机器人操作臂运动学PPT教案

1、会计学1机器人操作臂运动学机器人操作臂运动学3.1 连杆参数和连杆坐标系Denavit - Hartenberg Parameters第1页/共73页第2页/共73页通常为了能在三维空间定位末端执行器，最少要求有6个关节。第3页/共73页连杆坐标系 关节 1 是垂直于肩, 关节 2 经过肩水平线, 关节 3 是在肘部。关节 4, 5 & 6 是在手腕上，初始位置关节4 和关节6 共同沿着前臂,关节5 垂直于关节4 和关节6。第4页/共73页连杆坐标系Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i 111111 j

2、oint to1-joint from :direction and 1- axesbetween normalcommon arbitrary :direction 1,- axisjoint iiiiiixzyyiiiixiz第5页/共73页第6页/共73页Specification of Base & Final link frames 0 n第7页/共73页参数/变量: , a , d, 基本思想：每个关节分配一个坐标系。用D-H参数，描述框i相对于前一个框i-1的位姿需要4个参数D-H参数Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y

3、 i X i a i d i i 第8页/共73页1） ai-1 定义: ai-1 两个关节轴线公垂线的长度. 关节轴是围绕它发生旋转的有向空间直线，在图中是 Zi-1和 Zi 轴。Zi - 1Xi -1Yi -1 i - 1ai - 1Z i Y i X i a i d i i 第9页/共73页可视化方法：想象一个圆柱面围绕轴Z(i-1) 扩展 当圆柱面刚刚触及轴 i 时，圆柱的半径等于a(i-1)。图示方法： 若已经定义了坐标系, 公垂线通常是X(i-1) 轴.因此 a(i-1) 恰是沿着X(i-1)从框i-1 到框i 的位移如果连杆是移动关节, 那么 a(i-1) 是变量，而不是参数Z(

4、i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i 连杆参数连杆参数a(i-1) 的的识别方法：识别方法：第10页/共73页2) (i-1)定义: 使关节轴平行时，绕公垂线旋转的角度. 按右手规则确定正向旋转。绕X(i-1) 轴旋转使 Z(i-1) 指向Zi 轴的方向Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i 第11页/共73页3) di定义: 为了使公垂线a(i-1)和公垂线ai与Zi的交点对起，沿Zi 轴所需的位移。即，沿Zi 对准X(i-1) 和

5、 Xi 轴. Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i 第12页/共73页4) i 为了对准X(i-1) 轴和Xi 轴，绕Zi 轴所需转动的角度Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i 第13页/共73页1ia1in第14页/共73页idi第15页/共73页0 , 06060aad1和d6以及1和6的确定方法如下。 若关节1是转动关节，则1是可变的，称为关节变量，规定1 0为连杆1的零位。习惯约定d10若关节1是移动关节，则d1是可变的

6、，称为关节变量，规定d1=0为连杆1的零位。习惯约定10。 上面的约定对于关节6同样适用。第16页/共73页1ia1iidiidi1ia1iiid第17页/共73页第18页/共73页移动关节转动关节1ia1iidi连杆连杆i-1几何特征几何特征i-1从zi-1到zi沿xi-1旋转的角度ai-1 从zi-1到zi沿xi-1测量的距离di从xi-1到xi沿zi测量的距离i从xi-1到xi沿zi旋转的角度第19页/共73页3.1连杆变换和运动学方程第20页/共73页. axis along on translati)(; axisabout rotation )(; axis along on tr

7、anslati)(; axisabout rotation )(1111 -iiiiiiiizddzcxabxaT1ii连杆变换可以看成是坐标系i经以下四个子变换得到的：1ia1iidi用4个参数对准两个关节的轴线第21页/共73页Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i ),(),(),(),(111iiiiiidzTranszRotaxTransxRotT因为这些子变换都是相对于动坐标系描述的，按照“从左向右”的原则得到连杆变换矩阵第22页/共73页),(),(),(),(),(),(11111iiiiiiii

8、iidzScrewaxScrewdzTranszRotaxTransxRotTTii1（The Denavit-Hartenberg Matrix）10000111111111iiiiiiiiiiiiiiiiicdcscsssdscccsasc第23页/共73页D-H参数矩阵参数矩阵1000coscossincossinsinsinsincoscoscossin0sincosi1)(i1)(i1)(ii1)(iii1)(i1)(i1)(ii1)(ii1)(iiidda与齐次变换矩阵一样， D-H参数矩阵是从一个坐标系到下一个坐标系的变换。用一系列D-H参数矩阵相乘，最终的结果是从某个坐标系到初

9、始坐标系的变换。Z(i - 1)X(i -1)Y(i -1)( i - 1)a(i - 1 )Z i Y i X i a i d i i 第24页/共73页连杆变换依赖于四个参数，其中只有一个是变化的。以下用qi表示第i个关节变量TTTTnnn112010)()()(),(1212101210nnnnnqTqTqTqqqT10000pRTnnn)()()(1212101nnnqTqTqT手臂变换矩阵第25页/共73页Z0X0Y0Z1X2Y1Z2X1Y2d2a0a1Denavit-Hartenberg Link Parameter Table表的用途:1) 描述机器人的变量和参数2) 通过变量的

10、数值描述机器人的状态i-1从zi-1到zi沿xi-1旋转的角度ai-1从zi-1到zi沿xi-1测量的距离di从xi-1到xi沿zi测量的距离i从xi-1到xi沿zi旋转的角度第26页/共73页Z0X0Y0Z1X2Y1Z2X1Y2d2a0a1T)T)(T12011VVVTV222000ZYXZYX第27页/共73页100000000cossina0sincosT11011011100000cossind100a0sincosT22212212This is a translation by a0 followed by a rotation around the Z1 axisThis is

11、a translation by a1 and then d2 followed by a rotation around the X1 and Z2 axisTTT1201Z0X0Y0Z1X2Y1Z2X1Y2d2a0a1第28页/共73页i100029003001ia1iidi11l22l3100001000000111101csscTy3x3第29页/共73页100000010002212212cslscT100001000003323323cslscTTTTT23120103第30页/共73页The Situation:You have a robotic arm that starts

12、 out aligned with the xo-axis.You tell the first link to move by 1 and the second link to move by 2.The Quest:What is the position of the end of the robotic arm? 12两关节机器人第31页/共73页X2X3Y2Y3123123 Example Problem: You are have a three link arm that starts out aligned in the x-axis. Each link has length

13、s l1, l2, l3, respectively. You tell the first one to move by 1 , and so on as the diagram suggests. Find the Homogeneous matrix to get the position of the yellow dot in the X0Y0 frame.X1Y1X0Y0第32页/共73页The position of the yellow dot relative to the X3Y3 frame is (l1, 0). Multiplying H by that positi

14、on vector will give you the coordinates of the yellow point relative the the X0Y0 frame. X2X3Y2Y3123123X1Y1X0Y0H = Rz(1 ) * Tx1(l1) * Rz(2 ) * Tx2(l2) * Rz(3 ) i.e. Rotating by 1 will put you in the X1Y1 frame. Translate in the along the X1 axis by l1. Rotating by 2 will put you in the X2Y2 frame. a

15、nd so on until you are in the X3Y3 frame.第33页/共73页Slight variation on the last solution: Make the yellow dot the origin of a new coordinate X4Y4 frame X2X3Y2Y3123123X1Y1X0Y0X4Y4H = Rz(1 ) * Tx1(l1) * Rz(2 ) * Tx2(l2) * Rz(3 ) * Tx3(l3)This takes you from the X0Y0 frame to the X4Y4 frame.The position

16、 of the yellow dot relative to the X4Y4 frame is (0,0). 第34页/共73页We are interested in two kinematics topicsForward Kinematics (angles to position)What you are given: The length of each link The angle of each jointWhat you can find: The position of any point (i.e. its (x, y, z) coordinates)Inverse Ki

17、nematics (position to angles)What you are given:The length of each linkThe position of some point on the robotWhat you can find: The angles of each joint needed to obtain that position第35页/共73页PUMA560机器人关节空间运动第36页/共73页PUMA560连杆坐标系第37页/共73页第38页/共73页第39页/共73页第40页/共73页第41页/共73页第42页/共73页第43页/共73页则工具相对于工

18、作站的位姿为第44页/共73页I n v e r s e K i n e m a t i c sFrom Position to Angles 第45页/共73页A Simple Example 1XYSRevolute and Prismatic Joints Combined(x , y)Finding 1:)xyarctan( More Specifically:)xy(2arctan arctan2() specifies that its in the first quadrantFinding S:)y(xS22第46页/共73页 2 1(x , y)l2l1Inverse Kin

19、ematics of a Two Link ManipulatorGiven:l1, l2 , x , yFind: 1, 2Redundancy: A unique solution to this problem does not exist. Notice, that using the “givens” two solutions are possible. Sometimes no solution is possible.(x , y)l2l1l2l1第47页/共73页The Geometric Solutionl1l2 2 1 (x , y)Using the Law of Co

20、sines:2122212221222122212221222222arccos2)cos()cos()180cos()180cos(2)(cos2l lllyxl lllyxl lllyxCabbac22222Using the Law of Cosines:xy2arctanyx)sin(yx)sin(180sinsinsin1122222221lcCbBxy2arctanyx)sin(arcsin22221lRedundant since 2 could be in the first or fourth quadrant.Redundancy caused since 2 has tw

21、o possible values第48页/共73页 2122212222212221211211212221211212212221212112122122212122222yxarccosc2)(sins)(cc2)(sins2)(sins)(cc2)(ccyx)2(1)l llll llll llll llll lllThe Algebraic Solutionl1l2 2 1(x , y)212121121211122111(3)sinsy(2)ccx(1)cos( ccos cllllOnly Unknown)(sin(cos)(sin(cos)sin()(sin(sin)(cos(cos)cos(:abbababababaNote记：有第49页/共73页)(sin(cos)(sin(cos)sin()(sin(sin)(cos(cos)cos(:abbababababaNote)c(s)s(c cscss sinsy)()c(c ccc ccx2211221122212112121122122112122121121211llllllllslsllsslllllWe know what 2 is from the previous slide. We need to solve for 1 . N