1. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-Slider Introduction Acceleration Polygon for a Crank-Slider Mechanism This presentation shows how to construct the acceleration polygon for a crank-slider (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. Since the crank-slider has one degree-of-freedom, the angular velocity and acceleration of one of the links must be given as well. As an example, for the crank-slider shown, we first construct the velocity polygon. Then we will learn: 1. How to construct the acceleration polygon shown on the right 2. How to extract acceleration information from the polygon OAOA A n BA AnAAnA AtAAtA A A A BA AsBAsB A t BA B O2O2 A ω2ω2 α2α2

2. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-SliderInversion 1 Whether the crank-slider is offset or not, the process of constructing an acceleration polygon remains the same. Therefore, in this 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 and acceleration of the crank is given. B O2O2 A ω2ω2 α2α2 B O2O2 A ω2ω2 α2α2

3. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-SliderVector loop, differentiation Vector loop 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 first time derivative provides the velocity loop equation: V t AO 2 + V t BA = V s BO 4 Or, V t A + V t BA = V s B The second time derivative provides the acceleration loop equation: A A + A BA = A B We split the acceleration vectors on the left side of the equation into normal and a tangential components and express the acceleration vector on the right side of the equation as a slip acceleration: B O4O4 A t A + A n A + A t BA + A n BA = A s B We need the velocities to calculate some of the accelerations. Therefore we perform a velocity analysis first. RO4O2RO4O2 R BO 4 R BA R AO 2 O2O2 ► A

4. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-SliderVelocity analysis V t AO 2 + V t BA = V s BO 4 We calculate V t AO 2 : V t AO 2 = 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 and the direction of V s BO 4 is parallel to R BO 4 We draw the velocity polygon: We determine ω 3 as: 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 and ω 4 equals zero. B O4O4 A R BA R AO 2 O2O2 ω2ω2 V t AO 2 OVOV B A V t BA V s BO 4 ► ► ► ►►►► ► ω3ω3 RO4O2RO4O2 R BO 4

5. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-Slider Normal components We first calculate the magnitude of all the normal components: A n A = ω 2 2 ∙ R AO 2 A n BA = ω 3 2 ∙ R BA The direction of each normal component of acceleration is opposite to the corresponding position vector Normal components A t A + A n A + A t BA + A n BA = A s B A n BA AnAAnA ► B O4O4 A RO4O2RO4O2 R BO 4 R BA R AO 2 O2O2 ω2ω2 ω3ω3 B O2O2 A

6. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-Slider Tangential and slip components We first calculate the magnitude of A t A : A t A = α 2 ∙ R AO 2 The direction is found by rotating R AO 2 90° in the direction of α 2 We also know that A t BA is on an axis perpendicular to R BA Similarly we know that A s B is along the axis of R BO 4 Tangential components AtAAtA α2α2 ► ► ► A t A + A n A + A t BA + A n BA = A s B B O4O4 A R BO 4 R BA R AO 2 O2O2 A n BA AnAAnA B O2O2 A

7. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-Slider A n BA AnAAnA AtAAtA Acceleration polygon Now we are ready to draw the acceleration polygon. First we select the origin and add A t A and A n A to obtain A A A n BA is a added at A We also know the axis of A t BA which would be added to A n BA We also know the axis of A s B is added to the origin The two lines intersect at B. We complete the acceleration polygon by drawing the missing accelerations Next we determine the angular accelerations. Acceleration polygon ► ► ► ► ► A t A + A n A + A t BA + A n BA = A s B B O2O2 A AnAAnA A n BA OAOA A A AtAAtA AtAAtA AnAAnA B A BA AsBAsB A t BA

8. P. Nikravesh, AME, U of A Acceleration Polygon for a Crank-Slider Angular accelerations The absolute value of the angular acceleration of link 3 is computed as: α 3 = A t BA ∕ R BA R BA has to be rotated 90° cw to head in the direction of A t BA. Therefore α 3 is cw Since the slider does not rotate, α 4 is zero. This completes the acceleration analysis of this crank-slider. Angular accelerations α2α2 α3α3 ► OAOA A n BA AnAAnA AtAAtA A A A BA AsBAsB A t BA B O4O4 RO4O2RO4O2 R BO 4 R BA R AO 2 O2O2 A