Modeling and Optimization of a Highly Integrated

UNIVERSITY OF VERSAILLES SAINT-QUENTIN-EN-YVELINES

Modeling and Optimization of a Highly Integrated Electro-Hydraulic Energy Converter for Robotic Applications by Elmira AMROLLAH A thesis submitted in partial fulfillment for the degree of Doctor of Philosophy in the Subject Area: Robotics Department: S.T.V. Jury Advisor: Reviewers:

M. F.B.OUEZDOU M. Yannick AOUSTIN M. Gérard POISSON Examiners: M. Yasser ALAYLI M. Antoine DEQUIDT M. Fay¸cal NAMOUN M. Samer ALFAYAD

University of Versailles, LISV University of Nantes, IRCCyN University of Orléans, PRISME University of Versailles, LISV University of Valenciennes, LAMIH BIA Turnkey Test Systems University of Versailles, LISV

February 2013

I told my robot to my biddin’, He yawned and said, “You must be kiddin’.” I told my robot to cook me a stew. He said, “I got better things to do.” I told my robot to sweep my shack. He said. “You want me to strain my back?” I told my robot to answer the phone. He said, “I must make some calls of my own” I told my robot to brew me some tea. He said, “Why don’t you make tea for me?” I told my robot to boil me an egg. He said, “First lemme hear you beg” I told my robot, “There’s a song you can play me.” He said,“How much are you gonna pay me?” So I sold that robot, ’cause I never knew Exactly who belonged to who.

Shel Silverstein

Acknowledgements I would like to thank: Pr. Poisson and Dr. Aoustin for their helpful remarks on my thesis. Pr. Ben Ouezdou for being my supervisor and giving me the opportunity to join his research team and do this thesis. Samer Alfayad for all the discussions we had and his helpful advice. M.Namoun and M.Ramzi Sellaouti for their ongoing support and guidance. Thanks for always finding time to advice me on my thesis. Sylvain Bertrand for being a great friend and colleague at LISV. Thank you for making our ”nerd marathons” at the lab more fun! All my friends and colleagues at LISV and BIA. Specially Paul-Franois, JeanChristophe, Maja, and Marwa. Thank you for your kindness and helpful feedback. I’d also like to thank all my friends in Iran, France, and all over the world, for making the rough days seem not so bad after all. And last but not least, many thanks to my dearest parents and sister for their love and support all along the way.

iv

Contents Acknowledgements

iv

Introduction

1

R´ esum´ e

3

1 State of the Art

9

1.1

Electromagnetic actuators . . . . . . . . . . . . . . . . . . . . . . . 10

1.2

Pneumatic actuators . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3

Hydraulic actuators . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4

Actuator comparison . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Virtual Model

19

2.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.2

The Architecture of IEHA . . . . . . . . . . . . . . . . . . . . . . . 19

2.3

IEHA’s Dynamic Model . . . . . . . . . . . . . . . . . . . . . . . . 23 2.3.1

2.3.2

2.3.3

2.3.4

Fundamental Hydraulic Equations . . . . . . . . . . . . . . . 25 2.3.1.1

Rotary pump . . . . . . . . . . . . . . . . . . . . . 25

2.3.1.2

Servo valve . . . . . . . . . . . . . . . . . . . . . . 26

Hydraulic phenomena . . . . . . . . . . . . . . . . . . . . . . 27 2.3.2.1

Compressibility . . . . . . . . . . . . . . . . . . . . 27

2.3.2.2

Leakages

. . . . . . . . . . . . . . . . . . . . . . . 27

Motion Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3.3.1

Carriage eccentricity . . . . . . . . . . . . . . . . . 28

2.3.3.2

Fluid Pressure in the Rotary Pump . . . . . . . . . 28

2.3.3.3

The passive distributor . . . . . . . . . . . . . . . . 30

2.3.3.4

Linear Actuator

. . . . . . . . . . . . . . . . . . . 31

Input-Output Model . . . . . . . . . . . . . . . . . . . . . . 31 v

vi

CONTENTS 2.4

Simulink model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.5

Simulink Simulation Results . . . . . . . . . . . . . . . . . . . . . . 35

2.6

SimHydraulics Model . . . . . . . . . . . . . . . . . . . . . . . . . . 38 2.6.1

Servo-Valve & Carriage . . . . . . . . . . . . . . . . . . . . . 40

2.6.2

The Micro Pump . . . . . . . . . . . . . . . . . . . . . . . . 40

2.6.3

Passive distributor . . . . . . . . . . . . . . . . . . . . . . . 46 2.6.3.1

Linear passive distributor . . . . . . . . . . . . . . 47

2.6.3.2

Rotary passive distributor . . . . . . . . . . . . . . 49

2.7

SimHydraulics Simulation Results . . . . . . . . . . . . . . . . . . . 51

2.8

Comparison and conclusions . . . . . . . . . . . . . . . . . . . . . . 58

3 Parameter Optimization

61

3.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.2

Volumetric Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.3

3.4

3.5

3.2.1

Clearance and piston deviation . . . . . . . . . . . . . . . . 62

3.2.2

Leakage in pump pistons . . . . . . . . . . . . . . . . . . . . 62

3.2.3

Leakage in passive distributor . . . . . . . . . . . . . . . . . 66

Mechanical efficiency of the micro-pump . . . . . . . . . . . . . . . 66 3.3.1

Rolling motion . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.3.2

Vane dynamic model . . . . . . . . . . . . . . . . . . . . . . 67

3.3.3

Power loss due to piston-rotor friction . . . . . . . . . . . . . 71

3.3.4

Friction and the piston-carriage contact point . . . . . . . . 74

3.3.5

Friction and piston length . . . . . . . . . . . . . . . . . . . 75

3.3.6

Choosing the best length and tip parameters . . . . . . . . . 76

Ankle Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 3.4.1

Pump flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

3.4.2

Ankle pump efficiency . . . . . . . . . . . . . . . . . . . . . 81

Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4 Energy Storage

85

4.1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

4.2

Hopping monopod . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 4.2.1

4.2.2

Simulink model . . . . . . . . . . . . . . . . . . . . . . . . . 86 4.2.1.1

Hydraulic actuator in normal mode . . . . . . . . . 88

4.2.1.2

Hydraulic actuator in storage mode . . . . . . . . . 89

Simhydraulics model . . . . . . . . . . . . . . . . . . . . . . 96 4.2.2.1

Active distributor . . . . . . . . . . . . . . . . . . . 96

CONTENTS

vii 4.2.2.2

Active distributor in the monopod actuator . . . . 97

4.3

Walking biped robot . . . . . . . . . . . . . . . . . . . . . . . . . . 100

4.4

Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

5 Conclusion and perspective

109

5.1

conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

5.2

perspective

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

A Laminar flow in thin passages

111

Bibliography

115

List of Figures . . . . . . . . . . . . . . . . . . 11

1.1

Humanoid robots using electric actuators.

1.2

Life Magazine photo: Joseph L. McKibben with his wife, and paralyzed daugh-

. . . . . . . . . . . . . . . . . . . . . 12

ter wearing a pneumatic muscle, [1] 1.3

Hydraulically actuated robots: (a) BigDog, (b) HyQ, (c) PETMAN, (d) DB, (e) CBi.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 . . . . . . . . . 14

1.4

Hydraulic exoskeletons: (a) XOS 2, (b) HULC, (c) BLEEX

1.5

Classes of actuators in terms of performance criteria. (Images from [2]).

1.6

Hydrostatic actuation system consisting of a DC motor, pump, small accumu-

. . . . 16

lator, and rotary actuator. Fluid is pumped directly from one side of the rotary pump to the other. (Image from [3]).

. . . . . . . . . . . . . . . . . . . . 17 . . . . . . . . . . . 22

2.1

Schematic summary of the working mechanism of IEHA

2.2

Simplified model of the shaft with two pistons turning inside the housing [4].

2.3

Working schema of the passive distributor

2.4

Schema of the micro-valve mechanism, and the mechanical feedback of eccentricity (E).

. . . . . . . . . . . . . . . . . . 23

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 . . . . . . . . . . . . . . . . . . . . 25

2.5

Geometric study of the piston length

2.6

The 4-stage IEHA and the output cylinder

2.7

Output cylinder and the passive distributor model in Simlink.

2.8

Flow and pressure variation of the micro-piston chamber.

2.9

Internal variables of IEHA model in Simulink for a step input of X = −0.95 × Emax .

. 23

. . . . . . . . . . . . . . . . . 35 . . . . . . . . 36

. . . . . . . . . . . 37

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.10 Simscape model of micro-valve and carriage hydraulic chambers. 2.11 Micro-pump modeled as a combination of micro-pistons. 2.12 A piston of the micro-pump.

. . . . . . . 41

. . . . . . . . . . . 43

. . . . . . . . . . . . . . . . . . . . . . . . 43

2.13 Schematic model of micro-pump’s input/output channels, and the connection of a piston input orifice to these two.

. . . . . . . . . . . . . . . . . . . . 44

2.14 Intake and output channels drilled in the micro-pump shaft 2.15 SimHydraulic model of the passive distributor

viii

. . . . . . . . . . 45

. . . . . . . . . . . . . . . . 48

LIST OF FIGURES

ix

2.16 Pressure in the different lines of the system for a positive micro-valve displacement, X, E > 0.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.17 Circular overlap of orifices in the rotary passive distributor.

. . . . . . . . . 51

2.18 Nonlinear opening of the butterfly valve orifices w.r.t spool displacement. Valve shaft stroke ∈ [−22.5◦ , 22.5◦ ]

. . . . . . . . . . . . . . . . . . . . . . . . 52

2.19 Opening of triangular and rectangular orifices w.r.t valve displacement(h). Triangular orifice: Maximum opening section=ABC. Actual opening section=BCDE. Rectangular orifice: Maximum opening section=ABCD. Actual opening section=CDEF.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.20 Internal variables of IEHA model in SimHydraulics for a free movement, with step input of X = −0.95 × Emax . In the second curve: (dashed line) Channel A opening signal, (solid line) Channel B opening signal.

. . . . . . . . . . . 54

2.21 Variation of the internal variables of µp0 , for a free movement, and an eccentricity value going from 0 to E = −0.95 × Emax .

. . . . . . . . . . . . . . 55

2.22 Internal variables of IEHA model in SimHydraulics, for an output actuator pushing against a wall(spring-damper). In the second curve: (dashed line) Channel A opening signal, (solid line) Channel B opening signal.

. . . . . . . 56

2.23 Variation of the internal variables of µp0 , for an output actuator pushing against a wall(spring-damper).. 3.1

. . . . . . . . . . . . . . . . . . . . . . . . . . 57

(a) Deviation angle(α) from the central axes, causes by the clearance between the piston and its sleeve. (b) α depends on the clearance value(Cl), as well as the piston diameter/length ratio. (*) corresponds to the first prototype of IEHA micro-pistons.

3.2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Annular orifice created by the clearance between the piston and its slot. In the middle figure piston and slot are concerntric(e = 0). In the right figure they are at maximum eccentricity(e = cl).

3.3

. . . . . . . . . . . . . . . . . . . . 64

Passage length of annular leakage orifice(Lpass ), for E = Emax , and rotation angles 0 6 θ 6 π. Lpass =lp − ltip for 1 to 4, and Lpass =lp − lp Out(t) for 5 to 6.

3.4

64

Volumetric efficiency of the micro-pump w.r.t radial clearance and working pressure. (*) clearance=10µm, Pressure=50bar, in the first prototype of IEHA. (O)clearance=8µm, Pressure=50bar,

3.5

. . . . . . . . . . . . . . . . . . . . 65

Volumetric efficiency of the passive distributor vs. ∆P across the distributor. One side of the distributor is considered connected to pressure 1 bar, and the pressure on the other side variating.

3.6

. . . . . . . . . . . . . . . . . . . . . 66

Contact configuration of a round tip piston rolling inside an eccentric circle. β=angle between the contact point and piston center.

. . . . . . . . . . . . 68

x

LIST OF FIGURES 3.7

(a) Forces acting on a rotating piston. (b) Lever arms of the piston-carriage

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

contact force. 3.8

The value of the coefficient κ, depending on the direction of piston deviation.

3.9

Sum of torques resulting from pressure, inertial forces, and piston carriage

. 70

contact(Nc ). Positive torque turns the piston to the left, while negative torque turns the piston to the right.

. . . . . . . . . . . . . . . . . . . . . . . . 73

3.10 Instantaneous power loss due to the friction in contact points of the piston for half the cycle where the piston is discharging to the output channel of pressure 1 to 100bar.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 . . . . . . . . . . . . . . . . 74

3.11 Average power loss for a pump of 30 pistons.

3.12 Upper limit of the contact angle(β) for the contact point to remain on the tip. 3.13 Average power loss due to

N friction(W L ),

. 75

for different values of piston tip

radius(rpt ). By reducing the radius from 6mm(the actual value in IEHA) N

to 5mm, W L can be divided by half at 50bar. 3.14 Average power loss due to

N friction(W L ),

. . . . . . . . . . . . . . . 75

for different lengths of the piston.

By increasing the length from 2.7mm(the actual value in IEHA) to 4mm, the N

average power loss (W L ) can be divided by half at 50bar.

. . . . . . . . . . 76

3.15 Variation of mechanical and volumetric efficiency of the micro-pump for different combinations of piston length(lp ) and piston tip radius(rpt ): (a) Average N

power loss(W L ). (b) Volumetric efficiency. 3.16 Ankle pitch actuators in HYDRO¨ıD foot.

. . . . . . . . . . . . . . . . . 77 . . . . . . . . . . . . . . . . . . 78

3.17 Biomechanical data of ankle pitch for walking motion with speed of 1.17m.s−1 . 3.18 Pump and piston design parameters and chosen values.

79

. . . . . . . . . . . . 81

3.19 Instantaneous power loss of one piston and the total energy efficiency of the pump during the propulsion phase, for different piston tip values. Since the piston length is chosen high, the theoretical pump efficiency is also high.

. . . 82

3.20 Mechanical(a) and volumetric efficiency(b) of the micro-pump on ankle pitch during the propulsion phase, for different values of piston tip radius(rpt ) and piston length (lp ).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 . . . . . . . . . . . . . . . . . . . . . . 87

4.1

Monopod model in Simmechanics.

4.2

Jumping monopod actuated by a passive spring on an unideal ground with high damping.

4.3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Jumping monopod actuated by a passive spring on an unideal ground with high damping: (top) Position of center of mass of monopod body in the vertical axis. (bottom) variation of spring length.

. . . . . . . . . . . . . . . . . . . . . 88

LIST OF FIGURES 4.4

xi

Hopping monopod actuated by a hydraulic actuator and a serial passive spring. When the spring is compressed in contact with the ground the hydraulic actuator is activated to further compress the spring and compensate for the dissipated energy in contact.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 . . . . 90

4.5

Architecture of the monopod with an added hydraulic actuator IEHA.

4.6

Simulation results of the hopping monopod in Simulink. From top to bottom: Body position, spring length, spring velocity, hydraulic cylinder position, and contact force.

4.7

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Comparison of the contact force with only the passive spring (dashed red) & in the presence of the hydraulic actuator (blue). With the hydraulic actuator in the system, the robot pushes harder in the ground and take off faster.

4.8

The hydraulic actuator’s active(X 6= 0) and inactive phases(X = 0) during jump. The inactive phases(green arrows) are used for energy storage.

4.9

. . . . . 93

Simulink model of IEHA, including the energy storage function.

. . . . . 93

. . . . . . . 94

4.10 (top) Volume of fluid in the accumulator, (bottom ) the functioning mode of the pump: +1=storage, -1=normal.

. . . . . . . . . . . . . . . . . . . . . 95

4.11 Pressure in the accumulator, and the cylinder chambers: During the storage mode the accumulator is charged to 29bar. This stored pressure provides about 67% of the needed pressure in normal mode, demanding less power from the pump and the electric motor.

. . . . . . . . . . . . . . . . . . . . . . . . 95

4.12 The mechanism of the storage two-way valve.

. . . . . . . . . . . . . . . . 96

4.13 Combination of the passive and active distributor, modeled as variable orifices. ±P D:Opening signals of the passive distributor. ±ST : Opening signal of the active distributor.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

4.14 Pump direction from B to A:E < 0, P D > 0. (a)Normal mode:ST = −1. (b)Storage mode: ST = +1.

. . . . . . . . . . . . . . . . . . . . . . . . 99

4.15 Hydraulic actuator and the pump modeled in Simhydraulics and linked to the monopod architecture in Simmechanics.

. . . . . . . . . . . . . . . . . . . 100

4.16 Simulation results of the hopping monopod in Simhydraulics and Simmechanics. From top to bottom: Body position, spring length, spring velocity, hydraulic cylinder position, and contact force.

. . . . . . . . . . . . . . . . . . . . . 101

4.17 The variation of flow in the accumulators and cylinder chambers w.r.t functioning mode and the direction of movement.

. . . . . . . . . . . . . . . . . . 102

4.18 (a) Variation of flow in Rp accumulator as the pump switches between normal mode and storage mode. (b) Pressure in the accumulator, and cylinder chambers: During the storage mode the accumulator is charged to 28bar.

. . . 103

xii

LIST OF FIGURES 4.19 The six DoFs of the simulated robot, and the linear hydraulic actuator on the knees. Knee of the stance leg is mechanically blocked and in storage mode. 4.20 The biped robot simulator and the IEHA actuators of the robot knees.

. . 104

. . . . 105

4.21 The walking motion of the right leg. Storage mode of knee is on during stance, and off during swing.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

4.22 Right knee variables during walking. From top to bottom: angle, functioning mode, fluid volume in accumulator, pressure of the accumulator and the needed pressure during swing phase.

. . . . . . . . . . . . . . . . . . . . . . . . 107

A.1

Piston concentric in cylinder.

. . . . . . . . . . . . . . . . . . . . . . . . 111

A.2

Laminar flow of a thin layer of fluid

A.3

Laminar flow in annulus between eccentric shaft and cylinder.

. . . . . . . . . . . . . . . . . . . . . 112 . . . . . . . . 114

List of Tables 2.1

Value of parameters used in the virtual model of IEHA in Simulink. 38

3.1

Subject characteristics and walking speed. . . . . . . . . . . . . . . 79

4.1

IEHA parameter values of the actuator used in the monopod. . . . 91

4.2

Rp accumulator volume and pressure values. . . . . . . . . . . . . . 94

4.3

IEHA parameter values of the actuator used in the biped knees. . . 106

4.4

Rp accumulator volume and pressure values. . . . . . . . . . . . . . 106

xiii

xiv

Introduction Human body has inspired engineers and scientists throughout history. Humanoid robotics is all about creating robots that are directly inspired by human capabilities, that have similar kinematics, as well as similar shape and behaviour to humans. Humanoid robots are made for different purposes, such as general-purpose workers, home assistants, entertainers, and test-beds for neuroscience theories. Among all the components of a robot such as sensors, skeleton and so on, actuators are perhaps one of the most important component of any robot. The selection and design of a suitable actuator is a crucial part of any robot design process. An ideal actuator must have high power/mass ratio, and should be adapted to the performance specifications of its assigned tasks. It should also be energy efficient(ex. including an energy storage mechanism) and safe in applications involving humanrobot interaction. Hydraulic actuators have hardly been used in humanoid applications because a central hydraulic power unit is needed to supply high pressure fluid to all the actuators of the system which can be bulky. Another issue of this category of actuators is that the leakage from the hydraulic tube connections can cause safety issues, especially in human-robot interaction. In IEHA (Integrated Electro-Hydraulic Actuator), developed by Alfayad et al.[5],[6] there is no need for a central pressure source and each single actuator is autonomous. Thanks to IEHA’s compactness, it can be placed as near as possible to the output actuator (rotary or linear). Hence, the pressure drop is reduced and leakage may only exist inside the actuator. Hence, risk of polluting the working environment of the robot is drastically decreased. Therefore, we believe that this type of actuation is a competitive option for robotic actuation systems. The goal of the present work is to study the

1

2

Introduction

internal parameters of IEHA, as well as identifying the influence of each parameter on the behaviour of the system, depending on the specified task to accomplish. Compactness of IEHA makes it difficult to measure the internal variables of the system, such as micro-pump’s leakage, and internal pressure drop; moreover, it is important to study the effect of configuring the internal variables (ex. the dimensions of different mechanical parts) on the behaviour and energy efficiency of the system, before modifying the latter in future prototypes. Hence, development of a model of the actuator will help us identify the different properties of the system, and propose new solutions to improve its dynamic behaviour, which is very important in robotic applications. The effectiveness of the developed virtual model allows for quick and accurate evaluation of the internal parameters early in the design and development stage. This work is organized as follows: Chapter 1 gives a review of the three main types of actuation in robotics. It presents the pros and cons of each category. Most importantly it shows the advantages of using IEHA in robotic applications. In chapter 2 the mechanism and dynamic equations of IEHA are presented. Then, the actuator is simulated firstly by implementing the dynamic models in MATLABSimulink environment, and secondly in MATLAB-Simhydraulics. In the end of the chapter the advantages of each simulator are discussed. Chapter 3 discusses the different parameters and hydraulic phenomena that play a crucial role in the performance of the system. Optimizing these parameters improves the mechanical energy efficiency, volumetric efficiency, and response time of the overall system. Chapter 4 presents the concept of energy storage and how this function can be implemented in the architecture of IEHA. Examples of periodic tasks are used to show how a good energy storage strategy could make the actuator autonomous and smaller in terms of necessary hydraulic sources. Finally, the last chapter gives a summary of the contributions of this work and a list of future works is proposed.

R´ esum´ e A travers l’histoire le corps humain a été une grande inspiration pour les ingénieurs et les scientifiques. La robotique humano¨ıde a pour l’objectif créer des robots qui sont directement inspirés des capacités humaines; ils ont une cinématique similaire, sont anthropomorphiques, et ils ont le même comportement que l’être humain. Les robots humano¨ıdes sont créés dans des différents buts, tels que l’assistance, le divertissement, et servent de banc de test pour les neurosciences. Parmi tous les composants d’un robot comme les capteurs et le squelette, l’actionnement est sans doute l’élément le plus important. Le choix et la conception d’un actionneur approprié est une étape cruciale lors de la conception d’un robot. Un actionneur idéal doit avoir une puissance massique optimale, et doit être adapté aux performances exigées par les tâches pour lesquels il est destiné. Il faut aussi qu’il ait un bon rendement énergétique (par exemple avoir un mécanisme de stockage d’énergie), ainsi qu’il soit fiable d’un point de vue sécurité notamment pour les applications d’interaction homme-machine. Les actionneurs hydrauliques ont rarement été utilisés dans des applications robotique humano¨ıde, car ils nécessitent un système volumineux pour générer un fluide sous pression qu’il faut fournir a` tous les actionneurs du robot. Un autre problème majeur concerne les fuites des tubes de connection qui pourraient être dangereuses, surtout dans les applications d’interaction homme-machine. Avec l’IEHA (Integrated Electro-Hydraulic Actuator), développé par Alfayad et al., il n’est plus nécessaire d’avoir un réservoir central car chaque liaison est actionnée indépendamment. Grâce au faible encombrement du IEHA, il peut être placé près de l’actionneur (rotatif ou linéaire) de la liaison, il en résulte la réduction des pertes de pression et des fuites, diminuant la perte d’énergie ainsi que la pollution de l’espace de travail. C’est la raison pour laquelle, ce type d’actionnement paraˆıt 3

4

Résumé

être un choix adéquat pour l’actionnement des systèmes robotiques. Le but de ce travail est d’étudier et de modéliser les paramètres internes du IEHA, ainsi qu’identifier les paramètres qui influencent le comportement et le rendement du système, selon les tâches a` accomplir. Etant donné la conception compacte du IEHA, il est difficile de mesurer les variables internes du système tels que la perte de pression interne et les fuites. Cependant, il est nécessaire d’évaluer l’influence des paramètres géometriques, ainsi que les paramètres définis par la tâche sur ces variables internes et sur le comportement et le rendement du système, dans le but d’améliorer les futurs prototypes. Aussi, le développement d’un modèle de l’actionneur nous aidera à identifier les différentes caractèristiques du système, et proposer des solutions pour améliorer la dynamique de celui-ci. Nous avons choisi de développer un modèle numérique qui nous permettra l’évaluation des paramètres internes d’une manière rapide et précise, ce qui aura un impact sur les premières étapes de la conception. Ce travail est structuré de la manière suivante: Dans le premier chapitre les trois catégories les plus importantes d’actionnement dans la robotique sont présentées. La première concerne les actionneurs de type électromagnétiques. Ce type d’actionnement est le plus répondu, le rendant peu couteux et offrant une large diversité de ces actionneurs. Le premier inconvénient de ce type d’actionneur est une performance limitée; les actionneurs électriques sont rigides et ils ont une puissance massique faible. De plus, ces actionneurs produisent de faibles couples par rapport à leur taille et poids, c’est pourquoi ils ont besoin de réducteurs rendant souvent le système irréversible. Il est à noter que ceci pourrait être dangereux surtout dans les applications d’intéraction sécurisé entre l’homme et le système robotique. Une solution pour ce problème a été d’ajouter un actionneur élastique en série avec le moteur. Le robot ECCEROBOT(Embodied Cognition ub a Compliantly Engineered Robot), financé par l’UE est un exemple de ce type d’actionnement. Ce genre de système n’est pas robuste a` cause de la complexité des connections électriques. La deuxième catégorie englobe les actionneurs pneumatiques. Ces actionneurs sont basés sur la compressibilité de l’air, donc ils sont fiables dans les interactions

Résumé

5

homme-machine. Par contre, ils ont une faible bande passante, et la relation entre la pression et la force est non-linéaire, ce qui nécessitent des méthodes de contrôle très complexes. Un autre inconvénient des actionneurs pneumatiques est le besoin d’une source d’air comprimé qui peut être chère et bruyante. La troisième et la dernière catégorie d’actionneurs regroupe les actionneurs hydrauliques. Au début, les actionneurs hydrauliques étaient seulement utilisés dans l’industrie comme des sources de puissance, car ils offrent une puissance massique élevée. Les robots hydrauliques les plus connus de nos jours sont BigDog et son ”grand frère” PETMAN développés par Boston Dynamics pour des applications militaires, et un autre quadrupède hydraulique nommé HyQ développé a` l’IIT(Istituto Italiano di Tecnologia). Ce dernier utilise une combinaison de moteurs électriques et hydrauliques pour actionner ses liaisons. A part ces derniers, les actionneurs hydrauliques ont rarement été utilisés dans les applications humano¨ıdes, car ils ont besoin d’une centrale hydraulique. La solution EHA (ElectroHydraulic Actuator) introduite par Habibi et al. est un actionneur hydrostatique autonome qui nécessite sa propre pompe qui peut être dimensionner en fonction du besoin de la liaison. Cependant, cet actionneur a besoin d’un vérin symétrique spécialement conu cet effet, rendant ce système couteux. Dans IEHA développé par Alfayad et al. il n’y a pas besoin de centrale hydraulique de grande taille, car chaque actionneur est autonome, et il peut fonctionner avec des vérins asymétriques. En plus, cet actionneur peut être placé très proche des liaisons, diminuant les risques de fuites dans les tubes. Dans le deuxième chapitre, le mécanisme de IEHA et sa modélisation dans MATLAB et Simhydraulics sont présentés. L’idée de base de ce mécanisme est de transmettre l’énergie d’un moteur électrique a` un actionneur hydraulique. Ce système est composé de trois parties principales : 1)La micro-pompe; 2)la microvalve ; 3) le distributeur passif. La micro-pompe est une pompe radiale, composée de 15 a` 25 micro-pistons. Le rotor est connecté à un moteur électrique qui tourne a` vitesse constante. La pompe a un débit variable et transmet l’huile du côté A au côté B ou l’inverse. Le débit et la direction de la pompe sont contrôlés par l’excentricité du rotor par rapport au chariot. Le chariot est mobile et dispose de deux chambres hydrauliques, une sur chaque côté. L’huile est dirigée vers une de

6

Résumé

ces deux chambre via la micro-valve qui est une valve deux-lignes et trois positions. Un effort externe bouge la tige de la micro-valve dans un sens ou l’autre pour régler l’excentricité. Il existe une connection mécanique entre la partie fixe de la micro-valve et le corps du chariot. Cette connection physique crée une boucle fermée mécanique entre la position de la tige de la micro-valve et l’excentricité du chariot. Autrement dit, quand la tige est déplacée, la valve est ouverte et elle envoie l’huile dans une chambre du chariot ce qui le fera bouger dans la même direction que la tige. Une fois que le chariot est déplacé de la même valeur que la tige, puisqu’il bouge la partie fixe de la valve avec lui, la valve se referme et le chariot reste à cette position. Le débit à la sortie de la pompe est dirigé vers une des deux chambres du vérin de la sortie via le distributeur passif. Le distributeur passif est une valve rotative trois lignes trois positions qui s’ouvre dans un sens ou l’autre selon le sens du débit de la pompe et la différence de pression des deux côtés. Selon le sens de rotation du distributeur passif l’entrée de la pompe est connectée au réservoir sous pression, la sortie de la pompe à une des deux chambres du vérin de la sortie, et l’autre chambre du vérin au réservoir atmosphérique. En utilisant le distributeur passif la micro-pompe peut fournir de l’huile a` un coté du vérin ou l’autre sans changer la direction de rotation du moteur électrique. De plus, grâce au distributeur passif, puisque les deux cotés de la pompe ne sont pas connectés directement au vérin de la sortie, on n’est pas obligé d’utiliser des vérin symétriques. Autrement dit, ce système n’a nul besoin d’avoir le même débit a` l’entrée et la sortie de la micropompe. Aussi un vérin asymétrique qui a un volume différent de chaque coté peut être aussi bien utilisé qu’un vérin symétrique quelconque. Les différents composants de l’ensemble du système ont été modélisés sous MATLABSimulink ainsi que Simhydraulics. Le modèle sous Simhydraulics permet d’étudier les phénomènes liés à la dynamique de l’huile et la mécanique du système en détail. Ces phénomènes incluent les fuites de la micro-pompe, la variation de la pression dans les différentes parties, etc. Bien que le modèle Simhydraulics permet d’identifier en détail les variables internes du système et de les configurer, il coˆ ute cher en temps de calcul. Cependant, le modèle Simulink est plus rapide et permet de formuler les phénomènes macroscopiques ainsi que les phénomènes microscopiques sous forme mathématique. Ces phénomènes sont basés sur les théories

Résumé

7

mécaniques et hydrauliques, et sur les observation issues du modèle Simhydraulics. Le modèle Simulink est plus rapide, de plus il rend l’exportation et l’intégration aisées dans un modèle de robot complet destiné à la simulation des lois de commande dites ”haut niveau”. Dans le chapitre 3, le rendement mécanique et volumétrique de la pompe sont étudiés. Les fuites dans les pistons de la micro-pompe sont liées avec le jeu radial des pistons par une fonction cubique. Cependant diminuer les jeux nécessitent des procédés de fabrication onéreux. Ainsi, une autre solution pour diminuer les fuites serait d’augmenter le guidage des pistons. Par ailleurs, la perte de l’énergie dans la micro-pompe est liée aux frottements entre les pistons et le rotor. Pour calculer les forces d’interaction tangentielles et normales, l’ensemble des equations dynamiques des pistons ont été développées. Une version de la micro-pompe pour actionner la cheville du robot HYDRO¨ıD est proposée selon les données biomécaniques. L’étude sur les efforts présents dans les pistons (hydraulique, centrifuge, Coriolis, etc.) pour les différentes tailles de pistons et les différentes formes de la tête des pistons a été menée afin de trouver les meilleures dimensions de la micro-pompe ayant un rendement volumétrique et mécanique optimaux, tout en respectant les contraintes d’encombrement. Dans le chapitre 4 le mécanisme du stockage d’énergie est introduit dans l’architecture de l’actionneur. Il s’agit d’une valve trois lignes deux positions qui est placée entre le distributeur passif et le vérin. Quand la valve est activée le vérin est coupé de la pompe et le fluide du réservoir atmosphérique est restocké dans le réservoir sous pression via la micro-pompe. Pour montrer l’intérêt du stockage d’énergie, un monopode est modélisé. Ce monopode est actionné par un ressort linéaire intégré entre la jambe et le pied. Une partie de l’énergie du ressort est perdue lors de l’impact avec le sol du au contact modélisé par un ressort-amortisseur. Donc sans actionnement actif, après quelques sauts, le monopode arrête de sauter. Un actionneur de type IEHA, avec un accumulateur sous pression et un accumulateur atmosphérique sont intégrés en série avec le ressort, pour compenser l’énergie perdue a` l’impact en comprimant le ressort pendant son contact avec le sol. Le saut est une tâche périodique et il y a des phases pendant lesquelles l’actionneur est inactif. Donc avec une stratégie adéquate de stockage, ces phases de repos de l’actionneur sont utilisées pour stocker l’énergie de l’accumulateur atmosphérique

8

Résumé

dans celui sous pression. Avec cette méthode, le monopode peut continuer a` sauter pendant plusieurs secondes, avec des petits accumulateurs. En plus, tant que le volume de fluide est conservé quasi constant dans l’accumulateur sous pression, sa pression est aussi conservée. Ainsi la plus grande partie de l’effort nécessaire pour l’actionnement du monopode est fournie par la pression dans l’accumulateur, exigeant moins d’énergie du moteur électrique de la micro-pompe. On voit bien que ceci augmente le rendement énergétique de l’actionneur. Un deuxième exemple du stockage d’énergie est présenté sur un robot plan bipèd, dont les genoux sont acctionés par deux IEHA. Le genou est bloqué pendant la phase du support, et cette phase est utilisé pour le stockage, et l’énergie stock est utilisé pendant la phase de tranfert o` u l’actionneur de genou est en mode normal. Finalement, dans le chapitre 5, la contribution de ce travail est discutée et les perspectives sont présentées. Les frottements dans les différentes parties de IEHA seront identifiés en combinant la modélisation et les expérimentations sur le système réel. Selon les tâches désignées a` l’actionneur, les matériaux utilisés pour les composants pourront être modifiés afin d’augmenter le temps de réponse ou diminuer les coefficients de frottement. Une fois les paramètres de l’actionneur pour les liaisons du robot choisis, différentes méthodes de contrôle en position/force combinées avec la compliance seront testées sur le modèle développé afin de choisir la meilleure méthode adaptée a` notre système. L’objectif à terme est d’implémenter l’ensemble de ces résultas sur le robot HYDRO¨ıD en cours de développement.

Chapter 1 State of the Art Throughout history, the human body has inspired engineers and scientists. Humanoid robotics is all about creating robots that are directly inspired by human capabilities that have similar kinematics, as well as similar shape and behavior to humans. The motivation behind the existing humanoid robots varies from mechanical workers and assistance to test-beds for neuroscience, entertainment, etc. There is a long history of human looking robots performing human-like movements. Al-Jazari designed a humanoid automaton in the 13th century [7], and Leonardo da Vinci designed a humanoid in the 15th century [8]. Also in Japan, there is a tradition of making mechanical dolls called ”Karakuri ningyo” that dates back to at least the 18th century [9]. In the 20th century animatronics became very popular at theme parks, such as the Pirate’s of the Caribbean ride in Disneyland that opened in the 60s. However, these humanoids moved in an open-loop fashion without sensing their environment [10]. In the second half of the 20th century, thanks to the advances in digital computing, sensing, control, and actuation were included in the robots. Many roboticists developed systems for sensing, locomotion, and manipulation that were inspired by human capabilities. The first humanoid to integrate all these functions was WABOT-1 at Waseda University in Japan in 1973, [11][12]. In the 80s Honda began a confidential project to create a humanoid biped, and in 1996 it revealed the result of this project, Honda P2, [13], which had on-board power and processing, and was capable of stable bipedal walking. In parallel with these developments, 9

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Chapter 1 State of the Art

Cog project began in MIT, leading to an upper body humanoid that was mainly used for cognitive sciences. Ever since many companies and academic researchers have become involved with humanoid robots, and today there are numerous humanoid robots with different features across the world. Humanoid robots vary in form; some have focused only on the head and face, some have a head and two arms built on a stationary or mobile torso, others have an expressive face combined with legs, arms, and a torso. This variation impacts the uses of the robot in terms of mobility, manipulation, whole-body activities, and human-robot interaction [10]. Actuators are one of the main components of any robot, in order to achieve performance like humans(or animals). An ideal actuator must have a high power-to-mass ratio. It should also be energy efficient(including an energy storage mechanism), light-weight and safe in applications involving human-robot interaction. The three main types of actuators are electric, pneumatic, and hydraulic.

1.1

Electromagnetic actuators

Most common actuators used in robotics and other fields are electromagnetic actuators; therefore, they are inexpensive and available in various models and sizes.These actuators produce force and torque from magnetic fields and although there exists several types of them (solenoid, moving coil, etc.), the most common electromagnetic actuators are electric motors. The main inconvenience of this types of actuators is their limited performance; electric actuators are stiff and produce small torques w.r.t their size, which is why they need reduction gears to convert velocity to torque. Reduction gears introduce backlash and friction into the system, decreasing its backdrivability, which can cause safety issues, specially in human robot interaction. A solution to this problem has been associating a serial elastic actuator connected to the motor, as in [14] which uses a linear serial elastic spring between the motor and the output. The ECCEROBOT (Embodied Cognition in a Compliantly Engineered Robot), (Figure 1.1(a)), an EU-funded project, uses screwdriver motors, combined with a shock cord as the elastic component of the muscle[15]. In these systems the robustness is an issue because of complex electric connections. The famous electrically-driven lightweight DLR


11

arm has been made for aerospace application, (Figure 1.1(b)). Flexible harmonic drives have been specifically designed for this robot. Having flexible joints, a fine passivity-based Cartesian impedance controller is used for precise position control of this robotic arm [16] [17].

(a) The ECCE robot, a European Unionfunded project.

(b) Justin: A Humanoid Sporting Two DLRIII Lightweight Arms and Two DLR-II Hands

Figure 1.1: Humanoid robots using electric actuators.

1.2

Pneumatic actuators

The second category of actuators are pneumatic. The most common class of these actuators are pneumatic cylinders. The pneumatic cylindes are intrinsically compliant due to air compressibility, however one disadvantage compared to their hydraulic counterpart is that air, unlike oil, has no self-lubricating property; hence, friction in the piston seals is higher. Another class of pneumatic actuators is the M cKibben muscle, or the P M A(P neumaticM uscleActuator). An artificial actuator was designed to help grasping ability in 1957 at Rancho Los Amigos Hospital in collaboration with Dr. Joseph McKibben, a physicist whose daughter Karan had polio and was paralyzed since 1952, [1], (Figure 1.2). The pneumatically actuated muscle consisted of an inner rubber tube and a outer cord netting. The inner tube was expanding when inflated, and the cord which was inexpandable transferred this change of volume in a change of length. This type of actuator is an

12


inverse bellow, i.e. it contracts on inflation. The compressible air made the muscle elastic and safe for direct interaction with humans. In the 1960s, electrical motors replaced the Mc Kibben muscle as an orthodic device, because of the smaller design and the easier way to control. The interest in artificial muscles became higher when Bridgestone Company redesigned the Mc Kibben muscles and distributed them as Rubbertuators. Construction and materials differ from manufacturer to manufacturer, but the principle has remained the same. Pneumatic actuators have a low bandwidth and have a nonlinear pressure contraction force relation, (due to friction between the outer cord and the inner rubber tube), so higher control methods are needed to operate this type of actuators. Another disadvantage is that they need a costly dedicated compressed air source which makes them loud and noisy[18][19].

(a)

(b)

Figure 1.2: Life Magazine photo: Joseph L. McKibben with his wife, and paralyzed daughter wearing a pneumatic muscle, [1]

1.3

Hydraulic actuators

The third category is hydraulic actuators. Hydraulic actuators were chosen as power sources for earliest industrial robots, because they offer large force capacity, and they have a good power to weight ratio. Until recently, they were hardly used


13

for legged robotics. The most well-know robots of this category are the quadruped BigDog, and its ”big brother” PETMAN, developed by Boston Dynamics for military purposes[20][21]. Each leg in PETMAN has 4 hydraulic actuators that power the joints, plus 5 passive degrees of freedom, (Figure 1.3(a) 1.3(c)). Aerospacequality servovalves are used to regulate the low-friction hydraulic cylinders that actuate the joints in the legs. Another hydraulic quadruped inspired by BigDog is HyQ developed by Semini et al. at iit, (Figure 1.3(b)). This robot uses a combination of hydraulic and electrical motors to actuate the joints [22] [23]. There are a

(a)

(c)

(b)

(d)

(e)

Figure 1.3: Hydraulically actuated robots: (a) BigDog, (b) HyQ, (c) PETMAN, (d) DB, (e) CBi.

number of other interesting hydraulic legged robots, such as the robots produced by the Sarcos company: Sarcos Primus with 12 joints that control hand movement and use pneumatic actuators, and 34 joints actuated by linear hydraulic actuators

14


(a)

(b)

(c)

Figure 1.4: Hydraulic exoskeletons: (a) XOS 2, (b) HULC, (c) BLEEX

[24]. DB, which stands for Dynamic Brain, is a hydraulic anthropomorphic robot designed by the Sarcos company and the Kawato Dynamic Brain Project and was built by Sarcos [25],[26]. The robot is approximately 1.85 meters tall, weighs 80 kg, and contains 25 linear hydraulic actuators and five rotary hydraulic actuators, Figure 1.3(d). The other robot of this category is CB (Computational Brain) developed by Dr. Gordon Cheng and his colleagues,Figure 1.3(e). The biggest improvement of CB over DB is its autonomy. DB is powered by an external hydraulic pump, and its computer system was connected to it by wires. On the contrary, CB carries on-board power supplies and computer system, therefore it can function autonomously [27][28]. In terms of hydraulic exoskeletons, XOS 1 and XOS 2 are two exoskeletons developed by Raytheon for the US Army, Figure 1.4(a). The primary purpose of XOS is for work augmentation. HULC(Human Universal Load Carrier), is another hydraulic exoskeleton for military purposes developed by Lockheed Martin and Berkeley Bionics,[29]. It is a lower body exoskeleton design that the user puts on like a leg brace, extending to the hip, Figure 1.4(b). Berkeley’s lower Extremity Exoskeleton (BLEEX) at the University of Berkeley is a load-carrying, field-operational and energetically autonomous lower extremity exoskeleton, Figure 1.4(c). It consists of two powered anthropomorphic legs, a power unit, and a backpack-like frame on which heavy loads can be mounted, [30].


1.4

15

Actuator comparison

In this section we shall compare the actuation types mentioned above, plus some other types based on different performance criteria, extracted from [2]. Most actuators are characterized by their stress and strain. Stress(σ =

F ) A

is defined as

force(F ) per unit area, and like pressure it has the unit of Pascal[Pa]. Strain(ε = xl ) is displacement(x) divided by the actuator reference length(l), and is unit-less. The maximum stress and maximum strain are two important performance indexes of actuators. Figure 1.5(a) shows the maximum stress versus maximum strain of different classes of actuators. The graph shows that hydraulic and pneumatic actuators have high maximum strains, while in terms of maximum stress, hydraulics is stronger than pneumatics of about two orders of magnitude. In the field of robotics, maximum power density is a very important criteria in the comparison of actuator types, because it expresses the maximum power mass and maximum power volume ratio, Figure 1.5(c) and Figure 1.5(b). As can be seen in these two graphs, hydraulic actuators have the best volumetric power density and mass power density with respect to maximum strain. Thanks to technological advances in the field of electric motors, electric actuators have been the most widely used actuators in the robotic field during the past couple of decades, making them inexpensive and available in different sizes. However, their low power density and efficiency can be a big issue in dynamic robots. As we saw in Figure 1.5, hydraulic actuators have high power density values. The main drawback of hydraulics is that they need a central hydraulic power supply to provide high pressure fluid to all the actuators of the robot, which can be bulky. Another issue of this category of actuators is that the leakage from the hydraulic tube connections that connect the pump to the actuators can cause safety issues, specially in human-robot interaction. An alternative to using external pumps is to use hydrostatic transmission systems. These systems consist of a motor and a pump connected directly to a linear or rotary hydraulic actuator. An early example of this category of actuators for robotic applications was developed by J. Bobrow et al. [3] [31], and is presented in Figure 1.6. The pump can be constant speed with variable displacement, where the speed and the direction of the actuator is controlled by moving the swash plate of the

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pump. Another type of hydrostatic transmission systems have a fixed displacement pump with variable speed. EHA(ElectroHydraulic Actuator), developed by

Figure 1.6: Hydrostatic actuation system consisting of a DC motor, pump, small accumulator, and rotary actuator. Fluid is pumped directly from one side of the rotary pump to the other. (Image from [3]).

[32] is a well-known self-contained hydrostatic actuator that has an individual bi-directional pump that rotates with a variable speed. However, this hydraulic device needs a specifically designed symmetrical actuator, which can be costly. In IEHA (Integrated Electro-Hydraulic Actuator) developed by Alfayad et al. there is no need for a central pressure source and each single actuator is autonomous[5],[6]. Thanks to IEHA’s compactness, it can be placed as near as possible to the hydraulic actuator (rotary or linear). Hence, the pressure drop is reduced and leakage may only exist inside the actuator. Risk of polluting the working environment of the robot is drastically decreased since only four connectors are needed to actuate each arm joint. Therefore, we believe that this type of actuation is a good option for robotic applications. A first prototype of IEHA has been previously fabricated. However, the compactness of the system does not permit to have a detailed study of the internal parameters and phenomena that impact the system behaviour. On the other hand, it is important to determine the influence of each parameter on the performance of the system, as well as its energy efficiency, before modifying these parameters in the future prototypes. In the following chapters the analysis of IEHA through simulation and parameter optimization are detailed.

Chapter 2 Virtual Model

2.1

Introduction

Compactness of IEHA, makes it difficult to measure the internal variables of the system, such as micro-pump’s internal pressure drop and leakage; moreover, it is important to study the effect of configuring the internal variables (e.x. the dimensions of different mechanical parts) on the behavior of the system, before modifying the latter in future prototypes. Therefore, development of a model of the actuator would help us identify the different problems/issues of the system, and propose new solutions to improve its behavior. It also permits to have a virtual model to show externals users how the concept works. The effectiveness of the developed virtual model allows for quick and accurate evaluation of the internal parameters early in the design and development stages.

2.2

The Architecture of IEHA

IEHA is based on the power transmission from an electric motor to a hydraulic actuator. The basic idea consists of converting an electric power to a mechanical one by using a highly integrated micro-pump producing hydraulic energy. The three most important components of this energy converter are 1)the micro-pump; 2) the micro-valve; and 3) the passive distributor. In this section we will have a 19

20


look at the functioning principle of the components of IEHA. Figure 2.1 presents an overall schema of IEHA connected to an output hydraulic cylinder. The first part is an eccentric radial piston pump, composed of 15 or more micro-pistons, depending on the prototype version. The shaft is connected to an electrical motor, and rotates in one direction with constant speed. The flow discharge of the micro-pump is variable and from side A to B, or the inverse. For a given direction of rotation of shaft, fluid enters the pump, and the centrifugal force and hydraulic pressure push the pistons to the walls of the housing during half of the rotation cycle, [0, π], where the piston is connected to the intake port. As the rotor continues around, the vanes sweep the fluid to the opposite side. For the other half of the cycle, [π, 2π], the piston volume decreases and fluid exits the discharge port. Figure 2.2 presents a simplified IEHA model with two pistons. d is the distance between the bottom of chamber and the center of the shaft, while Rc is the radius of the carriage. Hp1 is the length of fluid volume in piston at the bottom of its stroke(piston on the left in figure),(2.1). In the same way, Hp2 is the length of fluid volume at top of its stoke(piston on the right in the figure), (2.2). In order to calculate the variation of the micro-pump stroke, defined by the volume of fluid produced during a rotation for a given eccentricity (E), the distance which the piston travels during half-rotation corresponding to either the intake or discharge is calculated. Hp1 = Rb − E − lp − d

(2.1)

Hp2 = Rb + E − lp − d

(2.2)

Hp2 − Hp1 = 2 × E

(2.3)

The eccentricity of the pump is controlled by tuning the displacement of the carriage, w.r.t the shaft. Moving the carriage from one position to another in between the two extreme values, ±Emax , changes the pump stroke and the flow produced by the micro-pump. The carriage is mobile and has two hydraulic chambers, one on each side. The 4-way, 3-position directional valve on the upper part in Figure 2.1, has the task of directing pressurized fluid to one of the two chambers of the carriage, thus moving the carriage and changing the eccentricity value and direction. An external force(produced by an electric magnet) actuates the spool of this micro-valve, and the value of eccentricity should follow that of the spool


21

displacement. In other words, once the eccentricity reaches the value of the spool displacement, the valve should be closed. In order to do so, the body of the carriage is mechanically connected to the micro-valve. Therefore, if the spool is displaced, say to +∆X, w.r.t the valve, the valve connects P to the left chamber, making the carriage to move to the right. Being connected to the carriage, the valve body also moves to the right, E approaches ∆X, until the carriage reaches the position E = ∆X and the valve closes completely. This mechanism is called a mechanical feedback loop built inside the IEHA and is explained in Figure 2.4. Pump’s output flow is directed to one of the two chambers of the output actuator through the passive distributor, according to the sign of eccentricity. The passive distributor is a 6-way, 3-position valve the working mechanism of which is shown in Figure 2.3. For E = 0, the passive distributor stays closed disconnecting the pump from the reservoirs and the output actuator. If E > 0, the pump intakes fluid from side A, and delivers it to side B. This creates a pressure differential across the spool of passive distributor sliding it to the left. In the left position, the passive distributor connects the supply reservoir(Rp ) to side A of the pump, so that pump aspires fluid from this reservoir. Also side B of pump is connected output actuator side B, while output actuator chamber A discharges into the atmospheric reservoir(Ratm ) and the output piston moves to the left. The scenario is inversed for E < 0 and the output piston moves in the right direction. We see that by using the passive distributor allows us to direct the flow produced by pump to either side of the output actuator without having to change the shaft rotation direction. Another advantage of the passive distributor is that despite of conventional hydrostatic actuators that need a specially made symmetric cylinder([32]), our system can work with asymmetric cylinders. In other words, in our system the in-going and out-going flows of the pump do not have to be the same, because the input and output of the pump are not directly connected to the actuator; the pumps aspires fluid from the supply chamber, discharges to one side of the output, while the other side of the output discharges into the atmospheric reservoir.

22

Chapter 2 Virtual Model T

P

Forcespool

micro-valve

shaft carriage

Pump Port A

Pump Port B Ratm

passive distributor (E>0)

Rp

Pa

Pb

actuator

Position Velocity Figure 2.1: Schematic summary of the working mechanism of IEHA


23

Figure 2.2: Simplified model of the shaft with two pistons turning inside the housing [4].

𝜇𝑃𝑢𝑚𝑝_A

𝜇𝑃𝑢𝑚𝑝_𝐵



Ratm

Rp

E>0

PDspool0



Ratm

Ratm

Rp

E=0

PDspool0

Rp

E 0. For zero eccentricity the volume of the micro-pistons is constant therefore there is no fluid transmission. As the pump eccentricity leaves zero value, it aspires fluid from one side pdA of the butterfly valve, causing a pressure drop(PpdA ↓), and pushes it out to side pdB , causing a pressure increase(PpdB ↑). The pressure difference created between the two sides of the valve, creates a torque(τpd ) that turns it to the left. As the valve turns it connects left channel of the micro-pump (µPA ) to the supply pressure(Rp), pump right channel(µPB ) to the output cylinder chamber cylB , and the cylinder chamber cylA to the atmospheric reservoir(T ). Neglecting friction forces, the dynamic equations ruling the passive distributor dynamics are as follows: τpd = Ipd .θ¨

(2.23)

τpd = Dpd .∆Ppd

(2.24)

, where PpdA − PpdB = ∆Ppd . Ipd and Dpd are passive distributor’s Inertia and displacement, respectively.

Chapter 2 Virtual Model 2.3.3.4

31

Linear Actuator

The dynamic equation of the hydraulic linear actuator is given by (2.25), where Pcyl Scyl and Fext are the hydraulic force and the external force exerted on the cylinder, respectively. Ff r represents all the friction forces including Coulomb and viscous friction. m and Y are the end-effector mass and position, respectively. Pcyl is the pressure difference between the two cylinder chambers A and B, and Scyl is the surface of the linear hydraulic cylinder’s piston. Pcyl Scyl − Fext − Ff r = mY¨

(2.25)

Pressure change due to compressibility and the input flow is defined as: Q = Scyl Y˙ +

νcyl ˙ Pcyl + Qleak β

(2.26)

where Q is the flow from the micro pump, defined by (2.4) or (2.9). Qleak is the internal leakage between the two chambers of the cylinder. νcyl is the volume of the cylinder chamber.

2.3.4

Input-Output Model

In this section, we will find a dynamic equation that best describes the relation between input and output of IEHA, while keeping a neat illustrative form. The ultimate goal of this section is to estimate the model order of the IEHA. The passive distributor is considered to be fast and that it does not effect the dynamics of the system. Since our goal is to find the simplest form of the equation corresponding to the lowest order equivalent of the system, leakage has been neglected in the following developments. Returning to the equations in subsection 2.3.3, if we derive and replace Pcyl from (2.25) into (2.26), and consider the macroscopic value of flow according to (2.9), we will have E in terms of Fext , Y , and their derivatives: E=

Scyl ˙ 1 νcyl 1 1 νcyl 1 ˙ Y + mY (3) + Fext 2N Sp ω 2N Sp ω β Scyl 2N Sp ω β Scyl

(2.27)

32


On the other hand, the force of the fluid pressure in the micro-pump pistons, exerted on the carriage chambers is given by (2.22). In order to simplify, we suppose that we have only two pistons as previously seen in Figure 2.2. Thus, ϕ = π, and (2.22) is simplified to: Fp = (Prp − Pcyl ) Sp cos (θ)

(2.28)

By replacing Pcyl in (2.28) from (2.25), the micro-pump’s force can also be written in terms of the output position, the external force, and their derivatives: 1 ¨ mY + Fext Sp cos (θ) Fp = Prp − Scyl

(2.29)

The pressure difference in carriage chambers can be written in terms of E¨ and Fp according to (2.13): ξ=

1 mcarr E¨ − Fp 2Scarr

(2.30)

Deriving E¨ from (2.27), ξ can also be written in terms of Y and Fext : ξ=

1 2Scarr

mcarr Scyl (3) Y 2N Sp ω

− Prp −

1

Scyl

+

mcarr νcyl 1 mY (5) 2N Sp ω β Scyl

mY¨ + Fext

Sp cos (ωt)

+

(3) mcarr νcyl 1 F 2N Sp ω β Scyl ext

(2.31)

, where Y (3) and Y (5) are, respectively, the payload jerk and the second derivative (3)

of the jerk. Fext is the third time derivative of the external force. Replacing the Bernoulli equation of flow of the servo-valve (2.10) in the dynamic equation of (2.14), we can write an expression which relates the input variable, X, to the internal variables E and ξ : r r 2 Ps νcarr ˙ C2πrtig (X − E) − ξ = Scarr E˙ + ξ % 2 β

(2.32)

Replacing E and ξ from (2.27), and (2.31) in the latter, we will obtain a nonlinear equation that represents the relationship between the system input, servovalve position X, and the measurable values at the end-effector, Y , Fext , and their derivatives, as shown in (2.33). This is a nonlinear equation of order 6, the coefficients of which, depend on the dimensions of the actuator, the properties of


33

the supply pressure and the fluid((2.3.4)). X =A+

B C

(2.33)

, where: A=(K8 Y˙ +K9m Y (3) +K9 F˙ext ) (4) B =K6 (K8 Y¨ +K9m Y (4) +K9 Fëxt )+K7 (K10 Y (4) +K11m Y (6) +K11 Fext )

−(K12 +K13m Y¨ +K13 Fext )sin(ωt)+K14m cos(ωt)Y (3) +K14 cos(ωt)F˙ext

C =K5

K1 = − K3 =

√

mcarr Scyl 2Ps Scarr Sp N ω

Sp Prp Scarr Ps s

K5 = C2πrtig K8 = K10 =

Scyl 2Sp N ω Scyl mcarr 2Sp N ω

K12 = −Prp Sp ω K14m =

2.4

(3) 1+K1 Y (3) +K2m Y (5) +K2 Fext +K3 cos(ωt)+K4m Y¨ cos(ωt)+K4 Fext cos(ωt)

Sp m Scyl

Ps β %

K2m = −

mcarr νcyl m 2Ps Scarr βScyl Sp N ω

K2 = −

mcarr νcyl 2Ps Scarr βScyl Sp N ω

(2.34)

K4m = −

Sp m Scarr Scyl Ps

K4 = −

Sp Scarr Scyl Ps

(2.35)

K6 = Scarr β

K7 =

νcarr 2Scarr

(2.36)

K9m =

νcyl m 2βScyl Sp N ω

K9 =

νcyl 2βScyl Sp N ω

(2.37)

K11 =

νcyl mcarr 2βScyl Sp N ω

K11m =

νcyl mcarr m 2βScyl Sp N ω

(2.38)

Sp ω Scyl

(2.39)

K13m = K14 =

Sp ωm Scyl Sp Scyl

K13 =

(2.40)

Simulink model

As a first attempt, we used MATLAB-Simulink to model the different stages of IEHA, using the dynamic equations of the system as well as fluid properties such as bulk modulus and internal leakage, as is illustrated in Figure 2.6. The two chambers of the cylinder are modeled separately, according to equation (2.26), as shown in Figure 2.7. The passive distributor is modeled as three switches driven by the sign of eccentricity(E), shown in dark green: One switch for each chamber of the output, connecting the latter to the flow from pump or to the atmospheric reservoir. A third switch that detects which cylinder chamber is receiving the

34


pump flow and connects the pressure of that chamber to the micro-pump. If E > 0, it means that chamber B is connected to the micro-pump’s output flow, QµP ump , while chamber A is connected to the atmosphere reservoir and its flow follows the Bernoulli equation (2.41). rcyl is the radius of the orifice connecting the output cylinder to IEHA. Inter-chamber leakage(Qf 12 ) depends on the pressure differential across the cylinder(Pa − Pb ). The pressure in each chamber of cylinder varies according to (2.26). The forces exerted on the output piston rod are the force resulting from the difference of hydraulic pressure in the two chambers, and an external force. Qcylλ =

2 Cπrcyl

r

2p Pcylλ − Patm %

(2.41)

The most import part of the system in Figure 2.6 is the micro-pump consisting of N micro-pistons, modeled as shown in Figure 2.8. The electric motor has been considered ideal, turning at speed ω, and is not modeled in the actual simulation. The embedded function in Figure 2.8 calculates the length and the speed of variation of oil volume in the chamber according to (2.5) (2.6) (2.7). Each piston is connected to supply pressure (Rp ) half of its cycle, where it aspires fluid, and its volume increases(H˙p > 0). The other half of the cycle the piston is connected to i

the output cylinder chamber(Pcyl ), where it pushes the fluid out to the cylinder and its volume decreases(H˙p < 0). As we saw in (2.4), the flow in the radial vane i

pump stays geometric, as is modeled in the bottom right part of Figure 2.8. The piston is only connected to the output cylinder when it is ejecting fluid and that its volume is decreasing, which is why a limiter has been used before the output port QPi . The upper part of the system consists of the micro-valve, and the two chambers on the sides of the carriage. The modeling of this part is very similar to the output cylinder in Figure 2.7, only that the input flow depends on the opening of the valve (X − E) as seen in (2.10). There is also no internal leakage in this part, since the two fluid chambers are on the two sides of the carriage and not connected. The input command, X, decides how much the servo-valve should be opened, and as the pressure in the chamber increases the carriage moves and E changes. The output of this block is the value of eccentricity and its variation, which are used to determine the pump flow and its direction(passive distributor) as described above.


35

P

1 Force

Servo & Carriage

du/dt

microPump

mPompe_A

mPompe_B

Ratm

P

Cyl_B

Cyl_A

Accumulator

Rp

Passive Distributor 2 External Force

Actuator

1 Position

2 Velocity

3 Pressure

Figure 2.6: The 4-stage IEHA and the output cylinder

2.5

Simulink Simulation Results

As a test for our system the values have been chosen according to Table 2.1. The actuator parameters are chosen for the toes of HYDRO¨ıD, [34]. An impulse force is applied to the micro-valve spool, moving it to −95% of Emax . The carriage

36


P

3

1

cyl

Qout

−1

Q

[dY]

[Y]

Y

Chamber Dynamics B Pa

P2

Chamber Dynamics A

Q

Qf12

−1

Qf12

dY

−K−

dY

Internal Leak Coeff.

[dY]

Y

Q

PD line: ChamberB to μPump/Ratm

[Pb]

[Pa]

[Y]

Qin

Qout

switch2

case

Q to atm

Pb

Pb

Pa

PD: Press to A/B μPump

Qout

case

Side to Pump

A/B

Q

Qin

case

Q to atm PD line: ChamberA to μPump/Ratm

E

Pump

Qout

Qμ

Pa

2

Scyl_A

Scyl_B

GotoA [Pa]

GotoB [Pb]

3 Fext

Piston Rod

[Y]

[dY]

1

Y

2

dY

Figure 2.7: Output cylinder and the passive distributor model in Simlink.

response time is fast (< 1ms), see Figure 2.9(a). In order to illustrate the pressure variation in the micro-pistons, we have shown the variation of pressure in the first and the forth micro-pistons, that have a phase difference of ∆φ = 4 ×

2π . 10

When a micro-piston is connected to the reservoir Rp its pressure decreases as it aspires fluid, and when it is connected to the output cylinder, as it pushes out its fluid into the cylinder, the pressure increases, (Figure 2.9(c)). The variations in piston pressures are linked to the oscillation of flow. Looking at pump flow curve (Figure 2.9(d)), we see that it oscillates with a frequency of 10ωHz. Number 10 corresponds to the radial distance between the pistons on the shaft( 2π ). There 10 is also a peak in the beginning, which corresponds to the peak in δE as shown in Figure 2.9(b). The faster the eccentricity changes the bigger this peak in flow


37

switch Rp

deltaP1 deltaP

dHpi

f(u) Qin/out

Q

case

Qf signQ deltaP2

1

2

P

dHpi

Sp

dVol

Hpi

Sp

Vol

PPi

cyl -K-

Patm

Piston Chamber Compression

Leakage Coeff

Sp

Sp W

W

dHpi t

E

Hpi

Rb

Rb

2

Hpi

Clock 3 dE

E

fcn

Sp

1 0 to -inf

dHpi

dHPi

Q

Pi

dE

Embedded MATLAB Function

Figure 2.8: Flow and pressure variation of the micro-piston chamber.

is. As for the output cylinder, in the beginning the pressure across the cylinder increases due to fluid compressibility, and as the load starts to move the pressure difference across the cylinder drops and the velocity reaches the stable value of 2.2 cm . Note that at steady speed, although the hydraulic force of the output drops s to zero(2.9(f)), the pressure difference across the output stays positive(' 0.45bar). This is because of the asymmetry of the output piston-ScylA is almost half ScylB . In these results the output flow of the pump seems to immediately increase the pressure force in the output cylinder and make the piston velocity increase to a stable value. This is because the dynamics of the passive distributor are not taken into account in this model in Simulink. In the following section the modelling of the system in SimHydraulics is presented. This model includes the passive distributor as a rotary valve, as well as a more detailed model of the pump taking into account the pump chambers and the pressure variation due to centrifugal forces.

38

Chapter 2 Virtual Model Parameter Ps rtig Mcarr Scarr Emax Vcarr β C % ω Prp N rpi Mp i Spi Lpi Rb d

m ScylA ScylB Strokecyl

Physical quantity Supply pressure of microvalve Micro-valve radius Carriage mass Carriage active surface Maximum eccentricity Chamber volume of carriage Bulk modulus Vena contracta coefficient Fluid density Electric motor rotational speed Supply pressure of micro-pump Number of pistons Micro-pump in-out radius Micro-piston mass Micro-piston surface Micro-piston length Interior ring radius distance between chamber bottom & shaft center Load mass Actuator piston surface side A Actuator piston surface side B Actuator piston stroke

Value 10bar 0, 25 cm 0, 091 kg 1, 644 cm2 0, 05 cm 0, 0822 cm3 800 M P a 0, 62 850 kg/m3 3000 rpm 1M P a 20 0, 15 cm 0, 45 gr 0, 197 cm2 0, 27 cm 0, 3 cm 0, 4 cm

10 kg 4.7124 cm2 9.6211 cm2 58 cm

Table 2.1: Value of parameters used in the virtual model of IEHA in Simulink.

2.6

SimHydraulics Model

IEHA has been simulated in four separate stages, Figure 2.6. These four stages are the micro-valve and the carriage eccentricity, micro-pump, passive distributor, and the output actuator.


39 2

0 X E −0.005

0

0 −0.01 −0.01

−2

−0.02 −0.03

−0.02

Δ E(cm/s)

X&E(cm)

−0.015

−0.04 −0.025

−0.05

0

0.005

0.01

0.015

0.02

−0.03

2 −4

0 −2 −4

−6

−6 −8

−0.035

−10

−8

−12 −0.04

0

0.005

0.01

0.015

0.02

−10 −0.045

−0.05

0

0.05

0.1

−12

0.15

0

0.02

0.04

0.06

time(s)

0.08

0.1 time(s)

0.12

0.14

0.16

0.18

0.2

˙ (b) Variation of the carrirage eccentricity, E.

(a) Displacement X of micro-valve, and carriage eccentricity E which follows X. 2.2

1.4 1.3 1.2

2

1.2

1.1 1 1.8

0.9 1

0.7

1.2

0.025

0.03

0.035

1.5

0.04

Q(l/min)

4

0.02 1.4

1

Pp & Pp (bar)

0.8 1.6

0.8 1 0.6 0.5

1

0.4 0

0

0.005

0.01

0.015

0.02

0.8 0.2 0.6 0

0.05

0.1

0

0.15

0

0.05

time(s)

1.4

60

1.2

50

1

40

0.8

0.6

30

20

0.4

10

0.2

0

0

0.05

0.15

(d) Flow produced by the rotary pump, Qmic .

Forcecyl(N)

Δ P(bar)

(c) Pressure variation in µ pistons Pp1 (blue) & Pp4 (dashed red)

0

0.1 time(s)

0.1

−10

0.15

0

0.05

time(s)

0.1

0.15

time(s)

(e) Pressure difference in the chambers of the output cylinder, PcylA − PcylB .

(f) Hydraulic force in the output actuator.

2.5

0.3

0.25

2

0.2

Pos(cm)

Vel(cm/s)

1.5

1

0.15

0.1

0.5 0.05

0

−0.5

0

0

0.05

0.1 time(s)

(g) Speed of the output piston, Y˙ .

0.15

−0.05

0

0.05

0.1 time(s)

(h) Position of the output piston, Y .

Figure 2.9: Internal variables of IEHA model in Simulink for a step input of X = −0.95 × Emax .

0.15

40

2.6.1


Servo-Valve & Carriage

The first stage consists of the carriage chambers and the servovalve commanding the eccentricity (Figure 2.10(a)). A control/external force moves the servovalve spool (Figure 2.10(c)). The latter is a mass (MSpool) the displacement of which is limited to maximum eccentricity value(±Emax ) by a translational hardstop. The carriage two chambers are simulated as a double-acting cylinder, and the mass of the piston is equal to the mass of the entire carriage (Figure 2.10(d)). The mechanical connection between the carriage and the servovalve body is simulated by connecting the spool reference to the carriage piston rod (physical connection Spool − Carriage). The servo-valve is modeled as four variable orifices (Figure 2.10(b)). Ports A and B are connected to the two chambers of the carriage, respectively. P is the connection to the supply reservoir, and T to the return one. The opening of each orifice depends on the sign and value of Spool displacement. For a positive displacement of the servo spool(P osSpool > 0), orifices P A and BT are opened, so carriage chamber A is connected to P , and its volume increases, increasing the pump eccentricity(E ↑).

2.6.2

The Micro Pump

The second and most complicated stage of the system is the micro pump, Figure 2.11, which is modeled as a combination of the rotating micro pistons. Each piston is modeled as a single-acting cylinder connected to two variable orifices connected to sides A and B of the passive distributor, Figure 2.12. These two variable orifices are used to model the connection between the piston and the intake and output chambers as shown in Figure 2.13(a). A and B side orifices open with positive and negative signals, respectively. However, there is an inter-chamber gap, noted by its corresponding angle, θg . The input orifice of each piston has a diameter of DpO and switches between the two channels passing by the inter-chamber gaps. When the piston orifice(pO) is against a gap, both opening signals are zero. As the rotation continues, piston orifice connects to one of the two channels gradually until it is completely connected to that channel. In Figure 2.13(b), in (i) position


41

Force Palette

1 Force

T A B

P

Q Sensor

A

2 P

2 T

B

1 P

Spool−Carriage

P_A

2 Edot

A

B

A

B

A_T

A

5 B

A

Vel_Carr

4 A

S

1 E

S

Pos_Carr A

B

P_B

B

Carriage

A

S

A

S

1 Orifice Opening Signal (Spool Position)

Servo Valve and Spool Actuator

B

B

B_T

Spool−Carriage B T 3

(a) Servovalve and Carriage

1 Force Palette

SPS

R

Ideal Force Source

C

Translational Damper

S C

(b) Servovalve

A

B

1

2

R R MSpool

C

Spool−Carriage C V P Position Sensor R

(c) Spool model

A

B

C

R

Pos Sensor Pos_Ecc

Translational Hard Stop

R

Vel_Ecc

Carriage Chambers

1 Pos_Carr 2 Vel_Carr

2 Pos Spool 1

Carriage Mass

3 Spool−Carriage

(d) Carriage

Figure 2.10: Simscape model of micro-valve and carriage hydraulic chambers.

the piston orifice is completely closed and starts to connect to the upper channel. In (ii) it is fully open and starting to close down. In (iii) it is against the wall and completely closed. In (iv) it is partially open to the lower channel and continues to open completely. In (v) it is completely open to the lower channel, and in (vi) it has just closed completely. In order to define the opening signals, lets see the opening as a function of the rotation angle for half a cylce (θ ∈ [0, π]), as described in (2.42). 0 and 1 mean fully closed and fully open, respectively. As for the model in Simhydraulics, the two variable orifices described above, each represents the connection of the piston orifice to one of the two chambers. One is chosen to open with positive signals, (OpeningA in (2.45)); while the other one opens with

42


negative signals,(OpeningB in (2.46)).   Opening = 0 if 0 6 θ < θk       θk 6 θ < θf  0 < Opening < 1 Opening = 1 if θf 6 θ < π − θf     0 < Opening < 1 if π − θf 6 θ < π − θk      Opening = 0 π − θk 6 θ < π

(2.42)

, with: θg + θpO 2 θg − θpO θk = 2

(2.43)

θf =

openingA =

     1

(2.44)

if θf < sin θpi (2.45)

]0, 1[ if θk < sin θpi < θf     0 else

   if sin θpi < −θf   −1 openingB = ]−1, 0[ if − θf < sin θpi < −θk     0 else

(2.46)

, where θpi = ωt + φpi , and φpi is the initial phase of piston pi . The forces on the pistons are considered in their translational axes. All the pistons slide inside the carriage wall, and have a sinusoidal translational movement: xpi = −Ecosθi

(2.47)

Hence, from newton’s second law the net force on the piston should be equal to: X

X

Fpi = mp · x¨pi

˙ ¨ Fpi = mp · Eω cos(θi ) + 2Eωsin(θ i ) − Ecos(θi ) 2

(2.48)

(2.49)


43

A B

2

B

A

A

B A

B

1

mPompeA

mPompeB

3 2

Eddot

Edot

E

1

E

Edot

A B

A B

E

P_B

Eddot

Edot Eddot

P_B

time

time

P_A

P_A

mPiston1

Clock

mPiston0

E

Edot P_B

Eddot

A B

A B

E

Edot Eddot

P_B

t

t

P_A

P_A

mPiston3

mPiston2

Figure 2.11: Micro-pump modeled as a combination of micro-pistons. P Sensor A

P

A

B

Annular Leakage

C

Ideal Force Source R

R

C

P pS

S

p

A

piston mass(m ) p

microPiston

[Opening_A]

m

x’’ p p PSS

S

SPS

piston section

B

A

Variable OrificeA

B

PSS

S A

2 Edot 3 Eddot

f(u)

Σ F

[Opening_B]

p

Variable OrificeB [Opening_A]

f(u) A Opening fcn

W 4 time

Rotor Speed 0 Piston#

B

B

1 E

1 P_A

A

Rotating PipeB

W

A W

Rotating PipeA

[Opening_B]

f(u) B Opening fcn

2 P_B W

Figure 2.12: A piston of the micro-pump.

For so, we apply an opposing force to the piston that is an image of the interaction force of the carriage wall on the piston. This force cancels out the hydraulic pressure force Ppi Sp and guarantees the sinusoidal movement of the piston as

44


(ii) (i)

 pO 2 

(iii)

g 2

(vi)

(iv) (v) (a) The intake and output chambers of the micropump, and the gap between(θg ).

(b) Different connections of the piston orifice to the intake/output channel.

Figure 2.13: Schematic model of micro-pump’s input/output channels, and the connection of a piston input orifice to these two.

defined in (2.47): X

Fpi = Finteraction + Ppi Sp

(2.50)

Finteraction = −Ppi Sp + mp · x¨pn

(2.51)

The pressure increase due to the rotation of the vane, is also calculated and accounted for with two rotating pipe elements of SimHydraulic library, inserted before the two variable orifices described above, as seen in Figure 2.12. Figure 2.14 shows a simplified schema of the rotating shaft with two fluid lines going from the two micro-pump in/out chambers to the micro-pistons. Because of rotational forces, the pressure at the port of each piston slot is bigger than the pressure in the micro-pump chamber. To formulate this pressure increase due to the centrifugal force, we calculate this force for each layer of fluid, x1 ..xn , at the slot entrance. For one layer of fluid, say xi , of section A, radius xi , and length ∂xi , the mass of fluid is: ∂mi = ρA∂xi

(2.52)


45

  rn r1

Micro-Pump Chamber A

x1

xi

xn

Micro-Pump Chamber B

Figure 2.14: Intake and output channels drilled in the micro-pump shaft

, where ρ is the density of the fluid. So the centrifugal force for the layer can be written as: ∂Fcentrif uge = ∂mi ω 2 xi = ρA∂xi ω 2 xi = ρAω 2 xi ∂xi

(2.53)

The force of fluid at the entrace of the slot is a combination of all the layers from r1 to rn : Z

rn

ρAω 2 xi dxi r1 Z rn 2 = ρAω xi .dxi

Fcentrif ugetotal =

r1

(2.54)

46


Pressure being the force per unit area, we find the pressure increase at the input of the piston slot due to rotation as: Pcentrif uge =

Fcentrif uge ZA

(2.55)

rn

= ρω 2

xi .dxi

(2.56)

r1

=

ρω 2 2 (rn − r12 ) 2

(2.57)

It is important to mention that this pressure increase in our actual prototype of IEHA is about 0.02bar and rather negligible. This is because the mass of fluid rotating around the axis is very small- rn − r1 = 1mm -, however this force could be non-negligible in future bigger prototypes. The last element of the micro-piston model in Figure 2.12 is the leakage. Leakage in the micro-piston is modeled as an annular orifice, resulting from the clearance between the piston and its slot. We will give a detailed study of these leaks in chapter 3.

2.6.3

Passive distributor

The passive distributor has the task of connecting the micro-pump’s lines to the output actuator and the supply and return reservoirs. The distributor can be linear or rotary. A rotary distributor is more size efficient than a linear one. However it requires higher design and assembly precision, and could be more costly. A rotary distributor consists of two concentric parts: a fixed rotating part including the input and output connections of the valve, and a rotating butterfly valve that opens orifices connecting the input/output lines as it rotates. If the valve is not well mounted on the fixed part, i.e. it is eccentric, it could drastically increase the inter-chamber leakage, as well as the friction. Therefore, if we do not have space limitations in the designated application of IEHA, it would be better to use a linear distributor, while in other application a rotary distributor is the better option. In the following we will show the modeling of both linear and rotary distributors.

Chapter 2 Virtual Model 2.6.3.1

47

Linear passive distributor

As a first approach to the modeling of the passive distributor, it was modeled to be linear, with the same characteristics as the rotary distributor. The linear passive distributor is modeled as a combination of a hydraulic cylinder and six variable orifices, corresponding to the moving spool of the distributor and a 6way -3position servo-valve, respectively (Figure 2.15). The six variable orifices represent the 6 possible fluid paths in the 6-way valve: • One from the source reservoir Rp to the micro-pump side A, one to side B: Rp mP A, Rp mP B • One from micro-pump side A/B to output cylinder chamber A/B: mP A CylA/mP B CylB • One from cylinder chamber A and one from chamber B to the atmospheric reservoir, Ratm : Cyl AT ,Cyl BT The opening of the each orifice depends on the sign and value of the opening signal. The opening signal is the position of the distributor spool. The two sides of the distributor spool (the cylinder chambers) are connected to the micro pump. Depending on the pressure difference on the two sides of the distributor spool, it slips to one side or the other, opening three of the six variable orifices. This will connect the micropump, output cylinder, and reservoirs Rp and Ratm , as shown in the hydraulic schema Figure 2.3. The stroke of the distributor spool is equivalent to the maximum opening of the valve orifices. The opening of the latter is linearly proportional to the position of the valve, i.e. the sections of the orifices are considered to be rectangular, which results in: Orif ice Opening = Orif ice AreaM ax ×

Spool Displacement Spool Stroke

(2.58)

As we described above, the movement of the passive distributor spool(linear or rotary) depends on the pressure differential on its two sides. If we neglect the pressure loss along tubes and lines, Figure 2.16 shows the pressure in different parts of the system, for a positive input signal,i.e. a positive value of eccentricty(see

48

Chapter 2 Virtual Model Passive Distributor Spool Pos Spool DistributorB DistributorA

1 mPompe_B

1

A

A

2 mPompe_A

Pos Spool

Pos Spool

R

Pos Sensor

Passive distributor B mPompe_B

B

A

A

B A internal leakage

B

Rp

Cyl_B

Cyl_A

6 Rp

R

A

MDistributor

6way Valve

4 Cyl_A

C

Cyl_B 5

2

B

Ratm

mPompe_A

S(t)

B

3 Ratm

DistributorB

DistributorA

(a) Spool and valve of the distributor.

3

(b) Distributor spool.

5 Rp

Rp_mPA

A

B

Rp_mPB

B

−

S

S

A

1 S(t)

+ 7 mPompe_B

A

mPA_CylB

B

mPA_CylA

+

B

A

P Sensor1

S

S

A

3 mPompe_A

−

A

A

CylB_T

B

CylA_T

B

−

S

6 Cyl_B

S

2 Cyl_A

A

P Sensor2

+

Ratm 4

(c) Distributor 6-way valve.

Figure 2.15: SimHydraulic model of the passive distributor

Figure 2.3. The pressure difference across the distributor is negative, ∆PP D = PRp −(PRp +∆PµP ump ) < 0, so the distributor spool moves to the left. The pressure supplied to the output actuator is the sum of PRp , and the pressure differential produced by the micro-pump∆PµP ump . By laws of energy conservation, working pressure demanded at the output actuator should always be superior to the supply pressure, otherwise the passive distributor will not move in the correct direction. This would make sense because we cannot produce negative energy with the micropump(The micro-pump is an energy ”source”, not a consumer)! This can also be


49

seen in the movement of the passive distributor spool: In the case of E > 0, the passive distributor spool should move to the left, therefore, the pressure on the right side of the distributor should be bigger that the pressure on the left side (PRp + ∆P > PRp ), otherwise the passive distributor will not move to its correct position(−stroke), and the produced flow by the pump will be lost in moving the passive distributor; in other words the produced energy by the system will be lost. This is why when the demanded output pressure is unknown, it is better to keep the reservoir pressure to the minimum (PRp = atm) to guarantee a correct displacement of the passive distributor.

 P Pump

PRP  PPump

PRP

Ratm

PRP

PRP  PPump

Rp

PcylA  atm

Pcyl B  PRP  PPump

Figure 2.16: Pressure in the different lines of the system for a positive microvalve displacement, X, E > 0.

2.6.3.2

Rotary passive distributor

In this version of the passive distributor, it is considered to be a butterfly valve which is the case in the current prototype of IEHA, Figure 2.17(a). The spool is modeled as a rotating hydraulic actuator with a shaft stroke of ±22.5◦ . The working mechanism is as explained in 2.6.3.1, only that the orifice area is no longer linear. The opening of each orifice is a function of the spool rotary displacement,

50


where the maximum displacement(stroke) is ±22.5◦ . The opening is the overlap of \ the circular section with the oblong hole connected to the fluid port, area ACBD \ is twice in Figure 2.17(b). The valve displacement is CD, and the area of ACBD \ as it is the intersection of two circules with the the area of the segment AHBD, \ same radius. This area is a a part of the circle segment AOBD: 4

\ area = AOBD \ area − AOB area AHBD

(2.59)

Having the valve displacement CD = 2 × DH, and the radius of the orifice r, we can calculate θ as: θ = 2 × acos(

r − DH OH ) = 2 × acos( ) r r

(2.60)

\ and the Which can be used to calculate the area of the circle segment AOBD, 4

triangle AOB , given in (2.61) and (2.62). \ area = θ × πr2 AOBD 2π

(2.61)

θ θ OH = r cos( ), AB = 2 × r sin( ) 2 2 4 OH × AB θ θ r2 AOB area = = r sin( ).r cos( ) = sin(θ) 2 2 2 2

(2.62)

Finally, using the equations above, we can calculate the area of the overlap: 4

\ area = 2 × (AOBD \ area − AOB area ) ACBD

(2.63)

\ area = r2 (θ − sin(θ)) ACBD

(2.64)

Figure 2.18 shows the nonlinear opening of the valve w.r.t spool displacement. The valve displacement goes from −stroke = −22.5 deg to stroke = 22.5 deg. We see that the valve opening speed around zero is small and it increases as the valve displacement moves away from the zero position and approaches its maximum stroke(±OpeningM ax ). Depending on the opening speed defined by the application this could slow that the passive distributor’s response time. A solution could be to have rectangular linear orifices, as explained in the linear modeling of the passive distributor, as in section 2.6.3.1. This would make the opening section linearly


51

proportional to the distributor displacement, (2.65). If for example we would like the movement of the spool to be fast at zero and damped at the end of its stroke, we can choose triangular orifices, given by equations (2.66) to (2.68). However, this could come at the price of making the mechanical design more complicated. See Figure 2.19 for the opening curves of rectangular and triangular orifices. h CDEFarea = ABCDarea AD

(2.65)

ADEarea AT AH − h = = ABCarea AH AH

(2.66)

BCDEarea = ABCarea − ADEarea 2 BCDEarea h =1− 1− ABCarea AH

(2.67) (2.68)

A r

O  C

H

D

r B (a) CATIA model of passive distributor

(b) Overlap of the butterfly valve and the circular orifices.

Figure 2.17: Circular overlap of orifices in the rotary passive distributor.

2.7

SimHydraulics Simulation Results

In this section we show some of the simulation results obtained by the model in section 2.6, and the component dimensions and supply pressure described in table 2.1. Here we have to mention that since the passive distributor dynamics are taken into account, for a system with these characteristics (input, load and no friction output actuator), PRp has been chosen as atmospheric, as described in section 2.6.3.1. In

52

Chapter 2 Virtual Model 1

0.8

Opening / Opening

Max

0.6

0.4

0.2

0

−0.2

−0.4

−0.6

−0.8

−1

−20

−15

−10

−5

0

5

10

15

20

Spool Displacement[°]

2.18: Nonlinear opening of the butterfly valve orifices w.r.t spool displacement. Valve shaft stroke ∈ [−22.5◦ , 22.5◦ ]

Figure

1 Rectangular Orifice Triangular Orifice

0.8

0.6

A E

D

E

B F

T h

h

Opening / OpeningMax

0.4

A

0.2

0

−0.2

−0.4

B

H

C

D

C −0.6

−0.8

−1

(a)

−20

−15

−10

−5 0 5 Spool Displacement[°]

10

15

20

(b)

Figure 2.19: Opening of triangular and rectangular orifices w.r.t valve displacement(h). Triangular orifice: Maximum opening section=ABC. Actual opening section=BCDE. Rectangular orifice: Maximum opening section=ABCD. Actual opening section=CDEF.

the first example, the load parameters are the same as the results presented in the Simulink section. The results are shown in Figure 2.20. The input force to the micro-valve spool is the same as in section 2.5. This moves the servo-valve spool to X = −0.95Emax , Figure 2.20(a). As the carriage eccentricity increases, the pump output flow increases until it reaches its stable value. The pump transfers flow from side B to A, causing the pressure on side B to drop, which makes the passive distributor rotate to the right until it reaches the positive stroke, Figure 2.20(c). The movement of the passive distributor opens the channels between the reservoirs, the micro-pump, and the output actuator. The micro-pump output flow(Figure 2.20(b)) is sent to side A of the output actuator(Figure 2.20(d)), and side B of the output actuator is connected to the atmospheric reservoir (Ratm ).


53

We can see that there is a difference in flow from the pump and the flow to the actuator up to t = 20ms. During this time the flow has been used to move the passive distributor to its end of stroke position. Once the passive distributor reaches its stroke, all the pump flow (minus leaks) is directed to the cylinder chamber A. The effect of this variation of flow at t = 20ms can also be seen in the velocity, pressure, and hydraulic force of the output, (Figure 2.20(g),2.20(e), 2.20(f)). To have a look at what actually goes on inside the micro-pump, the dynamics of the first micro-piston are shown in Figure 2.21. When the carriage becomes eccentric (E 6= 0), the micro-piston starts to move in its slot. The reference of the piston position has been fixed to its initial position in the sleeve, i.e. when xp = −Emax , the piston chamber is empty, and when xp = Emax it is completely filled. The piston is connected to channelA/B for positive/negative values of opening. For half a cycle, the piston is connected to channelB , fluid is sucked into it(QB ) and it is moved away from the shaft(xp ↑), until it reaches the gap between the two channels. In the second half of the cycle, the piston approaches the shaft(xp ↓), discharging into side A(QA ). We see that the inter-channel gaps barely have an effect on the pressure variation of the piston. This is because the gap diameter has been set equal to piston diameter(θg = θpO ),( seeFigure 2.13(a)). This means that the piston disconnection from both chambers is instantaneous. In the beginning of this phase as the piston is still partially against the gap, its pressure has a slight drop, as seen for example at t=0.03s in the figure. As it gets closer to the shaft it pushes out the fluid into channelA , (QA ). At the end of the cycle the piston reaches the gap between the two channels, and so on continues the cycle. Looking at the piston pressure(Pp0 ), we see that when the piston is aspiring fluid its pressure drops slightly below Rp = 1atm, this is due to the losses along the lines and orifices along the way from the reservoir to the piston input. When it is connected to the output cylinder its pressure increases to almost output cylinder pressure PcylA . The pressures and flows in Figure 2.20 and Figure 2.21 oscillate at a frequency of 10ωHz, same as previously seen in section 2.5. In another example the load mass here has been set to 1kg, and the output is , pushing against a wall through a spring-damper. The spring constant is 5000 N m and the damping is 20 N m , Figure 2.22. The flow to the output actuator, makes s

the hydraulic force increase, compressing the spring at the end-effector. Looking

54


0 X E −0.005

1

0 −0.01 −0.01 −0.015

−0.02

(l/min) 0

0.002 0.004 0.006 0.008 0.01 0.012

0.6

Q

−0.05

μ Pump

−0.04 −0.025

B

−0.03

−0.02

X&E(cm)

0.8

−0.03

0.4

−0.035

0.2 −0.04

−0.045

−0.05

0

0

0.05

0.1

0

0.15

0.05

0.1

0.15

time(s)

time(s)

(a) Displacement X of micro-valve, and carriage eccentricity E which follows X.

(b) Flow produced by the µP ump

1

0.9 1

1 0.8 0.8 0.6

0.8

0.4 0.2

0.5

0

0.01

0.02

0.6

cyl

0

A

0.6

Q (l/min)

displacementPD strokePD

0.7

0.03

0.4

0.4

0.3 0.2

0.2

0.1 0

0 0

0.05

0.1

0.15

0

0.05

time(s)

0.1

0.15

time(s)

(c) Passive distributor rotation.

(d) Flow into output actuator(cylA ).

0.8

30

0.7

25

20

0.6

15 0.5

Forcecyl(N)

Δ P(bar)

10 0.4

0.3

5

0 0.2 −5 0.1

−10

0

−0.1

−15

0

0.05

0.1

−20

0.15

0

0.05

time(s)

(e) Pressure difference in actuator(PcylA − PcylB ).

the

0.1

0.15

time(s)

chambers

of

output

(f) Hydraulic force in output actuator.

2.5

0.35

0.3 2 0.25 1.5

Pos(cm)

Vel(cm/s)

0.2

1

0.15

0.1 0.5 0.05 0 0

−0.5

0

0.05

0.1 time(s)

(g) Velocity of output actuator rod.

0.15

−0.05

0

0.05

0.1

0.15

time(s)

(h) Position of output actuator rod.

Figure 2.20: Internal variables of IEHA model in SimHydraulics for a free movement, with step input of X = −0.95 × Emax . In the second curve: (dashed line) Channel A opening signal, (solid line) Channel B opening signal.

at the pressure and flow variations in a single piston, Figure 2.23, we see that the pressure increases gradually from one rotation cycle to another, because the output


55

E(cm)

0 −0.02 −0.04 0

0.05

0.1

0.15

Op en Op enM a x

1 ChannelA 0

ChannelB

−1 0

0.05

0.1

0.15

0.05

0.1

0.15

0.05

0.1

0.15

0.05

0.1

0.15

0.05

0.1

0.15

0.1

0.15

xp 1(cm)

0.05 0

Pp 0(bar)

−0.05 0

1.2 1 0.8

0

l ) QA( min

0.1 0 −0.1 −0.2 0

l ) QB ( min

0.2 0.1 0 −0.1 0 −4

l ) Ql eak( min

x 10 4 2 0

0.05 time(s)

Figure 2.21: Variation of the internal variables of µp0 , for a free movement, and an eccentricity value going from 0 to E = −0.95 × Emax .

56


is compressing a spring as it moves and its force is increasing. The leakage(Qleak ) being directly proportional to the piston pressure it also increases at each rotation.

0 X E −0.005

1

0 −0.01 −0.01 −0.015

−0.02

Qμ Pump (l/min)

X&E(cm)

0.8

−0.03

−0.02

0.6

B

−0.04 −0.025

−0.05

0

0.002 0.004 0.006 0.008 0.01 0.012

−0.03

0.4

−0.035

0.2 −0.04

−0.045

−0.05

0

0

0.02

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

0.18

0

0.2

0.02

(a) Displacement X of micro-valve, and carriage eccentricity E which follows X.

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

0.18

0.2

(b) Flow produced by the µP ump

1

0.9 1

1 0.8

0.7

0.6

0.8

0.4

0.6

Qcyl (l/min)

0.2 0.5

0

0.6

A

d is p la c e m e n t P D s tr o k e P D

0.8

0

0.01

0.02

0.03

0.4

0.4

0.3 0.2

0.2

0.1 0

0 0

0.02

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

0.18

0.2

0

0.02

(c) Passive distributor rotation.

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

0.18

0.2

0.18

0.2

(d) Flow into output actuator(cylA ).

1.2

60

1

50

Forcecyl(N)

Δ P(bar)

70

0.8

0.6

40

30

0.4

20

0.2

10

0

0

0.02

0.04

0.06

0.08

0.1 time(s)

(e) Pressure difference in actuator(PcylA − PcylB ).

the

0.12

0.14

0.16

chambers

0.18

of

0

0.2

0

output

0.02

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

(f) Hydraulic force in output actuator.

4

0.7

3.5

0.6

3 0.5 2.5

Pos(cm)

Vel(cm/s)

0.4 2

1.5

0.3

0.2 1 0.1 0.5

0

0

−0.5

−0.1

0

0.02

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

(g) Velocity of output actuator rod.

0.18

0.2

0

0.02

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

(h) Position of output actuator rod.

Figure 2.22: Internal variables of IEHA model in SimHydraulics, for an output actuator pushing against a wall(spring-damper). In the second curve: (dashed line) Channel A opening signal, (solid line) Channel B opening signal.

0.18

0.2


57

E (cm)

0 −0.02 −0.04 0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

Op ening Op eningM a x

1 ChannelA 0

ChannelB

−1 0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

0.02

0.04

0.06

0.08

0.1 time(s)

0.12

0.14

0.16

0.18

0.2

xp 1 (cm)

0.05 0

Pp 0(bar)

−0.05 0

2 1 0

0

l ) QA( min

0.1 0 −0.1 −0.2 0

l ) QB ( min

0.2 0.1 0 −0.1 0 −4

l ) Qleak( min

x 10 6 4 2 0

0

Figure 2.23: Variation of the internal variables of µp0 , for an output actuator pushing against a wall(spring-damper)..

58

2.8


Comparison and conclusions

In this chapter we presented the steps of modeling IEHA using Simulink and SimHydraulics. Each of these simulators have their advantages and disadvantages. Using Simulink to model the system helped us have a better understanding of the mathematical equations underlying the system. The simulation is faster compared to the SimHydraulics model, but very sensitive to the simulation step time. The Simulink model is run with a fixed step time, so it can be compiled into C language and inserted easily in an external program. However, using Simulink on a complex system like ours has its own limits. For examples hard stops, (e.x. a piston at the end of its stroke), are modelled as saturation, and if we want to model the impact forces in these cases, we should add spring equations manually; this would increase the order of the system and the Simulink solver would be more sensitive to time step; however, decreasing the step time can make the simulation too slow. On the other hand the passive distributor is modelled as a sigmoid function, so its effects on the system can not be studied using this model. The Simscape library of MATLAB, including SimHydraulics, has a specific solver which uses a nonlinear solver to compute initial conditions and the transient initialization of the system. It has a modifiable ”consistency tolerance value” that determines how accurately the algebraic constraints are to be satisfied at the beginning of simulation and after every discrete event (for example, a discontinuity resulting from a valve opening, a hard stop, and so on). This adds to the reliability of the obtained results. In the SimHydraulic model of the micro-pistons, as we saw in section 2.7, the internal parameters change gradually and each slight phenomenon has an effect on the variation of the pressure and flow. This makes the model very ”stiff”, requiring an appropriate variable time solver. When the variations of the system states are smooth, a variable solver increases the step time making the simulation faster, and at sudden changes in the states of the system(like a hard stop), it tightens the step time computing a more accurate results for that transition phase. Also the dynamics of the passive distributor and its effect on the rest of the system are more realistic in the SimHydraulics system. Therefore, it is best to use the SimHydraulics model where we want to study and optimize the internal


59

parameters of the system. Once the results from the model are satisfying and that the transient phases can be neglected or identified, we can reconstruct the Simulink model by using mathematical equations with the determined values, and generate the C code of the system, which can be integrated in any other model, like the case of the biped model of our partners in laboratory of Nantes(IRCCyN). In the following chapter we analyse the efficiency of the system, and identify the most crucial parameters that have a direct impact on system losses in terms of leakage and friction.

Chapter 3 Parameter Optimization

3.1

Introduction

The performance of rotary vane machines is affected by interaction forces and friction losses resulting from vane motion. Mechanical friction is a major cause of power loss in machines of this type. On the other hand, an incompatible design choice of vanes and carriage can increase leakage in the pump and reduce the system’s volumetric efficiency to a high extent. In this chapter first we shall formulate the leakage in the micro-pump and the passive distributor. Then the dynamic equations of the forces and torques acting on the rotating pistons are presented. This will help us estimate the friction losses in the pump and identify the major parameters that could affect the system efficiency, such as piston length, etc. In the end of the chapter, using the developed mathematical models, a pump design for the ankle pitch actuator is proposed, based on volumetric and mechanical efficiency optimization criteria.

3.2

Volumetric Efficiency

We define volumetric efficiency as the ratio of the system output flow to the input flow. In other words, volumetric efficiency is a measure of the losses through

61

62


internal leakage. Leakage in the pump and the distributor are mainly a function of the clearance between the mechanical pieces, as we shall see in this section.

3.2.1

Clearance and piston deviation

Until now we have considered the pump pistons to always be perpendicular to the rotation axes. However, in reality there exists a gap between the piston and its sleeve, which can cause an angular deviation of piston from the center axes of the slot(α), as shown in Figure 3.1(a). lp refers to the piston length, and lp Out to the piston part outside the slot at each instant. The piston-carriage contact point is noted as C, and the top and bottom contact points with the slot are named t and b, respectively. The angular deviation is generally defined by equation (3.1). One way to reduce the angular deviation is to reduce the radial clearance, but this requires more accurate fabrication methods and can be costly. Nevertheless, given a specific radial clearance set by fabrication limits, the angular deviation can be reduced by adjusting the piston length w.r.t its diameter. Figure 3.1(b) shows the variation of the deviation angle vs. diameter/length ratio of a piston, for different clearance values. We see that for a ratio of length/diameter > 2.5, α stabilizes. This finding is consistent with the proposed ratio of length/diameter in the mechanical literature, which usually propose 2 6

length diameter

6 5, [35].

Even though the angular deviation can be quite small, it can have a major effect on system efficiency. In this study we have solely considered the deviation due to piston sleeve clearance. Nevertheless, this angle could be much larger than this theoretical value which considers the cylindrical form of the piston and the contact point to be ideal (no deformation). α = arctan(

3.2.2

2Cl ) lp − lpout

(3.1)

Leakage in pump pistons

This leakage is due to the annular orifices created by the radial clearance between the piston and its slot on the rotor. This type of flow is called annular flow and is give in (3.2), [33]. See Appendix A for the details of how this equation is obtained.


63

0.03

Cl=2µm Cl=5µm

c lp

lp Out



t b

Deviation Angle[rad]

0.025



Cl=10µm

Dp=5mm Lp=2.7mm

0.02

Cl=20µm Cl=24µm

∅ Cl=20µm

0.015

0.01

0.005

0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Length/Diameter

(a)

(b)

Figure 3.1: (a) Deviation angle(α) from the central axes, causes by the clearance between the piston and its sleeve. (b) α depends on the clearance value(Cl), as well as the piston diameter/length ratio. (*) corresponds to the first prototype of IEHA micro-pistons.

We see that this flow is a function of the geometrical values: slot radius(rslot ), radial clearance(Cl), passage length(Lpass ), and the piston-slot eccentricity(e) created by the clearance, (Figure 3.2). It is also a function of fluid dynamic viscosity(µ) and h 2 i pressure drop across the annular orifices(P1 − P2 ). The term 1 + 23 cle shows that the annular leakage increases quadratically with the vane-slot eccentricity, going up to 2.5 times the leakage value for a concentric vane, (3.3). In our analysis of the volumetric efficiency we consider the worst case scenario i.e. we consider the piston and slot to be fully eccentric(e = cl). Qannular

π.rslot .cl3 3 e 2 = 1+ (P1 − P2 ) 6.µ.Lpass 2 cl

Qannular (e = cl) = 2.5 × Qannular (e = 0)

(3.2) (3.3)

As we see in (3.2), annular leakage is inversely proportional to the length of the piston inside the slot(Lpass ). In our micro-pump this value changes with the radial position of the piston(xp ). When the piston is completely inside the slot (xp = −Emax , the annular orifice length is at its maximum and equals the piston length(lp ) minus its round shaped tip(ltip ). In this configuration the annular leak is minimum, while when the piston is at its maximum eccentricity(xp = Emax ), the annular orifices is the shortest and the annular leakage at its maximum value.

64


rslot

rslot

L pass

P1

P2

cl

+

+ + e

L pass

e 0

ecl

Figure 3.2: Annular orifice created by the clearance between the piston and its slot. In the middle figure piston and slot are concerntric(e = 0). In the right figure they are at maximum eccentricity(e = cl).

1

xp=

−Emax

2

−Emax/2

3

0

4

−Emax+ ltip

5

Emax/2

6

Emax

Figure 3.3: Passage length of annular leakage orifice(Lpass ), for E = Emax , and rotation angles 0 6 θ 6 π. Lpass =lp −ltip for 1 to 4, and Lpass =lp −lp Out(t) for 5 to 6.

Figure 3.3 shows the passage length for different values of xp and for an eccentricity of E = Emax = 0.5mm in the first prototype of the micro-pump. In the first prototype, at xp = −Emax , the piston is completely inside its slot. As it moves out the slot, the passage length stays constant, until the edge of the piston tip is at the same level as the vane edge, meaning lp Out(t) = ltip , for xp = −Emax + ltip . As the piston is pushed farther out, the passage length decreases and equals lp − lp Out(t). The annular leak has been implemented in the Simhydraulics model of the pistons (Figure 2.12) using the annular orifice block from the Simhydraulics library. This block calculates the flow through an annular orifice according to equation (3.2) and receives the insert displacement(∆Lpass ) as an input. ∆Lpass is calculated at each time step as a function of xp (t), lp , and ltip , and is passed to the annular orifice block. To calculate the volumetric efficiency of the micro-pump for different output pressures, it has been separated from the rest of the model and is connected to one constant pressure sources on each side. Pressure of the intake channel is set to 1 bar corresponding to an accumulator under atmospheric pressure, and


65

100

100

90

80

80

70

70

Efficiency[%]

Efficiency[%]

90

60

50

60

50

40

40

30

30

20

4

5

6

7

8

9

10

11

20

12

0

10

20

30

40

Clearance[μm]

50

60

70

80

90

100

Pressure[bar]

100 90

Efficiency[%]

80 70 60 50 40 30 20 0

20

4 40

6 60

Pressure[bar]

8 80

100

10 12

Clearance[μm]

Figure 3.4: Volumetric efficiency of the micro-pump w.r.t radial clearance and working pressure. (*) clearance=10µm, Pressure=50bar, in the first prototype of IEHA. (O)clearance=8µm, Pressure=50bar,

the pressure of the output channel is varied from 2 to 100 bar. Figure 3.4 shows the variation of the volumetric efficiency of the pump for radial clearance values going from 4µm to 12µm. The radial clearance in the actual model of IEHA is 10µm, shown with a black line in the figure. We see that for a working pressure of 50 bar, the efficiency of the pump can decrease to 80%. The first solution is to reduce the radial clearance, since annular leakage is cubically proportional to clearance; decreasing the clearance to 8µm can increase the efficiency up to 90% at this working pressure. The second solution would be to increase the piston length within the limits of the maximum allowed size of the pump.

66

Chapter 3 Parameter Optimization 100

99

PD Vol. Effic. [%]

98

97

96

95

94

93

10

20

30

40

50

ΔP [bar]

60

70

80

90

100

Figure 3.5: Volumetric efficiency of the passive distributor vs. ∆P across the distributor. One side of the distributor is considered connected to pressure 1 bar, and the pressure on the other side variating.

3.2.3

Leakage in passive distributor

Leaks in the passive distributor are both inter-chamber and external. This leak is created by the clearance between the distributor’s fixed part and the rotating butterfly valve. The leakage orifices in this component of IEHA are thin with rectangular cross sections. These leaks can be calculated by using equation (3.4), where wrectangle and Lrectangle , are the orifice width and depth, respectively. Calculating the leakage in the passive distributor, given in Figure 3.5, we see that for a working pressure of 50 bar , the volumetric efficiency of the passive distributor is about 97%. This shows that the leaks in the passive distributor are negligible compared to the micro-pump, therefore, in order to improve the volumetric efficiency of IEHA it is mostly important to modify the parameters of the micro-pump. Qannular =

3.3

wrectangle .cl3 (P1 − P2 ) 12.µ.drectangle

(3.4)

Mechanical efficiency of the micro-pump

Mechanical efficiency of the pump concerns the ratio of the output energy of the pump to the input energy. The difference between the input and output power of the pump are mainly due to power losses due to friction forces inside the pump. In order to calculate these friction forces we should evaluate the contact forces by


67

solving the dynamic equation of all the forces acting on the piston, including the pressure force, as well as axial and rotation forces.

3.3.1

Rolling motion

If the axial forces on a piston are large enough to guarantee its radial movement, the round tip of the piston will always be in contact with the inner wall of the carriage. The piston follows a rolling motion on the carriage wall which is due to the eccentricity between rotor and carriage. This rolling motion is illustrated in Figure 3.6(a), where the eccentricity value is fixed to Emax = 0.5mm. Red dashed lines show the vector from the rotor center to the center of the piston tip, while the blue lines are the vector from the carriage center to the contact point, noted by red cross marks. The rotation angle of the rotor is θ, and the angle between the contact point and piston center line is β. This latter is used to calculate the contact efforts as we shall see in the following section. We see that at the starting point(θ = 0), the contact point is at the tip center and moves away from the center as the rotation angle approaches θ = π2 . This angle decreases back to zero as the rotation angle moves towards π, and increases in the opposite direction. Figure 3.6(c) shows the variation of β for a full rotation of the rotor (θ = 0◦ to180◦ ). For IEHA pistons of diameter 0.5mm, tip radius of 6mm, and maximum eccentricity of 0.5mm, the contact angle,β, can increase up to 20◦ . The contact angle is calculated as follows: (Rc − rpt ) sin (β) = E sin (θ)

(3.5)

E sin (θ)) Rc − rpt

(3.6)

β = arcsin(

, where Rc is the carriage radius and rpt the radius of the piston tip sphere.

3.3.2

Vane dynamic model

Depending on the piston slot configuration, there can be 5 to 10 forces acting on the vane as shown in Figure 3.7(a). These forces are as follows:

68


Oc Emax Or

β

Oc

Oc

c

Or

Or

(a) Contact point(red cross) w.r.t rotor’s rotation angle(θ).

β[◦ ]



E

θ

R

rp

t



(b) Piston and carriage contact parameters.

20 0 −20 0

90

180

θ [°]

270

(c) Rolling motion of piston tip, and variation of the contact angle for IEHA piston of diameter 5mm, carriage radius Rc = 7.5mm, tip radius of rpt = 6mm, and eccentricity Emax = 0.5mm.

Figure 3.6: Contact configuration of a round tip piston rolling inside an eccentric circle. β=angle between the contact point and piston center.

• Fluid pressure force (Fpr ), centrifugal force (Fcg ), Coriolis force (Fco ), and the force resulting from the linear acceleration of the piston (Fma ) are given by equations (3.7) to (3.10), respectively. • Interaction force between the piston and the carriage (Nc ). If the piston slides on the inner surface of the carriage there is a friction force (Tc ), proportional to the friction coefficient between the piston tip and carriage material (µpc ), as shown in (3.11). • Interaction (Nt ) and friction (Tt ) forces of top edge of the slot exerted on the piston wall. Interaction (Nb ) and friction (Tb ) forces of the slot on the bottom


69

of the piston. The friction forces are proportional to the friction coefficient between the piston and rotor material (µpr ), and are given by equations (3.12), (3.13). The presence and direction of these forces depend on the piston placement in the slot, and its linear direction of movement(inwards or outwards).

Fpr = Sp Pp

(3.7)

Fcg = mp ω 2 OG

(3.8)

Fco = −2mp ω x˙p

(3.9)

Fma = mp x¨p

(3.10)

Tc = µpc Nc

(3.11)

Tt = µpr Nt

(3.12)

Tb = µpr Nb

(3.13)

The reference frame to calculate the forces and torques acting on the piston has been considered fixed to the piston bottom. The forces w.r.t the reference frame depend on the deviation angle, α, as well as the contact angle, β. In order to find the contact forces, we project the forces described above onto the x and y axis. We also calculate the torques exerted by these forces on the piston at the center of the reference frame(o). With the six equations, given in (3.11) to (3.18), we can find the unknown variables Nc , Nt , Nb , and their corresponding friction forces Tc , Tt , and Tb at each contact configuration. The sign coefficient κ, is defined by the piston slot configuration, i.e. whether the piston deviates from the central axes clockwise or counterclockwise, (Figure 3.8). ΣF · ~x = Fpr − Fma + (Fcg + Nc + Nt + Nb + Tc + Tb + Tt ) · ~x

(3.14)

70


Nc Tc

Fcg Fma x y

Nb

Nt

Fco

Tt

o

Tb Fpr

Oc

dY

x y



o

rpt





Nc Tc dX





Or

Figure 3.7: (a) Forces acting on a rotating piston. (b) Lever arms of the piston-carriage contact force.

 1 Figure

with:

  1

3.8: The value of the coefficient κ, depending on the direction of piston deviation.

                 

Fcg · ~x = Fcg cos(α) Nc · ~x = −Nc cos(β) Nt · ~x = −Nt sin(α) Nb · ~x = Nb sin(α)

(3.15)

Tc · ~x = −Tc sin(β) Tt · ~x = −Tt cos(α)sign(x˙p ) Tb · ~x = −Tb cos(α)sign(x˙p )

ΣF · ~y = Fco + (Fcg + Nc + Nt + Nb + Tc + Tb + Tt ) · ~y

(3.16)

Chapter 3 Parameter Optimization with:

                 

71

Fcg · ~y = −κFcg sin(α) Nc · ~y = Nc sin(β) Nt · ~y = −κNt cos(α) (3.17)

Nb · ~y = κNb cos(α) Tc · ~y = −Tc cos(β) Tt · ~y = κTt sin(α)sign(x˙p ) Tb · ~y = κTb sin(α)sign(x˙p )

Σtau = τco + τcg + τNc + τNt + τNb + τTc + τTt + τTb

(3.18)

with: 

τco    τcg    τNc     τNt   τ  Nb    τT  c    τTt   τTb

= Fco Or G = −κFcg sin(α)Or G = Nc sin(β)dX − Nc cos(β)dY = −κNt cos(α)(lp − lpOut ) − Nt sin(α) = κNb sin(α)

Dp 2

Dp 2

(3.19)

= −Tc cos(β)dX − Tc sin(β)dY Dp sign(x˙p ) + κTt sin(α)(lp − lpOut )sign(x˙p ) 2 Dp = −κTb cos(α) sign(x˙p ) 2

= κTt cos(α)

, where dX and dY are the lever arms for the contact forces of carriage-piston, i.e. Nc and Tc , as shown in Figure 3.7(b), and are calculated by (3.20) and (3.21).

3.3.3

dX = lp − rpt .(1 − cos(β))

(3.20)

dY = rpt sin(β)

(3.21)

Power loss due to piston-rotor friction

The eccentricity value is set to E = Emax , and the piston-rotor dynamic friction coefficient to µpr = 0.1. This value is an estimation of lubricated steel-steel contact friction coefficient[36]. In IEHA, the carriage inner ring is mounted on roller

72


bearings and follows the rotor motion. In this study, we have assumed the velocity of the carriage inner ring and the rotor to be sufficiently close, hence, we can assume the friction force between the piston tip and the carriage to be negligible, i.e. µpc = 0. In the first step, we should find the contact configuration of piston-sleeve during one complete rotation. For so, we calculate the sum of torques on piston resulting from pressure, inertial forces and only piston-carriage contact during one rotation. The result is shown in Figure 3.9. As we see in this figure, during the first half cycle the sum of these torques is negative (κ = −1), while in the second half it is positive (κ = +1). This means that in the first half, where piston intakes fluid, it tends to rotate counterclockwise, while in the second half, where it discharges oil and its chamber volume reduces, it tends to turn clockwise. It is important to note that if the contact configuration is not chosen correctly, the set of the system equations will not find a valid value for the contact forces, meaning that one or more of the interaction forces will give a negative value. This would mean that the sum of the forces and torques would not cancel out and that the chosen configuration is not stable. Once the contact configuration is determined, we solve the equations of the system for different working pressures to find the contact forces. We set the input pressure value to 1 bar and vary the output pressure from 1 to 100 bar. The instantaneous lost power due to friction between the piston and rotor, Tt and Tb , can be calculated with equation (3.22), [37] [38]. The lost power due to friction is shown in Figure 3.10 for different output pressures. The friction forces are rather important when the piston is being pushed in and discharging, mainly because that is where the hydraulic pressure force(Fp ) increases. WL = |Tt .x˙p | + |Tb .x˙p |

(3.22) N

The average power loss for N pistons in one rotation cycle(W L ) can be calculated by (3.23). For N = 30, this value is shown in Figure 3.11. The theoretical average lost power can increase up to 20W for a working pressure of 50bar. N WL

1 =N 2π

Z

2π

WL dθ 0

(3.23)


73

0.15

0.1

τ [N.m]

0.05

0

−0.05

−0.1 0

π 2

π

3π 2

2π

θ

Instantaneous Lost Power W L [W]

Figure 3.9: Sum of torques resulting from pressure, inertial forces, and piston carriage contact(Nc ). Positive torque turns the piston to the left, while negative torque turns the piston to the right.

6 5 4 3 2 1

0 100

2π 50

π

3 π2

π 2

Pressure [bar]

0 0

Angle of Rotation θ

Figure 3.10: Instantaneous power loss due to the friction in contact points of the piston for half the cycle where the piston is discharging to the output channel of pressure 1 to 100bar.

74

Chapter 3 Parameter Optimization 40

N

W L [W]

30

20

10

0

0

10

20

30

40

50 Pressure[bar]

60

70

80

90

100

Figure 3.11: Average power loss for a pump of 30 pistons.

3.3.4

Friction and the piston-carriage contact point

As we saw in equation (3.18), the piston-carriage contact point and its resultant torque are a function of β, which itself is a function of the piston tip radius rpt , as established by equation (3.6). Therefore, the first and cheapest solution to reducing the contact friction would be to only change the tip of the pistons, without changing the rest of the micro-pump design, such as piston length, rotor diameter. A study of the friction forces w.r.t rpt would lead to choosing the optimum value of the latter. In order to define the feasible values of rpt , we should first define its upper and lower bounds. It is clear that

Dp 2

< rptmin . rptmax should be chosen

so that the contact point always remains on the piston tip and not fall on the edge of the tip. The maximum angular deviation of the contact point happens for maximum eccentricity and rotation angles of θ = ± π2 , as we saw in Figure 3.6. βmax is shown in Figure 3.12, and is calculated as: β < βmax with: βmax = arcsin From equation (3.6) and for θ =

π 2

Dp /2 rpt

(3.24)

we can write:

Emax Dp arcsin < arcsin Rs − rpt 2rpt π Emax Dp 0