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1、. . . .ABB IRB 6600工业机器人运动学研究报告目录1机器人结构简介12机器人的运动学22.1、机器人正运动学22.2、机器人逆运动学82.2.1求各关节到末端的坐标变换矩阵82.2.2求Jacobian矩阵各列121机器人结构简介ABB工业机器人可以用于实现喷雾、涂胶、物料搬运、点焊等多种功能，是典型的机械臂，在网络上可以查找到较多的相关资料。本次作业就选取ABB IRB 6600机器人作为研究对象，首先对其结构进行简单简介。图 1图 2ABB IRB 6600是六自由度机器人，具有六个旋转关节，底座固定，通过各关节的旋转可以完成三维空间内的运动。图1是ABB IRB 6600
2、机器人的照片及工作范围图，图2是其结构简图和各轴的转动的参数。2机器人的运动学在这部分中运用所学知识对ABB IRB 6600 机器人进行D-H建模并求出对应的转换矩阵，并运用Jacobian 法进行逆运动学分析，求出Jacobian变换矩阵。2.1、机器人正运动学为了计算方便把机器人各关节前后两连杆共线作为初始状态，画出结构简图如图3图 3图3中的关节7实际上是末端执行机构。运用学过的D-H建模方法建立模型，建模过程中为了方便画出各关节坐标系，将部分连杆进行了拉长，且由于部分关节坐标的Z轴垂直于纸面，所以用X轴Y轴画出坐标系，用右手定则既得到对应的Z轴。最终建立模型如图4：图 4根据图4的可
3、以得到对应的D-H参数表：i变量范围100-180,1802900-65，80300-180，6040-90-300，3005-900-120，12060-90-300，3007（末端）00由此算出各关节变换矩阵：将这些关节坐标变换矩阵连乘就得到了由基坐标系到末端的坐标变换矩阵：但是由于矩阵规模较大，不便用矩阵形式写出，所以把malab计算得出的矩阵用分项的形式写出：sin(1)*sin(6) + cos(6)*(cos(5)*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) + sin(5)*(cos(1)*cos(2)*sin(3) + cos(1)
4、*cos(3)*sin(2)cos(6)*sin(1) - sin(6)*(cos(5)*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) + sin(5)*(cos(1)*cos(2)*sin(3) + cos(1)*cos(3)*sin(2)sin(5)*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) - cos(5)*(cos(1)*cos(2)*sin(3) + cos(1)*cos(3)*sin(2),h5*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) - h
5、6*(cos(5)*(cos(1)*cos(2)*sin(3) + cos(1)*cos(3)*sin(2) - sin(5)*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) - h7*(cos(5)*(cos(1)*cos(2)*sin(3) + cos(1)*cos(3)*sin(2) - sin(5)*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) + h2*cos(1) - (h4 + h5)*(cos(1)*cos(2)*sin(3) + cos(1)*cos(3)*sin(2) + h7*(sin(
6、1)*sin(6) + cos(6)*(cos(5)*(cos(1)*sin(2)*sin(3) - cos(1)*cos(2)*cos(3) + sin(5)*(cos(1)*cos(2)*sin(3) + cos(1)*cos(3)*sin(2) - h3*cos(1)*sin(2)cos(6)*(cos(5)*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) + sin(5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2) - cos(1)*sin(6),- cos(1)*cos(6) - sin(6)*(
7、cos(5)*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) + sin(5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2)sin(5)*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) - cos(5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2),sin(5)*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) - cos(5)*(cos(2)*sin(1)*sin(3) + cos(
8、3)*sin(1)*sin(2), h5*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) - h7*(cos(1)*sin(6) - cos(6)*(cos(5)*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) + sin(5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2) - h6*(cos(5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2) - sin(5)*(sin(1)*sin(2)*sin(3) - cos(2)*
9、cos(3)*sin(1) - h7*(cos(5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2) - sin(5)*(sin(1)*sin(2)*sin(3) - cos(2)*cos(3)*sin(1) + h2*sin(1) - (h4 + h5)*(cos(2)*sin(1)*sin(3) + cos(3)*sin(1)*sin(2) - h3*sin(1)*sin(2)-cos(6)*(cos(5)*(cos(2)*sin(3) + cos(3)*sin(2) + sin(5)*(cos(2)*cos(3) - sin(2)*sin(3),
10、sin(6)*(cos(5)*(cos(2)*sin(3) + cos(3)*sin(2) + sin(5)*(cos(2)*cos(3) - sin(2)*sin(3),cos(5)*(cos(2)*cos(3) - sin(2)*sin(3) - sin(5)*(cos(2)*sin(3) + cos(3)*sin(2),h1 + h2 + (h4 + h5)*(cos(2)*cos(3) - sin(2)*sin(3) - h5*(cos(2)*sin(3) + cos(3)*sin(2) + h3*cos(2) + h6*(cos(5)*(cos(2)*cos(3) - sin(2)*
11、sin(3) - sin(5)*(cos(2)*sin(3) + cos(3)*sin(2) + h7*(cos(5)*(cos(2)*cos(3) - sin(2)*sin(3) - sin(5)*(cos(2)*sin(3) + cos(3)*sin(2) - h7*cos(6)*(cos(5)*(cos(2)*sin(3) + cos(3)*sin(2) + sin(5)*(cos(2)*cos(3) - sin(2)*sin(3) 用各关节转角的初值来检查变换矩阵内的正确性：设：代入各关节变换矩阵可以求出各矩阵初值，由于中没有关节变量，所以保持不变记为矩阵g图 5将图5中的初值矩阵连乘
12、可以得到基坐标系到末端坐标系的变换矩阵：计算结果见图6：图 6观察图6中的矩阵，检查初始状态末端坐标系在基坐标系的位姿，很容易看出：在中的坐标为，并且与反向，与反向，与同向，这与D-H建模图（图4）中所得到的结果相同，可以确定计算过程是正确的。2.2、机器人逆运动学运用所学的Jacobian法进行机器人逆运动学的求解。2.2.1求各关节到末端的坐标变换矩阵首先已经列出的，可以依次求出使用matlab计算过程中：把依次记作a，b，c，d，e，f把记作（sita i）计算结果如下:由于矩阵元素表达式较复杂，不便列出整个矩阵，所以逐列写出：第一列：第二列：第三列：第四列：由于比的矩阵元素表达式更为复
13、杂，所以逐项列出：-cos(6)*(cos(5)*(cos(2)*cos(3) - sin(2)*sin(3) - sin(5)*(cos(2)*sin(3) + cos(3)*sin(2),sin(6)*(cos(5)*(cos(2)*cos(3) - sin(2)*sin(3) - sin(5)*(cos(2)*sin(3) + cos(3)*sin(2),- cos(5)*(cos(2)*sin(3) + cos(3)*sin(2) - sin(5)*(cos(2)*cos(3) - sin(2)*sin(3),h2 - (h4 + h5)*(cos(2)*sin(3) + cos(3)*sin(2) - h5*(cos(2)*cos(3) - sin(2)*sin(3) - h3*sin(2) - h6*(cos(5)*(cos(2)*sin(3) + cos(3)*sin(2) + sin(5)*(cos(2)*cos(3) - sin(2)*sin(3)-sin(6),-cos(6),-cos(6)*(cos(5)*(cos(2)*sin(3) + cos(3)*sin(2) + sin(5)*(cos(2)*cos(3) - sin(2)*sin(3)
