CA10B中間軸軸承支架加工工藝規(guī)程及車銑夾具設(shè)計(jì)【2套】【三維UG工件】【含全套CAD圖紙】

喜歡就充值下載吧。資源目錄里展示的全都有，下載后全都有，圖紙均為CAD原圖，有疑問咨詢QQ:414951605 或1304139763 山東華源萊動內(nèi)燃機(jī)有限公司專用夾具設(shè)計(jì)實(shí)習(xí)報(bào)告 姓 名： 學(xué) 號： 班 級： 指導(dǎo)教師： 機(jī)械工程學(xué)院青 島 理 工 大 學(xué)一、實(shí)習(xí)基本情況實(shí)習(xí)單位： 山東華源萊動內(nèi)燃機(jī)有限公司實(shí)習(xí)時(shí)間： 2010年3月08日至3月19日實(shí)習(xí)流程：一、進(jìn)行實(shí)習(xí)準(zhǔn)備二、前往實(shí)習(xí)公司三、了解公司的發(fā)展概況、運(yùn)營情況等，接受進(jìn)廠前安全教育四、學(xué)習(xí)、觀看萊動內(nèi)燃機(jī)零件專用夾具設(shè)計(jì)圖紙五、了解內(nèi)燃機(jī)各車間的整個(gè)產(chǎn)品加工，檢驗(yàn)，包裝等過程六、上午詳細(xì)了解幾種專用夾具在加工過程中的應(yīng)用七、返校八、實(shí)習(xí)總結(jié)，整理報(bào)告二、公司基本情況山東華源萊動內(nèi)燃機(jī)有限公司位于山東省萊陽市， 1943年建廠，至今已有60多年歷史，是中國北方重要的中小功率柴油機(jī)生產(chǎn)基地。公司主要從事發(fā)動機(jī)及零部件的開發(fā)設(shè)計(jì)、生產(chǎn)、銷售及相關(guān)進(jìn)出口貿(mào)易。 目前的主導(dǎo)產(chǎn)品有：為中小型拖拉機(jī)、低速載貨汽車配套的單缸系列柴油機(jī)；為低速載貨汽車、微型汽車、中型拖拉機(jī)、工程機(jī)械、收獲機(jī)械配套的L系列自然吸氣機(jī)型和增壓中冷機(jī)型等小缸徑多缸柴油機(jī)；為輕型卡車、工程機(jī)械、收獲機(jī)械配套的D系列自然吸氣機(jī)型和增壓中冷機(jī)型等高速車用柴油機(jī)。公司與國外先進(jìn)企業(yè)合資合作、聯(lián)合開發(fā)的具有國際一流水平，排放達(dá)到歐、歐標(biāo)準(zhǔn)的系列柴油發(fā)動機(jī)產(chǎn)品已經(jīng)實(shí)現(xiàn)批量生產(chǎn)。此外，公司還對外承攬各種鑄件、油泵、發(fā)電機(jī)、起動機(jī)等業(yè)務(wù)，從而逐步形成以發(fā)動機(jī)為主業(yè)，各種零部件、收獲機(jī)械、水泵機(jī)組、發(fā)電機(jī)組、船用機(jī)組等齊頭并進(jìn)的產(chǎn)業(yè)布局。多年來，“萊動牌”產(chǎn)品以其先進(jìn)的性能、優(yōu)良的質(zhì)量，多次榮獲國家權(quán)威部門的質(zhì)量管理獎和科技進(jìn)步獎。產(chǎn)品現(xiàn)與國內(nèi)主要的汽車、農(nóng)用車和拖拉機(jī)生產(chǎn)廠家建立了穩(wěn)定的配套關(guān)系，暢銷全國29個(gè)省、市、自治區(qū)，并遠(yuǎn)銷東南亞、中東、南美等20多個(gè)國家和地區(qū)。60多年的內(nèi)燃機(jī)生產(chǎn)歷史，萊動技術(shù)力量雄厚，工藝裝備先進(jìn)。近年來先后從德、美、英、意等國引進(jìn)了具有國際先進(jìn)水平的砂處理生產(chǎn)線、鑄造氣沖造型線、靜壓造型線、數(shù)控加工中心和其他加工、總裝和檢測設(shè)備。萊動極為重視產(chǎn)品開發(fā)工作，引進(jìn)了先進(jìn)的產(chǎn)品開發(fā)、排放檢測等試驗(yàn)設(shè)備，大大提高了產(chǎn)品的技術(shù)含量和技術(shù)保證能力。三、實(shí)習(xí)內(nèi)容在去萊動實(shí)習(xí)前，我對夾具的認(rèn)識只是通過書本了解到一些簡單的理論，理解不深刻。印象中的夾具應(yīng)該是體積不大，結(jié)構(gòu)簡單的小零件。在這次實(shí)習(xí)過程中，通過老師的講解，理論聯(lián)系實(shí)踐，把課本的知識與車間的實(shí)際應(yīng)用進(jìn)行對比、參考，充分顛覆了我原先的想法，讓我對夾具有了新的、更深入的認(rèn)識。機(jī)床夾具是在機(jī)械加工過程中，用來固定加工對象，使之占有正確位置，以接受加工或檢測并保證加工要求的機(jī)床附加裝置。鄭老師告訴我們按使用的機(jī)床不同可以把夾具分為車床夾具、銑床夾具、鉆床夾具、鏜床夾具、磨床夾具、齒輪機(jī)床夾具、數(shù)控機(jī)床夾具等。在萊動加工車間我們看到了銑床夾具（如右圖所示）、鉆床夾具和數(shù)控機(jī)床夾具。而按夾具夾緊動力源不同又可將夾具分為手動夾具和機(jī)動夾具兩大類。常用的機(jī)動夾具有氣動夾具、液壓夾具、氣液夾具、電動夾具、電磁夾具、真空夾具和離心力夾具等；在萊動公司我們見到的夾具大部分為液壓夾具，夾具體上帶有液壓缸，液壓夾具壓力穩(wěn)定、可靠，可承受較大載荷。在機(jī)床上加工工件時(shí)，必須用夾具裝好夾牢工件。將工件裝好，即定位，就是在機(jī)床上確定工件相對于刀具的正確位置。將工件夾牢，即夾緊，就是對工件施加作用力，使之在已經(jīng)定好的位置上將工件可靠地夾緊。從定位到夾緊的全過程，即為裝夾。工件裝夾情況的好壞，將直接影響工件的加工精度。零件加工時(shí)，先將夾具安置在工作臺上固定，對于大批量生產(chǎn)，夾具一經(jīng)固定，液壓管道連接好后一般就不再變動。當(dāng)加工一批零件時(shí)只需要對一個(gè)零件進(jìn)行一次調(diào)整，就可以對這同一批零件進(jìn)行連續(xù)加工。夾具的自動定位夾緊減輕了工人的勞動強(qiáng)度，大大提高勞動生產(chǎn)率，而且夾具定位準(zhǔn)確可靠，夾緊迅速，對工人技術(shù)要求不高。在不規(guī)則零件加工車間，我們見到了為不同零件設(shè)計(jì)的各種專用夾具。專用夾具是針對某一工件的某一工序的加工要求而專門設(shè)計(jì)和制造的夾具。其特點(diǎn)是針對性極強(qiáng)，沒有通用性。在產(chǎn)品相對穩(wěn)定、批量較大的生產(chǎn)中，常用各種專用夾具，可獲得較高的生產(chǎn)率和加工精度。但是專用夾具的設(shè)計(jì)制造周期較長。零件的定位一般采用“六點(diǎn)定位原則”。定位時(shí)欠定位是不允許的，過定位在一些精度要求較高，定位基準(zhǔn)定位精度也較高的情況下是允許的。在萊動我們見到了各種不同的實(shí)現(xiàn)定位夾緊的小零件，定位有：一面兩銷定位，長中心孔定位，圓錐定位等定位方式。有利用杠桿原理實(shí)現(xiàn)的快速夾緊，利用彈簧力的夾緊，。專用夾具的設(shè)計(jì)是為了提高加工速度，提高生產(chǎn)率，所以在車間見到的夾具都是能實(shí)現(xiàn)自動定位夾緊，機(jī)床一次調(diào)整后進(jìn)行連續(xù)加工，一般都是一個(gè)工人負(fù)責(zé)數(shù)臺加工機(jī)床，工人只需要把零件放于夾具上，然后啟動機(jī)床，機(jī)床便能實(shí)現(xiàn)定位夾緊加工加工完成，松開零件等一系列工作，加工完成后工人卸下零件，換上下一個(gè)待加工零件進(jìn)行切削加工。工作簡單又重復(fù)，讓我真實(shí)見到夾具在大批量自動化生產(chǎn)中的優(yōu)點(diǎn)。四、心得體會三年來第一次來到一個(gè)比較遠(yuǎn)的地方實(shí)習(xí)，真是一件令人興奮的事情。實(shí)習(xí)時(shí)間雖然很短，但在短暫的日子里我們卻學(xué)到了很多本專業(yè)實(shí)用的知識，我們從一個(gè)車間走到另一個(gè)車間，了解了各個(gè)車間的生產(chǎn)情況，與本專業(yè)有關(guān)的各種知識，各車間工人的工作情況等等。第一次親身感受了所學(xué)知識與實(shí)際的應(yīng)用，鑄造車間的砂型鑄造，加工車間的車、銑、刨、磨等工藝流程，裝配車間的流水線組裝，等等理論與實(shí)際的相結(jié)合，讓我們大開眼界，了解了大量一線運(yùn)作的流程，認(rèn)識到了很多基本的或者通用的器械，使我們對整個(gè)成品的加工過程有了系統(tǒng)的了解，對加工生產(chǎn)有了更深入的認(rèn)識，對機(jī)械這個(gè)行業(yè)有了更深刻的體會，這為我們以后走上工作崗位具有巨大的幫助。在專用夾具車間我們對銑床夾具進(jìn)行了深入的研究、探討，對每個(gè)部件進(jìn)行了分析，夾具并不像我想象中的那么簡單，要設(shè)計(jì)一個(gè)使用又簡便的夾具并不是一件容易的事情。我做的畢業(yè)設(shè)計(jì)就是有關(guān)夾具設(shè)計(jì)，這次實(shí)習(xí)對正在進(jìn)行的設(shè)計(jì)具有很大的幫助。從實(shí)習(xí)中體會最深的是機(jī)械專業(yè)類屬于重工型企業(yè)，在車間工作的工人有嚴(yán)格的上下班制度，每年有規(guī)定的產(chǎn)量要求，每天在噪音與空氣污染嚴(yán)重的環(huán)境中與冷冰冰的機(jī)器打交道，每天重復(fù)著同樣的動作，日復(fù)一日，年復(fù)一年。在鑄造車間鄭老師曾開玩笑說：“不好好學(xué)習(xí)畢業(yè)后就來干這個(gè)吧！”并不是說那個(gè)工作不好，而是老師看我們在學(xué)校整天懶懶散散也不專心對待本專業(yè)知識，借此機(jī)會提醒我們在學(xué)校要認(rèn)真對待學(xué)習(xí)，掌握好的知識，到企業(yè)后才能發(fā)揮出作為一名大學(xué)生應(yīng)有的實(shí)力。仔細(xì)一想我們在大學(xué)期間都學(xué)了些什么呢，對于本專業(yè)也只能算是初入門的了解，到了萊動加工車間才知道自己什么也不懂，理論知識與實(shí)踐應(yīng)用脫離得是那么大。此次實(shí)習(xí)讓我親身體會到即使是大學(xué)生，剛畢業(yè)去企業(yè)確實(shí)是什么也不會干的“小孩”，所以要從車間的基本工作干起，多向老師傅學(xué)習(xí)，自己什么也不會就該謙虛一點(diǎn)。機(jī)械類技術(shù)工人都需要時(shí)間與經(jīng)驗(yàn)的積累。此次實(shí)習(xí)為我即將走向的就業(yè)之路鋪墊下了一定的心理基礎(chǔ)。FP7 2:30 AN EFFICIENT ROBOT ARM CONTROL UNDER GEOMETRIC PATH CONSTRAINTS1 Kang G. Shin and Neil D. McKay Department of Electrical and Computer Engineering The University of Michigan Ann Arbor, Michigan 48109 ABSTRACT Conventionally, robot control algorithms are divided into two stages, namely, path planning and path tracking (or path con- tml). This division has been adopted mainly as a means of allevi- ating difficulties in dealing with complex, coupled robot arm dynamics. Unfortunately, the simplicity obtained from the division comes at the expense of efficiency in utilizing robots capabili- ties. To remove at least partially this inefficiency, this paper con- siders a so!ution to the problem of moving a robot arm in minimum time along a specified geometric path subject to input torque/force constraints. We first describe the robot arm dynam- ics using parametric functions which represent geometric path constraints to be honored for collision avoidance as well as task requirements. Secondly, constraints on input torques/forces are converted to those on the parameters. Finally, the minimum-time solution is deduced in an algorithm form using phase-plane tech- niques. 1. lntroduction During the past several years a great deal of attention has been focused on industrial automation techniques, especially the use of general-purpose robots. Since the purpose of industrial robots is to increase productivity, an obvious question to ask is how robots should be controlled so as to produce as many units as possible per dollar invested. The usual assumption is that fixed costs dominate the cost per item produced, so that it is desirable to produce as many units as possible in a given time. There are a variety of algorithms available for minimum-time or near-minimum-time robot arm controi. These algorithms usually assume that the control structure of the robot has been divided into two levels. The first level is called path planning, and the second level is called path cmfrol or path tracking The usual definition of path control is the act of attempting to make the robots actual position and velocity match desired values of posi- tion and velocity; the desired ,values are provided to the con- troller by the path planner. The path planner receives as input control 11, resolved acceleration control2, and various adaptive techniques3-5. Unfortunately, the simplicity obtained from the division into path planning and path tracking comes at the expense of effi- ciency. The source of the inefficiency is the path planner. In order to use the robot efficiently, the path planner must be aware of the robots dynamic properties, and the more accurate the dynamic model is, the better the robots capabilities can be used. However, most of the path planning algorithms presented to date assume very little about the robots dynamics. The usual assump- tion is that there are constant or piecewise constant bounds on the robots velocity and acceleration 6,7. In fact, these bounds vary with position, payload mass, and even with payload shape. Thus in order to make the constant-upper-bound scheme work, the upper bounds must be chosen to be global greatest lower bounds of the velocity and acceleration values; in other words, the worst case limits have to be used. Since the moments of inertia seen at the joints of the robot, and hence the acce;eration limits, may vary by a factor of three or more, such bounds can result in considerable inefficiency or under-utilization of the robot. To alleviate the inefficiency, this paper presents a solution to the minimum-time robot arm path control problem subject to constraints on its geometric path and input torques/forces. The solution will be in the form of a path planning algorithm, and will take into account the detai!s of the dynamics of the robot arm. The output of the path planner will be the true minimum-time solu- tion, and so will be useful as a standard against which the perfor- mance of other path planning algorlthms may be measured. Note that the problem and its solution considered in this paper are dif- ferent from the near minimum-time control methods in 8,9. The remainder of this paper is divided into five sections. Section 2 describe; a method for making the robot arm dynamic equations more tractable and a method for handling input torque constraints. Section 3 contains a detailed formulation of the minimum-time control problem. In Section 4, the form of the optimal solution is deduced using phase-plane techniques. Sec- tion 5 presents the highlight of this paper, that is, an a!gorithm for generating optima.l(i.e. minimum-time) trajectories. The final sec- tion is a discussion of the significance of the results. some sort of spatisl path descriptor from which it calculates a time historv of the desired Dositions and velocities. The oath 2. Robot ynamics with Constraints tracker thLn takes care of any deviations of the actual position Before delving headlong into the problem of minimum-time and velocity from the desired va:ues. control, consider the behavior of the system to be controlled, that The for dividing the control scheme in this way is that is, a dynamic model of the robot arm. There are a number of ways the process of robot control, if sonsidered in its entirety, is very of obtaining the dynamic equations of a robot arm, i.e. the equa- complicated, Since the dynamics of all but the simplest robots are tions which relate joint forces and torques to positions, veloci- highly and coupled, i,idi, the controller into the two ties, and accelerations. The two most common methods are the parts makes the whole process simpler. h path tracker is fie- Lagrange method and the Newton-mer method. The Newton-Euler quently a linear controller(e,g, a controller), While the method is cornputationally efficient, but is a recursive formulation linearities of robot arm dynamics frequently are not taken into which is hard to deal with in control problems. The Lagrange for- account at this level, such trackers can generally keep the robot mulation, while not computationally efficient, does yield a set of arm fairly to the desired More sophisticated differential equations which are easy to manipulate for robot con- methods can be used, though, such as resolved motion rate trol problems. Since the dynamic equations will be used here only to obtain analytical results, we have used Lagranges method to The work reported here IS supported 10 ,?art by the US AFOSR contract Nc. derive the following robot arm dynamic equations 23 3i F49E23-82-C-0089 and Robot Systems 31 vision, Center for Robotics and Integrat- ed h?aufactur,ng(CRIM), The Universttj of Michigan, Ann Arbor, Michigan. Any opinions, f .togs, and conclusions OT recommendations in this paper are those ot tb authors and do not necessvily reflect the view of the funding agencies. 1449 0191-2216183/0000-1449 S1.00 E 1983 IEEE Authorized licensed use limited to: Nanjing Southeast University. Downloaded on October 23, 2009 at 08:14 from IEEE Xplore. Restrictions apply. (1 a) =,(n)VJ+,j+(n)dvt+,(n) (1 b) where q! =it generalized coordinate v =ith generalized velocity Q =if generalized force Jij = the inertia matrix G, = gravitational force on the it joint ci, = Coriolis force array Kj = viscous friction matrix The Einstein summation convention has been used, and all indices run from one to n inclusive for an n-degree-of-freedom robot. The elements of the inertia matrix J,j are constants of pro- portionality which relate the torque/force exerted on the ifh joint to the acceleration of the jth joint. The entries ci,k of the Coriolis array describe the force felt at joint i due to the veloci- ties of joints j and k. The viscous friction matrix qj gives the frictional force felt at joint i due to the velocity at joint j. Note that this matrix is diagonal, and all the entries are non-negative. The motion of the robot arm will not, of course, be completely unconstrained. In fact, it will later be assumed that the robot arm must be constrained to a fixed path in joint space,2 and that the path is given as a pamrneterized cume. The curve is assumed to be given by a set of R. functions of a single parameter A, so that we are given q* = f i(h), 0 I x A, (2) where A= a parameter for describing the desired path, and it is assumed that the coordinates q vary continuously with A and that the path never retraces itself as X goes from 0 to A, i.e. It should be noted that in practice the spatial paths are given in Cartesian coordinates. While it is in general difficult to convert a curve in Cartesian coordinates to that in joint coordi- nates, it is relatively easy to perform the conversion for individual points. One can then pick a sufficiently large number of points on the Cartesian path, convert to joint coordinates, and use some sort of interpolation technique (e.g. cubic splines) to obtain a similar path in joint space (seelO for an example). Returning to the problem at hand, we may use the parame- terization of the q* and differentiate with respect to time, giving h(O)=O, h(tj)=hm,. where p=h. The equations of motion along the curve (Le. the geometric path) then become h=p (4a) dh dh dh Note that if h is used to represent arc length along the path, then /.L and p are the velocity and the acceleration along the path, respectively. With this parameterization, there are two state variables, i.e. h and p, but (n + 1) equations. One way to look at the system is to choose the equation h=p and one of the remaining equations as state equations, regarding the other equations as constraints on the inputs and on p. However, a better way of obtaining a sin- gle state equation from the n equations given is to multiply the i equation by and sum over i, giving dh Csrtesian space to joint space is needed here. Since any detailed treatment of the 2A path is normally described in Cartesian space and the conversion from mnversion is cut of the scope of this paper, oniy brief remarks on this aregiven in the next pararaph. This formulation has a distinct advantage. Note that the coeffi- cient of p is quadratic in the vector of derivatives of the con- straint functions. Since a smooth curve3 can always be parameterized in such a way that the first derivatives never all disappear simultaneously, and since the inertia matrix is positive definite, the whole equation can be divided by the non-zero, posi- tive coefficient of p, providing a solution for p in terms of h and p. Now there are only two state equations, and the original equations can be regarded as constraints on the inputs and on p (more on this will be discussed later). With this formulation, the state equations become h=p (6a) (6b) Consider now the constraints on the inputs, namely 1% 1 l*(tf)Pf (Sa) x:o)=G, h(tf )=A, (9b) 3.1. Application of the Maximum Principle In order to apply the state constraint Oshh, it is con- venient to add a third state equation. Call the third state v, and let the third equation have the form I/=X1(-h)+(h,-X)Z1:h-X,) (1 0) where , Note that bG, so that the boundary conditions v(O)=v(t,)=G force 1- to be identically zero. But the only way for v to be Identi- cally zero is to have OhIX, thus forcing h to remain in the desired interval. Before doing any further manipulations on the state equa- tions, define the functions U (A) =u, c dh S(A)=G,;h) q dh Again, for convenience the dependence of the above coefficients on h will be omitted in the sequel. Now rewrite the state equa- tions in the following form h=p (1 la) The .U term is a quadratic form reminiscent of the expression for the robot arms kinetic energy. In fact, if the parametric expressions for the Q are plugged into the formula for kinetic energy, one obtains the expression K=Mpz/2. The Q term represents the components of the Coriolis and centrifugal forces which act along the path plus the artificial forces of constraint generated by the parameterization. The R term represents fric- tional components, and S gives the gravitational force along the path. i/ is the weighted input term. Under the above setting the problem MTPP can be converted to the following. Problem MTPP: Find y = (h,u*,v*) and Li by minimizing (8) subject to (1 1 a), (1 1 b), (1 IC), (7d), (sa), and (Sb). meet task requirements 4This is done at the stage of task planning io avoid cclllsion as ne/ as to 1451 Authorized licensed use limited to: Nanjing Southeast University. Downloaded on October 23, 2009 at 08:14 from IEEE Xplore. Restrictions apply. For the MTPP problem the Hamiltonian now becomes 1 Or using the foregoing substitutions, the Hamiltonian is Differentiating with respect to p, we obtain Differentiating with respect to X, pi= - = - - -p aH PZ ai.: d -p- d d-c ah M ah dh dh dh M Finally, differentiating with respect to v, Thus if we are to apply the maximum principle to this problem, we must minimize H in (12b). In addition, constraints (1 la), (1 lb), (1 1 c), (gal, (9b), and (7d) must be met, and H must satisfy the boundary condition. Here, the state vector y is the vector (h,p,v), and we have a sin- gle input term U(tf)=q (Amzx) - df (Ama). In (1 4) we used the dh fact that H does not explicitly depend upon t. Also one can see y(tf) = (Amzx, pl, O)T from the boundary conditions (9), and the condition v(t,)=O. Note that the Hamiltonian functional (12b) is linear in lJ, and 1: is bounded because of the boundedness of u, and dfc in O, Amax. This implies that the optimal solution for I: should fol- low a bang-bang policy. That is, at any point on the optimal tra- jectory the quantity li in (1 2b) must assume its maximum or minimum. The extremum of I can be obtained by maximizing or minimizing itself with respect to the u,. S,ince we may write the equality constraints on the u, as q=g,(A)pth,(h,p), we have dh Because of the bang-bang nature of the control Ii , and because for a given state (A+) the quantity I; is linearly related to /. through the above equation, will also be bang-bang. Therefore p will be equal to either GLB(?,p) or LLB(h,p). Thinking of the three dimensional case again, p must put the input at one of the points where the line formed by the input equality constraints meets one of the sides of the prism formed by the inequalities. If the i-th joint input is on one of the sides of the constraint box, then it is driven at its maximum capability, while the others in general are not. The slowest joint, then, is pushed as hard as it will go while the others exert the proper amount of force to keep the robot arm on the right path. The above discussion is valid whenever the coefficient of the input term is not zero, i.e. when p2 in (1 3a) is not zero. If p2 is zero at isolated points only, then the optimal control is deter- mined almost everywhere. On the other hand, if pz is zero on some interval, we have the following theorem. Theorem 1: If pz is zero on some interval ;tl,t2 (t,S(O)li,(O) then p2(0)O. Proof: It is known that Olhlh, so that at t=tf we must have pl0. Otherwise, since p(t,)=O, we would have to have had p(t)O for some t hmax. But at tf we have ,u(t,)=M-(C;-S)O, this gives U-StO. Now the value of H at time tf is 0, so that If p2(tf)s0, then H(t,)O, so we must have pztZ)O. To determine the sign of p2(0), consider the quantity b(0). By an argument similar to that given above, p(O)O. This gives the result C:-SO. Since the control is bang-bang, we must have U=L:, since otherwise LJ= C7fin and U,-SO. But if U=Umax, then p2 must be less than zero. Therefore pz(0)O. Q.E.D. One consequence of these theorems is that the number of switching points is odd. If the number of switching points were even, then the sign of pz(tf) would be the same as that of p2(0), since s3n(p2:tf)=(-l)msgn(p2(0), where m is the number of slgn changes. 4. Phase Plane Interpretation At this point, it is instructive to look at the systems behavior in the phase plane. The equations of the phase plane trajectories can be obtained by dividing equation (1 1 b) by (1 1 a). This gives It is interesting to note that the total time T that it takes to go from initial to final states is The idea, then, is to minimize this integral subject to the given 1452 Authorized licensed use limited to: Nanjing Southeast University. Downloaded on October 23, 2009 at 08:14 from IEEE Xplore. Restrictions apply. constraints. We therefore want to make p as large as possible, a result which would be expected intuitively. The constraints on have two effects. One effect is to place limits on the slope of the phase trajectory. The other is to place limits on the value of p. To obtain the limits on $, one simply divicies the limits on ,u by b, since %h,gjO. Then the left-hand side of the inequality (1 7b) is a parabola which is concave upward, whereas for (1 7c) it is concave downward. When the parabola is concave downward, then the inequality holds when is between the two roots of the quadratic. If the parabola is concave upward, then the inequality holds outside of the region between the roots(Figure 1 ). Thus in one case p must lie within a closed inter- val and in the other it must lie outside an open interval, unless of course the open interval is of length zero. In that case, the ine- quality constraint is always satisfied and the roots of the qua- dratic will be complex. Since the admissible values of p are those which satisfy all of the inequalities, the admissible values must lie in the intersec- tion of all the regions determined by the inequalities. There are R (n -:)/ 2 inequalities which give closed intervals, so the inter- section of these regions is also a closed interval. The other n (%-:I/ 2 inequalities, when intersected with this closed inter- val, each may have the effect of punching a hole in the interval(Figure 2). It is thus possible to have, for any particular value of A, a set of admissible values for p which consists of as many as n(n+ I)/ 2 distinct intervals. When the phase portrait of the optimal path is drawn, it may be necessary to have the optimal trajectory dodge the little islands which can occur in the admissible region of the phase plane. (Hereafter, these inad- missible regions will be referred to as islands 3f inadr.LF.si.biiZty or just islands) It shoulc! be noted, though, that if there is no friction, then P,j=O. which means tha.7 in the concave upward case the inequality is satisfied for all values of k. Thus in this case there will be no islands in the admissible region. In addition to the constraints on ,LL described above, we must also have p 2 3. This can be shown as follows: if p 0, then the trajectory has passed below the line p = 0. Below this line, the trajectories always move to the left, since p = dX/ cit 9. Since the optimal trajsctorj must approach the desired final state through positive values of p, the trajectory would then have to pass through p = 0 again, and would pass from p 0 at a point to the left of where it had passed from p C to p h,X2 then there must be a zero. If g(h) is not continuously differentiable, assume that no zero exists. Then there are one or more points where g (x) has a discontinuity in its derivative, and a sign change must occur at one or more of these points. If this were not so, then there would have to be a sign change at a point where ?(A) is continu- ous, and hence there would be a zero. Call the first of these points Ad. This being the first (smallest) value of h where a sign change occurs, the change must be from positive to negative. Then at Ad two of the (continuously differentiable) constraint curves which generate g (A) are equal. Call the two active con- straints g, and Y, g1 being active for hhd. Then, smce llmv(X) 0 lim;s(h), we must have h4$ h+h$ But then, for some E we would have Except for higher order terms, this is just g,(Xdtc! hi, contrary to hypothesis. Therefore there must be at least one zero of g(h). The graphical meaning of this theorem is illustrated in Figure 7. Note from the figure that any cusps in 9 (A) must