# Chapter3 Kinematic Analysis of Mechanisms

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Chapter3 Kinematic Analysis of Mechanisms
§3－1 Tasks and Methods of Kinematic Analysis §3－2 Velocity Analysis by the Method of Instant Centers §3－3 Graphical Velocity and Acceleration Analysis

§3－1 Tasks and Methods of Kinematic Analysis

Methods of Kinematic Analysis：method of instant centres、 graphical method、analytical method. Kinematic analysis of mechanisms can be carried out by graphical or analytical or experimental methods. By geometric drawing, the positions of all links in a grade II linkage can be determined easily according to the assembly order of Assur groups, step by step. In this chapter, we will introduce one of the graphical methods, named the instant centers method, for velocity analysis. The analytical method has many advantages over graphical methods.

§3－2 Velocity Analysis by the Method of Instant Centers
1 2 A2(A1) vA2A1 Shown in Fig. are two bodies 1 and 2 having relative planar motion. At any instant there exists a pair of coincident points, e.g.P1 and P2, the absolute velocities of which are the same, in both magnitude and direction.. i. e. v P1 =V P2. At this instant, there is no relative velocity between this pair of coincident points, i.e. B2(B1) vB2B1 Bi(Bj) vBiBj P12 VP1P2= VP2P1=0. Thus, at this instant, either link will have pure rotation relative to the other link about the point. This pair of coincident points with the same velocities is defined as the instantaneous center of relative rotation, or more briefly the instant center, denoted as P 12 or P 21. 二、Number of Instant Centers of a Mechanism K＝ N ( N - 1) / 2

1 . Two Links Connected by a Kinematic Pair 1 2 P12 Revolute pair 1 2 P12 P12 1 2 P12 Sliding pair 1 2 P12 vM1M2 M 1 2 1 2 Higher pair 2 . Theorem of Three Centers Theorem of Three Centers——the three instant centers of any three independent links in general plane motion must lie on a common straight line.

Example 3-1 For the four-bar mechanism shown in figure, the angular velocity 2 of driver 2 is given. For the position shown, locate all instant centers for the mechanism and find the angular velocity 4 of the link 4. K＝N(N-1)/2＝6 Solution:Number of Instant Centers P13 P12 、P23 、P34、P14 P13 、P24 （Theorem of Three Centers） P13 P12 P23 P34 P14 ω4 P24 P12 P23 Instant Center P24

vP24 Solution: Determine the number and location of instant centers
Example In the slide-crank mechanism shown in Fig., the angular velocity of crank 2 is known. The velocity vc of slider 4 is to be found for the position shown.。 Solution: Determine the number and location of instant centers P24 vP24 instant center P24 P23 P12 P34 P14 The direction is to the left.

vP23 Solution: Determine the number and location of instant centers
Example 3-3 In the cam mechanism shown in Fig., supposing that the angular velocity 2 of the cam is known, the velocity v3 of the follower 3 is to be found for the position shown. Solution: Determine the number and location of instant centers 1 2 3 ω2 Three instant centers P12、P13、P23 K vP23 P12 P13 P23 Advantages and Disadvantages of the Method of Instant Centers: 1）To offer an excellent tool in the velocity analysis of simple mechanisms. 2）In a complex mechanism, some instant centres may be difficult to find. In some cases they will lie off the paper. 3）The instant center method can not be used in acceleration analysis.

§3－3 Graphical Velocity and Acceleration Analysis
This chapter presents a few simple principles of rigid body planar kinematics. These principles are useful in visualizing and understanding the motions of interconnected rigid bodies. Case 1: Two points velocity and acceleration in the same body Case 2: Coincidence point velocity and acceleration in two bodies 一、 Two points velocity and acceleration in the same link The lengths of all the links and the angular velocity 1 of driver 1 are known. We wish to solve for the linear velocities of points C and E(vc , vE), accelerations of points C and E(ac, aE)，and the angular velocities and accelerations of links 2 and 3 (2 , 3 , 2 and 3) velocity = translation component + rotation component

ω3 ω2 vc = vB + vCB 1 .Velocity analysis procedures：
4 B w1 1 D E 2 3 C ω2 vc = vB vCB direction CD AB CB magnitude ？ 1lAB ？ ω3 procedures： 1）Select a velocity scale b c 2) Select a point p draw pb // vB 3) Draw the directions of vCB and vc pbvB bc vCB p e vc = vpc vCB = v bc 2 = vCB/ lBC (clockwise) 3 = vC/ lCD (counter clockwise) vE = vB vEB = vC vEC direction ? √ BE √ CE magnitude ？ √ ？ √ ？

ω2 Velocity polygon properties：
A 4 B w1 1 D E 2 3 C ω2 1）The vector from zero velocity point p (starting point)to point b represents the absolute velocity VB. （ pbvB ） b 2）The line from point b to point c represents the velocity difference (relative velocity) VCB. Notice that VCB is represented by the vector from point b to point c .（ bc vCB ）. p e c 3) ∵ bce ∽BCE，bce is called the velocity image of BCE. Once velocities of two points on a link are calculated, the velocity of any other point can be obtained by similar triangles.

α3 ω2 α2 ω3 2. Acceleration analysis
p’ c’ e’ A 4 B w1 1 D E 2 3 C ω2 c’’ α2 α3 ω3 b’ c’’’ magnitude ω32 lCD ？ ω12 lAB ω22 lBC ？ direction CD CD BA C B CB Select a acceleration scale ，draw acceleration vector diagram. magnitude  ω22 lBE ？  ω22 lCE ？ direction  EB BE  EC CE

α3 Acceleration polygon properties： ω2 α2 ω3
4 B w1 1 D E 2 3 C ω2 α2 α3 ω3 b’ c’’’ 1） The vector from zero acceleration point p’ (starting point) to point b represents the absolute acceleration（ p ’b’ aB）. 2）The acceleration difference (relative acceleration) between two points, say B and C, is represented by the line point b’ to point c’. (b’c’ aCB ) 3）∵ b’ c’ e’ ∽BCE，b’ c’ e’ is called the acceleration image of BCE. Once accelerations of two points on a link are calculated, the acceleration of any other point can be obtained by similar triangles.

Given lAC , lBC , 1(counter clockwise)， Find 2, 3, 2 , 3. 1 2 ω1 3 A B C instantaneous absolute motion = frame of reference motion + relative motion 1 . Velocity analysis ω3 vB2 = vB vB2B1 magnitude ？ 1lAB ？ direction BC AB //AB p b2 vB2B1 = vb1b vB3 = vB2 = vpb2 2 = 3 = vB2 / lBC (counter clockwise) b1

2 . Acceleration analysis
p’ 1 2 ω1 3 A B C p ak B2B1 b2 b2’’ b1’ α3 b2’ ω3 b1 k’ magnitude ω22 lBC ？ ω12 lAB vB2B ？ direction B C BC BA AB // AB aB3 = aB2= ap 'b2' (counter clockwise)