ROBOTICS, Lecture, CHAP 4- Robotic Mechanics 1

ALPEN-ADRIA UNIVERSITÄT KLAGENFURT| WIEN GRAZ

ROBOTICS CHAP 4: Robotics Mechanics 1

by DI Ahmad Haj Mosa, Fadi Al Machot and Kyandoghere Kyamakya

Institute of Smart Systems Technologies, Transportation Informatics Group Alpen Adria

Universität Klagenfurt


Ahmad Haj Mosa

Introduction

• Machines have been evolving for several decades

• As the Ame is passing, as the machines become more and more important in our daily life

• The evoluAon of machines started from a pure mechanical machine unAl recent days where a single machine contains mechanical, electrical components integrated with a smart soFware (Mechatronics)

• RoboAcs is on applicaAon of Mechatronics where the mechanical design is significantly important

210/19/2015


Materials

3


Mechanical Design

• Mechanical design has three processes that collaborate together to end up with an optimum design:

1. Developing of the Mechanics (Joints, static and dynamic analysis, kinematics)

3. Mechanical Components design (safety, reliability, stress, fatigue, dimensions,cost)

5. Industrial design (shape, presentation, cost)

4

Industrial Design

ComponentDesign

Mechanics


Course Plan

• Mechanics Terminology, Definitions and Assumptions • Position and Displacement • Velocity Analysis • Acceleration • Static Force • Dynamic Force • Vibration • Balancing • Mechanical Component Design

5


Mechanics

• Mechanics deals with motions, time and forces

• It is made up of two parts: static and dynamic

• Static deals with the analysis of stationary systems

• Dynamic deals with systems that changes overtime which is also made up of two parts

6

Mechanics

MechanicsStatics Dynamics

KineticsKinematics


Ahmad Haj Mosa

Kinematics, Kinetics, Rigid and Soft Body

10/19/20157

Kinematics

• Is the study of motion, quite apart from the forces that produce the motion

• Study of Motion means the study of position, displacement, rotation,velocity and acceleration

Kinetics

• Is the study of motion under the action of external forces

• The study includes the study of force transition, dynamic stress, vibration and balancing

Both Kinematics & Kinetics study the rigid body For soft bodies it is not possible to separate the motion from the force

Rigid Body• Is a non-‐deformable body

• Regarding of the external force the distance between two points in the body remains the same

Soft Body

• Is deformable body

• External forces lead to change in shape/form of the body

https://en.wikipedia.org/wiki/Deformation_(engineering)


Ahmad Haj Mosa

Structure vs Mechanism

10/19/20158

Structure (Static Mechanics)

• Is a combination of resistant rigid bodies connected by static joints

• The purpose of a structure is be rigid and not to transform motion

• Has no internal mobility (degrees of freedom)

• Has no relative motions between its various members

Mechanisms (Kinematics)

• Is an assemble of rigid bodies connected by movable joints

• The purpose of a mechanisms is to transform motion

• Has an internal mobility (degrees of freedom)

• Has relative motions between its various members


Machines

• Machine is a system of mechanisms

• The purpose of machines is to do a work and to transform a force, torque and power (Kinetics analysis)

9

Mechanism 1 Mechanism 2

Mechanism 3

Robot Machine


Links, driver, follower and pairs

10

Driver, Follower and pairs/joints

• Links are connected together to transmit motion from the driver link to a the follower link

• The connection between the links is called pairs

Links

• Are the rigid connection between two or more joint elements

• Must be completely rigid (Springs are not links)

• Links connected to two pairs are a binary links

• link connected to there pairs are ternary links

Kinematic Chain

• Is a several links connected together by joints


Kinematic Higher and Lower Pairs

11

Lower Pairs

• When the two elements of a pair have a surface contact when relative motion takes place and the surface of one element slides over the surface of the other, the pair formed is known as lower pair

Higher Pairs

• When the two elements of a pair have a /ine or point contact when relative motion takes place and the motion between the two elements is partly turning and partly sliding, then the pair is known as higher pair


Planar, Spherical and Spatial Mechanism

12

Planar

• All particles move in a planar curves in space in which all curves lie in parallel planes

• A motion transformation of such mechanism is called coplanar

Planar 4 bar linkage mechanism

Spatial

• All particles move in a non parallel nor spherical planes

Spherical 4 bar linkage mechanism Spatial 4 bar linkage mechanism

Spherical

• All particles move in a Spherical curves in space in which all curves lie in parallel planes

• A motion transformation of such mechanism is called concentric

Animation source: https://www.youtube.com/channel/UCli_RJkGWfZvw4IlDLHNCQg

https://www.youtube.com/channel/UCli_RJkGWfZvw4IlDLHNCQg


Mobility or Degree of Freedom

13

• First important step in the design and the analysis of a mechanism is its mobility/degree of freedom

• The mobility of a mechanism is the number of pair variables that must be controlled independently to bring the device into a particular position or orientation

• In general It is possible to determine the mobility of a mechanism directly from the count of links and the number and types of joints

• In order to estimate the mobility of a planar mechanism we need to follow the following assumptions:

1. A free link of a planar mechanism has three degrees of freedom when moving relative to a fixed link

2. Then n-‐link planar mechanism has 3(n-‐1) degree of freedom

3. Connected two links by a joint that has a one degree of freedom (jp) (i.e. pin or slidining pair ) provide two constrains

4. Connected two links by a joint that has two degrees of freedom (jh) (i.e. cam or gear pair ) provide one constrains

• Using the previous four assumptions then the we can estimate the degree of freedom using the following:

8 CHAPTER ONE

(a) Single degree-of-freedom (M = 1) (b) Locked mechanism (M = 0) (c) Multi-degree-of-freedom (M = 2)

FIGURE 1.13 Mechanisms and structures with varying mobility.

5. Draw the Kinematic Diagram

The kinematic diagram is given in Figure 1.12.

1.6 KINEMATIC INVERSION

Absolute motion is measured with respect to a stationaryframe. Relative motion is measured for one point or linkwith respect to another link. As seen in the previous exam-ples, the first step in drawing a kinematic diagram isselecting a member to serve as the frame. In some cases,the selection of a frame is arbitrary, as in the vise gripsfrom Example Problem 1.2. As different links are chosen asa frame, the relative motion of the links is not altered, butthe absolute motion can be drastically different. Formachines without a stationary link, relative motion isoften the desired result of kinematic analysis.

In Example Problem 1.2, an important result of kine-matic analysis is the distance that the handle must be movedin order to open the jaw. This is a question of relative posi-tion of the links: the handle and jaw. Because the relativemotion of the links does not change with the selection of aframe, the choice of a frame link is often not important.Utilizing alternate links to serve as the fixed link is termedkinematic inversion.

1.7 MOBILITY

An important property in mechanism analysis is the number ofdegrees of freedom of the linkage. The degree of freedom is thenumber of independent inputs required to precisely positionall links of the mechanism with respect to the ground. It canalso be defined as the number of actuators needed to operatethe mechanism. A mechanism actuator could be manuallymoving one link to another position, connecting a motor to theshaft of one link, or pushing a piston of a hydraulic cylinder.

The number of degrees of freedom of a mechanism isalso called the mobility, and it is given the symbol . WhenM

the configuration of a mechanism is completely defined bypositioning one link, that system has one degree of freedom.Most commercially produced mechanisms have one degreeof freedom. In constrast, robotic arms can have three, ormore, degrees of freedom.

1.7.1 Gruebler’s EquationDegrees of freedom for planar linkages joined with commonjoints can be calculated through Gruebler’s equation:

where:

jh total number of higher-order joints (cam or gear joints)

As mentioned, most linkages used in machines have onedegree of freedom. A single degree-of-freedom linkage isshown in Figure 1.13a.

Linkages with zero, or negative, degrees of freedom aretermed locked mechanisms. These mechanisms are unableto move and form a structure. A truss is a structure com-posed of simple links and connected with pin joints andzero degrees of freedom. A locked mechanism is shown inFigure 1.13b.

Linkages with multiple degrees of freedom need morethan one driver to precisely operate them. Commonmulti-degree-of-freedom mechanisms are open-loopkinematic chains used for reaching and positioning, suchas robotic arms and backhoes. In general, multi-degree-of-freedom linkages offer greater ability to precisely positiona link. A multi-degree-of-freedom mechanism is shown inFigure 1.13c.

=jp = total number of primary joints (pins or sliding joints)

n = total number of links in the mechanism

M = degrees of freedom = 3(n - 1) - 2jp - jh

FIGURE 1.12 Kinematic diagram for Example Problem 1.2.

2

41

AD

Y

C B

X3


Mobility or Degree of Freedom

14

• Normally most joints used in machines have one degree of freedom. A single degree-‐of-‐freedom linkage is shown in Figure 1.13a.

• Joints with multiple degrees of freedom need more than one driver to precisely operate them. Common multi-‐degree-‐of-‐freedom mechanisms are open-‐loop kinematic chains used for reaching and positioning, such as robotic arms and backhoes. In general, multi-‐degree-‐of-‐ freedom linkages offer greater ability to precisely position a link

• A multi-‐degree-‐of-‐freedom mechanism is shown in Figure 1.13c.

8 CHAPTER ONE

(a) Single degree-of-freedom (M = 1) (b) Locked mechanism (M = 0) (c) Multi-degree-of-freedom (M = 2)

FIGURE 1.13 Mechanisms and structures with varying mobility.



1.6 KINEMATIC INVERSION

Absolute motion is measured with respect to a stationaryframe. Relative motion is measured for one point or linkwith respect to another link. As seen in the previous exam-ples, the first step in drawing a kinematic diagram isselecting a member to serve as the frame. In some cases,the selection of a frame is arbitrary, as in the vise gripsfrom Example Problem 1.2. As different links are chosen asa frame, the relative motion of the links is not altered, butthe absolute motion can be drastically different. Formachines without a stationary link, relative motion isoften the desired result of kinematic analysis.

In Example Problem 1.2, an important result of kine-matic analysis is the distance that the handle must be movedin order to open the jaw. This is a question of relative posi-tion of the links: the handle and jaw. Because the relativemotion of the links does not change with the selection of aframe, the choice of a frame link is often not important.Utilizing alternate links to serve as the fixed link is termedkinematic inversion.

1.7 MOBILITY

An important property in mechanism analysis is the number ofdegrees of freedom of the linkage. The degree of freedom is thenumber of independent inputs required to precisely positionall links of the mechanism with respect to the ground. It canalso be defined as the number of actuators needed to operatethe mechanism. A mechanism actuator could be manuallymoving one link to another position, connecting a motor to theshaft of one link, or pushing a piston of a hydraulic cylinder.

The number of degrees of freedom of a mechanism isalso called the mobility, and it is given the symbol . WhenM

the configuration of a mechanism is completely defined bypositioning one link, that system has one degree of freedom.Most commercially produced mechanisms have one degreeof freedom. In constrast, robotic arms can have three, ormore, degrees of freedom.

1.7.1 Gruebler’s EquationDegrees of freedom for planar linkages joined with commonjoints can be calculated through Gruebler’s equation:

where:

jh total number of higher-order joints (cam or gear joints)

As mentioned, most linkages used in machines have onedegree of freedom. A single degree-of-freedom linkage isshown in Figure 1.13a.

Linkages with zero, or negative, degrees of freedom aretermed locked mechanisms. These mechanisms are unableto move and form a structure. A truss is a structure com-posed of simple links and connected with pin joints andzero degrees of freedom. A locked mechanism is shown inFigure 1.13b.

Linkages with multiple degrees of freedom need morethan one driver to precisely operate them. Commonmulti-degree-of-freedom mechanisms are open-loopkinematic chains used for reaching and positioning, suchas robotic arms and backhoes. In general, multi-degree-of-freedom linkages offer greater ability to precisely positiona link. A multi-degree-of-freedom mechanism is shown inFigure 1.13c.

=jp = total number of primary joints (pins or sliding joints)

n = total number of links in the mechanism

M = degrees of freedom = 3(n - 1) - 2jp - jh


2

41

AD

Y

C B

X3


Mobility examples

• In the following figure we have a toggle clamp

• First draw the kinematic diagram

• Then Calculate mobility

15

Introduction to Mechanisms and Kinematics 9

FIGURE 1.14 Toggle clamp for Example Problem 1.3.

EXAMPLE PROBLEM 1.3

Figure 1.14 shows a toggle clamp. Draw a kinematic diagram, using the clamping jaw and the handle as points ofinterest. Also compute the degrees of freedom for the clamp.

SOLUTION: 1. Identify the Frame

The component that is bolted to the table is designated as the frame. The motion of all other links is determinedrelative to this frame. The frame is numbered as link 1.

2. Identify All Other Links

Careful observation reveals three other moving parts:Link 2: HandleLink 3: Arm that serves as the clamping jawLink 4: Bar that connects the clamping arm and handle

3. Identify the Joints

Four pin joints are used to connect these different links (link 1 to 2, 2 to 3, 3 to 4, and 4 to 1). These joints arelettered A through D.

4. Identify Any Points of Interest

The motion of the clamping jaw is desired. This is designated as point of interest X. Finally, the motion of theend of the handle is also desired. This is designated as point of interest Y.


The kinematic diagram is detailed in Figure 1.15.

1

4

3

X

A

BC

D

Y

2


6. Calculate Mobility

Having four links and four pin joints,

n = 4, jp = 4 pins, jh = 0




Toggle clamp

• Identify the links

• we have a handle denoted by link 2

• Arm denoted by link 3

• Bar denoted by link 4

• The frame denoted by link 1

16



EXAMPLE PROBLEM 1.3












1

4

3

X

A

BC

D

Y

2




n = 4, jp = 4 pins, jh = 0

Arm: Link 3Handle: Link 2

Bar: Link 4

Frame: Link 1


Toggle clamp Kinematic Diagram

17



EXAMPLE PROBLEM 1.3












1

4

3

X

A

BC

D

Y

2




n = 4, jp = 4 pins, jh = 0

Handle: Link 2

Arm: Link 3

Bar: Link 4

Frame: Link 1


Toggle clamp Mobility

• We have four links and four pin joints

• Then the mobility is given by

18



EXAMPLE PROBLEM 1.3












1

4

3

X

A

BC

D

Y

2




n = 4, jp = 4 pins, jh = 0



EXAMPLE PROBLEM 1.3












1

4

3

X

A

BC

D

Y

2




n = 4, jp = 4 pins, jh = 0

Handle: Link 2

Arm: Link 3

Bar: Link 4

Frame: Link 1

10 CHAPTER ONE

1

2

4

3C

A

B D

X


FIGURE 1.16 Can crusher for Example Problem 1.4.

and

With one degree of freedom, the clamp mechanism is constrained. Moving only one link, the handle, preciselypositions all other links in the clamp.

M = 3(n - 1) - 2jp - j h = 3(4 - 1) - 2(4) - 0 = 1

EXAMPLE PROBLEM 1.4

Figure 1.16 shows a beverage can crusher used to reduce the size of cans for easier storage prior to recycling. Draw akinematic diagram, using the end of the handle as a point of interest. Also compute the degrees of freedom for the device.


The back portion of the device serves as a base and can be attached to a wall. This component is designatedas the frame. The motion of all other links is determined relative to this frame. The frame is numbered as link 1.


Careful observation shows a planar mechanism with three other moving parts:Link 2: HandleLink 3: Block that serves as the crushing surfaceLink 4: Bar that connects the crushing block and handle


Three pin joints are used to connect these different parts. One pin connects the handle to the base. This joint islabeled as A. A second pin is used to connect link 4 to the handle. This joint is labeled B. The third pin connectsthe crushing block and link 4. This joint is labeled C.

The crushing block slides vertically during operation; therefore, a sliding joint connects the crushing blockto the base. This joint is labeled D.


The motion of the handle end is desired. This is designated as point of interest X.




Task 1

• For the following mechanism:


• Identify the joint type

• Draw the kinematic digram

• calculate the DOF or Mobility

19

10 CHAPTER ONE

1

2

4

3C

A

B D

X



and


M = 3(n - 1) - 2jp - j h = 3(4 - 1) - 2(4) - 0 = 1

EXAMPLE PROBLEM 1.4














Task 1 solution

• For the following mechanism:


• Identify the joint type

• Draw the kinematic digram

• calculate the DOF or Mobility

20

10 CHAPTER ONE

1

2

4

3C

A

B D

X



and


M = 3(n - 1) - 2jp - j h = 3(4 - 1) - 2(4) - 0 = 1

EXAMPLE PROBLEM 1.4













10 CHAPTER ONE

1

2

4

3C

A

B D

X



and


M = 3(n - 1) - 2jp - j h = 3(4 - 1) - 2(4) - 0 = 1

EXAMPLE PROBLEM 1.4














FIGURE 1.18 Shear press for Example Problem 1.5.


It was determined that there are four links in this mechanism. There are also three pin joints and one slider joint.Therefore,

and

With one degree of freedom, the can crusher mechanism is constrained. Moving only one link, the handle, preciselypositions all other links and crushes a beverage can placed under the crushing block.

M = 3(n - 1) - 2jp - j h = 3(4 - 1) - 2(4) - 0 = 1

n = 4, jp = (3 pins + 1 slider) = 4, jh = 0

EXAMPLE PROBLEM 1.5

Figure 1.18 shows another device that can be used to shear material. Draw a kinematic diagram, using the end of the handle and the cutting edge as points of interest. Also, compute the degrees of freedom for the shear press.


The base is bolted to a working surface and can be designated as the frame. The motion of all other links is de-termined relative to this frame. The frame is numbered as link 1.


Careful observation reveals two other moving parts:Link 2: Gear/handleLink 3: Cutting lever


Two pin joints are used to connect these different parts. One pin connects the cutting lever to the frame.This joint is labeled as A. A second pin is used to connect the gear/handle to the cutting lever. This joint islabeled B.

The gear/handle is also connected to the frame with a gear joint. This higher-order joint is labeled C.


The motion of the handle end is desired and is designated as point of interest X. The motion of the cutting surface isalso desired and is designated as point of interest Y.




FIGURE 1.18 Shear press for Example Problem 1.5.


It was determined that there are four links in this mechanism. There are also three pin joints and one slider joint.Therefore,

and

With one degree of freedom, the can crusher mechanism is constrained. Moving only one link, the handle, preciselypositions all other links and crushes a beverage can placed under the crushing block.

M = 3(n - 1) - 2jp - j h = 3(4 - 1) - 2(4) - 0 = 1

n = 4, jp = (3 pins + 1 slider) = 4, jh = 0

EXAMPLE PROBLEM 1.5

Figure 1.18 shows another device that can be used to shear material. Draw a kinematic diagram, using the end of the handle and the cutting edge as points of interest. Also, compute the degrees of freedom for the shear press.


The base is bolted to a working surface and can be designated as the frame. The motion of all other links is de-termined relative to this frame. The frame is numbered as link 1.


Careful observation reveals two other moving parts:Link 2: Gear/handleLink 3: Cutting lever


Two pin joints are used to connect these different parts. One pin connects the cutting lever to the frame.This joint is labeled as A. A second pin is used to connect the gear/handle to the cutting lever. This joint islabeled B.

The gear/handle is also connected to the frame with a gear joint. This higher-order joint is labeled C.


The motion of the handle end is desired and is designated as point of interest X. The motion of the cutting surface isalso desired and is designated as point of interest Y.




Course Plan

• Mechanics Terminology, Definitions and Assumptions • Position and Displacement • Velocity • Acceleration • Static Force • Dynamic Force • Vibration • Balancing • Mechanical Component Design

21


Position of a Point

• A position of a point is determined always using a reference point (Mostly the origin of the reference coordinates)

• For example, giving a reference point O, then the position of the point P is determined from the distance between O and P (based on the unit distance also called the magnitude) and the relative angular orientation of the line OP.

• Then the position of a point P is defined by a vector form the origin O to P

• In this course, a vector is denoted by

• In a 3D Cartesian coordinates a vector is represented by

• were are the unit vectors in the directions, respectively

22

The Magnitude of the vector

Position and Displacement Analysis 73

P

P'

RP∆

RP

P'

P

y

x

θ

θ

RP'

FIGURE 4.3 Displacement vector for point .P

two points on that link forms with a reference axis.Referring to Figure 4.2, line MN lies on link 4. The angularposition of link 4 is defined by , which is the anglebetween the -axis and line MN. For consistency, angularposition is defined as positive if the angle is measured coun-terclockwise from the reference axis and negative if it ismeasured clockwise.

4.2.3 Position of a MechanismThe primary purpose in analyzing a mechanism is to studyits motion. Motion occurs when the position of the links andthe reference points that comprise the mechanism arechanged. As the position of the links is altered, the mecha-nism is forced into a different configuration, and motionproceeds.

Recall from Chapter 1 that an important property ofa mechanism is the mobility or degrees of freedom. Forlinkages with one degree of freedom, the position of onelink or point can precisely determine the position of allother links or points. Likewise, for linkages with two degreesof freedom, the position of two links can precisely deter-mine the position of all other links.

Therefore, the position of all points and links in amechanism is not arbitrary and independent. The inde-pendent parameters are the positions of certain “driver”links or “driver” points. The primary goal of positionanalysis is to determine the resulting positions of thepoints on a mechanism as a function of the position ofthese “driver” links or points.

4.3 DISPLACEMENT

Displacement is the end product of motion. It is a vector thatrepresents the distance between the starting and endingpositions of a point or link. There are two types of displace-ments that will be considered: linear and angular.

4.3.1 Linear DisplacementLinear displacement, , is the straight line distance betweenthe starting and ending position of a point during a timeinterval under consideration. Figure 4.3 illustrates a point on a mechanism that is displaced to position Pœ.

P

¢R

xu4

The linear displacement of point is denoted as and is calculated as the vectoral difference between the initialposition and the final position. Given in equation form:

(4.1)

Notice that linear displacement is not the distancetraveled by the point during motion.

The magnitude of the displacement vector is thedistance between the initial and final position during aninterval. This magnitude will be in linear units (inches, feet,millimeters, etc.). The direction can be identified by an anglefrom a reference axis to the line that connects the two posi-tions. The sense of the vector is from the initial position andpointing toward the final position.

4.3.2 Angular DisplacementAngular displacement, , is the angular distance betweentwo configurations of a rotating link. It is the differencebetween the starting and ending angular positions of a link,as shown in Figure 4.4. While possessing a magnitudeand direction (clockwise or counterclockwise), angular dis-placement is not technically a vector since it does not adhereto commutative and associative laws of vector addition.

The angular displacement of a link, say link 3, isdenoted as and determined with equation (4.2).

(4.2)

The magnitude of the angular displacement is the anglebetween the initial and final configuration of a link duringan interval. This magnitude will be in rotational units (e.g.,degrees, radians, and revolutions), and denoting eitherclockwise or counterclockwise specifies the direction.

¢u3 = u3œ - u3

¢u3

¢u

¢RP = RP œ - 7 RP

¢RPP

RyP

RxP

RP

y

x

P

N

4

M

P

4θ

θ

FIGURE 4.2 Position vector for point .P

3'

B'

B

A

3

3'3∆θ

θ

θ3

FIGURE 4.4 Angular displacement.


The Angular Position of a Link

The angular posiAon of a link is also an important quanAty. An angular posi6on, is defined as the angle between a link and a reference axes

23


P

P'

RP∆

RP

P'

P

y

x

θ

θ

RP'






4.3 DISPLACEMENT



P

¢R

xu4


(4.1)





(4.2)


¢u3 = u3œ - u3

¢u3

¢u

¢RP = RP œ - 7 RP

¢RPP

RyP

RxP

RP

y

x

P

N

4

M

P

4θ

θ


3'

B'

B

A

3

3'3∆θ

θ

θ3



Ahmad Haj Mosa

Link Displacement Analysis

10/19/201524

Linear Displacement

• Linear displacement, is the he straight line distance between the starting and ending position and is given by

Angular Discplacement

• Angular displacement, is the angular distance between two configurations of a rotating link. It is the difference between the starting and ending angular positions of a link and given by


P

P'

RP∆

RP

P'

P

y

x

θ

θ

RP'






4.3 DISPLACEMENT



P

¢R

xu4


(4.1)





(4.2)


¢u3 = u3œ - u3

¢u3

¢u

¢RP = RP œ - 7 RP

¢RPP

RyP

RxP

RP

y

x

P

N

4

M

P

4θ

θ


3'

B'

B

A

3

3'3∆θ

θ

θ3



P

P'

RP∆

RP

P'

P

y

x

θ

θ

RP'






4.3 DISPLACEMENT



P

¢R

xu4


(4.1)





(4.2)


¢u3 = u3œ - u3

¢u3

¢u

¢RP = RP œ - 7 RP

¢RPP

RyP

RxP

RP

y

x

P

N

4

M

P

4θ

θ


3'

B'

B

A

3

3'3∆θ

θ

θ3



Graphical Displacement Analysis

• A mechanism must have a driving link and followers/slave links. First we need to analysis the displacement of the driving link.

25

74 CHAPTER FOUR

B′

B

A

1

15°

C3 = ?

∆θ

∆θ

4 = ?

4

3

2

FIGURE 4.5 Typical position analysis.

FIGURE 4.6 Two geometric inversions of a four-bar mechanism.

4.4 DISPLACEMENT ANALYSIS

A common kinematic investigation is locating the positionof all links in a mechanism as the driver link(s) is displaced.As stated in Section 4.2, the degrees of freedom of a mecha-nism determine the number of independent driver links. Forthe most common mechanisms, those with one degree offreedom, displacement analysis consists of determining theposition of all links as one link is displaced. The positions ofall links are called the configuration of the mechanism.

Figure 4.5 illustrates this investigation. The mechanismshown has four links, as numbered. Recall that the fixed link,or frame, must always be included as a link. The mechanismalso has four revolute, or pin, joints.

From equation (1.1), the degrees of freedom can becalculated as follows:

With one degree of freedom, moving one link precisely posi-tions all other links in the mechanism. Therefore, a typicaldisplacement analysis problem involves determining theposition of links 3 and 4 in Figure 4.5 as link 2 moves toa specified displacement. In this example, the drivingdisplacement is angular, clockwise.¢u2 = 15°

M = 3(4 - 1) - 2(4) = 1

disassembling the mechanism or traveling through deadpoints. Thus, when conducting a displacement analysis,inspecting the original configuration of the mechanism isnecessary to determine the geometric inversion of interest.

4.5 DISPLACEMENT: GRAPHICAL ANALYSIS

4.5.1 Displacement of a Single Driving LinkIn placing a mechanism in a new configuration, it is neces-sary to relocate the links in their respective new positions.Simple links that rotate about fixed centers can be relocatedby drawing arcs, centered at the fixed pivot, through themoving pivot, at the specified angular displacement. This wasillustrated in Figure 4.5 as link 2 was rotated 15° clockwise.

In some analyses, complex links that are attached to theframe also must be rotated. This can be done using severalmethods. In most cases, the simplest method begins byrelocating only one line of the link. The other geometry thatdescribes the link can then be reconstructed, based on theposition of the line that has already been relocated.

Figure 4.7 illustrates the process of rotating a complexlink. In Figure 4.7a, line AB of the link is displaced to itsdesired position, clockwise. Notice that therelocated position of point is designated as .BœB

¢u2 = 80°

Nearly all linkages exhibit alternate configurations for agiven position of the driver link(s). Two configurations forthe same crank position of a four-bar mechanism are shownin Figure 4.6. These alternate configurations are calledgeometric inversions. It is a rare instance when a linkage canmove from one geometric inversion to a second without

80°

B

C

A

B'

rBC

rAB

rBC

rAC

rAC2

B

C

A

C' B'

2 2

(b)(a)

FIGURE 4.7 Rotating a complex link.

74 CHAPTER FOUR

B′

B

A

1

15°

C3 = ?

∆θ

∆θ

4 = ?

4

3

2

FIGURE 4.5 Typical position analysis.

FIGURE 4.6 Two geometric inversions of a four-bar mechanism.

4.4 DISPLACEMENT ANALYSIS

A common kinematic investigation is locating the positionof all links in a mechanism as the driver link(s) is displaced.As stated in Section 4.2, the degrees of freedom of a mecha-nism determine the number of independent driver links. Forthe most common mechanisms, those with one degree offreedom, displacement analysis consists of determining theposition of all links as one link is displaced. The positions ofall links are called the configuration of the mechanism.

Figure 4.5 illustrates this investigation. The mechanismshown has four links, as numbered. Recall that the fixed link,or frame, must always be included as a link. The mechanismalso has four revolute, or pin, joints.

From equation (1.1), the degrees of freedom can becalculated as follows:

With one degree of freedom, moving one link precisely posi-tions all other links in the mechanism. Therefore, a typicaldisplacement analysis problem involves determining theposition of links 3 and 4 in Figure 4.5 as link 2 moves toa specified displacement. In this example, the drivingdisplacement is angular, clockwise.¢u2 = 15°

M = 3(4 - 1) - 2(4) = 1

disassembling the mechanism or traveling through deadpoints. Thus, when conducting a displacement analysis,inspecting the original configuration of the mechanism isnecessary to determine the geometric inversion of interest.

4.5 DISPLACEMENT: GRAPHICAL ANALYSIS

4.5.1 Displacement of a Single Driving LinkIn placing a mechanism in a new configuration, it is neces-sary to relocate the links in their respective new positions.Simple links that rotate about fixed centers can be relocatedby drawing arcs, centered at the fixed pivot, through themoving pivot, at the specified angular displacement. This wasillustrated in Figure 4.5 as link 2 was rotated 15° clockwise.

In some analyses, complex links that are attached to theframe also must be rotated. This can be done using severalmethods. In most cases, the simplest method begins byrelocating only one line of the link. The other geometry thatdescribes the link can then be reconstructed, based on theposition of the line that has already been relocated.

Figure 4.7 illustrates the process of rotating a complexlink. In Figure 4.7a, line AB of the link is displaced to itsdesired position, clockwise. Notice that therelocated position of point is designated as .BœB

¢u2 = 80°

Nearly all linkages exhibit alternate configurations for agiven position of the driver link(s). Two configurations forthe same crank position of a four-bar mechanism are shownin Figure 4.6. These alternate configurations are calledgeometric inversions. It is a rare instance when a linkage canmove from one geometric inversion to a second without

80°

B

C

A

B'

rBC

rAB

rBC

rAC

rAC2

B

C

A

C' B'

2 2

(b)(a)

FIGURE 4.7 Rotating a complex link.


1Constrainedpath of point A

Constrainedpath of point C

Constrainedpath of point EE

FB

C

DA

rAB

rCD

rEF

3

4

56

2

FIGURE 4.8 Constrained paths of points on a link pinned to the frame.

4

D


C

3

1

2

ArAB

rCD15°B

B′Constrainedpath of point B

FIGURE 4.9 Constructing the constrained path of .C

The next step is to determine the position of therelocated point , which is designated as . Because thecomplex link is rigid and does not change shape duringmovement, the lengths of lines AC and BC do not change.Therefore, point can be located by measuring the lengthsof AC and BC and striking arcs from points and , respec-tively (Figure 4.7b).

A second approach can be employed on a CAD system.The lines that comprise the link can be duplicated androtated to yield the relocated link. All CAD systems have acommand that can easily rotate and copy geometric entities.This command can be used to rotate all lines of a link abouta designated point, a specified angular displacement. It isconvenient to display the rotated link in another color andplace it on a different layer.

4.5.2 Displacement of the Remaining Slave LinksOnce a driver link is repositioned, the position of all otherlinks must be determined. To accomplish this, the possiblepaths of all links that are connected to the frame should beconstructed. For links that are pinned to the frame, all pointson the link can only rotate relative to the frame. Thus, thepossible paths of those points are circular arcs, centered atthe pin connecting the link to the frame.

Figure 4.8 illustrates a kinematic diagram of a mecha-nism. Links 2, 4, and 6 are all pinned to the frame. Becausepoints , and are located on links 2, 4, and 6, respec-tively, their constrained paths can be readily constructed.The constrained path of point is a circular arc, centered atpoint , which is the pin that connects link 2 to the frame.The constrained paths of and can be determined in asimilar manner.

The constrained path of a point on a link that isconnected to the frame with a slider joint can also be easilydetermined. All points on this link move in a straight line,parallel to the direction of the sliding surface.

After the constrained paths of all links joined to theframe are constructed, the positions of the connecting linkscan be determined. This is a logical process that stems fromthe fact that all links are rigid. Rigidity means that the linksdo not change length or shape during motion.

In Figure 4.5, the positions of links 3 and 4 are desired aslink 2 rotates 15° clockwise. Using the procedures described

ECA

B

EB,C

BœAC œ

C œCin Section 4.5.1, Figure 4.9 shows link 2 relocated to itsdisplaced location, which defines the position of point .The constrained path of point has also been constructedand shown in Figure 4.9.

Because of its rigidity, the length of link 3 does notchange during motion. Although link 2 has been reposi-tioned, the length between points and does notchange. To summarize the facts of this displacement analy-sis, the following is known:

1. Point has been moved to

2. Point must always lay on its constrained path (lengthfrom and

3. The length between and must stay constant (C!must be a length from .

From these facts, the new position of link 3 can beconstructed. The length of line BC should be measured.Because point has been moved to , an arc of length isconstructed with its center at . By sweeping this arc, thefeasible path of point has been determined. However,point must also lay on its constrained path, as shown inFigure 4.9. Therefore, point must be located at theintersection of the two arcs. This process is illustrated inFigure 4.10. Note that the two arcs will also intersect at asecond point. This second point of intersection is a consider-able distance from and represents a second geometricinversion for this linkage. The linkage must be disassembledand reassembled to achieve this alternate configuration, sothat intersection can be ignored.

It is possible that the two arcs do not intersect at all.Cases where the constrained path and feasible path do notintersect indicate that length of the individual linksprevents the driver link from achieving the specifieddisplacement.

C

C œC

C œBœ

rBCBœB

Bœ)rBC

CB

D)rCD

C

BœB

(rBC)CB

CBœ


Graphical Displacement of the slave links

• Once the driver link is repositioned, the position of all other links must be determined. To do that, one needs to draw the possible paths of thee links. For example, Links that are pinned to the frame have a circular path entered at the pin point on the frame. Also prism links has a linear path.

26





FB

C

DA

rAB

rCD

rEF

3

4

56

2


4

D


C

3

1

2

ArAB

rCD15°B










ECA

B

EB,C

BœAC œ








C

C œC

C œBœ

rBCBœB

Bœ)rBC

CB

D)rCD

C

BœB

(rBC)CB

CBœ

76 CHAPTER FOUR

4


Intersection representsprecise location of C′

Feasible path of C′

C

DA

3

1

2 15°B

B′

rBC

rAB

rCD

Second intersection representsanother geometricinversion

1

D

C

B

E

A90°

3

2

4

5

6

3.2''

.8''

5.3''

4.9''

10.1''

3.3''

3''


FIGURE 4.10 Locating the position of .C œ

Once has been located, the position of links 3 and 4can be drawn. Thus, the configuration of the mechanism asthe driver link was repositioned has been determined

This section presents the logic behind graphical posi-tion analysis—that is, locating a displaced point as theintersection of the constrained and feasible paths. Thislogic is merely repeated as the mechanisms become morecomplex. The actual solution can be completed using

C œ manual drawing techniques (using a protractor andcompass) or can be completed on a CAD system (using arotate and copy command). The logic is identical; however,the CAD solution is not susceptible to the limitations ofdrafting accuracy. Regardless of the method used, theunderlying concepts of graphical position analysis can befurther illustrated and expanded through the followingexample problems.

EXAMPLE PROBLEM 4.1

Figure 4.11 shows a kinematic diagram of a mechanism that is driven by moving link 2. Graphically reposition thelinks of the mechanism as link 2 is displaced 30° counterclockwise. Determine the resulting angular displacement oflink 4 and the linear displacement of point .

SOLUTION: 1. Calculate Mobility

To verify that the mechanism is uniquely positioned by moving one link, its mobility can be calculated. Six linksare labeled. Notice that three of these links are connected at point . Recall from Chapter 1 that this arrangementmust be counted as two pin joints. Therefore, a total of six pin joints are tallied. One sliding joint connects links1 and 6. No gear or cam joints exist:

and

With one degree of freedom, moving one link uniquely positions all other links of the mechanism.

M = 3(n - 1) - 2jp - jh = 3(6 - 1) - 2(7) - 0 = 15- 14= 1

n = 6 jp = (6 pins + 1 sliding) = 7 jh = 0

C

E





FB

C

DA

rAB

rCD

rEF

3

4

56

2


4

D


C

3

1

2

ArAB

rCD15°B










ECA

B

EB,C

BœAC œ








C

C œC

C œBœ

rBCBœB

Bœ)rBC

CB

D)rCD

C

BœB

(rBC)CB

CBœ





FB

C

DA

rAB

rCD

rEF

3

4

56

2


4

D


C

3

1

2

ArAB

rCD15°B










ECA

B

EB,C

BœAC œ








C

C œC

C œBœ

rBCBœB

Bœ)rBC

CB

D)rCD

C

BœB

(rBC)CB

CBœ


Task 2

• The attached figure shows a kinematic diagram of a mechanism driven by moving link 2. Graphically reposition the links of the mechanism as link 2 is displaced 30° counter clock wise

27

76 CHAPTER FOUR

4


Intersection representsprecise location of C′

Feasible path of C′

C

DA

3

1

2 15°B

B′

rBC

rAB

rCD

Second intersection representsanother geometricinversion

1

D

C

B

E

A90°

3

2

4

5

6

3.2''

.8''

5.3''

4.9''

10.1''

3.3''

3''


FIGURE 4.10 Locating the position of .C œ

Once has been located, the position of links 3 and 4can be drawn. Thus, the configuration of the mechanism asthe driver link was repositioned has been determined

This section presents the logic behind graphical posi-tion analysis—that is, locating a displaced point as theintersection of the constrained and feasible paths. Thislogic is merely repeated as the mechanisms become morecomplex. The actual solution can be completed using

C œ manual drawing techniques (using a protractor andcompass) or can be completed on a CAD system (using arotate and copy command). The logic is identical; however,the CAD solution is not susceptible to the limitations ofdrafting accuracy. Regardless of the method used, theunderlying concepts of graphical position analysis can befurther illustrated and expanded through the followingexample problems.

EXAMPLE PROBLEM 4.1

Figure 4.11 shows a kinematic diagram of a mechanism that is driven by moving link 2. Graphically reposition thelinks of the mechanism as link 2 is displaced 30° counterclockwise. Determine the resulting angular displacement oflink 4 and the linear displacement of point .

SOLUTION: 1. Calculate Mobility

To verify that the mechanism is uniquely positioned by moving one link, its mobility can be calculated. Six linksare labeled. Notice that three of these links are connected at point . Recall from Chapter 1 that this arrangementmust be counted as two pin joints. Therefore, a total of six pin joints are tallied. One sliding joint connects links1 and 6. No gear or cam joints exist:

and

With one degree of freedom, moving one link uniquely positions all other links of the mechanism.

M = 3(n - 1) - 2jp - jh = 3(6 - 1) - 2(7) - 0 = 15- 14= 1

n = 6 jp = (6 pins + 1 sliding) = 7 jh = 0

C

E


Task 2

• Link 2 and 4 are pinned to the frame therefore they have a circular constrained path centred at the points A and D respectively. Link 6 has a linear constrained path.

28


(a)

(b)

C

B

rBC rCE

B′

C′E

E′

Feasiblepath ofC′

1

D

C

BB′

30°

E

Constrained pathof point C

Possible pathof point D

4.0"

Constrained pathof point E

A

3

2

4 5 6

rAB

rCD

(c)

FIGURE 4.12 Displacement constructions for Example Problem 4.1.

2. Reposition the Driving Link

Link 2 is graphically rotated 30° counterclockwise, locating the position of point This is shown inFigure 4.12a

Bœ.

3. Determine the Paths of All Links Directly Connected to the Frame

To reposition the mechanism, the constrained paths of all the points on links that are connected to the frame, and are drawn. This is also shown in Figure 4.12a.

4. Determine the Precise Position of Point

Being rigid, the shape of link 3 cannot change, and the distance between points and remainsconstant. Because point has been moved to , an arc can be drawn of length , centered at . This arcBœrBCBœB

(rBC)CB

C œE)(B,C

Draw the constrained paths


Task 2

• The links 3 has a fixed length. therefore, we draw an arc (with radius equal to the length of the link 3) centred at the point B’. the intersection point between the arc and the constrained path of C is C’ the new position. We do the same for the link 5 to find the new position of E which is E’

29


(a)

(b)

C

B

rBC rCE

B′

C′E

E′

Feasiblepath ofC′

1

D

C

BB′

30°

E

Constrained pathof point C

Possible pathof point D

4.0"

Constrained pathof point E

A

3

2

4 5 6

rAB

rCD

(c)

FIGURE 4.12 Displacement constructions for Example Problem 4.1.

2. Reposition the Driving Link

Link 2 is graphically rotated 30° counterclockwise, locating the position of point This is shown inFigure 4.12a

Bœ.

3. Determine the Paths of All Links Directly Connected to the Frame

To reposition the mechanism, the constrained paths of all the points on links that are connected to the frame, and are drawn. This is also shown in Figure 4.12a.

4. Determine the Precise Position of Point

Being rigid, the shape of link 3 cannot change, and the distance between points and remainsconstant. Because point has been moved to , an arc can be drawn of length , centered at . This arcBœrBCBœB

(rBC)CB

C œE)(B,C

Find the intersection points


Analytical Displacement Analysis

30

• Analytical methods can be used in positions analysis to yield results with a high degree of accuracy, However the accuracy comes with a price in that the methods often become numerically intensive

• Many methods have been developed for analytical position analysis including simple trigonometric, complex algebra and more high-‐order math techniques

• For design situations, where kinematic analysis is not a daily task, the complex methods can be difficult to understand and implement, then a more straightforward method of position analysis involves using the trigonometric laws for triangles

• In this course we will cover the following two techniques:

1. Trigonometric Algebraic laws

2. Complex Polar Algebra


Trigonometric Algebraic laws

• In general, this method involves inserAng reference lines within a mechanism and analysing the triangles

• Laws of general and right triangles are then used to determine the lengths of the triangle sides and the magnitude of the interior angles

• We will explain the method through an example. The following figures shows the Toggle clamp mechanism. We want analysis the posiAons analyAcal

31

80 CHAPTER FOUR

30°

40Dimensions are in millimeters

50


EXAMPLE PROBLEM 4.3

Figure 4.15 shows a toggle clamp used to securely hold parts. Analytically determine the displacement of the clampsurface as the handle rotates downward, 15°.

efficient for those involved in kinematic research. However,for the typical design engineer, the simplicity far outweighsall inefficiencies. Thus, this triangle method of positionanalysis will be used in this text.

In general, this method involves inserting reference lineswithin a mechanism and analyzing the triangles. Laws of gen-eral and right triangles are then used to determine the lengthsof the triangle sides and the magnitude of the interior angles.As details about the geometry of the triangles are determined,this information is assembled to analyze the entire mechanism.

A substantial benefit of analytical analysis is the abilityto alter dimensions and quickly recalculate a solution.During the design stages, many machine configurations anddimensions are evaluated. Graphical analysis must be com-pletely repeated for each evaluation. Analytical methods,specifically when implemented with spreadsheets or othercomputer-based tools, can update solutions quickly.

The analytical method of position analysis can best beseen through the following examples.

SOLUTION: 1. Draw a Kinematic Diagram

The kinematic diagram is given in Figure 4.16a. The end of the handle was labeled as point of interest .X

A

B

(a)

(b)

X

C

A

B

40

30°

50

C

3

4

1

2

(c)

A

B′40 15°50

C

FIGURE 4.16 Mechanism for Example Problem 4.3.




Toggle clamp mechanism position analysis

• First, draw the kinematic diagram and then analyse the position in the original configuration

32

80 CHAPTER FOUR

30°


50


EXAMPLE PROBLEM 4.3








A

B

(a)

(b)

X

C

A

B

40

30°

50

C

3

4

1

2

(c)

A

B′40 15°50

C


80 CHAPTER FOUR

30°


50


EXAMPLE PROBLEM 4.3








A

B

(a)

(b)

X

C

A

B

40

30°

50

C

3

4

1

2

(c)

A

B′40 15°50

C


Original Configuration


γ

θ2 θ3

L2L3

L4

FIGURE 4.17 In-line slider-crank mechanism.

2. Analyze the Geometry in the Original Configuration

For this slider-crank mechanism, a triangle is naturally formed between pin joints , , and . This triangle isshown in Figure 4.16b.

Prior to observing the mechanism in a displaced configuration, all properties of the original configurationmust be determined. The internal angle at joint , , can be determined from the law of sines,equation (3.6):

The interior angle at joint , , can be found because the sum of all interior angles in any trianglemust total 180°:

The length side AC represents the original position of the slider and can be determined from the law ofcosines, equation (3.7):

3. Analyze the Geometry in the Displaced Configuration

The displaced configuration is shown in Figure 4.16c when the handle is rotated downward 15°. Note that thisdisplacement yields an interior angle at joint , , of 15°. The law of sines can be used to find the interiorangle at joint , :

Again, the interior angle at joint , , can be found because the sum of all interior angles in anytriangle must total 180°:

The length side represents the displaced position of the slider. As before, it can be determined from thelaw of cosines:

4. Calculate the Desired Displacement

The displacement of point during this motion can be found as the difference of the triangle sides and AC:

¢RC = AC œ - AC = 86.14 - 74.52 = 11.62 mm ;ACœC

= 86.14 mm

= 3(50 mm)2 + (40 mm)2 - 2(50 mm)(40 mm)cos(146.12 !) = 86.14 mm

AC œ = 3ABœ2 + B œCœ2 - 2(ABœ) (BœCœ) cos∠ABœC œ

AC œ

∠AB œC œ = 180° - (15° + 18.88°) = 146.12°

∠AB œC œB œ

∠BœC œA = sin-1 c a ABœ

B œC œ b sin∠C œABœ d = sin-1 c a 50 mm

40 mmb sin 15°d = 18.88°

∠B œC œAC œ∠C œABœA

= 74.52 mm

= 3(50 mm)2 + (40 mm)2 - 2(50 mm)(40 mm){cos 111.32°}

AC = 3AB2 + BC2 - 2(AB ) (BC )cos∠ABC

∠ABC = 180° - (30° + 38.68°) = 111.32°

∠ABCB

∠BCA = sin-1 c a AB

BCb sin∠B AC d = sin-1 c a 50 mm

40 mmb sin 30°d = 38.68°

sin∠BAC

(BC)= sin∠BCA

(AB)

∠BCAC

CBA

4.6.1 Closed-Form Position AnalysisEquations for an In-Line Slider-CrankThe clamp mechanism in Example Problem 4.3 is a slider-crank linkage. Specifically, it is termed an in-line slider-crankmechanism because the constrained path of the pin joint onthe slider extends through the center of the crank rotation.Figure 4.17 illustrates the basic configuration of an in-lineslider-crank linkage.


Toggle clamp mechanism position analysis

33

Displaced Configuration

80 CHAPTER FOUR

30°


50


EXAMPLE PROBLEM 4.3








A

B

(a)

(b)

X

C

A

B

40

30°

50

C

3

4

1

2

(c)

A

B′40 15°50

C



γ

θ2 θ3

L2L3

L4

FIGURE 4.17 In-line slider-crank mechanism.

2. Analyze the Geometry in the Original Configuration

For this slider-crank mechanism, a triangle is naturally formed between pin joints , , and . This triangle isshown in Figure 4.16b.

Prior to observing the mechanism in a displaced configuration, all properties of the original configurationmust be determined. The internal angle at joint , , can be determined from the law of sines,equation (3.6):

The interior angle at joint , , can be found because the sum of all interior angles in any trianglemust total 180°:

The length side AC represents the original position of the slider and can be determined from the law ofcosines, equation (3.7):

3. Analyze the Geometry in the Displaced Configuration

The displaced configuration is shown in Figure 4.16c when the handle is rotated downward 15°. Note that thisdisplacement yields an interior angle at joint , , of 15°. The law of sines can be used to find the interiorangle at joint , :

Again, the interior angle at joint , , can be found because the sum of all interior angles in anytriangle must total 180°:

The length side represents the displaced position of the slider. As before, it can be determined from thelaw of cosines:

4. Calculate the Desired Displacement

The displacement of point during this motion can be found as the difference of the triangle sides and AC:

¢RC = AC œ - AC = 86.14 - 74.52 = 11.62 mm ;ACœC

= 86.14 mm

= 3(50 mm)2 + (40 mm)2 - 2(50 mm)(40 mm)cos(146.12 !) = 86.14 mm

AC œ = 3ABœ2 + B œCœ2 - 2(ABœ) (BœCœ) cos∠ABœC œ

AC œ

∠AB œC œ = 180° - (15° + 18.88°) = 146.12°

∠AB œC œB œ

∠BœC œA = sin-1 c a ABœ

B œC œ b sin∠C œABœ d = sin-1 c a 50 mm

40 mmb sin 15°d = 18.88°

∠B œC œAC œ∠C œABœA

= 74.52 mm

= 3(50 mm)2 + (40 mm)2 - 2(50 mm)(40 mm){cos 111.32°}

AC = 3AB2 + BC2 - 2(AB ) (BC )cos∠ABC

∠ABC = 180° - (30° + 38.68°) = 111.32°

∠ABCB

∠BCA = sin-1 c a AB

BCb sin∠B AC d = sin-1 c a 50 mm

40 mmb sin 30°d = 38.68°

sin∠BAC

(BC)= sin∠BCA

(AB)

∠BCAC

CBA

4.6.1 Closed-Form Position AnalysisEquations for an In-Line Slider-CrankThe clamp mechanism in Example Problem 4.3 is a slider-crank linkage. Specifically, it is termed an in-line slider-crankmechanism because the constrained path of the pin joint onthe slider extends through the center of the crank rotation.Figure 4.17 illustrates the basic configuration of an in-lineslider-crank linkage.


Complex Algebra Position Analysis

• In planar mechanism it is often desirable to express a vector by specifying its magnitude and orientation in polar notation such as

• In the complex analysis this vector is then given by:

• where is the imaginary number

• The vector also can be described in the polar space as

34


Complex Algebra Position Analysis Example

• Perform a position analysis of the sliding block linkage by finding and the distance

35


Complex Algebra Position Analysis Example

• Perform a position analysis of the sliding block linkage by finding and the distance

36


Thank you for your attention !!

ROBOTICS, Lecture, CHAP 4- Robotic Mechanics 1

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