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连杆弹簧复位自动调偏装置设计,连杆,弹簧,复位,自动,装置,设计

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辽宁工程技术大学课程设计辽宁工程技术大学课 程 设 计题 目：连杆弹簧复位自动调偏装置班级：姓名：指导教师：完成日期：一、设计题目二、设计要求三、完成后应上交的材料四、进度安排五、指导教师评语成 绩： 指导教师日期摘要在带式输送机运转过程中，输送带的纵向中心线偏离输送机理论中心线的现象称输送带跑偏。它的表现是输送带边缘至托辊或滚筒边缘的距离与理论值相比或大或小。输送带的跑偏会使输送带与机架、托辊支架相摩擦，造成边胶磨损。严重的跑偏会使输送带翻边，若在滚筒表面边缘有凸起的螺丝头、覆盖胶层局部剥离、划伤等事故。跑偏会导致输送机的事故停机次数增多，影响生产；跑偏还可能引起物料外撒，使输送机系统的运营经济性下降。为此，曲柄连杆式自动调偏装置最突出的特色是曲柄连杆机构。调偏架通过滚动轴承安装在底座上，两端对称连接制作的连杆，连杆末端固定立辊。安装在曲柄上的两侧立辊靠近输送带边缘，一有跑偏出现，输送带就压向该侧立辊，使立辊随着曲柄向外转动，曲柄和连杆拉动调偏架按输送带运行方向旋转一定角度，从而产生调偏。SummaryWhile the bringing type conveyer operates, the phenomenon that the longitudinal centre line of the conveyer belt deviates from the mechanism of transporting and talks about the centre line claims the conveyer belt runs partially. Its behavior is either large or small on compared with theory value to holding the distance on the roller or edge of cylinder the edge of conveyer belt. The running and leaning towards and will enable conveyer belt and framework, ask the roller support to rub of conveyer belt, cause the glue is worn and torn. A serious one run, lean towards, can enable conveyer belt turn-ups, if surface edge have protruding screw head, cover glue layers of part strip, accident of scratching etc. in cylinder. Run, lean towards, can lead to the fact accident of conveyer shut down number of times increase, influence and produce; Run, may cause supplies let go also simply, make conveyer economic decline of operation of system. For this reason, it is the crank connecting rod organization that the crank connecting rod type adjusts and leans towards the most outstanding characteristic of the device automatically. Adjust and lean towards the shelf and install the connecting rod made of the symmetrical connection of both ends on the base through the rolling bearing, the connecting rod sets up the roller terminal fixedly. Both sides that are installed on the crank set up the roller close to the edge of conveyer belt, run and simply appear, conveyer belt press to should incline, set up roller, make, set up roller rotate outwards with the crank, the crank and connecting rod spur and adjust and lean towards the shelf to rotate certain angle according to the conveyer belt operation direction, thus produced and adjusted partially. 目 录1引言12系统分析22.1工作原理2 2.2初始条件22.3计算过程与分析22.3.1阻力计算22.3.2载荷计算32.3.3总阻力计算42.3.4弹簧弹力计算与选择42.3.5调偏架的固定52.3.6两连杆的固定及计算62.3.7其它杆件的安装62.3.8立辊的规格63系统说明74. 心得体会8参考文献8辽宁工程技术大学课程设计1引言皮带运输机在运输的过程中由于运输的东西沉重，和货物在往运输带上放的时候的方向方法不同，所以会导致运输带的跑偏，我们这里研究的就是防止运输带的跑偏过程。在带式输送机运转过程中，输送带的纵向中心线偏离输送机理论中心线的现象称输送带跑偏。它的表现是输送带边缘至托辊或滚筒边缘的距离与理论值相比或大或小。输送带的跑偏会使输送带与机架、托辊支架相摩擦，造成边胶磨损。严重的跑偏会使输送带翻边，若在滚筒表面边缘有凸起的螺丝头、覆盖胶层局部剥离、划伤等事故。跑偏会导致输送机的事故停机次数增多，影响生产；跑偏还可能引起物料外撒，使输送机系统的运营经济性下降。为此，曲柄连杆式自动调偏装置最突出的特色是曲柄连杆机构。调偏架通过滚动轴承安装在底座上，两端对称连接制作的连杆，连杆末端固定立辊。安装在曲柄上的两侧立辊靠近输送带边缘，一有跑偏出现，输送带就压向该侧立辊，使立辊随着曲柄向外转动，曲柄和连杆拉动调偏架按输送带运行方向旋转一定角度，从而产生调偏。特点：1 这个装置是无源装置，结构简单，经济实惠。2 通过连杆作用，比不使用连杆的调偏方式大，作用快。3 通过复位弹簧的作用，可以减少没有复位弹簧时，皮带在托辊上左右蛇形的运动状态。1立辊 2调偏架 3支架 4底座5中间转动部分 6连杆1 7连杆2整体调偏工作原理图2系统分析2.1工作原理连杆弹簧复位式自动调偏装置最突出的特色是连杆弹簧复位机构。由上图分析，调偏架通过滑动轴承固定在轨道上，两端对称连接连杆，连杆2末端固定立辊。安装在连杆上的两侧立辊靠近输送带边缘，一有跑偏出现，输送带就压向该侧立辊，使立辊立即随着连杆2向外转动，连杆拉动支架按输送带的运行方向旋转一定角度，从而通过受力分析，产生调偏作用，当皮带回到中心时，在一侧弹簧的作用下，支架又被拉回到了不偏转的位置上。从而也就完成了整个的调偏过程2.2初始条件：皮带带宽B1200mm 输送量Q2000t/h 带速V2.5m/s 托辊间距a=1.5m2.3计算过程与分析2.3.1阻力计算（1）输送机所受的主要阻力（ISO5048）： FHfLgqR0+qRu+(2qB+qG)cosf 模拟摩擦系数 f0.020 工作环境良好、制造、安装良好，带速低，物料内摩擦系数小。L 托辊长度g 重力加速度qR0输送机承载分支每米托辊旋转部分质量 查表取20kg/mqRu输送机回程分支每米托辊旋转部分质量qB 每米输送带质量 查表为23kg/mqG 物料每米质量 =222.22kg/m 机身倾角 取3最后求得FH26.57N(2)输送机所受的特殊阻力：FT=N= 皮带与托辊的摩擦系数 取0.4b1 前倾托辊与输送带接触长度 当跑偏0.075m时通过计算知b10.443m最后求得FT=359.66N2.3.2载荷计算（1）由CEMA所给出的托辊载荷的计算方法去计算承载托辊：由公式：K1物料系数 等长三辊槽形托辊组 K10.7K2输送带系数 等长三辊槽形托辊组 K20.4通过计算得P中1977.05NP侧423.65NCEMA方法中：作用在中间托辊上的力就70%，作用在侧托辊上的力为15%.2.3.3总阻力计算带在未跑偏时，所受的总的阻力为：F总FH+FT+(P中+ P侧*2)=2590.9N当带跑偏75mm时，额外增加的对侧托辊的载荷 F额= =318.88NF摩侧127.55NF总半1295.45N这里取1300N2.3.4弹簧弹力计算与选择故弹簧的最小应提供的力为1300N，才能保证弹簧能够将调心托辊拉回或推回到原位，这里取F弹1300N弹簧的选择：最小工作载荷： P11300N/m由表查得：弹簧截面圆直径 d=8mm中径 D=80mm工作极限载荷 Pj=1504N工作极限载荷下的变形量 fj=19.03N/mm 具体计算：工作行程先取 h=50mm最大工作载荷 Pn=1300N初算弹簧刚度 P=10.84N/mm由表通过弹簧截面直径和中径得到 单圈刚度 Pd=79N/mm，通过计算有效圈数 取标准值n=10总圈数 弹簧刚度 工作极限载荷下的变形量 节距 自由高度 弹簧外径 弹簧内经 螺旋角 展开长度 最小载荷时的高度 最大载荷时的高度 极限载荷时的高度 实际工作行程 故弹簧的总技术要求如下：1.总圈数： n1=122.旋向为右旋3.展开长度 L=3031.79mm4.硬度为 HRC=45505.弹簧固定在两杆中间，起到当支架跑偏时复位的作用。2.3.5调偏架的固定调偏架通过滑动轴承，固定在离支架500mm的位置上。调偏架的高度由余弦定理求出，调偏架的高度为424mm.通过滑动轴承固定在底座上。2.3.6两连杆的固定及计算由于支架只能旋转15，所以需保证两杆之间的夹角为105，由余弦定理c=500 105 a=b=314.98mm故杆应为314.98mm当连杆拉动支架旋转15时，支架摆过的距离为mm通过上面的计算，知弹簧被压缩了50mm，提供了1300N的力，这个力恰好能使偏转的支架回到平衡位置。由于弹簧的自由长度为 H0282.3mm由余弦定理算出固定位置为距离两连杆连接处的178.48mm 2.3.7其它杆件的安装两立辊与皮带接触就留用10间距。2.3.8立辊的规格立辊依据经验及所受的载荷和带宽确定为60mm.直径为25mm3系统说明安装调试完毕后，自动纠偏装置的维护与保养一、 紧固螺栓要每周检查一次，防止松动变位。二、 每周应将传动部分的灰尘进行一次清除。三、 每当设备检修时，应对自动纠偏装置进行除锈，并涂刷面漆。四、 使用过程中要经常检查，如发现有损坏部件，要及时更换。4. 心得体会 经过两个星期的努力,我终于将机械设计课程设计做完了.在这次作业过程中,我遇到了许多困难,一遍又一遍的计算,一次又一次的设计方案修改这都暴露出了前期我在这方面的知识欠缺和经验不足。至于画装配图和零件图,由于前期计算比较充分,整个过程用时不到一周,在此期间,我还得到了许多同学和老师的帮助.在此我要向他们表示最诚挚的谢意. 尽管这次作业的时间是漫长的,过程是曲折的,但我的收获还是很大的.不仅对制图有了更进一步的掌握;Auto CAD ,Word这些仅仅是工具软件,熟练掌握也是必需的.对我来说,收获最大的是方法和能力.那些分析和解决问题的方法与能力.在整个过程中,我发现像我们这些学生最最缺少的是经验,没有感性的认识,空有理论知识,有些东西很可能与实际脱节.总体来说,我觉得做这种类型的作业对我们的帮助还是很大的,它需要我们将学过的相关知识都系统地联系起来,从中暴露出自身的不足,以待改进.有时候,一个人的力量是有限的,合众人智慧,我相信我们的作品会更完美!参考文献1宋伟刚，通用带式输送机设计，机械工业出版社，2006.52张钺 ，新型带式输送机设计手册，冶金工业出版社，2001.23成大先，机械设计手册单行本弹簧起重运输件五金件，化学工业出版社，2004.18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。它是一给予很少关注的机械工程学的组成部分。联杆是具有两个或更多运动副元件的刚性机构，用它的连接是为了传递力或运动。在每个机器中，在运动期间，联杆或者占据一相对于地面的固定位置或者作为一个整体来承载机床。这些连杆是机器主体被称为固定连杆。基于由循环的或滑动的分界面的元件连接的布局被称作连接。这类旋转的和菱形的连接机构被称作低副。高副基于接触点或弯曲分界面的。低副的例子包括铰链循环的轴承和滑道以及万向接头。高副的例子包括通信区主站和齿轮。动力分析得到，根据机件几何学有利条件研究是一特别的机构,它是识别输入角速度和角加速度等等的运动。运动合成作用是处理机构设计到完成完成要求的任务。这里, 两者的选择类型和新的机制尺寸可能是运动学的综合部份。平面的、空间性的和球面运动机构平面的机构是其中全部的点描述平面曲线是间隔和全部平面是共面的， 大多数连杆和机构被设计成这样，例如刨床体系。主要的理由是这个平面的体系对工程师来说更方便。计算机综合法对工程师来说空间性的装置会有更多的麻烦。平面低副机构被称作二维的连接装置。平面的连接仅仅包括旋转的和一对等截面的使用。空间机构没有对相对运动的点的限制。平面的和球面运动机构是亚垫铁等锻工工具的空间机构。空间机构的连接不是被认为这时候被记录。球面运动机构有一接触点接通各个连杆，它是不动的并且平稳点在所有当中联杆场中工作。在所有机件当中，运动是同心并且由他们的盲区接通球面表现出来，它是集中于普遍的定位。空间机构的连接认为不是这时候被记录。可动性连杆在运动中所表现的自由度数是一个很重要的问题。为了使装置被送到指定位置应控制独立的活动自由度。它可能是由杆的数量和连接方式决定的。一自由连杆通常有3个自由度(x , y, )。由于自由度数的限制在n连杆装置中，通常把一个杆固定。自由度数=3(n-1).连接二连杆的机构有两个自由度约束的增加。有两个约束的二连杆连接，其中一个自由度是来约束这个系统的。有一个约束的连杆机构的自由度是j1，有两个约束的连杆机构的自由度是j2。这个系统的自由度数可表示为 m = 3 (n-1) - 2 j 1 - j 2以下为可动的连杆机构装置的示例0是这个体系中可动的机构。系统中仅仅由一连杆的位置固定可以将可动1安装在固定位置。系统中需要一个可动的2与两个连杆来确定连接位置。这是个一般的规则，但也存在例外，它可以作为一个可动性连杆布局的很有用的参考。格朗定律当设计一连接连杆时，在连续地旋转连杆处，例如由一马达输入时，连线可以自由地旋转完全运行驱动是很重要的。如果连杆锁在任一点则方案不会工作。四杆联动机构和grashof定律对这个情况进行提供了简单的测验。格朗的定律如下：b(短的链环)+c(长的链环)a+d四个典型的四连杆机构注意：如果非之上情况则只有连杆滑块机构可行。四连杆机构的优点四连杆机构按比例增大了施加在主动杆上的输入扭矩。可以证明其正比例系数是Sin( )其中是c、d 两杆之间的角度。反比例于sin( )。其中是b、c两杆之间的角度。这些角度不恒定，因此很明显，机构的优点是规律性的变幻。 如下图显示当角度=0 o或则=180 o时接近于无限增矩机构。这些位置是极限位置， 这些位置使四连杆机构可以用于夹具机构。角被叫做“传输角度”。当传输角度的sin值趋于无限小时，机构的增距接近于0。在这样的情况下连杆容易因为很小的摩擦而产生自锁。一般来说，当使用四连杆机构时，避免采用低于400到500的传输角度。弗洛伊德方程这些方程提供了确定内外连杆位置及连杆长度的简单代数学方法。假设四连杆机构如下所示：四连杆的位置矢量如下：l 1 + l 2 + l 3 + l 4 = 0 水平位移方程：l 1 cos 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 垂直位移方程：l 1 sin 1 + l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 假设 1 = 1800 then sin 1 = 0 and cos 1 = -1 Therefore 而l 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0方程两边同时消去l 3：l 32 cos 2 3 = (l 1 - l 2 cos 2 - l 4 cos 4 ) 2 l 32 sin 2 3 = ( - l 2 sin 2 - l 4 sin 4) 2由以上两式可得如下关系cos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 ) and sin2 + cos2 = 1结果如下所示弗洛伊德方程得出这样的参数关系结论K 1 cos 2 + K2 cos 4 + K 3 = cos ( 2 - 4 )K1 = l1 / l4 K2 = l 1 / l 2 K3 = ( l 32 - l 12 - l 22 - l 2 4 ) / 2 l 2 l 4 这个方程符合四连杆机构的有限元分析。如果外连杆机构中的三个参量已知，那么可以由公式得出其他连杆的位置与长度参数。连杆的速度矢量杆上一点的速度必须与杆的轴向垂直否则连杆的长度将产生变化。在B下的连杆速度为vAB = .AB，方向垂直于AB杆，速度矢量图如下： 考虑到下面四连杆机构的实例，速度矢量图表示如下：1）A和D相连并固定，相对加速度=0，A和D位于同一点2）B点相对A点加速是vAB = .AB垂直于AB杆。3）C点相对D点速度通过D点垂直于CD杆。4）P店读速度由速度矢量图和比例bp/bc = BP/BC获得。速度矢量简图如下所示：连杆上滑块的速度认为B滑块绕着A在连杆上滑动，滑块瞬间位移到B点。B点的速度为A = .AB并垂直于线的方向。其链接滑块和速度矢量图如下所示： 连杆的加速矢量杆上一点相对另一点的加速矢量由两部分组成：1）向心加速度由其角速度和连杆长度决定为 2.L2）角加速度由连杆角加速度度决定以下图表显示如何到构造一矢量图表下图显示如何构造单连杆机构的加速矢量向心加速度ab = 2.AB方向指向圆心，角加速度为bb = . AB方向垂直于杆。下图显示如何构造四连杆机构的加速矢量画法1）. A和D相连并固定，相对加速度=0(a,d同)2）. B点相对A点加速在上面的杆上画出3）. B点相对C点向心加速度为：B = v 2CB，方向指向B。4）. B点相对C点角加速度未知但是方向已知5）. C点相对D点向心加速度为：D = v 2CD， 与d( dc2)方向相同。6）. C点相对D点角加速度未知但是方向已知7）. 通过线c1 和c 2的交叉点找出cP点的速度由比例bp/bc=bp/bc获得，且其绝对加速度为P = ap。下面的图表显示其构造方式和转杆上滑块的加速矢量图。两个滑块之间呈dw角。连杆上点的速度与B点变化一致，变化范围为.r =a b 1 到 ( + d) (r +dr) = a b 2b1b2速度的变化分为沿杆方向的r d 及沿其切线方向的dr + r d。滑块上B点的速度与连杆上相关点的变化有关v = a b 3 to v + dv = a b 4.沿着dv与v d 方向速度的变化= b3b4 。在速度切线方向总变化= dv- r d 加速度 = dv / dt = r d / dt = a - 2 r 速度在正切方向总变化= v d + dr + r 正切加速= v d / dt + dr/dt + r d / dt = v + v + r = r + 2 v 加速矢量图表显示如下：注: 其中2 v代表块的正切加速度。每当链接滑通过一个旋转的块，相对一致点沿着一旋转链环一块滑动。- 7 -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 a spherical 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, ). 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.Grashofs 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 Grashofs law provides a simple test for this conditionGrashofs 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 be continuous relative rotation between two members.Referring to the 4 inversions of a four bar linkage shown below .Grashofs 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( ) the angle between the coupler link(c) and the driven link(d), and is inversely proportional to sin( ) 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 = 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 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 Freudensteins 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 follows l 1 + l 2 + l 3 + l 4 = 0 Equating horizontal distances l 1 cos 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 Equating Vertical distances l 1 sin 1 + l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 Assuming 1 = 1800 then sin 1 = 0 and cos 1 = -1 Therefore - l 1 + l 2 cos 2 + l 3 cos 3 + l 4 cos 4 = 0 and . l 2 sin 2 + l 3 sin 3 + l 4 sin 4 = 0 Moving all terms except those containing l 3 to the RHS and Squaring both sides l 32 cos 2 3 = (l 1 - l 2 cos 2 - l 4 cos 4 ) 2l 32 sin 2 3 = ( - l 2 sin 2 - l 4 sin 4) 2Adding the above 2 equations and using the relationshipscos ( 2 - 4 ) = cos 2 cos 4 + sin 2sin 4 ) and sin2 + cos2 = 1the following relationship results.Freudensteins Equation results from this relationship as K 1 cos 2 + K2 cos 4 + K 3 = cos ( 2 - 4 )K1 = l1 / l4 K2 = l 1 / l 2 K3 = ( l 32 - l 12 - l 22 - l 2 4 ) / 2 l 2 l 4 This 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 vAB = .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 vAB = .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/BC The 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 = .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 centripetal component due to the angular velocity of the link. 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 = 2.AB towards the centre of rotation. The tangential component bb = . AB in a direction perpendicular to the link. The diagram