Manipulators Robots2

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Master Thesis 2011-2012 Industrial Technology Department

Narvik University College

Design of a Hydraulic

Manipulator Arm

Rami Shafik Radwan

Supervisor: Professor Bjrn Solvang


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To my love ........................................Sohad

To charming smile.............. My daughter

To angle Face..................................My son


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Preface

This report is the documentation of master thesis in industrial technology

department at Narvik University College

This thesis divided into six main part: introduction part, a part concern on

hydraulic system design, a part concern on kinematic model, a part studying

dynamic model, a part concern on bond graph methodology, and a part for

control.

The report contains a literature survey over ideas of ways to design and control

hydraulic manipulator arm. The kinematic and dynamic model of the robot is

simulated by Matlab. Also the control of robot emerged into interpolator model

to check it.

Finally, I want to thank professor Bjrn solvang who support me and help me to

me to accomplish this thesis. Also the thank for Gabor Sziebig who help me.

The work in this report was done during 2011-2012


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List of Figures

Figure

Number

Description page

1-1 Schematic of robot architecture 2

1-2 a Redundant heavy manipulator 4

1-2 b Super Scorpio rov 4

1-3 Schematic of hydraulic system 4

1-4 Hydraulic system and its structure 6

1-5 The principle of speech interface system 91-6 Basic task to motion control 12

1-7 Typical robot joint control 13

1-8 The robot control architecture 14

2-1 The hydraulic manipulator arm 16

2-2 The hydraulic system for 3DOF manipulator arm 18

2-3 Density variation with temperature 19

2-4 Velocity variation for fluid at point 20

2-5 Variation of viscosity with oil temperature 21

2-6 Variation of bulk modulus with temperature for different fluids 22

2-7 schematic of an axial piston motor 24

2-8 The control signal sequence 26

2-9 The components of proportional direction valve 27

2-10 Spool overlap types for proportional direction valves 28

2-11 The components of pressure relief valve 29

2-12 Dual pilot-operated check valves symbol 30

2-13 Bladder accumulator type 38

2-14 Theoretical performance for valve CP722-11 43

3-1 The schematic representation of forward and inverse kinematic 52

3-2 Coordinate rotate around Z axis with angle c 54

3-3 Coordinate rotation around Y axis with angle B 55

3-4 Coordinate rotate around X axis with angle A 563-5 illustrate 3DOF manipulator arm , joint 1 rotate around Z, joint 2

and 3 rotate around Y

58

3-6 Forward Kinematic simulation for joint displacement (0,0,0) 61

3-7 Forward Kinematic simulation for joint displacement (30,30,30) 62

3-8 Projection of 3DOF arm on XY plane 63

3-9 Projection of 3DOF arm on XZ plane 64

3-10 Inverse kinematic simulation with tool center point pose [0.9232

0.5330 0.1536 0 60 30]

65

3-11 a large translation and short re-orientation 66

3-11b large re-orientation and short translation 663-12 Speed acceleration steps to reach PF and Retardation to stop 70


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3-13 adaptive the breaking distance 71

3-14 Retardation algorithm 73

3-15 Flow chart for the interpolator simulation 75

3-16 The changing of GLR,SL,SL0 during interpolator period 76

3-17 Changing actual speed during interpolator period 77

3-18 Changing remaining distance for tool center point parametersduring interpolator period

78

3-19 Changing tool center point parameters during interpolator period 79

4-1 Valve torque motor assembly 84

4-2 Valve responding to changing in electric input 85

4-3 The load flow through the valve is expressed by the equation 86

5-1 0-junction in bond graph 102

5-2 1-junction in bond graph 102

5-3 Representation of source of effort in bond graph 105

5-4 Representation of source of flow in bond graph 106

5-5 The representation for transform element in bond graph 1065-6 The representation for gyrator element in bond graph 107

5-7 Combination of n single bonds into multibond of dimension n 108

5-8 Array of 1-junction 108

5-9 Pump representation by bond graph 109

5-10 Bond graph representation for relief valve 111

5-11 Bond graph for pump and relief valve unit 112

5-12 Accumulators dynamic 113

5-13 Representation for accumulator by bond graph 114

5-14 Bond graph for filter 114

5-15 a Symbolic configuration of 4/3 proportional direction valve 115

5-15 b The controlling variable restriction for the valve 1155-16 pressure and flow resistant for 4/3 proportional direction valve 116

5-17 Bond graph for 4/3 proportional direction valve 117

5-18 A schematic diagram indicating the flow path in the valve and the

path resistances

118

5-19 A schematic diagram indicating the pressure and flow resistance

Circuit

119

5-20 Bond graph for 3/2 directional valve 119

5-21 Bond graph for hydraulic motor 120

5-22 Bond graph for the hydraulic system of robot 122

5-23 a Multibond for revolution joint 1235-23 b Multibond for prismatic joint 123

5-24 Multiband for 3degree of freedom robot arm 124

5-25 Bond graph for the arm with hydraulic actuators. 125

6-1 Tracking design 130

6-2 Joint 1trajectory result from interpolator procedure 134

6-3 Joint 1 trajectory result from insert the controller into interpolator

Procedure

135

6-4 Comparing the trajectory for joint 1with controller and without

controller

136



Procedure

138


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6-7 Comparing the trajectory for joint 2with controller and without

controller

139



Procedure

141

6-10 Comparing the trajectory for joint 3with controller and withoutcontroller

142


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List of Tables

Table

Number

Description Page

2-1 Determination of the pipeline friction cofficient 31

2-2 The local loss coefficient for some type of pipes 32

2-3 Example for x = 100 and the hydraulic system application suite 33

2-4 The methods used to install in hydraulic system applications 34

2-5 Comparing between different types of pumps 35

2-6 Comparing between different types of pump 36

2-7 The characters for AE16 motor From Duesterloh 40

2-8 Links parameters 45

5-1 The generalized variable in different domains 100

6-1 Parameters of the Hydraulic Actuator 132


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List of symbols

1. UPPERCASE

Symbol Description System matrix for the it actuatorAF Actual speed for TCP Input matrix for the it actuator viscous damping coefficient, Kg.m.s/radB(q) The coriolis matrix Isothermal tangent bulk modulus isentropic tangent bulk modulus

3n*n input matrix for the integrated hydraulic robot manipulator

C Capacitor elements Flow coefficientC(q) The centrifugal matrix

DF Modify the constant acceleration into integer

DFn Constant acceleration for TCP Load distribution matrix for the it actuatorG(q) The vector of gravitational forces is n*1 vector

GLR Breaking distance after modify

GY Gyrator element

I Inductor/Inertia elements

The inertia tensor matrix for linkiJ inertia moment Jacobin matrix for frame i according to the base Jacobin matrix for frame i according to the base Flow-pressure coefficient Pressure sensitivity Flow gainL Length of pipeline m

M(q) The inertia matrix is n*n matrix

Driven power for motor kw

Pe Linear momentum, N.sPF Program speed for TCP

Pp Pressure momentum N.s/m

Pt Angular momentum, Nms

Q flow rate, m /s maximum flow rates, m /sQt Theoretical flow rate, m /s

T The torque in the output shaft, N.m

Tc Cushion torque

TF Transformer element


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The load torque acting on the it actuator due to the manipulator Load torque, N.mT Acceleration torque, N.m

Transform matrixR Resistance element Rotation matrix around x Rotation matrix around y Rotation matrix around z Reynolds number 3n*3n system matrix for the integrated hydraulic robot manipulatorSe Source of effort

Sf Source of flow

SL Breaking distance

SL0 Breaking distance after reduce it one speed period Volume mVm Geometric volume of motor, m /rev The vector of coriolis and centrifugal forces is n*1 vector Scalar input to the it actuator Rate of load distribution matrix for the i t actuator 3 x 1 State vector of the ith actuator2. LOWERCASE

Symbol Description

A State space matrix

B Input coefficient vector Christoffel symbolsD Diameter m

E Effort

F Flow

fv Viscosity factor

G Gravity acceleration

K A coefficient vectorM Mass, Kg

nm Motor speed, r.p.m

P Pressure, bar

Q Displacement joint vector Fluid velocity, m/s Desired input.X The six parameters for tool center point

xv Spool displacement

Desired trajectory


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3. GREEK SYMBOLS

Symbol Description

Angular acceleration for motor shaft rad/s

Bulk modulus.

The number of particles size in upstream filte to the number of particlessize in downstream Motor torque efficiency h Motor hydraulic efficiency m Motor mechanical efficiency Volumetric efficiency Friction coefficient

e Eigenvalues

Angle displacement, rad

Angular velocity rad/s

Angular acceleration rad/s Angular velocity for rotary shaft rad/s Kinematic viscosity m /s Local loss coefficient Fluid density, kg/m Dynamic viscosity Ns/m Shear stress N/m


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List of Hydraulic Symbol

Pump unit

Variable displacement pump

Fixed displacement pump

Filter

Heater

Cooler

Accumulator

Double action piston

Motor

3/2 Directional valve

4/3 Proportional direction valve

Pressure relief valve


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List of Abbreviations

DOF Degree of freedom

TCP Tool center point

Notice

The bold letter using in symbols meaning that symbole is vector.


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Table of Contents

Chapter 1 Introduction ..................................................................................... 1

1-1 Manipulators Robots: .............................................................................. 2

1-2 Hydraulic Manipulator System.................................................................. 3

1-3 Human friendly interfaces..........................................................................6

1-3-1 Speech Interface ................................................................................. 7

1-3-2 Learning from demonstration...............................................................9

1-4 Robot Control Architecture......................................................................11

Chapter 2 Hydraulic system design ................................................................ 152-1 Introduction to Hydraulic system ............................................................ 17

2-2 Hydraulic System components ................................................................ 17

2-2-1 Hydraulic Oil ..................................................................................... 19

2-2-1-1 Oil density ................................................................................. 19

2-2-1-2 Viscosity .................................................................................... 20

2-2-1-3 Specific gravity .......................................................................... 22

2-2-1-4 Bulk Modulus ............................................................................ 22

2-2-1-5 Choosing hydraulic oil ............................................................... 23

2-2-2 Axial Piston Motors ........................................................................... 23

2-2-3 Proportional Direction Valves ........................................................... 27

2-2-4 Pressure Relief Valve ........................................................................ 28

2-2-5 Dual pilot-operated check valves ...................................................... 29

2-2-6 Pipelines ........................................................................................... 30

2-2-8 Filter ................................................................................................. 32

2-2-9 Pump ................................................................................................ 34

2-2-10 Accumulator ................................................................................... 37


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2-3 Circuit Design .......................................................................................... 38

2-4 Load Calculation ...................................................................................... 44

Chapter 3 Kinematic model ........................................................................... 51

3-1 Forward Kinematic .................................................................................. 53

3-1-1 Forward Kinematic for 3 DOF ............................................................ 58

3-1-2 Forward Kinematic Simulation .......................................................... 60

3-2 Inverse Kinematic .................................................................................... 62

3-2-1Inverse Kinematic Simulation ............................................................. 65

3-3 Interpolator ............................................................................................. 663-2-1 Speed Calculation ............................................................................. 69

3-3-2Breaking Procedure ........................................................................... 70

3-3-3 Interpolator Simulation ..................................................................... 74

Chapter 4 Dynamic Model ............................................................................... 80

4-1 Dynamic model for 3 DOF manipulator ................................................... 82

4-2Dynamic model for servo-hydro actuators ............................................... 84

4-3 Integration link dynamic models with hydraulic dynamic model ............ 93

Chapter 5 Bond Graph Methodology ............................................................. 98

5-1 Bond Graph Basics .................................................................................. 99

5-1-1 Bond variables ................................................................................ 100

5-1-2 Ports and Bonds .............................................................................. 101

5-1-3 Basic Elements ................................................................................ 103

5-1-3-1 Resistor Element ...................................................................... 103

5-1-3-2 Capacitor Elements .................................................................. 104

5-1-3-3 Inductor/Inertia Elements ........................................................ 104

5-1-3-4 Source of Effort ........................................................................ 105

5-1-3-5 Source of Flow: ........................................................................ 105


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5-1-3-6 Transformer Element ............................................................... 106

5-1-3-7 Gyrator Element ....................................................................... 106

5-1-4 Multibond Graphs ........................................................................... 107

5-2 Modeling Hydraulic System ................................................................... 109

5-2-1 Pump .............................................................................................. 109

5-2-2 Pressure Relief Valve ...................................................................... 111

5-2-3 Accumulator ................................................................................... 113

5-2-4 Filter ............................................................................................... 114

5-2-5 4/3 Proportional Direction valve ..................................................... 115

5-2-6 3/2 direction valve: ......................................................................... 118

5-2-7 Motor ............................................................................................. 119

5-3 Links Representation ............................................................................ 123

Chapter 6 Control System.............................................................................. 126

6-1 Control Method .................................................................................... 127

6-2 Simulation ............................................................................................. 132

Concolusion .................................................................................................. 143

Appendix ....................................................................................................... 144

Appendix A .................................................................................................... 14

A-1 Axial piston motor ............................................................................. 146

A-2 Proportional direction valve .............................................................. 149

A-3 3/2 Directional valve ......................................................................... 155

A-4 Inline filter ......................................................................................... 159

A-5 Accumulator ...................................................................................... 162

A-6 Pump ................................................................................................. 164


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Appendix B Matlab modeling and simulation .............................................. 168

B-1 Forward kinematic ............................................................................. 169

B-2 Inverse Kinematic .............................................................................. 171

B-3 Interpolator ....................................................................................... 172

B-4 Dynamic Model .................................................................................. 179

B-5 Finding poles for joint 1 ..................................................................... 183

B-6 Control joint 1 .................................................................................... 184

B-7 Finding poles for joint 2 ..................................................................... 189

B-8 Control joint 2 .................................................................................... 190

B-9 Control joint 3 .................................................................................... 192

References .................................................................................................... 194


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Chapter 1

Introduction

I n this chapter, we wil l discussion the industr ial robots and their application,

and then we wil l have a review of types of robot's actuator . From these types,hydraulic actuators will discussion in more details. A human friendly

interaction becomes more and more interesting f ield for robot technology.

Thus we will have a review of human friendly interaction. Also human

fr iendly inter face is very essential part f rom human f r iendly i nteraction. So we

have a look for two types from this interface, first speech interface, and second

learning fr om demonstration. The last paragraph about the basic architecture

control whi ch is a corner stone for buil ding a robot.


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1-1 Manipulators Robots:

Nowadays, the competitions between the manufacturing companies lead to more

automation in manufacturing systems, which requiring less operators

intervention, more flexible and reliability. Using robots in manufacturing systemis good example for flexible manufacturing. They capable to performed many

different tasks and precisely operations (Pires, 2007).

A robot can be define as device able to doing manipulative tasks with object,

tool or end effort, and has capability to reprogramming make it able to

performed different tasks and operations. Therefore a robot considers a complex

system which has a mechanical system providing the suit motion and force to

performing the manipulative operation. The parameters of mechanical systemand the information about the state of mechanical system can obtain by sensors

system, which consist of external and internal sensors. All controlled and saved

in memory by reprogrammed control unit, which gathered the information from

sensors and give the correct signal to the manipulators actuators (Ceccarelli and

Ottaviano, 2008). Figure 1-1 illustrates a schematic of a typical robot system.

Figure 1-1 Schematic of robot architecture

Industrial manipulators robots can classify by the means of driving the actuators

(joint drive systems). Thus we can have three kind of driving system for themanipulators robots:


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1_ Electrical drive system: The joint drive by electrical motors providing

enormous advantages such as simplicity of power supply, availability of

electrical motors, easy control by speed, low energy cost and losses, the ratio of

power producing according to the weight, and high precision engines. But the

electrical drive system is not suitable for explosion areas or in under waterapplication in additional in heavy tools (Morecki and Knapczyk, 1999).

2_Penumatic drive systems: The actuators in pneumatic system driven by using

compressible fluid, usually compressed air. Pneumatic drive systems have

advantages such as the system is very safe to use, availability of air and easy to

release when finished its work, and air has good dynamic properties. However

the most disadvantages is pneumatic drive system use in small and simpleapplication (Morecki and Knapczyk, 1999).

3_Hydraulic drive system: This system use compressed fluid to drive the joint. It

has some advantages like high precision components, very powerful components

compared with its size, easy to store power in accumulator, power transformmedium can use as lubrication and for cooling, and good for use in explosion

areas and underwater application. However it has some disadvantages like the

leakage, loss of power as result of the fluid viscosity, need power pack and the

noise from the power pack, and more safety procedures in the system and

surrounded environment.

1-2 Hydraulic Manipulator System:

Although the electrical drive system is more common in manipulator robots dueto simplicity, but the hydraulic drive system is still preferred in many industrial

tasks to use fluid power advantages. Thus there are no power source can

compete hydraulic drive system when heavy loads must be moved and position

precisely. We can find a hydraulic robot in construction, mining, forming,

forestry industry (Savela and Delta, 2011). Redundant heavy manipulator is an

example of hydraulic robot which was developed in Germany to make

maintenance process on tunnels roof and wall , it can load up to 500 kg in rangto 8 meter. Other famous example for hydraulic robot is 'Super Scorpio rov

which has saved Russian submarine Crew. The manipulator working by

hydraulic system with gripping force reach to 453 kg and the lifting capability at

full extension is 158 kg. Figure 1-2 shows picture for redundant and super

scorpio robots.

In general, the hydraulic system of the manipulator consists of hydraulic

actuators (linear or rotation actuators).


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( a) (b)

Figure 1-2 (a) Redundant heavy manipulator (youngester, 2011)(b) Super Scorpio rov (neptunems, 2012.)

The control of the direction and the displacement of actuators performed by

special kind of direction valve called proportional direction valve. Proportional

valves can control actuators by controlled with the fluid flow inter to the

actuators. Additional devices must be added to the system to ensure on thesecurity and the reliability, such as filters, pressure valves, accumulator, and

cooling/heating system. Figure 1-3 illustrates a schematic drawing of a typical

hydraulic system.

Figure 1-3 Schematic of hydraulic system


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A hydraulic system can be divided into two sections, a power section and a

power signal control.

The power section is consist of three main parts

1- Power supply section is section where the hydraulic power is generatedand convert the electrical power into mechanical and then into fluid

power. Thus electrical motor, pump, gages, reservoir, and protection

circuit are used. In additional some components are used to condition the

working fluid such as cooler, heater, thermometer, and filters.

2- Power control section is controlling with the direction of the fluid flow,the flow rate, the pressure of the fluid, and time of the flow. Generally

there are components perform this tasks such as direction control valves,

flow control valves, pressure valve and non-return valves.

3- Drive section is the part of the system where executes the work and themovement by moving the power from power supply section through

power control system to the components which consist the drive section.

The components formed this section can be hydraulic motors or cylinders

(Merle, Schrader, Thomes, 2003).

The signal control section is divided into two sections:

1- Signal input section is the kind of input signal which must processing inthe next section, so the input signal can be manually, mechanically, or

contactlessly.

2- Signal processing section where the input translate and transfer into topower control section. It can process by electrical means or mechanical

means, but generally in hydraulic robot we use electrical means (Merle,Schrader, Thomes, 2003).

Figure 1-4 shows a hydraulic system sample and the facing section in the

system.


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Figure 1-4 Hydraulic system and its structure

In robotics, the control of a precise position for the joints is more difficultcomparing with electrical drives, where the torque is proportional with electricalcurrent or voltage. While the power control section in the hydraulic manipulator

is proportional with pressure difference and flow rate, in additional some factors

can change according to the change of fluid properties such as viscosity or the

rate of impurities. All of these factors make the modeling and control of such

system a challenging task (Becker,Piestsch, and Hesselbach 2003).

1-3 Human friendly interfaces:

Human friendly interaction is define as the field which interesting to studydedicated to understand designing and evaluate robotic system for use by or

with human.

However, human friendly robots might has two properties; smart interfaces

facilities, easy for use, and safe mechanism. Human friendly robots must have

safety system to ensure that usage people are away from dangerous, thus all jointan movement part must controlled in the manner that prevent dangerous and

accidents ( Heinzmann and Zelinsky,1999).


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Designing the control architecture for human friendly robot might involve

decisions for the responsibility between robot control algorithms and the human

factors. These decisions must insure to minimize the possible damage to people.

However the defining of dangerous/safety situation is different from one

operation circumstances to another, which mean that the safety procedures insome case is safety in other case. Thus to make robot running in safety way the

robot must predicate of the dangerous and harmful situation and running in the

safety way ( Heinzmanny, Matsumotoy, Kieerz, and Zelinskyy,2007).

Human friendly robots will has widely applications such as health care, social

service, personal service, urban search and rescue, and in industrial field. Most

robots operate in manufacturing are manipulator arm, which need expertworkers to deal with their due to the difference control patterns to control with

the changing and various of operation situation, in additional that the robot is

very flexible device which mean that need mastering in setup its configuration

and teach it when there are a changing in operation. Therefore, it will be easy to

deal with in the case of presence a smart friendly human interface which allowsto workers to operate their robots by familiar language or through physical

movements. Smart human interface will remarkably reduce the difficulty of

teach the robot, and the manufacturing lead time ( Zhang, Ampornaramveth and

Ueno,2006)..

There are many software interface has been developed to operate industrial

robots. However, those software deal with control methodologies, such asspeech interface. Next paragraphs will discussion some human friendly

interfaces.

1-3-1 Speech Interface:

The interaction between the human and the machine through talking is idea

spreading in the science fiction where the machines can understand a complex

human sentences, and execute the order coming very accuracy . However this

technology is not very development yet, but of course it will be very useful forrobots and industrial robots. This technology will increase the production

efficiency and agility, faster and cheaper procedures for manufacturing. Thebenefits from using speech interface will performed by the following Speech

interface characters:

The similar between robot interface and the natural humancommunication.

Speech reduces the complexity of using and control with robots.


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Control and operation is possible from simple multirobot interfaces(Pires, 2007).

Speech interface system might be able to determine when the speaker starts

and finishes an utterance. Therefore there are different classes depending onthe type of utterance:

Isolated words: Required from the speaker to wait between utterances,and the processing performing during pause.

Connected words: Is similar to isolated words, but the pause betweenutterance is minimal.

Continuous speech: Allow to operator speak in nature way, and thecomputer determines the contents and the program keywords.

Spontaneous speech: System has ability to handle of changing ofnatural speech. Voice verification/identification: The program can identify specific

user (Pires, 2007).

In general, the user when talking the utterances transform to a software

platform via conductors or wireless. The software translates the utterances

into digital signal which create input signal to the controller. Controller send

electrical signal to motors/valves and operate the arm. Sensors detected the

arm parameters, and send feedback to the software which translated intoutterances user can understanding ( Zhang, Ampornaramveth and Ueno,

2006). Figure 1-6 explains the principle of the interaction between user and

robot through speech interface.

However the speech interface not common in industrial applications for some

reasons likes:

It considers a new technology and not strong enough to use inindustrial.

The industrial environment is noisy environment, which createadditional constrains to it

Industrial systems don't have powerful computer dedicated to humanmachine interface (Pires, 2007).


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Figure 1-5 The principle of speech interface system

1-3-2 Learning from demonstration:

Learning from demonstration is making a robot performing its tasks throughobservation the human movement by development algorithms. It makes

humans able to teach robots new tasks by showing the robot what to do

instead of by programming. The copy of the human movement done by using

special device such as data gloves (Skoglund, Iliev and Palm, 2010). This

technology allows to emulating the human movements and try reproduction

human movements, also allows to operate robots from distance by using

teleoperation system which aim to operate and control robots from distance

to avoid the dangerous situations or to perform work in inhumanly

environment like underwater or space (Hu, Li, Xie, and Wang , 2005).

The problem of imitate the human can divided into four problems or

questions:

What imitate: Here must determine what aspects of demonstration arerelevant to the task.

How to imitate: Meaning the methodologies applying to observationand encoding.

When to imitate: Meaning the agreement problem, which might be toobservation the sequence events then detected whether there aresomething related to the task.


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Who to imitate: Refer to identify to the operator (Billing andHellstrm, 2009)

To generate like this system we need powerful sensor data processing

techniques such as multiple vision systems, laser sensors, structured light,and in some cases especially designed input devices. Thus the demonstrationwill reduce a sequence of sensors reading which translated to mapping and

trajectory, which give the pose for the motor to doing the imitate (S.Ekvall,

2007). Vision system can be used to recognize the position of sensors and

precision control.

In general, the system consist of three main parts

Human operation input: Which allows following the humanmovement, it can be a vision system recording the changing of thesensors, or data gloves. Also input device can provided with additional

sensors to detect the forces, pressure...etc.

Demonstrator: Is a robot program which received data from humanoperation input and shows the robot what to do according to data

gathered. In addition to record the pose of movement, program can

also define the object which grasped, or the pressure, applied to the

object.

Robot system: Is the robot which connected with the demonstrator andemulate the human movement (Skoglund, Iliev and Palm, 2010).

Those parts can provide the sequence of learning process and motion, which

go through:

Sensing: Where sensors or vision system observed and capture themovement.

Recognition: The observed information interpreted and analysis intobasic movement skills.

Generation: The motion is generation from the last step. Simulation: This step is optional, but it offer useful information about

manipulator and correct it before transfer it to real robot (Billing,

2012).

Execute: Robot executes the imitation.Learning by demonstrate can reduce the time and the effort spending in the

traditional way in robot programming, also make robot has the ability to

build Hierarchical structures for knowledge , which open broad prospects of

artificial intelligent.


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1-4 Robot Control Architecture:

A robot control system is an electronic programmable system, responsible to

move and correctly pose to the joints, also control with joint velocity,

acceleration and declaration in accordance with the dynamic model.

In general, there are different ways to programming the robot:

Teach by leading: In this case the operator lead the arm, and the jointvariable is stored in program memory

Teach by pendant: The user move the robot by using joystick, and thepath and velocity is stored point by point. The movement by using

pendant can performing according to different coordinate systems :

First: According to joints coordinates the control performed forin

single joint.Second: According to the base coordinate where the base will be a

reference for end effort (tool center point).

Third: The joystick movement corresponds to end effort coordinate

systemFourth: According to external coordinate system.

Offline programming: A program can be writing in normal computer,after that the program upload to robot. The program informs the robot

about the joints parameters V and tool center point parameter X(Solvang, 2011).

To achieve motion user must inter basic parameters and final parameters, in

general case those parameters are the tool center point start Xi and final Xf

parameters in addition to program speed. Interpolator computes the intermediate

points X+dx and the velocity acceleration and declaration. Coordinate

transformation computes the joints variables according the interpolator outputs.

The outputs of coordinate transformation forming the basic inputs for a servocontrol system which controlling with the motor's joint. Figure 1-7 illustrate the

sequence of the motion tasks.


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Figure 1-6 Basic tasks to motion control

The servo controller considers very complex system which utilizes the data from

interpolator an coordinate transformer to drive the arm. There are many factors

might consider in servo controller such as the dynamic changing and the change

of electrical current. The servo controller connected with pose sensors and

velocity sensors to give the feedback to the controller. Figure 1-8 shows a

typical robot joint control (Pires, 2007).

Input parameters

Interpolation

Servo control

Coordinate

transformation


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Figure 1-7 Typical robot joint control (Pires, 2007)

Figure 1-9 illustrate the hole robot control architecture and which consists of

teaching methodologies which forming the input parameters. Then the motion

control tasks begin as explain above to send data to servo controller whichprovide the accuracy pose to the joints.


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Figure 1-8 The robot control architecture (Solvang, 2010)


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Chapter 2

Hydraulic system design

I n this chapter, we wil l have a looking on the basic components for the robot

hydraul ic system, character istics, and operation methods, how we can choosethe components, and the basic equations which al low fi nding the operation

parameters. Af ter that we wil l applying the information which we had have to

choosing the hydraul ic components for 3 degree of freedom robot arm, and

fi nd their parameters. Finall y we wil l f ind the maximum load can this robot

move.


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The thesis aim to studying and designing a hydraulic manipulator arm with three

degree of freedom by using hydraulic motors. This arm consists of 3 hydraulic

motors which form the arm joints. First joint will turn around Z axis and the

second and third joints will run around Y axis. Figure 1-2 shows the arm

Figure 2-1 The hydraulic manipulator arm

The first part in this chapter will begin with studying the hydraulic components,

characteristics, chosen method and the basic equations. Then we will choose thehydraulic components for the system. Finally we will find the load which arm

can moving.


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2-1 Introduction to Hydraulic system:

Fluid power is the use of fluids under pressure to generate, control, and transmit

power. Fluid power is subdivided into hydraulics using a liquid such as mineral

oil or water, and pneumatics using a gas such as air or other gases. Hydraulicfluid power character with high power transmission comparing with the other

method of power transmission. Thus hydraulic power has some advantages

comparing with other transmission medium:

High ratio of power to mass flow. Fast response and low installed weight. A hydraulic system is relatively simple to construct with fewer moving

parts.

Stepless speed control can be increased by little complex in the system. Force multiplication is possible by increasing actuator area or working

pressure.

In most cases the hydraulic fluid circulated will act as a lubricant andwill also carry away the heat generated by the system (James,John and.

Haberman, 1988), (Watton, 2009).

In general, the mechanical power converts in pump into hydraulic power, which

transmitting by oil to the actuator where the hydraulic power converts into force

or torque to produce work. There are many of control and regulation devices

through away the fluid pass which can control with flow and the pressure of the

oil, and keep oil in the perfect form

2-2 Hydraulic System components:

For 3DOF manipulator arm robot the hydraulic system consist of pump which

increase the oil pressure and deliver it to axial position motors to provide links

angular movement. Velocities and displacements links control can achieve by

proportional direction control valves. Additional components must be added to

the system such as pressure relief valve, and filter which providing the safety for

the system. To ensure that the arm stay in its position after the hydraulic system

stop and no return flow to the oil, check valve added after proportional direction

valve. An accumulator provides flow and pressure to the system when there is a

pressure drop in the system. More details for the systems components will

discussion later. Figure 2-2 shows a hydraulic system for 3DOF manipulator

arm.


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Figure 2-2 The hydraulic system for 3DOF manipulator arm


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2-2-1-2Viscosity:

Viscosity is the description the resistance to the laminar movement of two

neighboring fluid layers against each other. It results from the cohesion and

interaction between molecules. It caused by the generation of shear stress over awetted area, and can defined from Newtonian shear-stress equation at point on

the velocity profile (James, John and. Haberman, 1988), as shows in figure 2-4

(2-1)

Figure 2-4 Velocity variation for fluid at point

Where:: Shear stress N/m2: Dynamic viscosity Ns/m

2,

also it can expressed by poise (P), where 1 P = 0.1Ns/m: Fluid velocity m/s2y: Displacement perpendicular to the velocity vector, m

The ratio of dynamic viscosity to the mass density is called kinematic viscosity

(James,John and. Haberman, 1988).

(2-2)Where: Kinematic viscosity m2/s , can expressed in stokes (St) where 1 St =10-4 m2/s: Oil density Kg/m3The oil viscosity is affected by its temperature, as shown in Fig. 2-5. It

decreases with the increase in temperature. Therefore the viscosity is statedat a standard temperature (40C for the ISO specification).


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Figure 2-5 Variation of viscosity with oil temperature. (Shell Tellus,2009)

The oil viscosity has high influence on the hydraulic system performance, and

choosing the part of hydraulic system. It influences by:

Hydraulic losses in pipelines.

Resistance to fluid flow in narrow conduits.

Viscous friction forces and damping effect (Rabia, 2009).

Thus, its important during design considering keep the oil viscosity within a

certain range, otherwise the operation condition will change with temperature

(Watton, 2009).


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2-2-1-3 Specific gravity:

This is the ratio of the mass of a substance divided by the mass of an equal

volume of water at some specfied temperature, usually 20C. The unit is

therefore dimensionless and varies between 0.8 for some petroleumbased fluidsto as high as 1.5 for the chlorinated hydrocarbon (Akers, Gassman and Smith,

2006).

2-2-1-4 Bulk Modulus:

Is an elastic constant giving the amount that the oil volume is reduced for a

given application of pressure. Thus bulk modulus is a measure of a fluid

resistance to being compressed. It depends on pressure and temperature (Akers,

Gassman and Smith, 2006).

Bulk modulus is a measure of the compressibility of a fluid and will be required

when it is desired to calculate oil volume changes for high-pressure, large-

volume systems and dynamic model (Akers, Gassman and Smith, 2006). Bulk

modulus defines the compression of a fluid usually in one of two ways:

Isothermal tangent bulk modulus

(2-3)Isentropic tangent bulk modulus (2-4)Figure 2-6 shows the variation for a range of fluids under perfect conditions with

no pressure and dissolved air (Watton, 2009).

Figure 2-6 Variation of bulk modulus with temperature for different fluids(Watton,2009)


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2-2-1-5 choosing hydraulic oil:

Viscosity is the most important factor when selecting a hydraulic fluid for a

system, and when choosing an oil for system there are some design factors must

considering such as (Brendan, 2005):

starting viscosity at minimum ambient temperature

maximum expected operating temperature, which is influenced by maximum

ambient temperature

permissible and optimum viscosity range for the systems components

Thus the surrounding environment and climate changing during the year play a

critical factor to choosing the corrected oil.

The components in the system must considering when choosing oil. pump

design types and their required viscosity grades. Vane pumps require a viscosity

range of 14 to 160 cSt, while piston pumps require a viscosity range of 10 to 160

cSt. Gear pumps require a viscosity 300 cSt.

Hydraulic fluid has many roles in the operation. These roles range from a heat

transfer medium, power transfer medium and a lubrication medium. The

chemical makeup of a hydraulic fluid can take many forms when selecting it forspecific applications. It can range from full synthetic to water-based fluids used

in applications where there is a risk of fire or environment considerations

(Watton, 2009), (Sumerlin, 2009).

2-2-2 Axial Piston Motors:

The function of hydraulic motors is the reverse of that of the pump. Hydraulic

motors are displacement machines converting the supplied fluid power into

mechanical power

In many applications, the operation speed (e.g. robot, winches, vehicles ) range

between zero to hundreds revolution per minute. In robot application after

considering the rate of harmonic drive gear, the speed of motor must range

between 1 r.p.m to thousands r.p.m, to provide the accurate speed and pose.

There are number of low speed motor design includes axial piston motor, and

radial piston motor. However the motor's type is function for some variables

include: maximum and minimum speed, torque and pressure, shaft side load,fixed or variable displacement, and weight and cost (Chapple, 2003).


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In 3DOF robot, axial piston motor achieve the functions of robot like speed

range, torque weight and cost and provide the best sealing for high input

pressures and work best in high torque.

Figure 2-7 A schematic of an axial piston motor (DesignAerospace, 2011)

An axial piston motoris a piece of equipment that moves several pistons up anddown. The motor typically does this with the displacement motion of pistons,

and that slopes down and then up on one side, this moving rotate the swashplate

which connected with motor shaft.

A typical axial piston motor show in figure 2-7, which consists of a swashplate,

piston, piston shoe, cylinder block or barrel, and valve plate. The pistons are

placed vertically and at an equal distance from each other on top of the

swashplate.

In general, we can chose the motor from the maximum speed and calculate themaximum oil flow, pressure and pressure drop. The motor speed depends on

the flow rate, while the supply pressure depends mainly on the motor

loading torque. For ideal motor the relation between the speed and flow rate (2-5)Where

nm: Motor speed r.p.m
http://www.wisegeek.com/what-is-a-piston-pump.htmhttp://www.wisegeek.com/what-is-a-valve.htmhttp://www.wisegeek.com/what-is-a-valve.htmhttp://www.wisegeek.com/what-is-a-piston-pump.htm


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Vm: Geometric volume of motor, m3 /rev

Qt:Theoretical flow rate, m3

/s

Motor performance is function to the flow rate, since there are leakage the motor

speed is less than the theoretical speed. Thus we can define volumetricefficiency of the motor by: (2-6): Volumetric efficiencyQ: Real motor flow rate, m3 /s

Then:

(2-7)Because of the mechanical losses and the hydraulic losses, the output

mechanical power is less than the input hydraulic power. As pressure increases,

leakage increases, speed decreases, and thus the quantity of mechanical energy

delivered to the load decreases. Then the pressure differentiation is

(2-8)Where:: Motor mechanical efficiency h:Motor hydraulic efficiencyT: The torque in the output shaft N.m

p: pressure differentiation bar

The driven power can calculated from the equation

Kw (2-9)(Cundiff, 2002), (Rabie, 2009), (Chapple, 2003).


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2-2-3 Proportional Direction Valves:

Proportional valve is a heart of the hydraulic system, which control with fluid

flow direction. This technology combines the hydraulic transfer of force with

precision and flexibility of electronic control signal. The electrical signal caneffect on the flow rate or the flow direction by displacement the spool. The

change of the flow will influence in the actuators job ( Scholz, 1996). Figure 2-8

illustrate the signal sequence to control actuators by using proportional direction

valve.

An electrical input signal in form voltage converted by amplifier into current

corresponding with the input voltage. The electrical current form an input to

solenoid which produce variable magnetic force , which form an input to the

valve spool as variable displacement, producing proportional certain flow.

Therefore the actuators can run according different flow rate corresponding to

the electrical input signal (Schmitt, 1989).

Figure 2-8 The control signal sequence (Schmitt,1989).


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In the case of the spool shown in the figure 2-9 the spool control close the link

between the ports P, A, B and T. If we need to connected port P and B, port T

and A, Then the solenoid A is energized displacement the spool toward left. The

spring's force displacement the control spool into center position when the

actuator finish its work (Schmitt, 1989).

Figure 2-9 shows the components of proportional direction valve. The major

components are housing (1), proportional solenoid (2), inductive positional

transducer, control spool (4), return spring (5), fluid port (A, B, P, T ).

Figure 2-9 The components of proportional direction valve (Schmitt,1989).

The overlap of the control spool in its housing influences of the flow/ signal

function. Thus we have three types of overlap like in figure 2-10

1- Positive over small electrical signal displacements the spool control but the

flow is remains zero until certain value. The displacement correspond with zeroflow know as dead zone in the flow/signal function.

2- Zero overlap the function of flow/signal is linear function

3- Negative overlap: The flow /signal function in the small valve opening range

result in great shape.

The positive valves has more advantages than the other types like less in leakage

than the other types and the more tolerance in center position in power failure(Schmitt,1989), ( Rydberg, 2008).


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Figure 2-10 Spool overlap types for proportional direction valves ( Rydberg,

2008).

2-2-4 Pressure Relief Valve:

The relief valve (RV) is a type of valve used to control or limit the pressure in a

system which can build up by a process upset, instrument or equipment failure.

All hydraulic system must have at least one of this type. The function of this

valve is established the maximum pressure can developed in the hydraulic

system, which must large enough to overcome of pressure drop in hole system

and achieve the desired output pressure. Without using this type of the valves,

the pressure of the system could rise until damage the machine or the system

parts.


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Figure 2-11The components of pressure relief valve (Control open text

books, 2011)

Figure 2-10 shows schematic of typical pressure relief valve and its components.

When the pressure equal or under the pressure limitation, the spring force

prevent the seat disc from moving up and the forces are balancing in seat disc's

sides. If the pressure exceed, then the pressure force will overcome on spring

force and the seat disc opening allow to oil to flow and return to a tank. In this

design the maximum pressure or limitation can adjust by adjust a spring screwwhich increase or decrease the spring force (Chapple, 2003), (Akers, Gassman

and Smith, 2006).

2-2-5 Dual pilot-operated check valves:

In general, check valve allows oil to flow in one direction, that mean that it

works as electronic diode. A pilot-operated check is similar to a basic check

valve but can be held open permanently by application of an external pilot

pressure signal. Thus dual pilot -operated check valve is valve consists on its


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structure on two pilot operated check valve. There are two pilot pressures each

one work to open its check valve (Parr, 1999).

Figure 2-12 Dual pilot-operated check valves symbol

Check valves have excellent seal leakage in the closed position. So when wedesign a system to hold some load without using check valve, theoretical during

time the load might to stay on hold, but in practice the load tend to creep

because of leakage in the control valve (Parr, 1999).

In robot application it's important to keep the arm in its pose without any

changing pr creeping. So it's important to use check valve

2-2-6 Pipelines:

Pipelines are medium which transmission the hydraulic fluid into the different

parts of the system. The pipeline can effected the system by

Hydraulic friction losses; hydraulic resistance of lines Hydraulic local, or secondary, pressure losses Oil compressibility and elasticity of pipe material; hydraulic capacitance

of lines

Oil inertia; the hydraulic inertia of lines (Rabie, 2009).The proper tube diameter is determined according to the maximum flow rate and

selected fluid speed. Thus:

(2-10)Where:

: The inner pipe diameter

: Maximum flow rates, m3/s


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: Fluid mean velocity, m/s (Rabie, 2009).According to the type of fluid flow and the fabrication of pipe we can define two

type of pressure drop in pipeline.

A- Friction losses: Where the drop of the pressure depended in the surface

roughness in inner pipe, and fluid properties, thus we can define the drop of

pressure from friction losses in the equation:

(2-11)Where:

: Friction coefficientL: The length of pipeline m

d: The diameter of the pipe m: Fluid density kg/m3The friction coefficient is depend on the Reynolds number , the type of flow,

and the surface roughness by the following equations in table 1-2: (Rabie, 2009).

Table 2-1 Determination of the pipeline friction cofficient (Rabia, 2009).

B- Minor loss: It results from the rapid changing in the fluid velocity and the

direction of the flow. The pressure drop is calculation in the following equation:


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(2-12): Local loss coefficientThe local loss coefficient due to sudden expansion is calculated by the followingequation (Rabie, 2009):

2 (2-13)

Table 2-2 Local loss coefficient for some type of pipes (Merkle, Schrader,

Thomes, 2003).

Table 2-2 shows the local loss coefficient due to changing of the direction of

flow can determine for some frequency pipe types use in hydraulic system

(Merkle, Schrader, Thomes, 2003).

2-2-8 Filter:

Filters are an essential component and great significance of every hydraulic

system. Their function is to remove particle contaminants from the hydraulic

fluid, which reduce the service life of system components through abrasive

wear, or formed a suffocation point. Filters function is reducing the rate of

contaminants in acceptable level to protect the hydraulic systems components.

There are many distinction filters according to the grade, like absolute filtersfineness which indicates to the largest particle able to path, or average pore size

which indicate to the average size of particles path through the filters, other

important method is fvalue for given particle size x defined as the number of

particles size in upstream to the number of particles size in downstream (Merkle,

Schrader, Thomes, 2003), (Parr, 1999).

Example for x = 100 and the hydraulic system application suite.


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Table 2-4 The methods used to install in hydraulic system applications (Merkle,

Schrader, Thomes, 2003).

Every method will cause a drop in pressure. In pressure line filter the pressure

drop ranging between 1 ~ 1.5 bar. The pressure drop through a filter is the sum

of pressure drop from housing and from filter elements which specified by the

manufacturing, thus

(2-14)

Main flow filter By pass flow filter

Return flow

filter

Pump inlet

filter

Pressure line

filter

Circuit

Diagram

Advantages Economical.Simple

maint-

enance

Protectpump from

contam-

ination

Smaller poresize.

Possible for

valves

sensitive to

dirt

Smaller filterpossible as an

additional filter

Disadvantag

es

Contaminati

on can only

be checked

havingpassed

through the

hydraulic

components

Difficult

access, inlet

problems

with finepored filters.

Result: cavi-

tation

Expensive Lower dirt filtering

capacity

Remarks Frequently

used

Can also

used a head

of the pump

as a coarse

filter

Require

pressure tight

housing and

contaminatio

n indicator

Only part of the

delivery is filtered


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Where fv is viscosity factor.2-2-9 Pump:

A pump is a device which has ability to convert the mechanical energy into

hydraulic energy. There are a wide range of pumps in the market, but there

mechanical design can category into positive displacement pump and variable

and rotodynamic pumps. Also we can category pumps into variable

displacement pump, and fixed displacement pump (Cundiff, 2001). However,

the major aspects in selection pumps involve some factors summaries in the

follow points:

Cost. Pressure ripple and noise. Suction performance. Speed Weight. Displacement. Fluid type. Maximum pressure, flow or power (Chapple, 2003).

Table 5-2 comparing between operation parameters for different types of pumps,which can considering during the design.

Type Maximum

pressure (bar)

Maximum

flow (l/min)

Variable

displacement

Positive

displacement

Centrifugal 20 3000 No No

Gear 175 300 No Yes

Vane 175 500 Yes Yes

Axial piston

(port-plate)

300 650 Yes Yes

Axial piston

(valved)

700 650 Yes Yes

In line piston 1000 100 Yes Yes

Table 2-5 Comparing between different types of pumps (Parr, 1999)


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For comparing between the different type of pump, and the advantage and

disadvantages for each one and what is suitable for the application we can use

table 2-6

Gear Pump External type Internal type Other feature*Low cost

*Lowcontaminate

sensitivity

*Compact ,low

weight

*Good suction

performance

*Low noise

* Lowcontaminate

sensitivity

*In-line assembly

for multi pumpunits

Vane Fixed

displacement

Variable

displacement

*Low noise

*Good service

ability

*Low noise

*Low cost

*Goode service

ability*displacement

contols

*In-line assembly

for multi pump

units

Piston Fixed and variable displacement

*High efficiency

*Good service ability

Wide range of displacement control

*Integral boost

pump and multi

pump assemblies

*Can use mosttype in hydrostatic

transmission.

Table 2-6 Comparing between different types of pump (Rabia, 2009)

In robotic application, the pump will frequency operates under load or less than

nominal load. For avoid failed in the system or heating the oil which will reduce

the system reliability, we must choose variable displacement pump.

Variable displacement pumps provide wide range of control methods; include

load sensing, pressure compensation, power and torque sensing and limitation,

in additional control by electro-hydro device. In the term of profit, variable

displacement pump reduce the operation cost and reliability of the system, and


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reduce the system maintenance and increase the overall system efficiency

(Chapple, 2003).

The ideal flow from pump can achieve from the formula:

(2-15)Where:: The ideal flow from pump m3/sV: Volumetric displacement m

3/rev

: rotational speed rad/s

By considering the leakage from the pump, and is the volumetric efficiency (2-16)2-2-10 Accumulator:

Accumulator is a mean for storage energy, which can storage and release a

quantity of fluid at required system pressure. The storage of energy done when

the system is under load (Parr, 1999).

Often, accumulator installed in the system when:

Needing emergency supply. Leakage compensation. Required volume changes compensation due to temperature or pressure. Supposing system work frequency under load. Shock alleviation. There are simple suspension elements. Pulsation absorption. Supplementation of pump flow to meet high transient flow demand

(Chapple, 2003)

Figure 2-13 shows accumulator bladder type, where the bladder separates the

oil from pressure gas, usually is nitrogen. The gas supplied via valve with a pre-

charge pressure determine by the pressure range requirements for the system.

When the accumulator in storage phase. The gas does not flow in the bladderand the oil filling the accumulator. Where gas flow by rate proportional with the


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required oil pressure/flow in the system when we need to release oil from

accumulator, this gas push the oil to go out from the accumulator and feed the

system.

Figure 2-13 Bladder accumulator type (Brendan, 2009)

The accumulator is charged to a pressure p0 lower than the minimum pressure

operation p1. For gas has mass m and absolute temperature T and volume V, by

assuming the gas expansion polytropic, then:

(2-17)

(2-18)Where:

n: Polytropic index

R: Universal gas constant.

When the pressures fall in the system, the gas expansion and deliver V volume

of oil to the system


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(2-19)Where

V2: Maximum operation volume.

For comprising from V0 to V2: (2-20)Then

(2-21)Also for gas expansion from V1 to V2 with polytrophic index n2 (2-22) (2-23)This equation gives the conservative value for most application.

2-3 Circuit Design:

Figure 2-2 shows the hydraulic system for 3DOF manipulator arm. A pump

provides hydraulic fluid with operation pressure to the system. A pressure relief

valve and filter connecting to the system to safety insurance for the systems and

components. An accumulator storage the energy when the arm be under load

and compensates the system with fluid when it require that. A 3/2 direction

valve has on/off function for the system. The oil flow from 3/2 direction valve to

three 4/3 proportional direction valves, where each one control with the

direction and speed of one motor by controlling with the pass and flow quantity.

Motors:

The circuit design will begin by choosing the three motors and calculate their

parameters, after that finding the pressure drop in every part in the system to

choosing the correct pump.

Each joint is an axial piston motor provide the require rotation movement for the

link. The speed for robot tool center point might vary from very low speed toabout 1 m/s as maximum speed.


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Thus for choosing motor there are some aspects might be considering such as

the range of motor speed, the weight, the torque. For this reason, motors AE16

from Duesterloh has been chosen where this type of motor has speed range from

very low speed, and light weight with high torque. Table 2-7 shows the major

characters for AE16 motor, for more details see appendix A-1 (Duesterloh

Fluidtechnic,2005).

Table 2-7 The characters for AE16 motor From Duesterloh

The calculation of system parameters will be concentration on the maximum

flow/ speed/

From equation 2-7

Weight kg Kg 13

Displacement cm3/rev 16

Max.Operating pressure Bar 250

Max. Speed r.p.m 2500

Min. Speed r.p.m 5

Max. Torque N.m 57,2

max.Capacity Kw 9,6

Total efficiency 96%


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The maximum pressure drop in the motor can achieved from equation 2-8

From table the maximum operation pressure (inlet pressure) is:

P1= 250 bar

Then the outlet pressure:

p2= 16 bar

Each motor need 250 bar to operate at the maximum speed, and the high

pressure oil might deliver to motors from pump. According to the drop pressure

in the system parts the pump pressure must be higher than this value, to know

how much pump pressure we must find the drop of pressure in hole system:

1-The pipeline between the motors and proportional direction valves: If

considering the length of pipeline is d=5 meters with diameter L= 10 mm., the

density of oil 850 kg/m3 and kinematic viscosity= 1*10-4 m2/s. Then:

v=/2*pi*d) =1.32 m/sReynolds number can determine the type of flow if turbulence or laminar flow

66By using table 2-1 for laminar flow we find


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From equation finding the pressure drop

By assuming that, the system from proportional valves to motor needs 6

junctions forming in 90o

and according to table 2-2 we find From equation we can find the pressure drop on junctions

1-The pipeline between the pump and proportional direction valves

Between proportional valves and pump there is one line feed the three

proportional valves with oil, By assuming that the flow from each valve to the

motor are equal, then the flow in this line will be: 0,0416+0,0416+0,0416= 0,1248 m3/sThen to find the pressure drop in pipeline before the proportional valves, and by

assuming that pipelines length is 5 meter with diameter 10 mm. By flow thesame procedure above we find that

v= 3.97

Re=198.5

=0.332

Also we assume that the line consists of 6 junctions in 90o

form, then The total drop pressure in pipeline is 5.922 bar


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3-pressure drop in proportional valves:

We choosing Proportional direction valve from bosch Rexroth group with type

4WRPE 6with size 10 we find the pressure drop in the valve is 5 bar (bosch

Rexroth group, 2001). For more details you can see appendix A-2 bar3-The pressure drop in 3/2 direction valve:

We must considering during choosing the operation parameters for the valve the

nominal flow large or equal the total flow or:

Q>= Q1+Q2+Q3=125,9 l/min

Where:

Q1, Q2, Q3: The flow in motor 1, motor2 and motor 3

Also the pressure must be larger than the total pressure to the valve, which,

mean the nominal pressure for the valve >= the inlet pressure on the motor+ the

drop pressure after the 3/2 direction valve to the motor

Then we choose

Nominal pressure> 300 bar

We choose the valve CP722-11from (Sauer-Danfos). For more details see

Appendix A-3 (Sauer-Danfos, 2007)

In figure 2-14 we can find the pressure drop in the valve according to maximum

operation parameters

Figure 2-14 Theoretical performance for valve CP722-11(Sauer-Danfos, 2007)


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From figure we can see that the pressure drop in the valve 4- Pressure drop in the filter:

According to table 2-3, the suitable filter must has = 2-5. And according to

table 2-4 we need pressure inline filter. The operation pressure must be larger

than the pressure in the rest of the system. So we choose the filter350LEN0160-

H3XLA00-V5, 0-M -.which allow to maximum pressure until 350 bar and has =

2. The drop in pressure which happen through the filter according to appendix

A-4 (Bosch Rexroth group, 2011)

1bar .PUMP

From last discussion, we can choose the pump which has outlet pressure equal to

the motor inlet pressure+ drop pressure in the system

250+27.922=277.922 barAlso the outlet flow must equal to the total flow Q=124.9 l/min.Thus we choose the variable displacement pump A4VSOfrom Bosch Rexroth

group see appendix A-6 which has outlet pressure 300 bar, and flow rate 225

l/min (Bosch Rexroth group, 2002.)

Accumulator:

We choose accumulator which performance the maximum pressure we need.

Thus we choose accumulator can give operation pressure 350 bar after punch,

which is HAD..-1X/2X from bosch Rexroth group (Bosch Rexroth group ,

2012). For more details see appendix A-5

2-4 Load Calculation:

The hydraulic manipulator arm which will studying and designing consist is 3

degree of freedom (3DOF), consist of three rotation joints. Base joint rotate

around z axis and the second and third joint rotate around y axis. Each joint


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rotate by axial piston motor, the rotation speed and the direction of rotation for

the motors will controlled by proportional direction valves. Figure 2.2 shows the

hole hydraulic system for the arm. However this motor moved three rigid

aluminum cylindrical links. Table 2-8 shows the properties for the link.

Link number Length ; m Diameter ; m Weight ;kg

1 1 0.2 85

2 1 0.2 85

3 0.4 0.2 34

Table 2-8 Links parameters

In general the speed of robot's tool center point must be not high and might

average between 0,001 m/s to 1 m/s

To calculate the maximum load for the robot we must calculate the maximum

load for each motor then chose the minimum load. In order to reduce the

revolution speed for the motor we will use harmonic drive systems with rate

1/160.

For motor 3:

The maximum revolution speed for the link after harmonic drive system(Hibbeler, 2010):

r.p.m (2-24)Then the angular velocity for link 3

rad/s (2-25)Then the maximum velocity for tool center point is: m/s (2-26)The minimum revolution speed for the link after harmonic drive system:

r.p.mThen the angular velocity for link 3

rad/s


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Then the minimum velocity for tool center point is: m/sThe total torque applied in motor 3 is a sum of three types of torques

TL : Load torque which is the torque required to support the weight or force of

the load (Watford, 2005).

(2-27)Maximum load torque done when cos =1. Then equation 2-11 will be

(2-28)Where

mlmax : The maximum load mass

g: gravity acceleration

ml3: the mass for link 3

T: Acceleration torque is the torque required to overcome the inertia of the load

in order to provide a required acceleration or deceleration (Watford, 2005). (2-29)Where

JL3: The inertia moment for link 3 around connection point with motor 3

(2-30)Jload: The inertia moment for load around connection point with motor 3 (2-31)By compensation (2-30), (2-31) in (2-29) :

(2-32)


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Tc: Cushion torque is the torque that the actuator must apply to provide a

required angular deceleration, *. Cushion torque is generated by restricting the

flow out of the actuator in order to create a back pressure which decelerates the

load (Watford, 2005).

(2-33) (2-34)Tf: Friction torque is the torque required to overcome any friction between

moving parts, especially bearing surfaces. We will consider the friction torque

equal 0

Thus the total torque is (Watford, 2005): (2-35)By compensation equations (2-28), (2-32) and (2-34) in (2-35) find:

(2-36)The maximum torque for the motor is 57,3 N.m then, the maximum torque for

link 3 is. Then the maximum torque for links by respect the gear ratio:

Ttmax= 160*57,3= 9168 N.m

By applied this value in equation 2-20 we can find the maximum load for link 3

m=2624 kg

Motor 2:

The maximum revolution speed for the link after harmonic drive system:

r.p.m

Then the angular velocity for link 3


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rad/sThen the maximum velocity for tool center point is:

m/sThe minimum revolution speed for the link after harmonic drive system: r.p.mThen the angular velocity for link 3

rad/s

Then the minimum velocity for tool center point is: m/sIn the same way we can find the torques for motor 2

Then

mlmax2= 398 kg

Motor 1:


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r.p.mThen the angular velocity for link 3

rad/sThen the maximum velocity for tool center point is: m/sThe minimum revolution speed for the link after harmonic drive system:

r.p.m

Then the angular velocity for link 3

rad/sThen the minimum velocity for tool center point is: m/sThe torques:

mlmax= 442 kg


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the maximum load for the arm

mlmax= min(mlmax1,mlmax2,mlmax3) = 398 kg


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Chapter 3

Kinematic model

The kinematic model is essential in robot studying and for path planning. By

kinematic model the coordinates and orientations for tool center point can bedetermining by knowing the displacement of joint, which know the forward

kinematic. While by applying inverse kinematic, the join ts displacement can be

known according to tool center point coordinates and orientations. Last part

from this chapter wil l studying the interpolator, which can planning the path

for tool center point f rom one point to another by dividing it into small steps

and give the sui table velocity and acceleration for each step.


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The term of kinematic model for manipulator arm refer to the studying of the

geometry of manipulator. The kinematic model is a fundamental tool in design

and control with manipulator robot.

The purpose of a manipulator is to manipulate the tool center point for the robot,to performance that the robot must to know where the object to work on is

located, and what is located of tool center point. Thus robot kinematics studies

the relationship between the dimensions and connectivity of kinematic chains. In

other word, kinematic model describes the position and the orientation (pose),

velocity, acceleration and all higher order derivative of the pose.

The robot kinematics can be divided into forward kinematics and inverse

kinematics. Forward kinematics problem is straightforward and there is no

complexity deriving the equations. Hence, there is always a forward kinematics

solution of a manipulator. Inverse kinematics is a much more difficult problem

than forward kinematics. The solution of the inverse kinematics problem is

computationally expansive and generally takes a very long time in the real time

control of manipulators.

The relationship between forward and inverse kinematics is illustrated in Figure

3-1

Figure 3-1 The schematic representation of forward and inverse kinematic

The studying of dynamic model will involve:

Forward kinematic: Inverse kinematic:

Interpolator:


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3-1 Forward Kinematic:

A manipulator robot consists of a serial number of links which are affixed to

each other by revolution joints or prismatic joints. This serial begin from the

base of robot to the end effort (tool center point). The function of forwardkinematic is finding the pose of tool center point according to joints

displacement in the Euclidean space.

The minimum number of coordinates required to locate a body in Euclidean

space is six, three coordinates for position, and three for orientation. Thus we

can find the location of tool center point in the space by finding the six

coordinates. By considering the joints which connected each point from base to

the tool center point, then we can define the forward kinematic problem as:

if we have the joint displacement q , then the pose of tool center point (Solvang,

2010):

x=f(q, L) (3-1)

Where:

L: The length of link

q=[q1 q2 q3...qn]

x=[X Y Z A B C]

Where

q1, q2...qn : Displacement of joint 1, joint 2..

x: The six parameters for tool center point.

A: The rotation around x axis.

B: The rotation around y axis.

C: The rotation around z axis.


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The displacement of tool center point is result from the joint displacement, and

to determine the change of orientation we have to select a convention of

orientation. This can done by Euler angles. Which are three angles introduced by

Leonhard Euler to describe the orientation of a rigid body.

We can find the relationship between the rotation frame simply by finding the

rotation around X,Y,Z (Solvang,2010).

Figure 3-2 shows the coordinate rotation around Z or C angle

Figure 3-2 Coordinate rotate around Z axis with angle c

Then we can write the rotation matrix as (Solvang, 2010):

(3-2)

Where:

n: Describe X axis for frame 0 in frame 1


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s: Describe Y axis for frame 0 in frame 1

u: Describe Z axis for frame 0 in frame 1

In the same way figure 3-3 shows the coordinate rotation around Z or C angle

Figure 3-3 Coordinate rotation around Y axis with angle B

Then the rotation matrix around Y can write as (solvang, 2010):

(3-3)

Figure 3-4 shows the rotation around X


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Figure 3-4 Coordinate rotate around X axis with angle A

Then rotation matrix:

(3-5)

By multiply the matrixes we get the general rotation matrix from frame (i-1) toframe (i)

(3-6)By considering the three coordinates for position we can write the general

transform matrix from frame i-1 to i


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(3-7)

The last column of the last matrix was added to the general form to has

symmetric matrix (Solvang, 2010)

Thus, the transform matrix can be writing in the form

(3-8)

So we can move frame one frame to another by multiply the transformation

matrix to the frame. Which meaning, the transform matrix to robot's tool center

point has n link (3-9)This matrix can write as following:

(3-10)

From this matrix we can drive the six parameters [X Y Z A B C], or in other

world solve the forward kinematic problem (Solvang, 2010)

X=r41 (3-11)

Y=r42 (3-12)

Z=r43

(3-13)

(3-14)

(3-15)


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3-1-1 Forward Kinematic for 3 DOF:

The manipulator arm consist of 3 joint first joint rotate around z axis with

displacement q1 and the second and third rotate around y axis with

displacements q2, q3. There are three links in the arm, first link has length L1connecting joint 1 and joint 2, the second joint has length L2 and connecting the

second joint with third joint, where the third joint which has length L3

connecting the joint 3 with tool center point.

.Figure 3-5 shows the orientation of manipulator arm consist of three joint, and

how the coordinates changing according to the joints revolution.

Figure 3-5 illustrate 3DOF manipulator arm, joint 1 rotate around Z, joint 2

and 3 rotate around Y

From equation we can write:

(3-16)


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Where:: The transform matrix from joint 1 to the base

: The transform matrix from joint 2 to joint 1

: The transform matrix to joint 3 according joint 2

(3-17)Where

c1=cos(q1)

s1=sin(q1)

(3-18)

Where:

c2= cos(q2)

s2=sin(q2)

(3-19)c3=cos(q3)

s3=sin(q3)


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(3-20)

Then from equation (3-16)

(3-21)

Where:

c23=cos(q2+q3)

s23=sin(q2+q3)

3-1-2 Forward Kinematic Simulation:

Appendix B-1 consist Matalb function to simulate 3DOFmanipulator arm

by displacement (q1,q2,q3)= (0, 0 ,0). In simulation we suggested that

L1=L2= 1 m, L3=0.4 m

we will have result in figure


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Figure 3-6 Forward Kinematic simulation for joint displacement (0,0,0)

By joints displacements (q1,q2,q3)=(30,30,30), then figure shows the result


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Figure 3-7 Forward Kinematic simulation for joint displacement (30,30,30)

3-2 Inverse Kinematic:

From its name, the inverse kinematic is the opposite problem for forward

kinematic. Which mean that the inverse kinematic problem is centering about

finding the joint displacements from knowing tool center point coordinates and

orientation. In other world:

(q)=f(X, L) (3-22)

The inverse kinematics problem of the serial manipulators has been

studied for many decades. Solving the inverse kinematics is computationallyexpansive and generally takes a very long time in the real time control of

manipulators.

Cartesian space includes orientation matrix and position vector. However, joint

space is represented by joint angles. The conversion of the position and

orientation of a manipulator tool center point from Cartesian space to joint space

is called as inverse kinematics problem. There are about three methods to solve

this problem, but in this repor

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