Chapter 5 Synthesis of Planar Linkages

含有萘酰亚胺的菁染料太阳能电池敏化剂的合成詹文海 花建丽 金樱华 武文俊 田禾*(华东理工大学精细化工研究所结构可控先进功能材料及其制备教育部重点实验室 上海 200237)摘 要 通过Click反应把萘酰亚胺化合物连接到含有醛基的吲哚上,再通过醛基和吲哚碘盐的Knoevenagel缩合反应合成出带有萘酰亚胺部分的菁染料,用NMR、MS、元素分析、UV_Vis等方法对其结构和性能进行了表征和测试。

关键词 菁染料 萘酰亚胺 太阳能电池敏化染料 合成Synthesis of Novel Naphthalimide_containing Cyanine Dye Utilized as Solar Cell SensitizerZhan Wenhai,Hua Jianli,Jin Yinghua,Wu Wenjun,Tian He*(Laboratory for Advanced Materials and Institute of Fine Chemicals,East China University of Science&T echnology,Shanghai200237)Abstract Two novel cyanine dyes containing naphthali mide have been syn thesized by Knoevenagel condensation reaction between i ndole iodide and naphthali mide_indoles generated from naphthalimides and indole via Click reaction andcharacterized by NMR,MS,elemental analysis,UV_Vis absorption spectra etc.Key words Naphthalimide,Cyanine dye,Solar cell sensitizer,Synthesis目前,染料敏化太阳能电池的光电转换效率已经超过了10%,达到了实用的要求。

英文文献中的关键词以及重要句子关键词：Deployable structure 可展结构Scissor-like element 剪式单元Mechanism theory 机构理论Structure unit 结构单元重要句子：1.Hachem et al. surveyed biological systems in terms of their morphology, kinematics and structural characteristics, and presented the similarity of the biological systems with existing deployable structures and potential applications.哈希姆（Hachem）等人调查了生物系统的形态学、运动学以及结构特性，从生物系统中的共性发现了其中存在的可展开结构以及可能的应用。

2. On the basis of the basic module, various deployable trusses can be developed, such as equilateral triangles, squares, or normal hexagons.在基本模块的基础上的，开发出了各种可展开桁架，如等边三角形、正方形或正六边形。

3. This paper, therefore, addresses the mechanism theory that can guarantee the zero stress both at the two extreme configurations and throughout the whole deployment process. To begin with the mobility analysis of a rigid SLE, it deals with the rigid body model and kinematic principle of deployable scissor-like mechanisms.本文基于相关机械原理理论可以保证在机构的两个极端位置以及整个展开过程中到达零压力。

Chapter 5 Lineation and foliationFoliations and lineations are mesoscopic (handspecimen to outcrop) penetrative (occur throughout)features that characterize volumes of rocks and are related to larger-scale mappable features such as folds and faults. They are clues to large scale geometry, kinematics, strain at the microscopic level, and conditions of deformation.Part one: LineationLineation is parallel alignment of elongate linear fabric elements in a rock body. We use these features because they contain the most kinematic information. In other words, the most strain and stress history of the rock is most recorded in these structures. According its causes, lineation can be classified into primary and secondary lineation. Primary lineation is those formed during the formation of rocks, such as ropy structure in lava flows and linear alignment of phenocrysts and xenoliths (flow lineation) in plutonic rocks. Secondary lineations are those formed after the formation of rocks. The lineations discussed here are secondary lineations. From the viewpoint of observation, lineation can be classified as small scale and large-scale lineations.1.Small scale lineation●Stretching lineations: An important type of lineation is formed by the parallel alignmentof a set of objects that have acquired an elongate shape as a result of deformation.Individual detrital grains or fragments of any size may be deformed and/or rotated, to define a lineation. Extension lineations are generally inclined to the related fold axes at an angle close to 90° (transverse lineations) but locally may be parallel to the fold axis. Ina given area it is commonly one or the other.●Crenulation Lineation: is expressed as bundles of very small, closely spaced fold hinges(crests and troughs). Best developed in strongly anisotropic rocks, such as phyllites and schists.●Intersection Lineation: The intersection of any two planar structures forms a line. If manysuch intersections are present in a rock they constitute an intersection lineation. Commonintersections include bedding-cleavage intersections and intersections between two foliations of different generation. In cylindrically folded rocks, intersection lineation are subparallel to the fold axis.●Mineral Lineation: Metamorphic minerals often grow with a preferred crystallographicand dimensional orientation, i.e. with their long axes in parallel alignment. Mineral lineations are delineated by the long axes of individual, elongate or platy crystals (for example amphibole crystals) or mineral aggregates aligned and sub-parallel within a foliation plane. They are a penetrative element of the rock fabric, commonly parallel to other types of lineation, and serve to reinforce them. Mineral lineations may be parallel or inclined to the axes of related folds.rge scale lineations●Boudins -sausage-shaped segments of extended competent layers surrounded by lesscompetent matrix. If ductility contrast is large, boudins will be subangular withrectangular forms. As ductility contrasts diminish, boudins become more lens-shaped inprofile.'Boudin', 'boudinage', from a French word for sausage, describes the way that layers of rock break up under extension. Imagine the hand, fingers together, flat on thetable, encased in soft clay and being squeezed from above, as being like a layer ofrock. As the spreading clay moves the fingers (sausages) apart, the most mobile rockfractions (ore-making fluids) are drawn or squeezed into the developing gaps. In thatsimple analogy for rock deformation for which the term was coined almost a hundred years ago, is a model for mineral exploration which is as simple as it is controversial, and which is ignored by the consensusBoudinage depicts the periodical segmentation of pre-existing bodies, generally more competent than the rock surrounding them, when they are inhomogeneously stretched during deformation. Typically, a strong layer or dyke is broken up into a series of elongate and aligned blocks (whose cylinder-like shape motivated the name boudin).Boudin profiles are variable, with rectangular, rhomboidal shapes being common. In low-grade rocks, boudins are usually separated and form a pull-apart structure or gap, which is generally mineralised.. At higher grades, and in unconsolidated rocks, the competent layers have generally not broken through; narrow, thinned necks separate and alternate with boudins of relatively still, thick layers and the resulting structure is known as pinch-and-swell. Pinch-and-swell and pull-apart structures may be combined at any level since they really depend on the ductility contrast between the strong bed and its matrix.Boudins are commonly linear and aligned parallel to the axes of related folds.However, stretching may take place in two directions in the plane of layering.Segmentation in these two directions produces nearly equidimensional boudins rather than the elongate forms. This process is referred to as chocolate-tablet boudinage.Structures similar to boudins and pinch-and-swells may occur in certain zones of homogeneous strongly foliated rocks with no apparent lithological contrast between the boudins and the host rocks. These generally long lens-shaped structures are described as foliation boudinage.Boudins (and mullions) tend to be large in size and are commonly restricted to certain layers or, in the case of mullions, are restricted even to certain surfaces in a deformed sequence. Thus at outcrop scale they are a non-penetrative feature.A boudin axis can be measured, like a fold axis, as the nearest approximation to aline that, if moved parallel to itself, generates the boudin form. The neckline connects points of minimum layer-thickness. The length of a boudin is measured parallel to the boudin axis. The width and the thickness are dimensions orthogonal to this axis.●Pencil structure - formed by the intersection of bedding parallel foliationand cleavage,such that the rock breaks up into elongated, square sided "pencils". Pencil structure forms in weakly deformed shales or mudstone, representing an early stage in the development of slaty cleavage●Mullions are coarse structures formed in the original rock material as opposed tosegregated or introduced material. The mullion is a columnar corrugation of the surface ofa competent layer, at any size. These long features are remarkably cylindrical. They havea ribbed or grooved appearance, often cuspate in shape with broad smoothly curvedconvex surfaces separated by narrow, sharp, inward-closing hinges. The individual surface features are very persistent along the length of the mullion.●Rodding: is a morphological term for elongate, cylindrical and monomineralic bodies ofsome segregated mineral (quartz, calcite, pyrite, etc.) enclosed in metamorphic rocks ofall grades. In profile, rods may have any outline, from elliptical to irregular to that of a dismembered fold.3.Observation and measurement of lineations(1)Differentiate the primary and secondary lineation(2)Determine the types of lineation.(3)Measure its orientation. And find out its relationship with associated large scalestructures. Be sure that only those measured on the foliation plane is the reallineation.(4)As one of the important kinematic marker, lineation can show the movementdirection of materials during deformation as well as the strain state. In most cases,the directions of stretching lineation and mineral lineation are parallel to the X axisof strain ellipsoid. And the directions of boudins ,mullions, intersectionlineations ,pencil lineation and crenulation lineation are parallel to the Y-axis ofstrain ellipsoid.Part two: foliationFoliation are any type of planar fabric in rock, including bedding, cleavage, schistosity. Foliations are penetrative (occur throughout) in samples at 10's of cm scale. Thus faults are not foliations, nor are fractures and joints because the latter are simply fractures and not related to internal structure of rock.According to the formation and evolution procedure, foliation can be divided into two basic types:●Primary foliation includes layers in sedimentary rocks and flow banding and flowfoliations in igneous rocks.●Secondary foliations are usually associated with deformed metamorphic rocks andinclude (in increasing grade and grain size) slaty cleavage, phyllitic structure,schistosity and gneissic foliation.Cleavage is a secondary foliation formed under low grade metamorphic conditions (or less) that allows the rock to split along planes. Here ,we will discuss mainly on cleavage. Other types of foliation, such as phyllitic structure and schistosity will be introduced in petrology course.1. The structure of cleavageCleavage is a foliation that forms in relatively low-grade metamorphic rocks. One of the most important feature of cleavage is its domain structure. That is, the cleavage consists of cleavage domains and microlithons, each of which has a unique composition and geometry●Cleavage domains (M-domains) are thin zones of concentrated, strongly aligned, platyminerals(mica) or insoluble oxide and clay residue.●Microlithons(Q-domains) are lenticular or tabluar zones with less abundant platyminerals that exhibit a weak to strong alignment. It is usually the result of concentrationof quartz.2. Types of cleavageThere are many different kinds of classification of cleavages. Here we will only introduce the Powell’s classification. According to Powell’s classification scheme, cleavage is broken out into two main categories: continuous and spaced, depending on whether there are distinguishable, distinct cleavage domains in the rock: if no distinct domains, if distinct domains are present spaced. Scales of cleavage development from about .01 mm to approaching 1 m ,∙< 1 mm = Continuous∙> 1 mm = Spaced(1)Continuous cleavage is where the domains (microlithons) between the cleavage surfaces are too fine to observe with the naked eye. These are further broken down into fine and coarse. (2)Disjunctive cleavage obviously, is the opposite where the microlithons between the cleavage surfaces are large enough to be seen plainly. However, there are two general categories of spaced cleavage: spaced and crenulation.●Spaced cleavage consists of an array of parallel to anastomosing, stylolitic to smooth,fracture-like partings common in slightly deformed sedimentary rocks.●Crenulation spaced cleavage cuts a preexisting continuous cleavage inherent to the rock.Crenulation cleavage is characterized by microscale kinking of an earlier fabric. In symmetric crenulation cleavage the boundaries between adjacent zones are approximate planes of symmetry. In sigmoidal or asymmetric crenulation cleavage, the relict earlier foliation is bent into a sigmoidal shape. Pressure solution（压溶）can also enhance crenulation cleavage, resulting in a concentration of mica in the cleavage domains (M-domains) and a concentration of quartz in the hinge zones of microlithons (Q-domains).Fig: Fold with axial plane foliation and boudins developed on limbs3. Observation and measurement of cleavage● Distinguish the layer and cleavage● Observe carefully the geometrical pattern and its structure.● Measure its orientation.● Determine the strain state of deformed rocks. In general, cleavage is perpendicular todirection of maximum shortening, or in other words , parallel to the maximum flattening of strain ellipsoid, that is , the XY plane. For example, the axial planar cleavage is always approximately parallel to axial surfaces of folds, showing that the cleavage is perpendicular to direction of maximum shortening.Determine the generations of different phases of cleavage. Usually, we use the S O to stand for the sedimentary bedding , S1 the earliest developed cleavage, and so on.。

Link mechanismLinkages include garage door mechanisms, car wiper mechanisms, gear shift mechanisms. They are a very important part of mechanical engineering which is given very little attention...A link is defined as a rigid body having two or more pairing elements which connect it to other bodies for the purpose of transmitting force or motion . In every machine, at least one link either occupies a fixed position relative to the earth or carries the machine as a whole along with it during motion. This link is the frame of the machine and is called the fixed link.An arrangement based on components connected by rotary or sliding interfaces only is called a linkage. These type of connections, revolute and prismatic, are called lower pairs. Higher pairs are based on point line or curve interfaces.Examples of lower pairs include hinges rotary bearings, slideways , universal couplings. Examples of higher pairs include cams and gears.Kinematic analysis, a particular given mechanism is investigated based on the mechanism geometry plus factors which identify the motion such as input angular velocity, angular acceleration, etc. Kinematic synthesis is the process of designing a mechanism to accomplish a desired task. Here, both choosing the types as well as the dimensions of the new mechanism can be part of kinematic synthesis.Planar, Spatial and Spherical MechanismsA planar mechanism is one in which all particles describe plane curves is space and all of the planes are co-planar.. The majority of linkages and mechanisms are designed as planer systems. The main reason for this is that planar systems are more convenient to engineer. Spatial mechanisma are far more complicated to engineer requiring computer synthesis. Planar mechanisms ultilising only lower pairs are called planar linkages. Planar linkages only involve the use of revolute and prismatic pairsA spatial mechanism has no restrictions on the relative movement of the particles. Planar and spherical mechanisms are sub-sets of spatial mechanisms..Spatial mechanisms / linkages are not considered on this pageSpherical mechanisms has one point on each linkage which is stationary and the stationary point of all the links is at the same location. The motions of all of the particles in the mechanism are concentric and can be repesented by their shadow on aspherical surface which is centered on the common location..Spherical mechanisms/linkages are not considered on this pageMobilityAn important factor is considering a linkage is the mobility expressed as the number of degrees of freedom. The mobility of a linkage is the number of input parameters which must be controlled independently in order to bring the device to a set position. It is possible to determine this from the number of links and the number and types of joints which connect the links...A free planar link generally has 3 degrees of freedom (x , y, θ ). One link is always fixed so before any joints are attached the number of degrees of freedom of a linkage assembly with n links = DOF = 3 (n-1)Connecting two links using a joint which has only on degree of freedom adds two constraints. Connecting two links with a joint which has two degrees of freedom include 1 restraint to the systems. The number of 1 DOF joints = say j 1 and the number of joints with two degrees of freedom = say j 2.. The Mobility of a system is therefore expressed as mobility = m = 3 (n-1) - 2 j 1 - j 2Examples linkages showing the mobility are shown below..A system with a mobility of 0 is a structure. A system with a mobility of 1 can be fixed in position my positioning only one link. A system with a mobility of 2 requires two links to be positioned to fix the linkage position..This rule is general in nature and there are exceptions but it can provide a very useful initial guide as the the mobility of an arrangement of links...Grashof's LawWhen designing a linkage where the input linkage is continuously rotated e.g. driven by a motor it is important that the input link can freely rotate through complete revolutions. The arrangement would not work if the linkage locks at any point. For the four bar linkage Grashof's law provides a simple test for this conditionGrashof's law is as follows:For a planar four bar linkage, the sum of the shortest and longest links cannot be greater than the sum of the remaining links if there is to becontinuous relative rotation between two members.Referring to the 4 inversions of a four bar linkage shown below ..Grashof's law states that one of the links (generally the shortest link) will be able to rotate continuously if the following condition is met...b (shortest link ) + c(longest link) < a + dFour Inversions of a typical Four Bar LinkageNote: If the above condition was not met then only rocking motion would be possible for any link..Mechanical Advantage of 4 bar linkageThe mechanical advantage of a linkage is the ratio of the output torque exerted by the driven link to the required input torque at the driver link. It can be proved that the mechanical advantage is directly proportional to Sin( β ) the angle between the coupler link(c) and the driven link(d), and is inversely proportional to sin( α ) the angle between the driver link (b) and the coupler (c) . These angles are not constant so it is clear that the mechanical advantage is constantly changing.The linkage positions shown below with an angle α = 0 o and 180 o has a near infinite mechanical advantage. These positions are referred to as toggle positions. These positions allow the 4 bar linkage to be used a clamping tools.The angle β is called the "transmission angle". As the value sin(transmission angle) becomes small the mechanical advantage of the linkage approaches zero. In these region the linkage is very liable to lock up with very small amounts of friction. When using four bar linkages to transfer torque it is generally considered prudent to avoid transmission angles below 450 and 500.In the figure above if link (d) is made the driver the system shown is in a locked position. The system has no toggle positions and the linkage is a poor design Freudenstein's EquationThis equation provides a simple algebraic method of determining the position of an output lever knowing the four link lengths and the position of the input lever. Consider the 4 -bar linkage chain as shown below..The position vector of the links are related as followsl1 + l2 + l3 + l4 = 0Equating horizontal distancesl 1cos θ 1 + l 2cos θ 2 + l 3cos θ 3 + l 4cos θ 4 = 0Equating Vertical distancesl 1sin θ 1 + l 2sin θ 2 + l 3sin θ 3 + l 4sin θ 4 = 0Assuming θ 1 = 1800then sin θ 1= 0 and cosθ 1 = -1 Therefore- l 1 + l 2cosθ 2 + l 3cosθ 3 + l 4cos θ 4 = 0and .. l 2sin θ 2 + l 3sin θ 3 + l 4sin θ 4 = 0Moving all terms except those containing l 3 to the RHS and Squaring both sidesl 32 cos 2θ 3 = (l 1 - l 2cos θ 2 - l 4cos θ 4 ) 2l 32 sin 2θ 3 = ( - l 2sin θ 2 - l 4sin θ 4) 2Adding the above 2 equations and using the relationshipscos ( θ 2 - θ 4) = cos θ 2cos θ 4+ sin θ 2sin θ 4 ) and sin2θ + cos2θ = 1the following relationship results..Freudenstein's Equation results from this relationship asK 1cos θ 2 + K2cos θ 4 + K 3= cos ( θ 2 - θ 4 )K1 = l1 / l4K2 = l 1 / l 2K3 = ( l 32 - l 12 - l 22 - l 2 4 ) / 2 l 2 l 4This equation enables the analytic synthesis of a 4 bar linkage. If three position of the output lever are required corresponding to the angular position of the input lever at three positions then this equation can be used to determine the appropriate lever lengths using three simultaneous equations...Velocity Vectors for LinksThe velocity of one point on a link must be perpendicular to the axis of the link, otherwise there would be a change in length of the link.On the link shown below B has a velocity of v AB= ω.AB perpendicular to A-B." The velocity vector is shown...Considering the four bar arrangement shown below. The velocity vector diagram is built up as follows:∙As A and D are fixed then the velocity of D relative to A = 0 a and d are located at the same point∙The velocity of B relative to a is v AB= ω.AB perpendicular to A-B. This is drawn to scale as shown∙The velocity of C relative to B is perpedicular to CB and passes through b∙The velocity of C relative to D is perpedicular to CD and passes through d∙The velocity of P is obtained from the vector diagram by using the relationship bp/bc = BP/BCThe velocity vector diagram is easily drawn as shown...Velocity of sliding Block on Rotating LinkConsider a block B sliding on a link rotating about A. The block is instantaneously located at B' on the link..The velocity of B' relative to A = ω.AB perpendicular to the line. The velocity of B relative to B' = v. The link block and the associated vector diagram is shown below..Acceleration Vectors for LinksThe acceleration of a point on a link relative to another has two components:∙1) the cen tripetal component due to the angular velocity of the link.ω 2.Length∙2) the tangential component due to the angular acceleration of the link....∙The diagram below shows how to to construct a vector diagram for the acceleration components on a single link.The centripetal acceleration ab' = ω 2.AB towards the centre of rotation. The tangential component b'b = α. AB in a direction perpendicular to the link..The diagram below shows how to construct an acceleration vector drawing for a four bar linkage.∙For A and D are fixed relative to each other and the relative acceleration = 0 ( a,d are together )∙The acceleration of B relative to A are drawn as for the above link∙The centripetal acceleration of C relative to B = v 2CB and is directed towards B ( bc1 )∙The tangential acceleration of C relative to B is unknown but its direction is known∙The centripetal acceleration of C relative to D = v 2CD and is directed towards d( dc2)∙The tangential acceleration of C relative to D is unknown but its direction is known.∙The intersection of the lines through c1 and c 2 locates cThe location of the acceleration of point p is obtained by proportion bp/bc =BP/BC and the absolute acceleration of P = apThe diagram below shows how to construct and acceleration vector diagram for a sliding block on a rotating link..The link with the sliding block is drawn in two positions..at an angle dωThe velocity of the point on the link coincident with B changes from ω.r =a b 1to ( ω + dω) (r +dr) = a b 2The change in velocity b1b2has a radial component ωr d θ and a tangential component ωdr + r dωThe velocity of B on the sliding block relative to the coincident point on the link changes from v = a b 3 to v + dv = a b 4.The change in velocity = b3b4 which has radial components dv and tangential components v d θThe total change in velocity in the radial direction = dv- ω r d θRadial acceleration = dv / dt = ω r d θ / dt = a - ω2 rThe total change in velocity in the tangential direction = v dθ + ω dr + r αTang ential acceleration = v dθ / dt + ω dr/dt + r d ω / dt= v ω + ω v + r α = α r + 2 v ωThe acceleration vector diagram for the block is shown belowNote : The term 2 v ω representing the tangential acceleration of the block relative to the coincident point on the link is called the coriolis component and results whenever a block slides along a rotating link and whenever a link slides through a swivelling block连杆机构连杆存在于车库门装置，汽车擦装置，齿轮移动装置中。

Lesson 1 Basic Concepts in Mechanics1、The branch of scientific analysis which deals with motions, time, and forces is called mechanics and is made up of two parts statics and dynamics对运动、时间和作用力作出科学分析的分支称为力学。

它由静力学和动力学两部分组成。

2、Our intuitive concept of force includes such ideas as place of application, direction,and magnitudeand these are called the characteristics of a force.直观概念包括作用点、方向和大小。

这些称为力的要素3、Inertia is the property of mass that causes it to resist any effort to change its motion.惯性是质量所具有的抵抗任何外力改变其本身运动状态的性质.4、A particle is a body whose dimensions are so small that they may be neglected.当一个物体的尺寸特别小，可以忽略不计时，该物体可以被称为质点。

5、Law 1 If all the forces acting on a particle are balanced, the particle will either remain at rest or will continue to move in a straight line at a uniform velocity.第一定律如果作用在一个质点上的所有外力平衡，那么，这个质点保持原来的静止或者匀速直线运动状态不变。

Lesson 16 Basic Concepts of MechanismsA combination of interrelated parts(相关零件的组合体) having definite（确定的）motions and capable of performing useful work may be called machine.具有确定的运动轨迹,并且能做有用的功联接在一起的零件的组合称为机器.A mechanism is a component of a machine consisting of two or more bodies arranged so that the motion of one compels the motion of the others.机构是机器的组成要素.机器由二个及以上的机构组合而成.因此其中一个构件的运动会带动其它的构件的连锁运动.Kinematics is the study of motion in mechanisms without reference to the forces that act on the mechanism运动学仅研究机构的运动而不考虑作用在机构上的力.Dynamics is the study of motion individual bodies and mechanisms under the influence of forces and torques.动力学研究独立构件的运动及力和扭矩作用下对机构的影响.The study of forces and torques in stationary systems (and systems with negligible inertial effects) is called statics研究平衡中的力和扭矩,该系统忽略惯性影响,称之为静力学.Synthesis is the procedure by which a product (a mechanism, for example) is developed to satisfy a set of performance requirements合成是一个程序,通过该程序能够不断改进一个产品(例如一个机构)直到满足一系列的性能要求.If a product configuration is tentatively specified and then examined to determine whether the performance requirements are met, the process is called analysis.如果需要探索性的详细阐述产品的构造结构,然后检验,判断该产品是否满足性能要求,这个过程称为分解.The design of mechanisms involves both synthesis and analysis.机构设计包括合成和分解.The design process begins with the recognition of a need. A set of requirements is then listed设计从对于需求的确认开始,罗列出一系列的要求.Detailed analyses of displacements, velocities, and accelerations are usually required通常需要对位移,速度,加速度进行具体分析.This part of the design process is followed by an analysis of forces and torques. 设计过程中的这个部分是紧跟在对力和扭矩的分析之后.The design process may continue long after（很久以后）the first models have been produced and may include redesigns of components that affect velocities, accelerations, forces, and torques在制造完毕第一个模型之后,该设计过程仍有可能持续很长一段时间. 并且有可能会重新设计这些会影响速度,加速度及所知扭矩的零部件In order to successfully compete from year to year, most manufactures must continuously modify their product and their methods of production为了在年复一年的技术中取得优势,大多数制造商必须不断调整修改他们的产品及加工方式.Increases in the production rate, upgrading of product performance, redesign for cost and weight reduction are frequently required,为了增加出产率,提高产品的性能,经常需要重新设计来降低成本以及减轻重量. Success may hinge on（决定于） the accuracy of the kinematic and dynamic analysis 对问题进行正确的运动学和动力学分析是成功的关键.Many of the basic linkage configurations have been incorporated into（成为……的一部分）machines designed centuries ago, and the terms we use to describe them have changed over the years很久以前,许多基本的连杆构造就已经运用在机器设计中去了,并且我们用来称呼它们的术语在不断的变化,.Thus, definitions and terminology are not consistent throughout thetechnical literature.因此技术供应文献中的定义和说明不是一成不变的In most cases, however, meanings will be clear from the context of the descriptive matter.在大多数的情况下,然而定义,义意可以在的解释性的描述内容中清晰的获得,A few terms of particular interest to the study of kinematics and dynamics of machines are defined below下面定义了在学习,研究机构运动学和动力学中一些特别重要的术语.Link A link is one of the rigid bodies or members joined together to form a kinematic chain.杆件是一个刚体或者连接在一起形成运动链的零部件,因此被称为刚性构件或者是简单构件.The term rigid link, or sometimes simply link, is an idealization used in the study of mechanisms that does not consider small deflections due to strains in machine members刚性构件是在机械装置中的理想化模型.这样就不无需考虑由机器零部件所产生的应力造成的微小偏移量.A perfectly rigid or inextensible link can exist only as a textbook type of model of a real machine member.一个绝对刚性或不可延展的构件只能存在于教科书上的机械零部件的模型.For typical machine parts, the maximum changes in dimension are on the order of only a one-thousandth of the part length.对于典型的零部件,最大的尺寸变化量仅仅是沿着该部件长度方向的千分之一.We are justified（合理的） in neglecting this small motion when considering the much greater motion characteristic of most mechanisms当考虑到大多数的机构相对较为显着的运动特征,忽略这点微量的偏移是合情合理的. The word link is used in a general sense to include cams, gears, and other machine members in addition to cranks, connecting rods and other pin-connected（铰接） components通常意义所说的杆件包括凸轮,齿轮以及它的机器零件的曲柄,连杆机构和其它的销件连接部分.Frame The fixed or stationary link in a mechanism is called the frame机架机械装置中固定或静止不动的构件称为机架.When there is no link that is actually fixed, we may consider one as being fixed and determine the motion of the other links relative to it如果实际中确实没有构件是固定的,我们可以假设其中某个是固定的,这样可以判断出其它与之相连构件的运动.In an automotive engine, for example, the engine block (see Fig.14.1) isconsidered the frame, even though the automobile may be moving例如汽车发动机,我们通常是把发动机气缸视为机架,让它保持静止不动,实际上整部汽车者是运动的.Degree of Freedom The number of degrees of freedom of a linkageis the number of independent parameters required to specify the position of every link relative to the frame or fixed link自由度连杆机构的自由度是给定独立参数的数目，需要自由度来确定某一个与机架或固定杆件相连接的构件的位置，If the instantaneous configuration of a system may be completely defined bespecifying one independent variable, that system has one degree of freedom如果系统的瞬间结构完全可以由一个独立的变量来确定，这样系统就具有一个自由度，An unconstrained rigid body has six degrees of freedom:大多数实际上运用的机构装置只有一个自由度。

⽣物化学5-10章练习题英⽂版Exercises of Chapter 5-10A1.Which can transport acyl-CoA into the Mitochondria? ( )A .apolipoprotein B. lipoprotein C. albumin D. citrate E. carnitine2. In which following conditions are ketone bodies synthesis greatly accelerated? ( )A .glycolysis is enhanced B. fat mobilization is enhanced C. fatty acid synthesis is enhancedD. malonyl CoA is reducedE. aerobic oxidation of glucose is enhanced3. How many moles of ATP are yielded when 1 mol glycerol is oxidized to CO2 and H2O ? ( )A .19 B. 16.5/18.5 C. 20/22 D.23 E. 174. which is the key enzyme in the regulation of FA synthesis? ( )A .fatty acid synthase B. acetyl-CoA carboxylase C. keto-acyl-ACP synthaseD.enoyl-ACP reductaseE. thioesterase5. NADPH in FA synthesis comes from ( ):A .aerobic oxidation of glucose B. glycolytic pathway C. gluconeogenesisD.pentose phosphate pathwayE. glucuronate pathway6. which is not concerned with the cholesterol biosynthesis?A .acetyl-CoA B. HMGCoA C. NADPH +H+ D.seualene E. glycerol7. the active form of choline in phospholipid biosynthesis is:A .ADP-choline B. UDP-choline C. CDP-choline D. GDP-choline E. CMP-choline8. which is not concerned with the glycerophospholipid biosynthesis?A .serine B. GTP C. ATP D. CTP E. choline9.which is specific function of apoAⅠ?A .activates LPL B. inhibits HL C. activates LCAT D. inhibits LCAT E. activates ACAT10.which is not required in β-Oxidation of Fatty AcidsA .CoASH B. NADP+ C. carnitine D. FAD E. NAD+11. which is the primary carrier of cholesterol in the blood of humans?A .CM B. HDL C.VLDL D. LDL E. IDL12. How many moles of ATP are yielded when 1 mol stearic acid (18 C) is oxidized to CO2 and H2O ? ( )A .146 B. 120 C. 165 D.182 E. 19913. which is an intermediate on the pathway for synthesis of ketone bodies and cholesterol?A .malonyl CoA B. acetoacetyl-CoA C. HMGCoA D. acetyl-CoA E. acetoacetate14. which is prosthetic group of acetyl-CoA carboxylase?A . NAD+ B. biotin C. NADP+ D. FAD E. FMN15. which is the key enzyme in the regulation of cholesterol synthesis? ( )A . HMGCoA synthase B. squalene monooxygenase C. HMGCoA reductaseD. squalene synthaseE. HMGCoA thiolase16.which is the correct sequence of 4 reactions in fatty acids oxidation?A . dehydrogenation, redegydrogenation, hydration, thiolysisB. thiolysis, dehydrogenation, redegydrogenation, hydrationC. dehydrogenation, hydration, redegydrogenation, thiolysisD. hydration, dehydrogenation, thiolysis, redegydrogenationE. dehydrogenation, thiolysis, redegydrogenation, hydration17. which is the major form that transports endogenous TAGs from the liver to the tissues?A .HDL B. LDL C.VLDL D.CM E. IDL18. which lipoprotein contains apoB100 mostly?A .HDL B. LDL C.VLDL D.CM E. CM remnants19. which lipoprotein contains apoB48 mostly?A .HDL B. LDL C.VLDL D.CM E. IDL20. which enzyme ‘s activator is apo CⅡ?A .LPL B. LCAT C.ACAT D.HL E. CETP21. the net result of the oxidation of 1 mol palmitic acid (16C) will be ( ) mol ATP?A .129 B. 96 C. 165 D.182 E. 10622.which is the major form that participates in the reverse cholesterol transport?A . IDL B. HDL C. LDL D. VLDL E. CM23.which is the rate –limiting enzyme in the TAG degradation?A .HSL B. LPL C.DAG lipase D.HL E. MAG lipase24.which hormone can accelerate fat mobilization?A .epinephrine B.glucagons C.TSH D.ACTH E. all above25.which is anti-lipolytic hormone ?A .epinephrine B.glucagons C.TSH D.ACTH E. prostaglandin E226.which is essential fatty acid?A . stearic acid B. oleic acid C.linoleic acid D. palmitic acid E. oxaloacetic acid27.which is the pathway that acetyl-CoA is transported from the mitochondria into the cytosol to allow FA synthesis to occur?A . TAC cycle B. malate-asparate shuttle C.glucose pyruvate cycleD. citrate pyruvate cycleE. α-glycerophosphate shuttle28. which is the key e nzyme in β-Oxidation of Fatty Acids? ( )A . carnitine acyl transferaseⅠ B. carnitine acyl transferaseⅡ C. acyl-CoA synthetaseD. carnitine-acylcaritine translocaseE. acetyl-CoA carboxylase29. the lack of ( ) means that ketone bodies are synthesized exckusively in the liver and must be used elsewhere.A . HMGCoA synthase B. succinyl CoA transsulfurase C. HMGCoA reductaseD. acetoacetyl-CoA thiolaseE. HMGCoA thiolase30. which agent is an inhibitor of electron transport at Complex Ⅲ?A .rotenone B. antimycin A C. piericidin A D. amobarbital E. DNP31. which agent is not an inhibitor of electron transport at Complex Ⅳ?A .carbon monoxide B. azide C. sulfureted hydrogen D. amobarbital E. cyanide32. which is not the energy-rich compound?A .CP B. ATP C. PEP D. 1,3-bisphoglycerate E. 2,3-bisphoglycerate33.the sequence of cytochrome in respiratory chain is :A. c----c1----b----aa3----O2B. c1---- c ----b----aa3----O2C. b ----c1---- c ----aa3----O2D. b ---- c ---- c1 ----aa3----O2E. c---- b ---- c1 ----aa3----O234. Which of the following state is nitrogen equilibrium?A. advanced cancer patientB. pregnant womenC. patient recovering from illnessD. people suffer form long-time hungerE. normal adult35. Which of the following state that nutrition sufficicent infant in?A.nitrogen balanceB.positive nitrogen balance/doc/4c28100b79563c1ec5da7190.html ative nitrotgen balanceD. nitrogen equilibriumE.none of the above36. Which of the following state is positive nitrogen balance?A.advanced cancerB. pregnant womenC.people suffer from feverD. people suffer form long-time hungerE.normal adult37. Which of the following is not one carbon unit?CH3; CH2; CO2; CH; CHO38. Which of the following is not produced by Tyr?norepinephrine;epinephrine;dopamine; Phe;melanin39. Which of the following is the main deamination style in muscle? Transamination;ornithine cycle;L-glutamate oxidation deamination;purine nucleotide cycle;γ-glutamyl cycle40. Which of the following is the storage and transportation style of ammonia?Glu; Tyr;GSH; Asn; Gln41. Which of the following is the main outlet of ammonia in mammalian? infiltrate to intestinaltract;synthesis of Gln;synthesis of urea in liver;synthesis of amino acid again; excrete out of the body in kidney42. Which of the following amino acid can not produce one carbon unit?Serine;Valine ;Tryptophan;Histidine ;Glycine43.Which of the following ammino acids take part in the synthesis of creatine? Glycine,mehionine,arginine; Methionine,ornithine,citrulline; Glycine,methionine,citrulline; Glycine,arginine,ornithine;Ornithine,citrulline, arginine44. Which of the following is the carrier of one carbon unit ?biotin; SAM; FH4; FH2; folic acid45. Which of the following is ketogenic amino acid only?Glycine; Leucine; Threonine; Isoleucine; Tyrosine46. Which of the following are the nitrogen source of urea ?carbomoyl phosphate and aspartate ;carbomoyl phosphate and ornithine;α-amino group andγ-amino group of ornithine;α-amino group of citrulline andα-amino group of arginine;γ-amino group of ornithine and glycine47. Which of the following amino acid can produce PAPS?taurine; methionine; cysteine; homocysteine; glutamate48. Which of the following is the main deamination style in the body?oxidation deamination; reduction deamination;direct deamination; transamination; union deamination49. Which of the following is the process that transport ammonia from muscle to liver? TAC cycle;ornithine cycle;alanine-glucose cycle;methionine cycle ; γ-glutamyl cycle 50. Which of the following vitamin that the coenzyme of amino acid decarboxylase contain?C; B2; B12;B6;B151. Which of the following is the direct donor of CH3?N5-CH3-FH4; SAM; N5,N10- CH2-FH4; N5,N10=CH-FH4; FAD52. Which of the following is ketogenic and glucogenic amino acid ?Ala; Phe; His; Pro; Leu53. Which of the following is the main source of ammonia in kidney?union deamination of amino acid; hydrolysis of Gln;hydrolysis of urea ; nonoxidative deamination of amino aicd;oxidation of amine54. In the synthesis of urea, one N come from NH3, which of the following amino acid provide another N for urea?Pro; Asp; Glu; Phe; Lys55. Which of the following is first synthesized in the process of de novo synthesis of purine nucleotide?GMP; AMP; IMP; ATP; GTP56. Which of the following is the end product of purine nucleotide catabolic metabolism?urea; creatine; creatinine; uric acid; β-alanine57. Which of the following substance directly link nucleotide synthesis and glucose metabolism?glucose; glucose-6-phosphate ; glucose-1-phosphate ; glucose-1,6-diphosphate; ribose-5-phosphate58. Which of the following is not the direct material of de nove synthesis of purine nucleotide ?glycine; aspartate ; glutamate ; CO2 ; one carbon unit59. Which of the following material could provide C4 and C5 for purine ring? alanine; glycine; glutamate ; aspartate ; glutamine60. Which of the following provide only one C atom for purine synthesis?CO2; glutamine; glycine; aspartate; formic acid61. Which of the following provide two N atoms for pyrimidine ring?glutamine and ammonia ; glutamine and glutamateglutamine and carbomoyl phosphate; aspartate and carbomoyl phosphate; aspartate and glutamate62.Which of the following is the common material for the synthesis of pyrimidine and purine nucleotide?fumarate; glutamine ; formic acid ; asparagine; glycine63. Which of the following amino acid totally participate in the synthesis of purine nucleotide?glycine ; aspartate; glutamate ; glutamine ; asparagines64.Which of the following provide C2 for purine ring?N5-CH3FH4; N10-CHO-FH4;N5，N10=CH-FH4; N5，N10-CH2-FH4; CO2 65. Which of the following is the energy source of brain for long-term starvation patient ?glucose; amino acid; glycerol; acetone body; glucogen66. Which of the following is not correct about chemical modification of key enzyme? the enzyme have two style: active and inactive ; the active style and inactive style can be interconverted by the other enzyme ; the enzyme that catalyze the interconversion can be controlled by the hormon; do not need energy; phosphorylation and dephosphorylation is the main style67. Which of the following is the main style of chemical modification?methylation and demethylation ; acetylation and deacetylation; phosphorylation and dephosphorylation; polymerization and depolymerization ; synthesis and decomposition of enzyme 68. Which of the following reaction take place in plasma?TAC ; oxidative phosphorylation ; pyruvate carboxylation; β-oxidation of fatty acid ; synthesis of fatty acid69. In the condition of starvation, which of the following is not correct? insulin secretion is enhanced ; glucagon secretion is enhanced;fat mobilization is enhanced ; synthesis of acetone body is enhanced ; glyconeogenesis is enhanced70. Which of the following is not correct about the substance metabolism in different tissues and organs ?liver is the hinge of the substance metabolism;glucose decomposition in heart is mainly aerobic oxidation ;the energy of brain usually come from glucose ;the energy of red cell usually come from glycolytic pathway;liver is the only organ that have the reaction of glyconeogenesis71. Which of the following is not correct about the allosteric regulation ?the allosteric enzyme usually has two subunits ;the allosteric effectors usually are small materials ;the allosteric effectors usually bind to the site out of the active center of the enzyme ; the end production of the pathway usually is the allosteric inhibitors of the enzyme that catalyze the initiation reaction of the pathway ; can be amplified72. Which of the following is not correct about the key enzyme?the key enzyme usually catalyze the first reaciton of the metabolic pathway ; the key enzyme has the highest activity; the key enzyme usually is the allosteric enzyme ; the key enzyme usually is regulated by hormone; the reaction that catalyzed by the key enzyme usually is the one-way reaction or equilibrium reaction 73. Which of the following enzyme do not belong to chemical modification enzyme ? phosphorylase; glycogen synthetase; pyruvate kinase; malate dehydrogenase ; phosphorylase b kinase74. Which of the following metabolic pathway take place in the mitochondira? glycolytic pathway; the synthesis of acetone body; pentose phosphate pathway ; the synthesis of the fatty acid ; the synthesis of the glycogen75. Which of the following is the material of glyconeogenesis for short time starvation? glycerol ; glycerol-3-phosphate; alanine; pyruvate; lactate76. In the oxidation decomposition of glucose ，fat and amino acid , which is the main compound that go into the TAC cycle? isocitrate ; pyruvate ; α-ketoglutarate ; α-keto acid ; acetyl CoA77. Which of the following is not the outlet of Acetyl CoA?go into the TAC cycle ; use to synthesize fat ; transition ot ketogenic amino acid ; transition ot acetone body; transition to cholesterol78. Which of the following is the rate-limiting enzyme of glyconeogenesis ?Acetyl CoA carboxylase ; 7-α-hydroxylase; HMGCoA reductase; pyruvate carboxylase; 6-phosphofructokinase-179. Which of the following does not participate in DNA replication?helicase; single strand binding protein; transciption factor; ligase ; primase 80. DNA replication is a ( ) process. conservative; destructive; non-conservative; semi- conservative, random dipersive81. he primary structure of DNA is the squence of its nucletide residues connected by ( )?3’,5’-phosphodiester bonds; hydrogen bonds; peptide bonds; salt bonds; disulfide bonds82. When the two daughter strands of DNA replicated , ( )one strand in the 5’ to 3’ and the other in the 3’ to 5’direction; the both strands are all in the 3’ to 5’ direction; the both strands are all in the 5’ to 3’ direction; the b oth strands are all conservative replicated; the DNAsynthesis of leading strand is continuous83．Which one can separate the parental DNA strands?DnaC; DNA-pol; topoisomerase; primase ; helicase84．Human telomerase RNA is one of the ( ).reverse transcriptase ; DNA-pol; RNA-pol; DNA ligase; DNA topoisomerase 85．Which of these enzyme is not concerned with DNA replication？DNA dependent DNA polymerase; DNA dependent RNA polymerase; RNA dependent DNA polymerase; topoisomerase; DNA ligase86．In the initiation of DNA replication, the proper order that following enzyme and protein acting is? 1, DNA-polⅢ 2，SSB 3，primase 4, helicase1,2,3,4; 4,2,3,1; 3,1,2,4; 1,4,2,3; 2,3,4,187．Which of the following could proofreading, DNA repairing, filling of gaps? DNA polα; DNA polγ ; DNA pol δ; DNA polⅠ; DNA polⅢ88．Which of the following have primase activity?DNA pol α; DNA polγ; DNA pol δ; DNA polⅠ; DNA polⅢ89．Which of the following serve as DNA replication in mitochondria ?DNA polα; DNA polγ; DNA pol δ; DNA polⅠ ; DNA polⅢ90．In DNA replication, what is the reason of Okazaki fragment generated?the speed of DNA replication is too fast ; bidirectional replication; come fromprimer; the direction of replication is different with that of unwinding; DNA be coiled and tied knots in replication 91．Which of the following is the product of the replication discontinuously? Klenow fragment; primer; leading strand ; lagging strand, Okazaki fragment92．Which of the following is a short RNA?Klenow fragment; primer ;leading strand; lagging strand; Okazaki fragment. 93．Which of the following could contain the length of eukaryotic chromosome? telomerase；primase；SSB；DNA ligase；topoiserase94．The template DNA is 5′-ATAGC-3′,and the daughter DNA is：5′-GCTAG-3′；5′-GCTAT-3′；5′-GCATA-3′；5′-GATAT-3′；5′-TATCG-3′. 95．Which of the following is the function of SSBunwingding；to identify the ori of replication；synthetic primer；to stabilize the single strand；filling the gapX:which are correlated with FA synthesis?A . acetyl-CoA carboxylase B.ACP C.biotin D. NADPH +H+ E. acyl-CoA synthetase2. which is essential fatty acid?A . oleic acid B. linolenic acid C.linoleic acid D. arachidonate E. palmitic acid3. which materials can not be transported from the mitochondria into the cytosol?A . acetyl-CoA B.pyruvate C. acyl-CoA D. oxaloacetic acid E. malate4. the function of cholesterol is:A . conversion into vitamin D B. conversion into bile acid C. as a biological membraneD. providing energyE. conversion into cholesterol hormones5. which material synthesis requires acetyl-CoA?A . fatty acid B.ketone bodies C. cholesterol D. lipoprotein E. sugar6. which does not exist in hepatic cells?A . succinyl CoA transsulfurase B. acetyl-CoA carboxylase C. HMGCoA reductaseD. acetoacetyl-CoA thiolaseE. acetoacetyl-CoA thiokinase7. utilization of ketone bodies in ( )A . heart B. brain C. kidney D. liver E. skeletal8.proton pumps are ( ) where protons flow from the matrix to the intermembrane space and a membrane potential is formed.A . ComplexⅠ B. ComplexⅡ C. Complex Ⅲ D. Complex Ⅳ E. ATP Synthase9. which agents are inhibitors of electron transport at ComplexⅠ?A .rotenone B. antimycin A C. piericidin A D. amobarbital E. BAL10. Which of the following are essential amino acids?phenylalanine；tyrosine；serine；threonine；leucine11. Which of the following substance need one carbon unit to synthesize?AMP; cholesterol; TMP; heme; fatty acid12. Which of the following are produced by Cysteine ?5-HT ; γ-GABA ; taurine ; GSH ; PAPS13. Which of the following are the source of ammonia?deamination of amino acid ;absorption of ammonia in digestive canal; secretory ammonia in kidney; decomposion of purine and pyrimidine nucleotide; decomposition of urea14. Which of the following α-keto acid can be directly converted to amino acid by transamination ?oxaloacetate;pyruvate; α-ketoglutarate;α-isobutyric acid ; hydroxyproline15. Which of the following are the main transportation style of ammonia between tissues?NH4CL; alanine; urea; glutamine; oxaloacetate16. Which of the following are one carbon unit?-CH=NH;-CHO;-CH2-;-CH3;CO217. Which of the following are glucogenic aminio acid?arginine;lysine;histidine;alanine; leucine18. Which of the following are ketogenic amino acid ?tyrosine; phenylalanine;ornithine;leucine ;lysine19. Which of the following are ketogenic and glucogenic amino acid?isoleucine; tyrosine; tryptophan; phenylalanine; serine20. Which of the following amino aicds take part in the synthesis of creatine ? glycine; arginine; methionine; cysteine; ornithine21. Which of the following substance can produce uric acid ?AMP ; UMP ; IMP ; TMP ; GMP22. Which of the following are the products of pyrimidine nucleotide decomposition? NH3; uric acid; CO2; β-alanine;β-aminoisobutyrate23.Which of the following provide C and N atoms for purine nucleotide ? glutamine; glycine; aspartate ; glutamate; alanine24. Which of the following are the materials for de novo synthesis of pyrimidine nucleotide?glutamine and aspartate; one carbon unit ; ribose phosphate; CO2 ; glycine 25. Which of the following are the compound that have the high-energy bound ? phosphoenolpyruvic acid ; Acetyl CoA; creatine phosphate ; AMP; 3-phosphoglycerate26. Which of the following are right about the allosteric enzyme ?all the reaction catalyzed by the allosteric enzyme is reversible ; the binding to the allosteric effectors is reversible ; all the enzyme have the regulatory subunit and catalytic subunit; the allosterism of the enzyme ofen accompany with the polymerization and depolymerization of the subunit; allosteric effectors can be the substrate or the production27. The central dogma of biological genetic information contain：DNA→DNA；DNA→RNA；RNA→DNA；RNA→R NA ；RNA→Potein 28．Which of the following could catalize the synthesis of phosphodiester bond? DNA-pol；topoismerase；telomerase；reverse transcriptase；DNA ligase.29。

ANALYTICAL LINKAGESYNTHESISImagination is more important than knowledgeALBERT EINSTEIN5.0 INTRODUCTIONWith the fundamentals of position analysis established. We can now use these techniques to synthesize linkages for specified output positions analytically .The synthesis tech-niques presented in Chapter 3 were strictly graphical and somewhat intuitive .The analytical synthesis procedure is algebraic rather than graphical and is less intuitive .However,its algebraic nature makes it quite suitable for computerization.These analytical synthesis methods were origiated by Sandor and further developed by his students.Erdman,Kaufman,and Loerch et al.5.1TYPE OF KINEMATIC SYNTHESISErdman and Sandor define three types of kinematic synthesis ,function ,path,and motion generation,which were discussed in Section 3.2.Brief definitions are repeated here for your convenience,FUNCTION GENERATION is defined as the correlation of an input function with an output function in a mechanism.Typically. a double –rocker or crank-rocker is the result ,with pure rotation input and pure rotation output. A slider-crank linkage can be a function generator as well ,driven form either end ,i.e,rotation in and translation out or vice versa.PATH GENERATION is defined as the control of a point in the plane such that it fllows some prescribed path .This is typically accomplished with a fourbar crank-rocker or double-rocker,wherein a point on the coupler traces the desired output path .No attempt is made in path generation to control thelink which contains the point of interest.The coupler curve is made to pass through a set of desired output point. However,it is common for for the timing of the arrival of the coupler point at particular locations along the parth to be defined.5.4 COMPARISION OF ANALYTICAL AND GRAPHICAL TWO-POSITION SYNTHESISNote that in the geaphical solution to this two-position synthesis problem(in Example 3-3 and Figure 3-6,p.87).we also had to make three free choices to solve the problem.The identical two-position synthesis problem from Figure 3-6 is reproduced in Figure 5-3.The approach taken in Example 3-3 used the two points A and B as the attachments for the moving pivots O2 and O4.For the analytical solution we will use those points A and B as the joints of the joints of the two dyads WZ and US,These dyads meet at point P.which is the precision point.The relative position vector P21 defines the displacement of the precision point.Note that in the graphical solution, we implicitly defined the left dyad vector Z by locating attachment points A and B on link 3 as shown in Figure 5-3a,This defined the two variables.z andφWe also implicitly chose the value of w by selecting an arbitrary location for pivor O2 on the perpendicular bisector.When that thire choice was made. the remaining two unknowns,anglesβ2andθ,were solved for graphically at the same time,because the geometric construction was in fact a graphical “computation” for the solution of the two simultaneous equations 5.8.The graphical and analytical methods represent two alternate solutions to the same problem. All of these problems can be solved both analyticlally and graphically.One method can be provide a good check for the other,we will now solve this problemanalytically and correlate the results with the graphical solution from Chapter 3. EXAMPLE 5-1Two-position Analytical Motion Synthesis.Problem: Design a fourbar linkage to move the link APB shown from position A1P1B1 to A2P2.B2.Solution: (see Figure 5-3)1 Draw the link APB in its two desired positions, A1P2B2,to scale in the plane as shown.2 Measure or calculate the values of the magnitude and angle of vector P21, namely,P21 andδ2In this example they are:P21=2.416; δ2=165.2°3 Measure or calculate the values of the change in angle, α2,of vector Z from position 1 to position 2 .In this example it is;α2=43.3°4 The three values in steps 2 and 3 are the only ones defined in the problem statement.We must assume three additional ”free choices” to solve the problem..Method two(see equa-tions 5.8) chooses the length z and angleφof vector Z andβ2,the change in angle of vector W. In order to obtain the same solution as the graphical method ptoduced in Figure 5-3a(form the infinities of solutions available),we will choose those values consistent with the graphical solution.Z=1.298; φ=26.5°; β2=38.4°5 Substitute these six values in equations 5.8 and obtin:w=2.467 θ=71.6°6Compare these to the graphical solution;W=2.48 θ=71°Which is reasonable match given the graphical accuracy.This vector W1 is link 2 of the fourbar.7Repeat the procedure for the link-4 side of the linkage.The free choices will now be:S=1.035; ψ=104.1°; γ2=85.6°8Substitute these three values with the original three values from steps 2 and 3 in equations 5.8 and obtain;u=1.486 σ=15.4°9Compare these to the graphical solution:U=1.53 σ=14°These are a reasonable match for graphical accuracy. Vector U1 is link 4 of the fourbar.10Line A1B1 is link 3 and can be found from equation 5.2a. Line O2O4 is link 1 and can be found from equation 5.2b.11Check the Grashof condition, and repeat saeps 4 to 7 if unsatisfied. Note that any Grashof condition is potentially acceptable in this case.12Construct a cardboard model and check its function to be sure it can get from initial to final position without encountering any limit(toggle) positions.13Check transmission angles.5.5SIMULTANEOUS EQUATION SOLUTIONThese methods of analytical synthesis lead to sets of linear simultaneous equations. The two position synthesis problem results in two simultaneous equations which can be solved by direct substitution. The three-position synthesis problem will lead to a system of four simultaneous linear equations and will require a more complicated method of solution.A convenient approach to the solution of sets of linear simultaneous equations is to put them in a standard matrix from and use a numerical matrix solver to obtain the answers.Matrix solvers are built into most engineering and scientific pocket calculators. Some spreadsheet packages and equation solvers will also do a matrix solution.As an example of this general approach,consider the following of simultaneous equations:-2x1-x2+x3=-1X1+x2+x3=63x1+x2-x3=2A system this small can be solved lolnghand by the elimination method,but we will put it in matrix from to show the general approach which will work regardless of the number of equations. The equations 5.13a can be written as the product of two matrices set equal to as third matrix.[]We will refer to these matrices as A,B,and C,[A] * [B] = [C]Where A is the matrix of coeffcients of the unknowns,B is a column vector of the unknown terms and C is a column vector of the constant terms.When matrix A is munltiplied by B, the result will be the same as the left sides of equation 5.13a.See any text on linear algebra such as reference 7 for a discussion of the procedure for matrix mnltiplication.If equation 5.13c were a scalar equation,ab=crather than a vector(matrix) equation,it would be very easy to solve it for the unknown b when a and c are known, We would simply divide c by a to find b.b=c/aUnfortunately, division is not defined for msatrices,so another approach must be used .Note that we could also express the division in equation 5.14b as:B=(1/a)*cIf the equations to be solved are linearly independent. Then we can find the inverse of matrix A and multiply it by matrix C to find B .The inverse of a matrix is defined as that matrix which when multiplied by the original matrix yields the identity matrix. The identity matrix is a square matrix is denoted by adding a superscript of negative one to the symbol for the original matrix.。

优秀设计平面四杆机构的运动性能研究摘要：平面四杆机构是主要的常用基本机构之一,应用十分广泛,也是其他多杆机构的基础。

由于连杆机构的性能受机构上繁多的几何参数的影响,呈复杂的非线性关系,无论从性能分析上还是性能综合上都是一个比较困难的工作,尚需作进一步深入研究。

本文基于平面四杆机构的空间模型,将机构实际尺寸转化为相对尺寸,在有限的空间内表示出无限多的机构尺寸类型,从而建立起全部机构尺寸类型和空间点位的一一对应关系,为深入研究平面四杆机构的运动性能与构件尺寸之间的关系提供了基础。

根据曲柄摇杆机构、双曲柄机构、双摇杆机构、单滑块四杆机构的不同特点,详细分析各类机构的运动性能参数与构件尺寸之间的关系,指出构件尺寸的变化对机构运动性能的影响,并绘制相关的运动性能图谱。

针对具有急回特性的Ⅰ、Ⅱ型曲柄摇杆机构,通过深入分析极位夹角与构件尺寸之间的内在关系,获得了Ⅰ型曲柄摇杆机构极位夹角分别小于、等于或大于90°的几何条件以及Ⅱ型曲柄摇杆机构极位夹角一定小于90°的结论,揭示了曲柄摇杆机构设计时作为已知条件的极位夹角和摇杆摆角之间应满足的要求。

本文得出的图谱和相关结论,为工程应用中机构性能分析和机构综合提供了理论依据。

关键词：平面四杆机构空间模型运动性能Plane four clubs institutions of Sports performance research Abstract:The planar four-bar linkages are one type of basic mechanisms, and they are applied very extensively. The performances of the linkages depend on their geometrical parameters and present the complicated non-linear relations. It is necessary to make the further research on them for analysis, synthesis and application of linkages.By using of the three-dimensional models of the planar four-bar mechanisms, the actual sizes of mechanisms are transformed relative ones, and all size types of mechanisms can be figured by spatial coordinates. It is the foundation for research on the relations between the link dimensions and kinematic capability parameters.Aimed at the different characteristics of crank-rocker mechanism, double-crank mechanism, double-rocker mechanism and single-slider mechanism, some inherent relations between the link dimensions and the kinematic capability parameters are deeply analyzed, then the relative kinematic capability diagrams are obtained.Based on deeply analysis of inherent relations between the extreme position angle and the link dimensions of typeⅠand typeⅡcrank-rocker mechanisms with quick return characteristics, the geometrical conditions are put forward in this paper, by which we can judge whether the extreme position angle of typeⅠcrank-rocker mechanisms is less than, equal to or lager than 90°. It is proved that the extreme position angle of typeⅡcrank-rocker mechanism is certainly less than 90°. The relations between the extreme position angle and the angular stroke of the rocker are brought to light, which should be satisfied during the kinematic design of crank-rocker mechanisms.The diagrams and conclusions obtained in this paper provide theoretic foundation for the capability analysis and synthesis of mechanisms.Keyword:Planar four-bar linkage Space model Sports Performance如需源程序联系扣扣 194535455目录1 序言1.1 连杆机构 (1)1.2 平面连杆机构运动学分析 (2)1.3 本论文所作的主要工作 (3)2 平面四连杆机构的类型2.1 分类概念 (3)2.2 分类 (4)3 平面四杆机构运动分析3.1.1 连杆上任意点的轨迹分析 (6)3.1.2 Non-grashof机构的运动分析 (8)3.2 速度分析 (9)3.3 加速度分析 (10)4 平面连杆机构曲线分类基准及分类4.1 曲率 (11)4.2 弧长 (12)4.3 回转数 (12)4.4 结点 (13)4.5 变曲点、曲率极大点与极小点 (19)4.6 机构数据库的建立 (20)4.7 连杆曲线的分类结果 (20)5 平面连杆机构的仿真设计5.1 初始运行界面及程序 (23)5.2 部分仿真结果 (42)结论 (49)参考文献 (51)致谢 (52)1 序言连杆机构，是由许多刚性构件通过低副联结而成，也称低副机构。

机械原理英文词汇表周传喜编长江大学机械学院Chapter 1 Introduction第一章 绪论1.mechanism 机构2.kinematical element 运动学元件3.link 构件4.cam 凸轮5.gear 齿轮6.belt 带7.chain 链8.internal-combustion engine 内燃机9.slider-crank mechanism 曲柄滑块机构10.piston 活塞11.connecting rod 连杆12.crankshaft 曲轴13.frame 机架14.pinion 小齿轮15.cam mechanism 凸轮机构16.linkage 连杆机构17.synthesis 综合Chapter 2 Structure analysis of mechanisms 第二章 机构的结构分析1. structural analysis 结构分析2. planar mechanisms 平面机构3. planar kinematical pairs 平面运动副4. mobile connection 可动连接5. transmit 传输6. transform 转换7. pair element 运动副元素8. higher pair 高副9. revolute pair 转动副10. sliding pair ,prismatic pair 移动副11. gear pair 齿轮副12. cam pair 凸轮副13. screw pair 螺旋副14. spherical pair 球面副15. surface contact 面接触16. kinematical chain 运动链17. closed chain 闭式链18. open chain 开式链19. driving links 驱动件20. driven links 从动件21. planar mechanism. 平面机构22. spatial mechanism 空间机构23. The kinematical diagram of a mechanism机构运动简图24. schematic diagram 草图25. kinematical dimensions 运动学尺寸26. fixed pivot 固定铰链27. pathway 导路28. guide bar 导杆29. profiles 轮廓30. the actual cam contour 凸轮实际廓线.31. polygon 多边形32. route of transmission 传递路线33. structural block diagram 结构框图34. Degree of Freedom (DOF) 自由度35. constraints 约束36. common normal 公法线37. compound hinge 复合铰链38. gear-linkage mechanism 齿轮连杆机构40. passive DOF 局部自由度41. redundant constraint 虚约束42. The composition principle and structural analysis组成原理与结构分析43. the basic mechanism 基本机构44. Assur groups 阿苏尔杆组45. inner pair 内副46. outer pairs 外副.47. composition principle of mechanism 机构组成原理48. kinematical determination 运动确定性Chapter 3 kinematic analysis of mechanicsms 第三章 机构的运动分析1. velocity 速度2. acceleration 加速度3. parameter 参数3. graphical method 图解法4. analytical method 解析法5. experimental method 实验法6. instant center 瞬心7. classification of instant centers 瞬心的分类8. absolute instantaneous center 绝对瞬心9. relative instantaneous center 相对瞬心10. the method of instantaneous center 瞬心法11. the Aronhold-Kennedy theorem阿朗浩尔特-肯尼迪定理(即三心定理)12. the four-bar linkage 四杆机构13. inversion of the slider-crank导杆机构(曲柄滑块机构的倒置机构) 14. complex mechanism 复杂机构，多杆机构Chapter 4 Planar Linkage Mechanicsms 第四章 平面连杆机构1.four-bar linkage 四杆机构2.crank-rocker mechanism 曲柄摇杆机构3.double-crank mechanism 双曲柄机构4.double-rocker mechanism 双摇杆机构5.Crashof’s criterion 格拉索夫判据6.Condition for having a crank 有曲柄的条件7.slider-crank mechanism 曲柄滑块机构8.offset distance 偏距9.offset slider-crank mechanism 偏置曲柄滑块机构10.in-line slider-crank mechanism 对心曲柄滑块机构11.rotating guide-bar mechanism. 转动导杆机构12.oscillating guide-bar mechanism 摆动导杆机构13.double rotating block mechanism 双转块机构14.crank and oscillating block mechanism 曲柄摇块机构15.variations 变异16.inversions 倒置17.transmission angle 传动角18.dead point 死点19.imbalance angle 极位夹角20.time ratio 行程速比系数21.quick-return mechanism 急回机构22.pressure angle 压力角23.toggle positions 肘节位置24.oldham coupling 联轴器25.flywheel 飞轮26.clamping device 夹具27.dimensional synthesis 尺度综合28.function generation 函数发生器29.body guidance 刚体导引30.path generation 轨迹发生器Chapter 5 Cam Mechanisms第五章凸轮机构1. contour 轮廓2. Follower 从动件3. Plate cam(or disc cam) 盘形凸轮4. Translating cam 移动凸轮5. Three-dimensional cam 空间凸轮6. cylindrical cam 圆柱凸轮7. Translating follower 直动从动件8. Oscillating follower 摆动从动件9. Camshaft 凸轮轴10.in-line translating follower 对心直动从动件11.offset translating follower 偏置直动从动件12.Knife-edge follower 尖底从动件13.Roller follower 滚子从动件14.Flat-faced follower 平底从动件15.Force-closed cam mechanism 力封闭凸轮机构16.Form-closed cam mechanism 形封闭凸轮机构17.Lift 行程18.cam angle for rise 推程角19.cam angle for outer dwell 远休止角20.cam angle for return 回程角21.cam angle for inner dwell 近休止角22.the quasi-velocity 类速度23.the quasi-acceleration 类加速度24.Constant Velocity Motion Curve 等速运动规律25.rigid impulse 刚性冲击26.Constant Acceleration and Deceleration Motion Curve等加速等减速运动规律27.soft impulse 柔性冲击29.Cosine Acceleration Motion Curve (Simple Harmonic MotionCurve) 余弦加速度运动规律（简谐运动规律） 30.Sine Acceleration Motion Curve (Cycloid Motion Curve)正弦加速度运动规律（摆线运动规律）31.3-4-5 Polynomial Motion Curve 3-4-5多项式运动规律bined Motion Curves 组合运动规律33.the cam contour 实际廓线34.the pitch curve 理论廓线35.prime circle 基圆36.the common normal 公法线37.positive offset 正偏置38.negative offset 负偏置39.outer envelope 外包络线40.inner envelope 内包络线41.The locus of the centre of the milling cutter铣刀中心轨迹42.Pressure Angle 压力角43.acute angle 锐角44.the normal 法线45.The allowable pressure angle 许用压力角46.Radius of Curvature 曲率半径47.Cusp 尖点48.Undercutting 根切49.The angular lift 角行程50.interference 干涉Chapter 6 Gear Mechanisms第六章 齿轮机构1. constant transmission ratio 定传动比2. planar gear mechanisms 平面齿轮机构3. spatial gear mechanisms 空间齿轮机构4. external gear pair 外齿轮副5. internal gear pair 内齿轮副6. rack and pinion 齿条和齿轮7. spur gear 直齿轮8. helical gear 斜齿轮9. double helical gear 人字齿轮10.spur rack 直齿条11.helical rack 斜齿条12.bevel gear mechanism 圆锥齿轮机构13.crossed helical gears mechanism 螺旋齿轮机构14.worm and worm wheel mechanism 蜗杆蜗轮机构15.Fundamentals of Engagement of Tooth Profiles齿廓啮合基本定律16.the pitch point 节点17.the pitch circle 节圆18.conjugate profiles 共轭齿廓19.transmission ratio 传动比20.involute gear 渐开线齿轮21.the radius of base circle 基圆半径22.generating line 发生线23.unfolding angle 展角24.table of involute function 渐开线函数表25.gearing 啮合26.standard involute spur gears 标准渐开线直齿轮27.the facewidth 齿宽28.addendum circle (or tip circle) 齿顶圆29.dedendum circle (or root circle) 齿根圆30.arbitrary circle 任意圆31.the tooth space 齿槽32.the spacewidth 齿槽宽33.the pitch 齿距，周节34.the reference circle 分度圆35.module 模数36.addendum 齿顶高37.dedendum 齿根高38.tooth depth 齿全高39.the coefficient of addendum 齿顶高系数40.the coefficient of bottom clearance 顶隙系数41.bottom clearance 顶隙42.the normal tooth 正常齿43.the shorter tooth 短齿44.base pitch 基圆齿距，基节45.normal pitch 法向齿距，法节46.conjugated point 共轭点47.proper meshing conditions 正确啮合条件48.working pressure angle 啮合角49.the backlash 齿侧间隙50.the bottom clearance 顶隙51.the reference centre distance 标准中心距52.contact ratio 重合度53.the actual working profile 实际工作齿廓54.the actual line of action 实际啮合线55.manufacturing methods of involute profiles渐开线齿廓的加工方法56.form cutting 仿形法57.generating cutting 展成法或范成法58.disk milling cutter 盘形铣刀59.end milling cutter 指状铣刀60.broach 拉刀ling machines. 铣床62.rack-shaped shaper cutter 齿条插刀63.shaping 插齿64.hobbing 滚齿65.rack-shaped cutter 齿条型刀具the 车床67.cutter Interference 根切68.corrected gears 变位齿轮69.the modification coefficient 变位系数70.positively modified 正变位71.negatively modified 负变位72.the gearing equation without backlash 无侧隙啮合方程73.involute helicoids 渐开线螺旋面74.the transverse plane 端面75.the normal plane 法面78.the transverse contact ratio 端面重合度79.the overlap ratio 轴向重合度80.the virtual gear 当量齿轮81.the virtual number of teeth 当量齿数82.axial thrust 轴向推力83.worm gearing 蜗杆传动84.righthanded 右旋85.lefthanded 左旋86.ZA-worm 阿基米德蜗杆87.involute helicoid worms ----ZI-worm 渐开线蜗杆88.arc-contact worms -----ZC-worm 圆弧齿蜗杆89.enveloping worm 包络蜗杆90.The number of threads 头数91.bevel gears 圆锥齿轮92.back cone 背锥93.virtual gear 当量齿轮94.the reference cone 分度圆锥95.sector gear 扇形齿轮96.the outer cone distance 外锥距97.the reference cone angle 分度圆锥角98.The apexes 锥顶99.The dedendum angle 齿根角100.dedendum cone angle 齿根圆锥角Chapter 7 Gear Trains第七章 轮系1. gear train with fixed axes 定轴轮系2. epicyclical gear train 周转轮系3. elementary epicyclical gear trains 基本周转轮系4. combined gear trains 复合轮系5. planet gear 行星轮6. planet carrier 行星架,系杆,转臂7. sun gears 太阳轮,中心轮8. differential gear train 差动轮系9. the train ratio of a gear train 轮系传动比10.idle wheels 惰轮11.converted gear train 转化轮系12.the efficiency of the gear train 轮系效率13.branching transmission 分路传动14.the brake 刹车片15.the clutch 离合器16.negative mechanism 负号机构17.positive mechanism 正号机构18.train ratio condition 传动比条件19.concentric condition 同心条件20.assembly condition 装配条件21.planetary reducer with small tooth difference少齿差行星减速器 22.cycloidal-pin wheel planetary gearing摆线针轮行星传动 23.harmonic drive gearing 谐波传动Chapter 8 Other Common Mechanisms 第八章 其它常用机构1. ratchet mechanism 棘轮机构2. pawl 棘爪3. intermittent motion 间歇运动4. geneva mechanism 槽轮机构5. external geneva mechanism 外槽轮机构6. internal geneva mechanism 内槽轮机构7. geneva rack mechanism 齿条槽轮机构8. spherical geneva mechanisms 球面槽轮机构9. the ratio k between motion time and dwell time运动与停歇时间比k10.Cam-Type Index Mechanisms 凸轮式间歇运动机构11.Cylindrical Cam Index Mechanisms 圆柱凸轮式间歇运动机构12.Universal Joints 万向联轴节13.The Single Universal Joint 单万向联轴节14.The Double Universal Joint 双万向联轴节15.Screw Mechanisms 螺旋机构16.single-threadscrew mechanisms 单螺旋机构17.Double-thread screw mechanisms 复式螺旋机构18.Index cam mechanism 分度凸轮机构19.Geared linkages 齿轮连杆机构Chapter 10 Balancing of Machinery第十章 机械的平衡1. Vibration 振动2. Frequency 频率3. Resonant 共振4. Amplitudes 振幅5. Balancing of rotors 转子6. Rigid rotors 刚性转子7. Flexible rotors 柔性转子8. Balancing of mechanisms 机构的平衡9. Disk-like rotor 盘状转子10.Non-disk rigid rotor 非盘状转子11.the shaking force 振动力12.the shaking moment 振动力矩13.Balancing of Disk-like Rotors 盘状转子的平衡14.static imbalance 静不平衡15.static balancing machine 静平衡机16.the mass-radius product 质径积17.dynamically unbalanced 动不平衡18.balance planes 平衡基面19.Dynamic balancing machine 动平衡机20.Unbalancing Allowance 许用不平衡量Chapter 11 Motion of Mechanical Systemsand Its Regulation第十一章 机械系统的运转及其调节1. Periodic speed fluctuation 周期性波动2. punching machine 冲床3. Motion Equation of a Mechanical System机械系统的运动方程4. General Expression of the Equation of Motion运动方程的一般表达式5. the kinetic energy 动能6. the moment of inertia 转动惯量7. Dynamically Equivalent Model of a Mechanical System等效动力学模型8. the equivalent moment of inertia 等效转动惯量9. the equivalent moment of force 等效力矩10.the equivalent link 等效构件11.Pump 泵12.Blower 鼓风机13.Flywheel 飞轮Chapter 12 Creative Design of Mechanism Systems 第十二章 机械系统的创新设计1. prototype machine 样机2. Working cycle diagrams 工作循环图3. reference link 定标件4. Circular working cycle diagram 圆工作循环图5. Rectilinear working cycle diagram 矩形工作循环图6. Rectangular coordinate working cycle diagram直角坐标式工作循环图。

机械基础》课程标准)机械基础》课程标准一、课程概述为适应机械产业升级和企业高速发展技能应用型人才的需要，本课程旨在加强高职机械类专业学生的机械设计、机械结构认识与应用能力培养。

机械是人类在长期的生产实践中创造出来的技术装置，在现代生产和日常生活中起着非常重要的作用。

对于现代工程技术人员，研究和掌握一定的机械基础知识是必要的而且是必需的。

本课程是机械类专业的一门重要的专业基础课程。

以实用为原则，以会用为目标。

重点研究齿轮传动、带传动和链传动等常用的机械传动装置；铰链四杆机构、凸轮机构等常用机构。

熟悉蜗杆传动、轮系等传动，了解螺旋机构、间歇运动机构等机构形式及键、销等连接方式。

熟悉轴、轴承及联轴器、离合器等结构、类型与应用。

通过一体化教学，使学生对机械知识有一定的认识与掌握，并能合理应用机械结构等。

二、培养目标1．方法能力目标通过项目任务和工程案例，激发学生积极性，培养学生勤奋好学的惯，让学生在研究中享受成功的喜悦。

同时，通过课后拓展训练，巩固课堂项目研究效果，培养学生独立自主研究的良好惯。

目的是培养学生应用机械基础知识解决实际工程计算的能力。

本课程适用于数控技术专业、模具设计与制造专业、机械制造与自动化专业、机电一体化技术专业和工业机器人技术专业。

学时为48~56学时，学分为3~4学分。

开设时间在第一学期和第二学期。

为了提高学生的研究效果和信息处理能力，本课程将搭建课程网络研究平台和建立课程网络资源库，实现资源共享。

在社会能力目标方面，本课程通过项目任务交流与互助，培养学生团队协作精神，并锻炼他们的沟通交流能力。

此外，课后训练将拓展研究项目，培养学生自主研究能力。

通过项目训练，学生将掌握机械工程手册查阅能力，并培养爱岗敬业、精准求精的基本素养。

在专业能力目标方面，本课程将使学生掌握齿轮、带、链等机械传动的类型、原理及应用，并具备机械传动装置的分析与选用能力。

此外，学生还将熟悉轮系的类型、作用，以及定轴轮系的有关计算；掌握铰链四杆机构、凸轮机构等的结构、原理及性质，并具备常用机构的应用能力；了解棘轮机构等间歇运动等机构的结构、原理及应用；熟悉螺纹、键、销等连接件的类型、特点及应用；熟悉轴及轴承类型、应用及特点，并具备分析轴结构的能力；了解联轴器、离合器等连接件的类型、特点及应用场合。