygon shown on the right 2. How to extract velocity information from the polygon A O4O4 B O2O2 ω2ω2 V t A V t B V t BA A B OVOV

2. P. Nikravesh, AME, U of A Velocity Polygon for a Four-barExample 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). Introduction A O4O4 B O2O2 ω2ω2

3. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar Velocity analysis We define four position vectors to obtain a vector loop equation: R AO 2 + R BA = R O 4 O 2 + R BO 4 The time derivative of this equation yields the velocity loop equation: V AO 2 + V BA = V O 4 O 2 + V BO 4 Since R O 4 O 2 is fixed to the ground, V O 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: A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 Vector loop O2O2 ω2ω2 V A + V BA = V B For clarification purposes we assign superscript “t” to these vectors indicating they are tangential : V t A + V t BA = V t B ►

4. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar We can calculate V t A : V t A = 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 BA is perpendicular to R BA : The direction of V t B is perpendicular to R BO 4 : V A and lines of action A O4O4 B R BO 4 RO4O2RO4O2 R BA V t A O2O2 V t A + V t BA = V t B R AO 2 ω2ω2 ► ► ►

5. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar To construct the velocity polygon we select an origin and draw V t A : V t B starts at the origin. We know the line of action: V t BA 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! Velocity polygon A O4O4 B R BO 4 RO4O2RO4O2 V t A V t B V t BA V t A O2O2 V t A + V t BA = V t B A B OVOV R AO 2 ω2ω2 R BA ► ► ► ►

6. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar We can determine ω 3 : ω 3 = V t BA / R BA R BA has to be rotated 90° clockwise to point in the same direction as V t BA. Therefore ω 3 is clockwise: To determine ω 4 : ω 4 = V t B / R BO 4 R BO 4 has to be rotated 90° counter-clockwise to point in the same direction as V t B. Therefore ω 4 is ccw: ω4ω4 ω3ω3 Angular velocities A O4O4 B R BO 4 RO4O2RO4O2 R BA O2O2 V t A V t B V t BA V t A R AO 2 ω2ω2 A B OVOV ► ►

7. P. Nikravesh, AME, U of A Velocity Polygon for a Four-barExample 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). P A O4O4 B O2O2 ω2ω2

8. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar P Velocity analysis We define four position vectors to obtain a vector loop equation: R AO 2 + R BA = R O 4 O 2 + R BO 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: 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 A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 O2O2 ω2ω2 Vector loop ►

9. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar A O4O4 R BO 4 RO4O2RO4O2 R AO 2 V t A O2O2 ω2ω2 V A and lines of action We can calculate V t A : V t A = 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 BA is perpendicular to R BA : The direction of V t B is perpendicular to R BO 4 : P B R BA ► ► ► V t A + V t BA = V t B

10. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar V t A V t BA V t B OVOV A B Velocity polygon A O4O4 R BO 4 RO4O2RO4O2 R AO 2 V t A O2O2 ω2ω2 To construct the velocity polygon we select the origin and draw V t A : V t B starts at the origin. We know the line of action: V t BA 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! P B R BA ► ► ► ► V t A + V t BA = V t B

11. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar V t A V t BA V t B OVOV A B Velocity of coupler point P A O4O4 B R BO 4 RO4O2RO4O2 R AO 2 V t A O2O2 ω2ω2 P 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 V P : P VPVP R BA ► ► ►

12. P. Nikravesh, AME, U of A Velocity Polygon for a Four-bar V t A V t BA V t B OVOV A B Angular velocities A O4O4 B R BO 4 RO4O2RO4O2 R BA R AO 2 V t A O2O2 ω2ω2 P P V t P We can determine ω 3 : ω 3 = V t BA / R BA R BA has to be rotated 90° clockwise to point in the same direction as V t BA. Therefore ω 3 is clockwise: To determine ω 4 : ω 4 = V t B / R BO 4 R BO 4 has to be rotated 90° clockwise to point in the same direction as V t B. Therefore ω 4 is clockwise: ω3ω3 ω4ω4 ► ►