挖掘机臂液压系统的模型化参量估计外文翻译、中英文翻译、外文文献翻译

重庆交通大学二 OO 九届毕业设计（论文） 译文 外文原文： Modeling and parameter estimation for hydraulic system of excavators arm HE Qing-hua， HAO Peng， ZHANG Da-qing Abstract A retrofitted electro-hydraulic proportional system for hydraulic excavator was introduced firstly. According to the principle and characteristic of load independent flow distribution (LUDV) system, taking boom hydraulic system as an example and ignoring the leakage of hydraulic cylinder and the mass of oil in it ,a force equilibrium equation and a continuous equation of hydraulic cylinder were set up. Based on the flow equation of electro-hydraulic proportional valve, the pressure passing through the valve and the difference pressure were tested and analyzed. The results show that the difference of pressure does not change with load and it approximates to 2.0MPa. And then, assume the flow across the valve id directly proportional to spool displacement and is not influenced by load, a simplified model of electro-hydraulic system was put forward. At the same time, by analyzing the structure and load-bearing of boom instrument, and combining moment equivalent equation of manipulator with rotating law, the estimation methods and equations for such parameters as equivalent mass and bearing force of hydraulic- cylinder were set up. Finally, the step response of flow of boom cylinder was tested when the electro-hydraulic proportional valve was controlled by the step current. Based on the experiment curve, the flow gain coefficient of valve unidentified as 2.82510-4m3/(sA) and the mode is verified. Key words: Excavator, Hydraulic-cylinder proportional system, Load independent flow distribution (LUDV) system, Modeling, Parameter estimation 重庆交通大学二 OO 九届毕业设计（论文） 译文 1 Introduction For its high efficiency and multifunction, hydraulic excavator is widely used in mines，road building, civil and military construction， and hazardous waste cleanup areas The hydraulic excavator also plays an important role in construction machines Nowadays, macaronis and mobilization have been the latest trend for the construction machines So， the automatic excavator gradually becomes popular in many countries and is considered a focus Many control methods can be used to automatically control the manipulator of excavator Whichever method is used, the researchers must know the structure of manipulator and the dynamic and static characteristics of hydraulic system That is, the exact mathematical models are helpful to design controller However, it is difficult to model on time-variable parameters in mechanical structures and various nonlinearities in hydraulic actuators， and disturbance from outside Researches on time delay control for excavator were carried out in Refs NGUYE used fuzzy sliding mode control and impedance control to automate the motion of excavators manipulator SHAHRAM et al adopted impedance control to the teleported excavator Nonlinear models of hydraulic system were developed by some researchers. However, it is complicated and expensive to design controller, which 1imits its application In this paper, based on the proposed model， the model of boom hydraulic system of excavator was simplified according to engineering and by considering the force equilibrium， continuous equation of hydraulic cylinder and flow equation of electro-hydraulic proportional valve； at the same time， the estimation methods and equations for the parameters of model were developed 2 Overview of robotic excavator The backhoe hydraulic excavator studied is shown in Fig.1 In Fig.1， Fc presents the resultant force of hydraulic cylinder, gravity of boom， dipper, bucket and so on at point B，whose direction is along cylinder AB； Fc can be decomposed into Fcl and Fc2， and their directions are vertical and parallel to that of O1B， respectively； ac is the acceleration whose 重庆交通大学二 OO 九届毕业设计（论文） 译文 direction is same to that of Fc， and ac can be decomposed into acl an d ac2 too； G1， G2 and G3 are the gravity centers of boom， dipper and bucket， respectively； ml， m2 and m3 are the masses of them， and their values can be given by experiment( m1=868.136kg， m2=357.115kg and m3=210.736kg)； Ol， O2 and O3 are the hinged points； G1， G2and G3are projections of Gl， G2 and G3 on x axis， respectively The arm of excavator was considered a manipulator with three degrees of freedom (three inclinometers were set on the boom， dipper and bucket， respectively) In tracking control experiment， the objective trajectories were planed based on the kinematic equation of excavators manipulator Then， the motion of boom， dipper an d bucket was set by the controller In order to suit for automatic contro1 the normal hydraulic control excavator should be retrofitted to electro-hydraulic controller Based on original hydraulic system of SW E-85 The hydraulic pilot control system was replaced by an electro-hydraulic pilot control system The retrofitted hydraulic system is shown in Fig.2 In this work， because boom， dipper an d bucket are of the same characteristics ， the hydraulic system of boom was taken as an example In the electro-hydraulic pilot control system， the pilot electro hydraulic proportional valves were derived from adding proportional relief valves on the original SX-l4 main valve， and hydraulic pilot handle was substituted by electrical one The retrofitted system of excavator was still the LUDV system (Fig.3)of Rexroth with good controllability In Fig.3， y is the displacement of piston； Q1 and Q2 are the flows in and out to the cylinder respectively； pl，p2， ps and pr are the pressures of head and rod sides of cylinder, system and return oil，respectively； A1 and A2 are the areas of piston in the head and rod sides of cylinder, respectively； xv is the displacement of spool； m is the equivalent mass of load. Flg.1 Schematic diagtam of excavators arm 重庆交通大学二 OO 九届毕业设计（论文） 译文 Flg.2 Schematic diagram of retrofitted electro-hydraulic system of excavator Flg.3 Schematic diagram of LUDV hydraulic system after retrofitting 重庆交通大学二 OO 九届毕业设计（论文） 译文 3 Model of electro-hydraulic proportional system 3.1 Dynamics of electro hydraulic proportional valve In this work， the electro-hydraulic proportional valve consists of proportional relief valves and SX-14 main valve A transfer function from input current to the displacement of spool can be obtained as follows： Xv(s) Iv(s)=KI (1+bs) (1) where Xv is the Laplace transform of xv， m； KI is the current gain of electro-hydraulic proportional valves， m A； b is the time constant of the first order system， s： Iv=I(t)-Id，I(t)and Id are respectively the control current of proportional valve and the current to overcome dead band， A 3.2 Flow equation of electro-hydraulic proportional valve In this work ， LUDV system was adopted in the experimental robotic excavator According to the theory of LUDV system， the flow equation can be gotten： 11 2 pwxcQ vd 222 pwxcQvd = where p is the spring-setting pressure of load sense valve， MPa； cd is the flow coefficient m5 (Ns)； w is the area gradient of orifice， m2 m； is the oil density, kg m3； 1p and 2p are the two orifices pressure， respectively, M Pa When the flow of excavator is not saturated， p is a nearly constant In this work， the value was tested and gotten by experiment In Fig.4， ps， p1s,and p represent the system pressure， the load sense valve pressure and the diference of pressure, respectively. The pressure experiment curves of the system show the variation of three kinds of pressures Although Ps and pls change with load， their difference does not change with load， the value approximates to 2.0MPa.So， the effect of on the flow across the valve can be neglected It is assumed that the flow across the valve is proportional to the size of orifice valve， and the flow is not influenced by 0)(/2 1 tIppwxc rvd ， (2) 0)(2 tIpwxc vd ， 0)(/2 2 tIppwxc rvd ， 0)(2 tIpwxc vd ， (3) 重庆交通大学二 OO 九届毕业设计（论文） 译文 load Then， Eqn (2) can be simplified as Q1=Kqxv(t),I(t)0 (4) where is the flow gain coefficient of valve, m2 s, and /2 pwcKdq Flg.4 Curves of pressure experiment under boom moving condition 3.3 Continuity equation of hydraulic cylinder Generally speaking， construction machine does not permit external leakage At present， the external leakage can be controlled by sealing technology On the other hand，it has been proven that the internal leakage of excavator is quite little by experiments So, the influence of internal and external leakage of hydraulic system can be ignored When the oil flows into head side of cylinder and discharges from rod side， the continuity equation can be written as cpVyAQ /1111 cpVyAQ /2222 where V1 and V2 are the volumes of fluid flowing into and out the hydraulic cylinder, m3 ；c is the effective bulk modulus(including liquid， air in oil and so on)， N/m2 3.4 Force equilibrium equation of hydraulic cylinder It is assumed that the mass of oil in hydraulic cylinder is negligible,and the load is rigid. Then the force equilibrium equation of hydraulic cylinder can be calculated from the Newtons second law： (5) 重庆交通大学二 OO 九届毕业设计（论文） 译文 cc FyBymApAp 2211 (6) where Bc is the viscous damping coefficient， Ns m 3.5 Simplified model of electro hydraulic proportional system After the Laplace transform of Eqns (4) (6)， the simplified model can be expressed as 21201 asasasssFbsXbsY cfv (7) where Y(s) is the Laplace transform of y； 122211 / AAVAKb qc ；b1=V1V2；a0=V1V2m；a1=BcV1V2； 2212122 AVAVa c . 4 Parameters estimation From the process of modeling and Eqn (7)， it is clear that all parameters in the simplified model are related to the structure。 the motional situation and the posture of excavators arm Moreover， these parameters are time variable. So it is quite difficult to get accurate values and mathematic equations of these parameters. To solve this problem， those important parameters of model were estimated approximately by the estimation equation and method proposed in this work 4.1 Equivalent mass estimation for load on hydraulic cylinder The load of boom hydraulic cylinder(it is assumed there is no external load)consists of boom， dipper and bucket In Fig.1， boom， dipper and bucket rotate around points O1，O2 and O3， respectively So their motions are not straight line motions about the cylinders, that is to say, their motion directions are different from Y in Eqn.(5) So， m in Eqn.(6)cannot be simply regarded as the sum mass of boom， dipper and bucket Considering O1 at an axis of manipulator, the torque and angular acceleration can begiven as follows： s in111 BOcBOc lFlFM BOcBOc lala 11 s in1 where M and are the torque and angular acceleration of manipulator to O1，respectively；BQl1 is the length from point O1 to point B According to the rotating law： M=J ,we get (8) 重庆交通大学二 OO 九届毕业设计（论文） 译文 BOcBOc lJalF 11 /s ins in that is BOcc lJaF 12/ （ 9） where J is the equivalent moment inertia of manipulator to point O1,kgm2， and it can be written as follows： 312111 233222211 GOGOGO lmJlmJlmJJ (10) J1, J2 and J3 are the moment inertia of boom， dipper and bucket to their own bary center respectively The values of them can be obtained by dynamic simulation based on the dynamic mode, J1=450.9Nm, J2=240.2Nm, J3=94.9Nm. Comparing Eqn (9)with Fc=mac， the equivalent mass at point B can be given： BOlJm 12/ (11) 4.2 Estimation for load on hydraulic cylinder The equivalent moment equation of manipulator to O1 is 3121111 321s i n GOGOGOBOc glmglmglmlF (12) where 2111 , GOGO lland 31GOl are the length from pointO1 to point G1 ， G2and G3；，respectively Then， the counter force of load is s in1312111 321 BOGOGOGOc lglmglmglmF (13) 4.3 Estimation for flow gain coefficient of valve The flow of pump can be measured by flow transducer The instrument used in this work was Multi system 5050 The step response of flow of boom cylinder under the electro hydraulic proportional valve controlled by the step curent is shown in Fig.5 At the same time， the curve verifies Eqn 11 Based on the experiment curve， the range of KqKl can be identified according to Eqns.(1)and(4) And then， according to data in Fig.4， we can get： KqKl=2.82510-4m3/(sA) 重庆交通大学二 OO 九届毕业设计（论文） 译文 Flg.4 Flow of boom cylinder under electro-hydeaulic proportional value controlled by step current 5 Conclusions （ 1） The mathematic model of electro hydraulic system is developed according to the characteristics of excavator It is assumed that the flow across the valve is directly proportional to the size of valve orifice， and the influence of intemal and extemal leakage of hydraulic system is ignored The simplified model can be obtained： )(/)()()( 212011 asasasssFbsXbsY cv where represent the displacement of piston and the displacement of spool （ 2） From the model of electro hydraulic system， we can obtain the equivalent mass BOlJm 12/ ， bearing force )(31211 3121 GOGOGOl lmglmglmF , flow gain coefficient of value KqKl=2.82510-4m3/(sA) ， where KI is the current gain of electro hydraulic proportional valves From: Journal of Central South University (English) 2008 Vol 15 No. 3 pages 382-386 重庆交通大学二 OO 九届毕业设计（论文） 译文 译文： 挖掘机臂液压系统的模型化参量估计 何清华，赫鹏，张大庆 摘 要 首先介绍了液压挖掘机的一个改装的电动液压的比例系统。根据负载独立流量分配（ LUDV ）系统的原则和特点，以动臂液压系统为例并忽略液压缸中的油大量泄漏，建立一个 力平衡方程和一个液压缸的连续性方程。基于电动液压的比例阀门的流体运动方程，测试的分析穿过阀门的压力的不同。结果显示压力的差异并不会改变负载，此时负载接近 2.0MPa。然后假设穿过阀门的液压油与阀芯的位移成正比并且不受负载影响，提出了一个电液控制系统的简化模型。同时通过分析结构和承重的动臂装置，并将机械臂的力矩等效方程与旋转法、参数估计估计法结合起来建立了液压缸以等质量等为参数的受力平衡参数方程。最后用阶跃电流控制电液比例阀来测试动臂液压缸中液压油的阶跃响应。根据实验曲线，阀门的流量增益系数被确定为2.82510-4m3/(sA)，并验证了该模型。 关键词 ： 挖掘机，电液比例系统，负载独立流量分配（ LUDV ）系统，建模，参数估计 重庆交通大学二 OO 九届毕业设计（论文） 译文 1 引言 由于液压挖掘机具有高效率、多功能的优点，所以被广泛应用于矿山，道路建设，民事和军事建设，危险废物清理领域。液压挖掘机在施工机械领域中也发挥了重要作用。目前，机电一体化和自动化已成为施工机械发展的最新趋势。因此，自动挖掘机在许多国家逐渐变得普遍并被认为重点。挖掘机可以用许多控制方法自动地控制操作器。 每种使用方法，研究员必须知道操作器结构和液压机构的动 态和静态特征。即确切的数学模型有利于控制器的设计。然而，来自外部的干扰使得机械结构模型和各种非线性液压制动器的时变参数很难确定。关于挖掘机时滞控制的研究已经有人在研究了。 NGUYEN 利用模糊的滑动方式和阻抗来控制挖掘机动臂的运动， SHAHRAM 等采取了阻抗对挖掘机远距传物的控制。液压机构非线性模型已经由研究员开发出来了。 然而，复杂和昂贵的设计控制器限制了它的应用。在本文，根据提出的模型，根据工程学和受力平衡，挖掘机臂液压机构模型简化为连续均衡的液压缸和流动均衡的电液比例阀；同时，确定了模型的参量的估计方法 和等式。 2 挖掘机机械臂概述 液压挖掘机的挖掘研究结果如图 1。在图中， Fc 表示液压缸，动臂的重力，斗杆，铲斗的重力等在 B 点合力，其方向是沿着液压缸 AB 方向 ； Fc 可分解成 Fc1 和 Fc2 ，他们的方向分别为垂直于和平行于 O1B ，加速度 ac 的方向与 Fc 是相同的，并且 ac 也可以分解成 ac1 和 ac2； G1 ， G2 和 G3 分别是动臂，斗杆和铲斗的重心 ； m1， m2， m3是 它 们 各 自 的 质 量 且 能 通 过 实 验 给 定 （ m1=868.136kg ， m2=357.115kg and m3=210.736kg） ； Ol， O2 和 O3 是铰接点 ； G1， G2和 G3分别是 G1 ， G2 和 G3 在 X轴上的投影。 挖掘机的臂被认为是一个三个自由度的的机械手（三个 测斜仪 分别装在动臂，斗杆和铲斗上）。在跟踪控制实验中，其目标轨迹是根据挖掘机机械手运动学方程确定的。然后，动臂，斗杆和 铲斗的动作有操作员控制。为了适应自动控制，普通液压控制挖掘机应改造电动液压控制挖掘机。 基于 SW E-85 型原有的液压系统，把先导液压控制系统更换为先导电液控制系统。重庆交通大学二 OO 九届毕业设计（论文） 译文 新改进的液压系统如图 2 所示。在这系统中，因为动臂，斗杆和铲斗具有相同的特点，将动臂的液压系统作为一个例子。在先导电液控制系统中，先导电液比例阀是在原始的 SX-l4 主要阀门基础上增加比例泄压阀衍生出的并且用电子手柄替代液压手柄。 挖掘机的改装系统仍是具有良好的可控性的 LUDV 系统（图 3 ）。在图 3 中 ， y是可移动的活塞的位移； Q1 和 Q2 分别 代表流进和流出液压缸的流量； pl， p2， ps 和pr 分别表示汽缸的有杆腔和无杆腔，系统和回油路的压力； A1 和 A2 分别表示汽缸的有杆腔和无杆腔的面积； xv 代表阀芯的位移； m代表加载的负载； 图 1 挖掘机工作装示意图 图 2 挖掘机液压系统示意图 重庆交通大学二 OO 九届毕业设计（论文） 译文 图 3 改造后 LUDV 液压系统示意图 3 模型的电液比例系统 3.1 电动液压的比例阀门 动力学特性 在本文中，电液比例阀包括比例减压阀和 SX-14 主要阀 .传递功能从输入液流的阀芯位移可如下 ： Xv(s) Iv(s)=KI (1+bs) (1) 其中 Xv 是 xv 的 拉 普拉斯变换 值，单位为 m； KI 是电液比例阀获得的液流，单位为 m/A； b 是一阶系统的时间常数，单位为 s； Iv=I(t)-Id， I(t)和 Id 分别表示比例阀门的控制潮流和克服静带的各自潮流，单位为 A 。 3.2 电动液压的比例阀门的流体运动方程 在 本文中，实验性机器人挖掘机采取了 LUDV 系统。根据 LUDV 系统的理论，可以得到流体运动方程 ： 重庆交通大学二 OO 九届毕业设计（论文） 译文 11 2 pwxcQ vd 222 pwxcQvd = 其中 p 是负荷传感阀门的压力差，单位为 MPa； cd是径流系数，单位为 m5 (Ns)；w 是管口的面积梯度，单位为 m2 m； 是油密度，单位为 kg/m3； 1p 和 2p 分别为二个管口压力，单位为 MPa；当挖掘机流程没有饱和时， p 是一几乎恒定。在本文中，其值由实验测试得到。 在图 4 中， ps， p1s,和 p 分别表示系统压力、负荷传感阀门压力和它们的压力差；压力系统的实验曲线显示三种不同的压力值。虽然 ps 和 p1s随着荷载而改变，但是他们的区别不会随着荷载而改变，其值接近对 2.0MPa。因此，对横跨阀门的流量的作用 p可以被忽略。假设，流 过阀门的流量与管口阀门的大小成比例，并且荷载不影响流量。那么方程 (2)能被 简化为 ： Q1=Kqxv(t),I(t)0 (4) 其中 Kq是阀门流量系数，单位为 m2/s；并且 /2 pwcKdq 图 4 动臂移动压力曲线图 3.3 液压缸的连续性方程 一般来说，工程机械不允许外泄。当前，外在泄漏可以通过密封技术控制。另 一 0)(/2 1 tIppwxc rvd ， （ 2） 压力 0)(2 tIpwxc vd ， 0)(/2 2 tIppwxc rvd ， 0)(2 tIpwxc vd ， (3) 时间 重庆交通大学二 OO 九届毕业设计（论文） 译文 方面，由实验证明了挖掘机内部泄漏是相当小的。因此，液压机构内部和外在泄漏的影响可以被忽略。当油流进汽缸无杆腔并且进入到有杆腔内时，连续性方程可以写成 ： cpVyAQ /1111 cpVyAQ /2222 其中 V1 和 V2 分别表示流入及流出的液压缸液体的体积，单位是 m3；c是