# Velocity Analysis with Instant Centers for a Four-bar Mechanism

## Presentation on theme: "Velocity Analysis with Instant Centers for a Four-bar Mechanism"— Presentation transcript:

Velocity Analysis with Instant Centers for a Four-bar Mechanism
Introduction Instant Centers for a Four-bar Velocity Analysis with Instant Centers for a Four-bar Mechanism This presentation shows how to perform velocity analysis for a four-bar mechanism with the method of instant centers. 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 (or one other velocity information) must be given as well. For a given four-bar mechanism the velocity analysis consists of two steps: Finding the instant centers Finding velocities Note that the instant center method is a graphical method.

Four-bar mechanism Instant Centers for a Four-bar Four-bar mechanism Assume that for this four-bar mechanism all the link lengths are known and the angular velocity of the crank is given as ω2 ccw. In the configuration shown we can perform a velocity analysis with the instant center method. B A ω2 O2 O4

Number of instant centers
Instant Centers for a Four-bar Number of instant centers The first task is to determine how many instant centers exist for a four-bar. The number of links in a four-bar 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. B A ω2 O2 1 2 O4 4 3

Finding the instant centers
Instant Centers for a Four-bar Finding the instant centers Four of the centers are already known: They are the four pin joints. We don’t know I1,3 but we know that it lies on the same line as I4,1 and I3,4. I1,3 also lies on the same line as I2,1 and I3,2. The point of intersection is I1,3. I2,4 is also unknown but it lies on the same line as I3,4 and I2,3. I2,4 also lies on the same line as I4,1 and I1,2. The point of intersection is I2,4. Now we have found all 6 centers B = I3,4 A = I2,3 I2,4 ω2 O2 = I1,2 1 2 O4 = I4,1 4 3 I1,3

A (or I2,3) is a point on link 2, therefore:
Finding ω3, knowing ω2 Instant Centers for a Four-bar Finding velocities A (or I2,3) is a point on link 2, therefore: VA = ω2 ∙ RAI1,2 Its direction is obtained by rotating RAI1,2 90° in the direction of ω2. A (or I2,3) is also a point on link 3, which rotates around I1,3, this means: VA = ω3 ∙ RAI1,3 Since we already know VA, we can solve for ω3: ω3 = VA ∕ RAI1,3 ω3 VA B = I3,4 A = I2,3 I2,4 RAI1,2 ω2 O2 = I1,2 RAI1,3 O4= I4,1 I1,3

Since we know ω3, we can find VB. B is a point on link 3:
Finding ω4, knowing ω3 Instant Centers for a Four-bar Finding velocities Since we know ω3, we can find VB. B is a point on link 3: VB = ω3 ∙ RBI1,3 Its direction is obtained by rotating RBI1,3 90° in the direction of ω3. B is also a point on link 4 which yields: VB = ω4 ∙ RBI1,4 We already know VB so we can solve for ω4: ω4 = VB ∕ RBI1,4 VB ω3 VA B = I3,4 A = I2,3 I2,4 RBI4,1 ω2 O2 = I1,2 ω4 O4= I4,1 RBI1,3 I1,3

We could have determined ω4 without knowing ω3:
Finding ω4, knowing ω2 Instant Centers for a Four-bar Another approach We could have determined ω4 without knowing ω3: I2,4 is a point on link 2, therefore: VI2,4 = ω2 ∙ RI2,4 I1,2 Its direction is obtained by rotating RI2,4 I1,2 90° in the direction of ω2. I2,4 is also a point on link 4, which rotates around I4,1. This means: VI2,4 = ω4 ∙ RI2,4 I4,1 We already know VI2,4 so we can solve for ω4: ω4 = VI2,4 ∕ RI2,4 I4,1 B = I3,4 A = I2,3 I2,4 VI2,4 RI2,4 I1,2 ω2 O2 = I1,2 ω4 RI2,4 I4,1 O4= I4,1 I1,3

Velocity of a coupler point
Velocity of coupler point Instant Centers for a Four-bar VP Velocity of a coupler point The instant center method makes it easy to find the velocities of additional points. For example suppose link 3 is a triangular plate and we want to determine the velocity of P. Since P is a point on link 3, we have: VP = ω3 ∙ RPI1,3 The direction is found by rotating RPI1,3 90° in the direction of ω3. P B = I3,4 A = I2,3 I2,4 RPI4,1 ω2 O2 = I1,2 O4= I4,1 I1,3