1. P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider Introduction Velocity Analysis with Instant Centers for a Crank-Slider Mechanism (Inversion 2) This presentation shows how to perform velocity analysis for a crank-slider mechanism (inversion 2) with the instant center method. 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 of the crank (or one other velocity information) must be given as well. For a given crank-slider mechanism the velocity analysis consists of two steps: 1. Finding the instant centers 2. Finding velocities O4O4 A O2O2 ω2ω2 P

2. P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider Crank-slider mechanism Assume that for this crank-slider mechanism all the link lengths are known and the angular velocity of the crank is given as ω 2 ccw. In the configuration shown we can perform a velocity analysis with the instant center method. Crank-slider mechanism (inversion 2) O4O4 A O2O2 ω2ω2 P

3. P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider Number of instant centers The first task is to determine how many instant centers exist for a crank-slider. The number of links in a crank-slider is n = 4 Between n links, there are n (n − 1) ∕ 2 instant centers. That means in a fourbar there are 4 (4 − 1) ∕ 2 = 6 instant centers. A small circle will help us keep track of locating each center. On the circumference of the circle we put as many marks as the number of links. Each time we find a center between two links, we draw a line between the corresponding marks on the circle. Number of instant centers 12 43 ► O4O4 A O2O2 (2) (3) (4) (1)

4. P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider O 2 = I 1,2 = I 3,4 A = I 2,3 O 4 = I 4,1 Finding the instant centers Four of the centers are already known: They are the three pin joints and the sliding joint. Note: The instant center between two bodies connected by a slider is located in infinity on any axis perpendicular to the axis of sliding. We don’t know I 1,3 but we know that it lies on the same line as I 4,1 and I 3,4. I 1,3 also lies on the same line as I 1,2 and I 2,3. The point of intersection is I 1,3. I 2,4 is also unknown but it lies on the same line as I 3,4 and I 2,3. I 2,4 also lies on the same line as I 4,1 and I 1,2. The point of intersection is I 2,4. Now we have found all 6 centers. Finding the instant centers 12 43 I 1,3 I 2,4 ► ► ► ► ► (2) (3) (4) (1) ► ►

5. P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider R AI 1,3 Finding velocities A (or I 2,3 ) is a point on link 2, therefore: V A = ω 2 ∙ R AI 1,2 Its direction is obtained by rotating R AI 1,2 90° in the direction of ω 2. A (or I 2,3 ) is also a point on link 3, which rotates around I 1,3. This means: V A = ω 3 ∙ R AI 1,3 Since we already know V A, we can solve for ω 3 : ω 3 = V A ∕ R AI 1,3 Since links 3 and 4 are connected by a sliding joint, their angular velocities are the same; i.e., ω 4 = ω 3 Finding ω 3 and ω 4, knowing ω 2 ω3ω3 VAVA R AI 1,2 ► ► ω2ω2 I 1,2 A = I 2,3 I 1,3 ω4ω4 ►

6. P. Nikravesh, AME, U of A Instant Centers for a Crank-Slider R I 2,4 I 4,1 Finding velocities We could have determined ω 4 directly from ω 2 : I 2,4 is a point on link 2, therefore: V I 2,4 = ω 2 ∙ R I 2,4 I 1,2 Its direction is obtained by rotating R I 2,4 I 1,2 90° in the direction of ω 2. I 2,4 is also a point on link 4, which rotates around I 4,1. This means: ω 4 = V I 2,4 ∕ R I 2,4 I 4,1 Since links 3 and 4 are connected by a sliding joint, their angular velocities are the same; i.e., ω 3 = ω 4 Finding ω 4 and ω 3, knowing ω 2 ω4ω4 V I 2,4 R I 2,4 I 1,2 ► ► ω2ω2 I 1,2 I 4,1 I 2,4 ω3ω3 ►