Abstract: In miniature aircraft research and manufacture process, flying imitate experiment is essential , flying imitate experiment primarily includes experiment for check the airplane motive function and check airplane dynamic control fuction.

A semi-substance simulation test system is developed for the text and experiment of micro sensors and micro measuring units used in micro aerial vehicles. The simulator is designed with double functions. The bottom mechanism provides yawing angular motions. The middle mechanism provides pitching angular motion. And the top mechanism provides rolling angular motion. When the bottom mechanism is used individually, the system works as a rotating table. The mechanical system is constructed with three independent axes to fulfill the corresponding three angular motions. It adopted gear organization, worm, soft stalk organization to imitate three movement in the design.

The main function of this product is under the control system, can do movement accurately as follows :step forward as hand control, step backward as hand control, step forward automatically, step backward automatically. The designing turns have advantages such as constructed tightly ,can be manipulated conveniently, is well controled , and have powerful fuction etc.

Key words: Flight simulator Semi-substance simulator

1 前言………………………………………………………………………………….............. ( 1)

1.1 模拟转台的简介........................................................................................ ( 1)

1.2 研究概况及发展趋势.......................................................................... .( 2)

1.3 研究内容……………………………………………………………………………... .( 2)

2 转台机械系统方案的选择价…………………………………………………... ( 3)

2.1 控制台的功能分析………………………………………………………... ( 3)

2.2 方案的选择和评价……………………………………………………………................ ( 3)

3 原理设计及评价………………………………………………………............ (13)

3.1 原理一……………………………………………………………......................(13)

3.2 原理二……………………………………………………………......................(15)

3.3 原理三……………………………………………………………......................(17)

3.4 总述方案……………………………………………………………..................(18)

3.5 方案二的介绍……………………………………………………………...........(19)

4 齿轮机构设计……………………………………………………………..........(21)

4.1 失效形式…………………………………………………………....................(21)

4.2 设计准则……………………………………………………………...................(22)

4.3 齿轮设计……………………………………………………………...................(23)

5 蜗杆蜗轮设计…………………………………………………………..........(28)

5.1 蜗杆传动简介…………………………………………………………........... (28)

5.2 蜗杆传动的失效形式…………………………………………………….....(29)

5.3 蜗杆传动的材料选择……………………………………………….......... .(29)

5.4 蜗杆计算…………………………………………………………….................(30)

6 软轴设计与计算………………………………………………………........ .(34)

6.1 钢丝软轴的简介…………………………………………………………....(34)

6.2 钢丝软轴的选择与算…………………………………………………...... (36)

7 软轴的校核…………………………………………………………................(37)

7.1 按扭矩强度校核……………………………………………………….....(37)

7.2 按弯扭合成强度条核…………………………………………………...(37)

8 总结……………………………………………………..........................................(39)

参考文献………………………………………………………………...................... (41)

致谢…………………………………………………………………………………………… (42)

1 前言

§1.1模拟转台的简介

微型飞行器（Micro Aerial Vehicles 或Micro Air Vehicles,简称MAVs）的概念最初是美国科学家布鲁诺 •W•奥根斯坦在1992年美国国防高级研究计划局（Defense Advanced Research Projects Agency简称DARPA）主持的一次未来军事会议上提出的。现代的微型飞行器具有尺寸小、重量轻、隐身性能好、自主飞行，实时传输图象等突出优点，因而在军事、消防、交通等领域有独特的应用。

随着社会的发展，工业水平的提高，逐步的由大型化转向小型化，转化成微型化。在此发展的过程中，机械制造技术的水平与制造业也得到了有个较快的发展，它反映的是一个国家的经济实力和科学技术水平。新技术的推广和应用，使得新产品的不断涌现，与原有的事物相比，其体积是越来越小。机械工业历来都是发达国家的重要支柱产业，是一个国家的工业基础。从70年代开始，由于各国政府重视和发展高新技术，特别是微电子技术，微机技术的引入，使传统的机械工业在产品结构和生产系统结构等方面发生了质的变化，使其焕发了新的生命，形成了一个崭新的现代机械工业。

一个国家需安全、稳定，其国防技术就不能落后于其它国家。目前，各国在国防建设的投入是越来越多，逐年增加，为的就是研制出新的产品，使自己有优越于其他国家。现代军队的装备越来越重视其轻型化、技术化、现代化等。如：微型冲锋枪、微型机器人等。飞机在现代战争是必不可少的，其发展趋势也是向无人驾驶，微型化发展。

在微型飞行器的研制过程中，飞行仿真实验是必不可少的重要步骤。飞行仿真实验主要分为测试飞机飞行动力学性能的风洞实验和测试飞机动态控制性能以及各种机载传感器性能的地面半实物飞行仿真实验。只有通过仿真实验，获取足够多的飞行器性能数据，确证飞行器的外形设计符合动力学要求，控制系统具有足够的稳定性以及各种机载稳定性以及各种机载仪器能够在模拟工作环境中正常工作，才能进行飞行器的试飞工作。

三轴摇摆台是半实物飞行仿真实验系统中的一个关键设备。它可以按照实验要求，提供飞行器飞行时的航向角、俯仰角、横滚角以及飞行扰动，实时模拟飞行器在空气飞行的姿态。通过模拟飞行器的飞行姿态，测试飞行器控制系统能否在飞行器受到外界扰动时控制飞行器调整到安全飞行姿态。同时，还可以测试飞行器携带的机载传感器在模拟飞行条件下的工作状况。

§1.2 研究概况及发展趋势

目前，在微型飞行器模拟实验转台的研究开发方面，多采用齿轮传动，用步进电机驱动。一般是三轴飞行模拟转台，实验时能提供飞行器飞行时的航向角（偏转角）、俯仰角、横滚角（滚转角），即只有三个自由度。这种三轴飞行模拟转台，并不能完全模拟飞行器在空中的姿态。它采用三层转台提供三个方向的转动。在实际使用时，由三个独立的电机驱动。下层转台模拟飞行时的航向角，中层转台模拟飞行时的俯仰角，上层转台模拟飞行时的横滚角。

飞行器在空中飞行时，有六自由度，所以在风洞实验时，模拟转台的发展趋势是具有五个或六个自由度，要能真实的模拟飞行器在空中受到的各种力以及影响，并且使其智能化，即由计算机控制，以增加模拟的准确性。

§1.3 研究内容

1．搜集毕业设计相关资料，包括参考图纸、技术论文及外文资料。

2．对相关类型的模拟转台进行分析比较，并确定出新的传动方案，绘制出相应传动系统图。

3．绘制结构图，包括展开图和剖截图，并进行相关设计的计算。

4．综合计算结果及图，进行合理性检验。

5．确定方案并进性设计记录的修改、整理。

6．绘制总装配图。

7．确定驱动方式，并确定驱动动力来源。

8．撰写技术论文及设计说明书。

9．翻译外文资料。

2 转台机械系统方案的选择及评价

§2.1 控制台的功能分析：

微型飞行器模拟转台的有效载荷重量为300—500克，有效载荷空间为Φ150mm，能够模拟微型飞行器飞行时的偏转、俯仰、滚转，以及飞行扰动，实时模拟微型飞行器在空中的姿态。通过模拟飞行器的飞行姿态，测试飞行器控制系统能否在飞行器受到外界扰动时控制飞行器调整到安全飞行姿态。同时还可以测试飞行器携带的机载传感器在模拟飞行条件下的工作状况。

由于在微型飞行器模拟实验时，模拟转台要根据飞行器在空中飞行的实际情况进行模拟，所以，在传动方面要比较精确。设计要求中，模拟转台具有三个自由度。要实现三个自由度大致把它分为三个转台（上、中、下）实现，控制飞行器的偏转、俯仰、滚转。由于转动精度要较高，转动的角位移分辨率要低，连续转动速度范围要较大，运动角度范围要符合设计要求。所以要求转动的三轴必须共点，同时要求中层转台和上层转台有较高的同轴度、等高度。对要求转动的在以各种速度转动的过程中，转台运动要平稳，过渡过程要较迅速。直线运动的速度范围要较大，运动要平稳。直线运动与转动过渡过程要迅速、平稳。

全部文件.jpg

总图.jpg

字数统计.jpg

内容简介：: 南昌航空大学科技学院学士学位论文中文译文液压支架的最优化设计摘要：本文介绍了从两组不同参数的采矿工程所使用的液压支架（如图1）中选优的流程。这种流程建立在一定的数学模型之上。第一步，寻找四连杆机构的最理想的结构参数以便确保支架的理想的运动轨迹有最小的横向位移。第二步，计算出四连杆有最理想的参数时的最大误差，以便得出最理想的、最满意的液压支架。图1 液压支架关键词：四连杆机构；优化设计；精确设计；模糊设计；误差 1.前言：设计者的目的时寻找机械系统的最优设计。导致的结果是一个系统所选择的参数是最优的。一个数学函数伴随着一个合适的系统的数学模型的出现而出现。当然这数学函数建立在这种类型的系统上。有了这种数学函数模型，加上一台好的计算机的支持，一定能找出系统最优的参数。Harl描述的液压支架是斯洛文尼亚的Velenje矿场的采煤设备的一个组成部分，它用来支护采煤工作面的巷道。它由两组四连杆机构组成，如图2所示.四连杆机构AEDB控制绞结点C的运动轨迹，四连杆机构FEDG通过液压泵来驱动液压支架。图2中，支架的运动，确切的说，支架上绞结点C点竖向的双纽线的运动轨迹要求横向位移最小。如果不是这种情况，液压支架将不能很好的工作，因为支架工作在运动的地层上。实验室测试了一液压支架的原型。支架表现出大的双纽线位移，这种双纽线位移的方式回见少支架的承受能力。因此，重新设计很有必要。如果允许的话，这会减少支架的承受能力。因此，重新设计很有必要。如果允许的话，这种设计还可以在最少的成本上下文章。它能决定去怎样寻找最主要的图2 两四连杆机构四连杆机构数学模型AEDB的最有问题的参数。否则的话这将有必要在最小的机构AEDB改变这种设计方案。上面所罗列出的所有问题的解决方案将告诉我们关于最理想的液压支架的答案。真正的答案将是不同的，因为系统有各种不同的参数的误差，那就是为什么在数学模型的帮助下，参数允许的最大的误差将被计算出来。2.液压支架的确定性模型首先，有必要进一步研究适当的液压支架的机械模型。它有可能建立在下面所列假设之上：（1）连接体是刚性的，（2）单个独立的连接体的运动是相对缓慢的.液压支架是只有一个方向自由度的机械装置。它的运动学规律可以通过同步的两个四连杆机构FEDG和AEDB的运动来模拟。最主要的四连杆机构对液压支架的运动规律有决定性的影响。机构2只是被用来通过液压泵来驱动液压支架。绞结点C的运动轨迹L可以很好地来描述液压支架的运动规律。因此，设计任务就是通过使点C的轨迹尽可能地接近轨迹K来找到机构1的最理想的连接长度值。四连杆机构1的综合可以通过 Rao 和 Dukkipati给出运动的运动学方程式的帮助来完成。图3 点C轨迹L图3描述了一般的情况。点C的轨迹L的方程式将在同一框架下被打印出来。点C的相对应的坐标x和y随着四连杆机构的独有的参数一起被打印出来。点B和D的坐标分别是xB=x -cos (1)yB=y -sin (2)xD=x -cos() (3)yD=y -sin() (4) 参数也彼此相关xB2 +yB2= (5)(xD-1)2+ yD2= (6)把(1) (4)代入（5）（6）即可获得支架的最终方程式(x-cos)2+ (y- sin)2- =0 (7)x- cos()-2+ y- sin()2- =0 (8)此方程式描述了计算参数的理想值的最基本的数学模型。2.1数学模型Haug和Arora提议，系统的数学模型可以用下面形式的公式表示min f(u,v), (9)约束于gi(u,v)0, i=1,2,l, (10)和响应函数hi(u,v)=0, j=1,2,m. (11) 向量 u=u1,u2,unT 响应设计时的变量, v=v1,v2,vmT是可变响应向量，(9)式中的f是目标函数。为了使设计的主导四连杆机构AEDB达到最佳，设计时的变量可被定义为u= T, (12)可变响应向量可被定义为v=x yT. (13)相应复数3，5，6的尺寸是确定的。目标函数被定义为理想轨迹K和实际轨迹L之间的一些“有差异的尺寸”f(u,v) =maxg0(y)-f0(y)2, (14)式中x= g0(y) 是曲线K的函数，x= f0(y)是曲线L的函数。我们将为系统挑选一定局限性。这种系统必须满足众所周知的最一般的情况。 (15) (16)不等式表达了四连杆机构这样的特性：复数只可能只振荡的。这种情况： (17)给出了设计变量的上下约束条件。用基于梯度的最优化式方法不能直接的解决(9)(11)的问题。min un+1 (18)从属于gi(u,v) 0, i=1,2,l, (19)f(u,v)- un+10, (20)并响应函数hj(u,v)=0, j=1,2,m, (21)式中： u=u1 un un+1T v=v1 vn vn+1T因此，主导四连杆机构AEDB的一个非线性设计问题可以被描述为：min7, (22)从属于约束 (23) (24) , (25) (26)并响应函数： (27) (28) 有了上面的公式，使得点C的横向位移和轨迹K之间的有最微小的差别变得可能。结果是参数有最理想的值。3.液压支架的随机模型数学模型可以用来计算比如参数确保轨迹 L 和 K 之间的距离保持最小。然而端点C的计算轨迹L可能有些偏离，因为在运动中存在一些干扰因数。看这些偏离到底合时与否关键在于这个偏差是否在参数容许的公差范围内。响应函数（27）（28）允许我们考虑响应变量v的矢量，这个矢量依赖设计变量v的矢量。这就意味着vh (v)，函数h是数学模型（22）（28）的基础，因为它描述出了响应变量v的矢量和设计变量v的矢量以及和数学模型中v的关系。同样，函数h用来考虑参数的误差值的最大允许值。在随机模型中，设计变量的矢量u=u1,unT可以被看作U=U1,UnT的随机矢量，也就是意味着响应变量的矢量v=v1,vnT也是一个随机矢量V=V1,V2,VnT v=h(u) (29)假设设计变量 U1,Un 从概率论的观点以及正常的分类函数Uk (k=1,2,n)中独立出来。主要参数和 (k=1,2,n)可以与如测量这类科学概念和公差联系起来，比如=,。所以只要选择合适的存在概率， k=1,2,n (30)式（30）就计算出结果。随机矢量 V 的概率分布函数被探求依赖随机矢量 U 概率分布函数及它实际不可计算性。因此，随意矢量 V 被描述借助于数学特性，而这个特性被确定是利用Taylor的有关点 u=u1,unT 的函数h逼近描述，或者借助被Oblak和Harl在论文提出的Monte Carlo 的方法。3.1 数学模型用来计算液压支架最优化的容许误差的数学模型将会以非线性问题的独立的变量 w= (31)和目标函数 (32)的型式描述出来。约束条件 (33) , (34)在式（33）中，E是是坐标C点的x 值的最大允许偏差，其中 A=1,2,4 (35)非线性工程问题的计算公差定义式如下： (36)它服从以下条件： (37) , (38) (39)4.有数字的实列液压支架的工作阻力为1600kN。以及四连杆机构AEDB及FEDG 必须符合以下要求：它们必须确保铰接点C 的横向位移控制在最小的范围内，它们必须提供充分的运动稳定性图2中的液压支架的有关参数列在表1 中。支撑四杆机构 FEDG 可以由矢量 (mm) (40)来确定。四连杆AEDB 可以通过下面矢量关系来确定。 (mm) 在方程(39)中，参数d是液压支架的移动步距，为925mm .四连杆AEDA的杆系的有关参数列于表2中。表 1 液压支架的参数表 2 四连杆AEDA的参数4.1四连杆AEDA的优化四连杆的数学模型AEDA的相关数据在方程(22)-(28)中都有表述。(图3)铰接点C双纽线的横向最大偏距为65mm。那就是为什么式(26)为 (41)杆AA与杆AE之间的角度范围在76.8o和94.8o之间，将数依次导入公式(41)中所得结果列于表3中。这些点所对应的角都在角度范围76.8o,94.8o内而且它们每个角度之差为1o设计变量的最小和最大范围是 (mm) (42) (mm) (43)非线性设计问题以方程(22)与(28)的形式表述出来。这个问题通过Kegl et al(1991)提出的基于近似值逼近的优化方法来解决。通过用直接的区分方法来计算出设计派生数据。设计变量的初始值为 (mm) (44)优化设计的参数经过25次反复计算后是表3 绞结点C对应的x与y 的值角度x初值(mm)y初值(mm)x终值(mm)y终值(mm)76.866.781784.8769.471787.5077.865.911817.6768.741820.4078.864.951850.0967.931852.9279.863.921882.1567.041885.0780.862.841913.8566.121916.8781.861.751945.2065.201948.3282.860.671976.2264.291979.4483.859.652006.9163.462010.4384.858.722037.2862.722040.7085.857.922067.3562.132070.8786.857.302097.1161.732100.7487.856.912126.5961.572130.3288.856.812155.8061.722159.6389.857.062184.7462.242188.6790.857.732213.4263.212217.4691.858.912241.8764.712246.0192.860.712270.0866.852274.3393.863.212298.0969.732302.4494.866.562325.8970.502330.36 (mm) (45) 在表3中C点x值与y 值分别对应开始设计变量和优化设计变量。图 4 用图表示了端点 C开始的双纽线轨迹 L(虚线)和垂直的理想轨迹K(实线)。图4 绞结点C 的轨迹4.2 四连杆机构AEDA的最优误差在非线性问题(36)-(38),选择的独立变量的最小值和最大值为 (mm) (46) (mm) (47)独立变量的初始值为 (mm) (48)轨迹偏离选择了两种情况E=0.01和E=0.05。在第一种情况，设计变量的理想公差经过9次反复的计算，已初结果。第二种情况也在7次的反复计算后得到了理想值。这些结果列在表 4和表5 中。图 5和图 6的标准偏差已经由Monte Carlo方法计算出来并表示在图中(图中双点划线示)同时比较泰勒近似法的曲线(实线)。图5 E=0.01时的标准误差图6 E=0.05时的标准误差5.结论通过选用系统的合适的数学模型以及采用数学函数，让液压支架的设计得到改良，而且产品的性能更加可靠。然而，由于理想误差的结果的出现，将有理由再考虑一个新的问题。这个问题在四连杆的问题上表现的尤为突出，因为一个公差变化稍微都能导致产品成本的升高。10南昌航空大学科技学院学士学位论文 Optimal design of hydraulic supportm. oblak. Harl and b. butinarAbstract :This paper describes a procedure for optimal determination of two groups of parameters of a hydraulic support employed in the mining industry. The procedure is based on mathematical programming methods . In the first step, the optimal values of some parameters of the leading four-bar mechanism are found in order to ensure the desired motion of the support with minimal transversal displacements. In the second step, maximal tolerances of the optimal values of the response of hydraulic support wil be satisfying.Keywords: four-bar mechanism, optimal design, mathematical programming , approximation method, tolerance1 IntroductionThe designer aims to find the best design for the mechanical system considered. Part of thie effort is the optimal choice of some selected parameters of a system. Methods of mathematical programming can be used, Of course, it depends on the type of the systemWith this foemulation, good computer support is assured to look for optimal parameters of the system. The hydraulic support (Fig.1) described by Harl (1998) is a part of the mining industry equipmenr port in the mine Velenje-Slovenia, used for protection of working environment in the gallery. It consists of four-bar mechanisms FEDG and AEDB as shown in Fig.2. The mechanism AEDB defines the path of coupler point C and the mechanism FEDG is used to drive the support by a hydraulic actuator。Fig. 1 Hydraulic supportIt is required that the motion of the support,more precisely, the motion of the point C in Fig.2, is vertical with minimal transversal displacements. If this is not the case, the hydraulic support will not work properly because it is stranded on removal of the earth machine.A prototype of the hydraulic support was tested in a laboratory (Grm 1992). The support exhibited large transversal displacements, which reduce its employability. Thetefore, a redesign was necessary. The project should be improved with minimal cost if possible. It was decided to find the best values for the most problematicFig.2 Two four-bar mechanismsParameters of the leading four-bar me AEDB with methods of mathematical programming. Otherwise it chanisms would be necessary to change the project, at least mechanism AEDB. The solution of above problem will give us the response of hydranlic support for the ideal system. Real response will be different because of tolerances of various parmeters of the system, which is why the maximal allowed tolerances of paramentsa1,a2,a3,a4 will be calculated support. ,with help of mathematical programming.2 The deterministic model of the hydraulic supportAt fist it is necessary to develop an appropriate metical model of the hydraulic support.It could be based on the following assumptions:- the links are rigid bodies,- the motion of individual is relatively slow.The hydraulic support is a mechanism with one degree of freedom. Its kinematics can be model consists of four-bar mechanisms FEDG and AEDB (Oblak et al. 1998).The leading four-bar mechanisms AEDB with methods of mathematical programming. Otherwise it would be necessary to change the project, at least mechanism AEDB. It is required that the motion of the support,more precisely, the motion of the poit C. Therefore, the path of coupler point C is as near as possible to the desired trajectory k.The synthesis of the four-bar mechanism 1 has been performed with help of motion given by Rao Dukkipati(1989). The general situation is depicted in Fig,3. Fig.3 Trajectory L of the point CEquations of trajectory L of the point C will be written in the coordinate frame considered. Coordinates x and y of the point C will be written with the typical parameters of a four-bar mechanism a1,a2,.a6.The coordinates of points B and D are xBcos (1)yB=sin (2)xD=cos() (3)yD=sin() (4) The parameters are related to each other byxB2+ (5)1)2+ yD2= (6)By substituting (1) (4) into （5）（6）the response equations of the support are obtained as(xcos)2+ (y- sin)2- =0 (7)x- cos()-2+ y- sin()2- =0 (8) This equation represents the mathematical model for calculating the optimal values of paramerters a1,a2,a4.2.1 Mathematical model The mathemtial model of the system will be formulated in the from proposed by Haug and Arora (1979):gi(u,v)0, i=1,2,l, (10)and response equationshi(u,v)=0, j=1,2,m. (11) The vector u=u1,u2,unT is called the vector of design variables, v=v1,v2,vmTis the vector of response variables and f in(9)is the objective function. Tobperform the optimal design of the leading four-bar mechanism AEDB,the vector of design variables is defined asu= T, (12)and the vector of response variables asv=xyT. (13)The dimensions 3，5，6 of the corresponding links are kept fixed. Thef(u,v) =maxg0(y)-f0(y)2, (14)where x= g0(y) is the equation of the curve K and x= f0(y) is the equation of the curve L.Suitable limitations for our system will be chosen.The system must satisfy the well-known Grasshoff conditions (15) (16)Inequalities (15) and (16) express the property of a four-bar mechanism, where the links may only oscillate.The condition： (17)Prescribes the lower upper bounds of the design variables.The problem (9)(11)is not dirrctly solvable with the usual gradient-based optimization methods. This could be cirumvented by int express the property of the objective function be written with the typical parameters be written asminun+1 (18)sobject togi(u,v) 0, i=1,2,l, (19)f(u,v)- un+10, (20)and response equationshj(u,v)=0, j=1,2,m, (21)where： u=u1 un un+1T v=v1 vn vn+1TA nonlinear programming problem of the leading four-bar mechanism AEDB can therefore be difined asmina7, (22)sobject to constraints (23) (24) , (25) (26)And respose equationt (27) (28)3.The stochastic model of the hydraulic support The mathematical model can be used to calculate the parameters of L and K to ensure that the track such as to maintain the distance between the minimum. However the endpoint C calculation L may deviate from the track, because of the movements in the presence of some interference factor. Look at these deviations from what should or not lies in the deviation is in the parametric tolerance tolerance range.Response function (27) - (28) allows us to consider the response variable V vector, the vector of dependent variable V vector design. This means that v = H ( V, H ) function is a mathematical model (22) - (28) foundation, because it describes a response variable V vector and V vector as well as design variables and the mathematical model of the relationship between v. Similarly, the function H used to consider the parameter errors in the value of the maximum permissible value.In the stochastic model, design variable vector u=u1, . , unT can be viewed in U=U1, . , UnT random vector, which means the response variable vector v=v1, . , vnT is a random vector V=V1, V2, . , VnTV=h ( U ) (29)Suppose design variable U1, . , Un from probability theory and the classification of normal function of Uk ( k=1,2, . , n ) of independence. The main parameters and ( k=1,2, . , n ) can be associated with such as the measurement of this kind of scientific concepts and tolerance to link, such as a =,. So as long as the choice of suitable existence probability, k=1,2, . , n (30)Type (30) is calculated the results of.Random vector V probability distribution function is search for dependent random vector U probability distribution function and its actual computability. Therefore, random vector V is described by mathematical properties, and the properties were identified using Taylor on u=u1, . , unT h approximation function description, or with the aid of Oblak and Harl in the Monte Carlo method.3.1 The mathematical modelUsed to calculate the allowable error of hydraulic support optimization mathematical model will be nonlinear problem of independent variablew= (31)and objective function (32)With conditions (33) , (34)In（33），E is the maximal allowed standard deviation of coordinate x of the point C and A=1,2,4 (35)The nonlinear programming problem for calculating the optmal tolerances could be therefore defined as ： (36)Subject to constraints (37) , (38)4.Numerical examply The carrying of the hydraulic support is 1600kN. Both four-bar AEDB andFEDG must fulfill the following demand：they must allow minimal transversal displacements of the point C, and，they must provide sufficient side stability. The parameters of the hydraulic support (Fig.2) are given in Table 1. The drive mechanism FEDG is specified by the vector (mm) (39)And the mechanism AEDB by (mm) In(39),the parameter d is a walk of the support with maximal value of 925 mm. Parameters for the shaft of the mechanism AEDB are given in Table 2. 4.1Four connecting rod AEDA optimizationFour link model AEDA related data in equation (22) - (28) are expressed. ( Fig 3). C lemniscate of maximum horizontal offset for65mm. That is why type (26) for the (41)Rod and bar between AE AA angle in the range of 76.8o and 94.8o, will be a number successively introduced formula (41) obtained results are listed in table 3.These points corresponding to the angle ofin the range of 76.8o,94.8o and they each angle difference of1The design variables of the minimum and maximum range is (mm) (42) (mm) (43)Nonlinear design problems in equation (22) and (28) in the form of. This problem byKegl et al (1991) based on the approximation of the optimal approximation solution. By using the direct method of distinguishing to calculate design derived data.Design variables for the initial value (mm) (44)Optimization of design parameters through calculation is repeated 25 timesTable3 node C corresponding to the X and Y values角度x初值(mm)y初值(mm)x终值(mm)y终值(mm)76.866.781784.8769.471787.5077.865.911817.6768.741820.4078.864.951850.0967.931852.9279.863.921882.1567.041885.0780.862.841913.8566.121916.8781.861.751945.2065.201948.3282.860.671976.2264.291979.4483.859.652006.9163.462010.4384.858.722037.2862.722040.7085.857.922067.3562.132070.8786.857.302097.1161.732100.7487.856.912126.5961.572130.3288.856.812155.8061.722159.6389.857.062184.7462.242188.6790.857.732213.4263.212217.4691.858.912241.8764.712246.0192.860.712270.0866.852274.3393.863.212298.0969.732302.4494.866.562325.8970.502330.36 (mm)In Table 3 of C x values and Y values respectively corresponding to the design variables and the variables of optimization design line )。Fig 4diagram represents the endpoint C started the lemniscate locus L ( dotted line) and perpendicular to the ideal trajectory K ( solid line).Lateral displacement and the trajectory of Figure 4graph represents the endpoint C started the le

温馨提示:
1: 本站所有资源如无特殊说明，都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
2: 本站的文档不包含任何第三方提供的附件图纸等，如果需要附件，请联系上传者。文件的所有权益归上传用户所有。
3.本站RAR压缩包中若带图纸，网页内容里面会有图纸预览，若没有图纸预览就没有图纸。
4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
5. 人人文库网仅提供信息存储空间，仅对用户上传内容的表现方式做保护处理，对用户上传分享的文档内容本身不做任何修改或编辑，并不能对任何下载内容负责。
6. 下载文件中如有侵权或不适当内容，请与我们联系，我们立即纠正。
7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

人人文库网所有资源均是用户自行上传分享，仅供网友学习交流，未经上传用户书面授权，请勿作他用。