1. 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

2. §3－1 Tasks and Methods of Kinematic Analysis
一、Tasks of Kinematic Analysis
The tasks of kinematic analysis are to find angular positions, angular velocities and angular accelerations of driven links and/or positions, linear velocities and linear accelerations of points on driven links, according to input parameters of driving link (s) and the dimensions of all links.
In order to determine whether or not all links will interfere with each other, or to determine the stroke of a driven link, or to find the locus of a point, we must analyze positions of the links and/or the points of interest.

3. 二、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.

4. §3－2 Velocity Analysis by the Method of Instant Centers1 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

5. 三、 Location of the Instant Center
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.

6. 四、Applications of Instant Centers
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

7. 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.

8. 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.

9. §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

10. 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 ？ √ ？ √ ？

11. 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.

12. α3 ω2 α2 ω3 2. Acceleration analysisp’ 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

13. α3 Acceleration polygon properties： ω2 α2 ω34 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.

14. 二、Coincidence point velocity and acceleration in two bodies
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

15. 2 . Acceleration analysisp’ 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)