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挖掘机臂液压系统的模型化参量估计摘 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2.825104m3/（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 E85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。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SX14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： （2）(3)=其中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 m2m；是油密度，单位为 kg/m3；和分别为二个管口压力，单位为 MPa;当挖掘机流程没有饱和时，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；并且压力时间图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:(5)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；是有效体积模量(包括液体,油中的空气等）,单位是N/m2。3.4 液压缸力的平衡方程据推测，液压缸中油的质量可以忽略，而且负载是刚性的。那么可以根据牛顿的法律得到液压缸的力量平衡等式： (6）其中Bc是黏阻止的系数，单位是 Ns/m。3。5 电动液压的比例系统简化的模型 方程(4)(6）在拉伯拉斯变换以后，简化的模型可以表达为： （7)其中Y是y拉伯拉斯变换得到的；;bf=V1V2；a0=V1V2m；a1=BcV1V2；。4 参量估计 从塑造的过程和方程（7）中可以得到在确切的简化的模型中与结构，运动情况以及挖掘机动臂的体位有关的所有参量.而且,这些参量是时变。因此要得到这些参量的准确值和数学等式是相当难的。要解决这个问题，本文提出了估计方程和方法来估算模型中的这些重要参数。4。1 估算液压缸负载液压缸臂上的负载（假定没有外部负载）由动臂，斗杆和铲斗上的负载组成。在图1中，动臂,斗杆和铲斗分别绕着各自的铰接点旋转。因此他们的运动不是沿着汽缸的直线运动，也就是说他们的运动方向与方程（5）中的y的方向是不同的。因此方程（6）中的m不能简单的认为是动臂，斗杆和铲斗质量的总和。考虑到机械手的坐标轴心O1，机械手的转矩和角加速度可考虑如下：(8)其中的M 和 分别是工作装置对O1的转矩和角加速度.是点O1到点B的长度；由转动定律M=J可得：，即： (9）其中的J是工作装置指向O1的等效转动惯量，单位是kgm2;并且写成如下式子： (10）J1， J2 和 J3分别是动臂,斗杆和铲斗对各自的中心的惯性力矩；它们的值可以通过模拟动态模型得出J1=450.9Nm，J2=240.2Nm,J3=94。9Nm。比较方程（9)和Fc=mac，可以得出点B的等效质量： （11)4。2 液压缸负载的估算 工作装置对于O1等效力矩等式为： （12)其中和分别表示O1点到 G1 ，G2和 G3三点的距离；那么反力负荷为： (13）4。3增益系数阀流量的估计流量传感器可以测量泵的流量。用于这项工作的仪器为多系统5050型。动臂液压缸流量的阶跃响应在电液比例阀控制下的结果如图5所示。同时，该曲线验证等式(11) 。根据实验曲线和等式（1）和（4)可确定KqKl的范围。那么根据图4中的数据我们可得出：KqKl=2。825104m3/(sA) .流量（L/min）时间图5 动臂液压缸流量的阶跃响应在电液比例阀控制下的曲线图5 结论（1）电液控制系统的数学模型是根据挖掘机的特点发展起来的.假定流过阀的流量与阀口大小成正比,并忽略液压系统的内部和外部泄漏影响。简化模型可以得到： ，其中Y（s）和Xv（s)分别是活塞和阀芯的位移.（2）从电液控制系统的模型中，我们可以得到等效的质量，承载力，流量增益系数的值KqKl=2.82510-4m3/（sA）,其中KI 是电液比例阀的增益系数.出自:中南大学学报(英文版）2008年第15卷第3期382-386页Modeling and parameter estimation for hydraulic system of excavators armHE Qinghua，HAO Peng,ZHANG Da-qingAbstractA retrofitted electrohydraulic 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 electrohydraulic 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 electrohydraulic 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 estimation1 IntroductionFor its high efficiency and multifunction， hydraulic excavator is widely used in mines，road building, civil and military construction，and hazardous waste cleanup areasThe hydraulic excavator also plays an important role in construction machinesNowadays, macaronis and mobilization have been the latest trend for the construction machinesSo，the automatic excavator gradually becomes popular in many countries and is considered a focusMany control methods can be used to automatically control the manipulator of excavatorWhichever method is used， the researchers must know the structure of manipulator and the dynamic and static characteristics of hydraulic systemThat 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 outsideResearches on time delay control for excavator were carried out in RefsNGUYE used fuzzy sliding mode control and impedance control to automate the motion of excavators manipulator SHAHRAM et al adopted impedance control to the teleported excavatorNonlinear models of hydraulic system were developed by some researchers. However, it is complicated and expensive to design controller, which 1imits its applicationIn 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 electrohydraulic proportional valve;at the same time，the estimation methods and equations for the parameters of model were developed2 Overview of robotic excavatorThe backhoe hydraulic excavator studied is shown in Fig.1In 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 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 manipulatorThen，the motion of boom,dipper an d bucket was set by the controllerIn order to suit for automatic contro1the normal hydraulic control excavator should be retrofitted to electrohydraulic controllerBased on original hydraulic system of SW E85The hydraulic pilot control system was replaced by an electrohydraulic pilot control systemThe retrofitted hydraulic system is shown in Fig.2In this work，because boom，dipper an d bucket are of the same characteristics，the hydraulic system of boom was taken as an exampleIn 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 oneThe retrofitted system of excavator was still the LUDV system (Fig。3）of Rexroth with good controllabilityIn 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 armFlg。2 Schematic diagram of retrofitted electro-hydraulic system of excavatorFlg.3 Schematic diagram of LUDV hydraulic system after retrofitting3 Model of electro-hydraulic proportional system3。1 Dynamics of electrohydraulic proportional valveIn this work， the electro-hydraulic proportional valve consists of proportional relief valves and SX-14 main valveA 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 electrohydraulic proportional valves，mA；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，A3.2 Flow equation of electrohydraulic proportional valveIn this work，LUDV system was adopted in the experimental robotic excavatorAccording to the theory of LUDV system，the flow equation can be gotten: (2)(3)=where is the springsetting pressure of load sense valve，MPa；cd is the flow coefficient m5（Ns）；w is the area gradient of orifice，m2m； is the oil density， kgm3;and are the two orifices pressure，respectively， M PaWhen the flow of excavator is not saturated，is a nearly constantIn this work,the value was tested and gotten by experimentIn Fig。4，ps,p1s，andrepresent 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 pressuresAlthough 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 neglectedIt is assumed that the flow across the valve is proportional to the size of orifice valve，and the flow is not influenced by loadThen，Eqn(2） can be simplified as Q1=Kqxv（t),I(t)0 （4)where is the flow gain coefficient of valve， m2s, and Flg.4 Curves of pressure experiment under boom moving condition3。3 Continuity equation of hydraulic cylinder Generally speaking，construction machine does not permit external leakageAt present，the external leakage can be controlled by sealing technologyOn the other hand，it has been proven that the internal leakage of excavator is quite little by experimentsSo， the influence of internal and external leakage of hydraulic system can be ignoredWhen the oil flows into head side of cylinder and discharges from rod side， the continuity equation can be written as (5)where V1 and V2 are the volumes of fluid flowing into and out the hydraulic cylinder, m3 ； is the effective bulk modulus（including liquid，air in oil and so on)，N/m23.4 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: （6）where Bc is the viscous damping coefficient，Nsm3.5 Simplified model of electro-hydraulic proportional systemAfter the Laplace transform of Eqns(4）（6)，the simplified model can be expressed as （7）where Y（s） is the Laplace transform of y；;b1=V1V2；a0=V1V2m;a1=BcV1V2;.4 Parameters estimationFrom 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 armMoreover，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 work4.1 Equivalent mass estimation for load on hydraulic cylinderThe load of boom hydraulic cylinder（it is assumed there is no external load）consists of boom，dipper and bucketIn Fig。1，boom，dipper and bucket rotate around points O1，O2 and O3,respectivelySo 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 bucketConsidering O1 at an axis of manipulator， the torque and angular acceleration can begiven as follows:(8)where M and are the torque and angular acceleration of manipulator to O1，respectively; is the length from point O1 to point BAccording to the rotating law：M=J,we get that is （9） where J is the equivalent moment inertia of manipulator to point O1,kgm2,and it can be written as follows: (10）J1， J2 and J3 are the moment inertia of boom，dipper and bucket to their own bary center respectivelyThe 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： （11)4.2 Estimation for load on hydraulic cylinderThe equivalent moment equation of manipulator to O1 is （12)where and are the length from pointO1 to point G1 ，G2and G3；，respectivelyThen，the counter force of load is （13）4。3 Estimation for flow gain coefficient of valveThe flow of pump can be measured by flow transducer The instrument used in this work was Multisystem 5050The step response of flow of boom cylinder under the electrohydraulic proportional valve controlled by the step curent is shown in Fig。5At the same time，the curve verifies Eqn11Based 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。825104m3/(sA）Flg.4 Flow of boom cylinder under electro-hydeaulic proportional value controlled by step current5 Conclusions（1)The mathematic model of electrohydraulic system is developed according to the characteristics of excavatorIt 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 ignoredThe simplified model can be obtained：where represent the displacement of piston and the displacement of spool（2）From the model of electrohydraulic system,we can obtain the equivalent mass ,bearing force , flow gain coefficient of value KqKl=2。82510-4m3/(sA） ，where KI is the current gain of electrohydraulic proportional valvesFrom： Journal of Central South University (English) 2008 Vol 15 No. 3 pages 382-386
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