挖掘机外文翻译外文文献中英翻译

挖掘机外文翻译外文文献中英翻译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 2.0MP&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2.825 X10- 4m3/(s A),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。NGUYE利用模糊的滑动方式和阻抗来控制挖掘机动臂的 运动，SHAHRAM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Fci和 Fc2 ,他们的方向分别为垂直于和平行于 OB ,加速度ac 的方向与Fc是相同的，并且ac也可以分解成aci和ac2； G , G2和G分别是动臂，斗杆和铲斗的重心；m, m,m3是它们各自的质量且能通过实验给定(m=868.136kg , m2=357.115kg and m 3=210.736kg) ; Oi, Q 和 Q是较接 点；G ,G和G3分别是G , G2和G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(三个分别装在动臂，斗杆和铲 斗上)0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。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-14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；Qi和Q2分别代表流进和流出液压缸的流量；pi, 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；Ai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尸Ki/(1+bs)(1)其中Xv是Xv的值，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 Ao3.2 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：QiCdWXv2piCdWXvj2 pj , I 0CdWXv J2 PiPr / , I (t) 0CdWXv J2 P2Pr / , I (t) 0Q2CdWXv - P2CdWXvj2 p/ , I(t) 0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/ (N s)； w是管口的面积梯度，单位为 m2/m； p是油密度，单位为kg/m3; Pi和P2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是一几乎包定。在本文中，其值由实验测试得到。在图4中，Ps, Pis,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pis随着荷载而改变，但是他们的区别不会随着荷载而改变，具值接近对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：Qi = KqXv(t),I(t) 0(4)其中Kq是阀门流量系数，单位为 m2/s；并且Kqcdwj2 p/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：?-Qi Ai y Vi pl cT?LQ2 A2 y V 2 P2/ c其中Vi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(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：?PiAi p2A2 my Bc y Fc(6)其中Bc是黏阻止的系数，单位是N s/m。3.5 电动液压的比例系统简化的模型方程一(6)在拉伯拉斯变换以后，简化的模型可以表达为：Y sbiXv s bfsFc s sa0s2 a1s a22其中Y是y拉伯拉斯变换得到的；bicKq?AV2 4 /A ； bf=ViV2；ao=ViV2m；22ai=BcViV2； a2cV2A ViA,。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较接点旋转。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Oi,机械手的转矩和角加速度可考虑如下：Fclo1BSin ac sin &(8)其中的M和分别是工作装置对Oi的转矩和角加速度。Qb是点Oi到点B的长度；由转动定律M=J可得：FclOiBsinJacsin /loiB ,即：Fc aJ/l2OiB(9)其中的J是工作装置指向Oi的等效转动惯量，单位是kg m2;并且写成如下式子：222J J imlOiGiJ 2rm21O2J 3rm31O3(i0)Ji, 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 Ji=450.9Nm, J2=240.2N m, J3=94.9N m。比较方程(9)和Fc=mac,可以得出点B的等效质量：2m J/12OiB(ii)4.2 液压缸负载的估算工作装置对于Oi等效力矩等式为：FclOiB sinmiglOiGi 仁m3glOiG3(i2)其中I。，L 和he分别表示Oi点至1J Gi，G2和G3三点的距离；那么反力负 OiGi O|G2OiG 3荷为：FcmiglOiGi m2glOiG2m3glO lOiB sin(i3)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KqKi的范围。那么根据图4中的数据我们 可得出：KqKi=2.825 10-4m3/(s A)。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：Y(s) bXv(s) b1sFc(s)/s(%s2 as a2),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m J/12of,承载力Fl(mMoGm2g16Gm31GJ ,流量增益系数的值 KqKi=2.825 10-4m3/(s A),其中O1G1O1G2QG3Ki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 excavator s armHE Qing-hua, HAO Peng, ZHANG Da-qingAbstractA 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-hydraulicproportional valve was controlled by the step current. Based on the experiment curve, the flow gain coefficient of valve unidentified as 2.825 10-4m3/(s A) and I shutt】。valvtXad Sen sc I vaIvc .p Jshiltiiw) -IRcxrolh SX1-1 valve，r=。Flg.3 Schematic diagram of LUDV hydraulic system after retrofitting3 Model of electro-hydraulic proportional system3.1 Dynamics of electro hydraulic proportional valveIn 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尸KI/(1+bs)(1)where Xv is the Laplace transform of x/, 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 valveIn this work, LUDV system was adopted in the experimental robotic excavator. According to the theory of LUDV system , the flow equation can be gotten, cdwxv2 p/ , I(t) 0Qi cd wxvPi -J(2)Icdwxvj2 phpr / , I(t) 02cd WXv 72Pr / ，I (t) 0Q2cdwXvj p2 =YIcdw.j2 p, I(t) 0where p is the spring-setting pressure of load sense valveMPa； cd is the flow coefficient m5/(N s)； w is the area gradient of orifice, m2/m； p is the oil densitykg/ 3m ; p1 and p2 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 pressure sAlthough 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 load. Then, Eqn. (2) can be simplified asQ=Kqxv(t),I(t) 0(4)where is the flow gain coefficient of valve, m2/s, and Kqow 2 p/04 H 1216Tims Flg.4 Curves of pressure experiment under boom moving condition3.3 Continuity equation of hydraulic cylinderGenerally 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 experiment tsSo, the influence of internal and external leakage of hydraulic system can be ignore d When the oil flows into head side of cylinder and discharges from rod side the continuity equation can be written as?-QiAi y Vi Pi/ cn?Q Q2 A2 y V 2 P2/ cwhere Vi 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 cylinderIt 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 theNewton s second la w ?P1A1P2 A2my Bcy Fc(6)where Bc is the viscous damping coefficient, N s/m.3.5 Simplified model of electro hydraulic proportional systemAfter the Laplace transform of Eqns. (4) (6), the simplified model can be expressed asY sb1Xv s bfs艮 s s a0s2 a1s a2(7)where Y(s) is the Laplace transform of y 222b(cKq?AV2A2/ A1;bi=ViV2；ao=ViV2m;ai=BcViV2； a2cv2AViA24 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 ofexcavators arrMfloreover, these parameters are time variable. So it is quite difficult toget accurate values and mathematic equations of these parameters. To solve this prob |em 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 bucket In Fig.1, boom, dipper and bucket rotate around points O, 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 boq mdipper and bucketConsidering Oi at an axis of manipulator, the torque and angular acceleration can begiven as follows：1 MFciloiBFcMsin1-11(8)I _aci lo1Bac sinlo1Bwhere M and are the torque and angular acceleration of manipulator to O respectively;lQB is the length from point Oi to point B. According to the rotating law: M=J ,we getFclOi B 1 nvJ c 1/ lO|B-2，一、that isFc acJ/l OiB(9)where J is the equivalent moment inertia of manipulator to point O,kg m2, and it can be written as follows：222J JimJ01GlJ2m2l01G2J3m3l01G3(i0)Ji, J2 and S are the moment inertia of boon dipper and bucket to their own bary center respectively The values of them can be obtained by dynamic simulation based on the dynamic mode, J=450.9N m, J2=240.2N m, J3=94.9N m.Comparing Eqn. (9)with Fc=mac, the equivalent mass at point B can be given2m J/l2OiB(11)4.2 Estimation for load on hydraulic cylinderThe equivalent moment equation of manipulator to O isFclOiB sinmlglOiGim2glOiG2m3glOiG3(12)where ,l” and l” are the length from point。to point Gi, G2 and , 。1 Gi OG 2OG 3respectively. Then, the counter force of load isFCm1glQGm2gloGm3glQGlO1Bsin(13)GiO1G2G314.3 Estimation for flow gain coefficient of valveThe 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 timq 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=2.825 10-4m3/(s A).Flg.4 Flow of boom cylinder under electro-hydeaulic proportional value controlled by stepcurrent5 Conclusions(1) The mathematic model of electro hydraulic system is developed according tothe 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 ofhydraulic system is ignored. The simplified model can be obtainedY(s) hXv(s) hsFc(s)/s(a0s2 &s a?)where represent the displacement of piston and the displacement of spo ol(2) From the model of electro hydraulic system, we can obtain the equivalent2mass m J /l qb , bearing forceFl(m1g10Gm2g101G m3lOG3) , flow gain coefficientof value KqKl=2.825 10-4m3/(s A) , 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