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1、The movement simulation and trajectory planning of PUMA560 robot Shibo zhao Abstract: In this essay, we adopt modeling method to study PUMA560 robot in the use of Robotics Toolbox based on MATLAB. We mainly focus on three problems include: the forward kinematics, inverse kinematics and trajectory pl
2、anning. At the same time, we simulate each problem above, observe the movement of each joint and explain the reason for the selection of some parameters. Finally, we verify the feasibility of the modeling method.Key words: PUMA560 robot; kinematics; Robotics Toolbox; The simulation; I.IntroductionAs
3、 automation becomes more prevalent in peoples life, robot begins more further to change peoples world. Therefore, we are obliged to study the mechanism of robot. How to move, how to determine the position of target and the robot itself, and how to determine the angles of each point needed to obtain
4、the position. In order to study robot more validly, we adopt robot simulation and object-oriented method to simulate the robot kinematic characteristics. We help researchers understand the configuration and limit of the robots working space and reveal the mechanism of reasonable movement and control
5、 algorithm. We can let the user to see the effect of the design, and timely find out the shortcomings and the insufficiency, which help us avoid the accident and unnecessary losses on operating entity. This paper establishes a model for Robot PUMA560 by using Robotics Toolbox, and study the forward
6、kinematics and inverse kinematics of the robot and trajectory planning problem. II.The introduction of the parameters for the PUMA560 robotPUMA560 robot is produced by Unimation Company and is defined as 6 degrees of freedom robot. It consists 6 degrees of freedom rotary joints (The structure diagra
7、m is shown in figure 1). Referring to the human body structure, the first joint（J1） called waist joints. The second joint（J2）called shoulder joint. The third joint （J3）called elbow joints. The joints J4 J5, J6, are called wrist joints. Where, the first three joints determine the position of wrist re
8、ference point. The latter three joints determine the orientation of the wrist. The axis of the joint J1 located vertical direction. The axis direction of joint J2, J3 is horizontal and parallel, a3 meters apart. Joint J1, J2 axis are vertical intersection and joint J3, J4 axis are vertical crisscros
9、s, distance of a4. The latter three joints axes have an intersection point which is also origin point for 4, 5, 6 coordinate. (Each link coordinate system is shown in figure 2)Fig1 the structure of puma560 Fig2 the links coordinate of puma 560 When PUMA560 Robot is in the initial state, the correspo
10、nding link parameters are showed in table 1. The expression of parameters:Let length of the bar represent the distance between and along.Torsion angle denote the angle revolving from to.The measuring distance between and along is. Joint angle is the angle revolving from to along.Table 1 the paramete
11、rs of puma560 link Range100900-1601602-90000.1491-22545300.4318-900-452254-90-0.021300.4331-110170590000-1001006-90000-266266III.The movement analysis of Puma560 robot3.1 Forward kinematicDefinition: Forward kinematics problem is to solve the pose of end-effecter coordinate relative to the base coor
12、dinate when given the geometric parameters of link and the translation of joint. Let make things clearly: What you are given: the length of each link and the angle of each jointWhat you can find: the position of any point (i.e. its coordinate)3.2 The solution of forward kinematicsMethod: Algebraic s
13、olution Principal: The kinematic model of a robot can be written like this, where denotes the vector of joint variable, denotes the vector of task variable,is the direct kinematic function that can be derived for any robot structure . The origin of Each joint is assigned a coordinate frame. Using th
14、e Denavit-Hartenberg notation, you need 4 parameters () to describe how a frame () relates to a previous frame (). For two frames positioned in space, the first can be moved into coincidence with the second by a sequence of 4 operations:1. Rotate around the axis by an angle.2. Translate along the ax
15、is by a distance.3. Rotate around the new z axis by an angle.4. Translate along the new z axis by a distance. (1.1) (1.2) Therefore, according to the theory above the final homogeneous transform corresponding to the last link of the manipulator: (1.3) 3.3Inverse kinematicDefinition: Robot inverse ki
16、nematics problem is that resolve each joint variables of the robot based on given the position and direction of the end-effecter or of the link (It can show as position matrix T). As for PUMA560 Robot, variable need to be resolved.Let make things clearly: What you are given: The length of each link
17、and the position of some point on the robot.What you can find: The angles of each joint needed to obtain that position. 3.4 The solution of inverse kinematics Method: Algebraic solution Principal: Where is the robot Jacobian. Jacobian can be seen as a mapping from Joint velocity space to Operational
18、 velocity space. 3.5 The trajectory planning of robot kinematicsThe trajectory planning of robot kinematics mainly studies the movement of robot. Our goal is to let robot moves along given path. We can divide the trajectory of robots into two kinds. One is point to point while the other is trajector
19、y tracking. The former is only focus on specific location point. The latter cares the whole path.Trajectory tracking is based on point to point, but the route is not determined. So, trajectory tracking only can ensure the robots arrives the desired pose in the end position, but can not ensure in the
20、 whole trajectory. In order to let the end-effecter arriving desired path, we try to let the distance between two paths as small as possible when we plan Cartesian space path. In addition, in order to eliminate pose and positions uncertainty between two path points, we usually do motivation plan amo
21、ng every joints under gang control. In a word, let each joint has same run duration when we do trajectory planning in joint space.At same time, in order to make the trajectory planning more smoothly, we need to apply the interpolating method. Method: polynomial interpolating 1 Given: boundary condit
22、ion (1.3) (1.4) Output: joint space trajectory between two points = （1.5）Polynomial coefficient can be computed as follows: (1.6)IV. Kinematic simulation based on MATLABHow to use linkIn Robotics Toolbox, function link is used to create a bar. There are two methods. One is to adopt standard D-H para
23、meters and the other is to adopt modified D-H parameters, which correspond to two coordinate systems. We adopt modified D-H parameters in our paper. The first 4 elements in Function link are , a, , d. The last element is 0 (represent Rotational joint) or 1 (represent translation joint). The final pa
24、rameter of link is mod, which means standard or modified. The default is standard.Therefore, if you want to build your own robot, you may use function link. You can call it like this: L1=link(0 0 pi 0 0,modified); The step of simulation is:Step1: First of all, according to the data from Table 1, we
25、build simulation program of the robot (shown in Appendix rob1.m).Step2: Present 3D figure of the robot (shown in Fig4). This is a three-dimensional figure when the robot located the initial position (). We can adjust the position of the slider in control panel to make the joint rotation (in Fig 5),
26、just like controlling real robot. Step3:Point A located at initial position. It can de described as . The target point is Point B. The joint rotation angle can be expressed as. We can achieve the solution of forward kinematics and obtain the end-effecter pose relative to the base coordinate system i
27、s (0.737, 0.149, 0.326) , relative to the three axes of rotation angle is the (0, 0, -1). The robots three-dimensional pose inis shown in Fig 6.Step4: According to the homogeneous transformation matrix, we can obtain each joint variable from the initial position to the specified location Step5:Simul
28、ate trajectory from point A to point B. The simulation time is 10s. Time interval is 0.1s. Then, we can picture location image, the angular velocity and angular acceleration image (shown as Fig 8) which describe each joint transforms over time from Point A to Point B. In this paper, we only present
29、the picture of joint 3. By using the function T=fkine(r,q), we obtain T a three-dimensional matrix. The first two dimensional matrix represent the coordinate change while the last dimension is time t. Fig 4 Fig 5Fig 6 Fig 7 Fig8V The problem during the simulationThe reason for selection of some para
30、meter The parameter of link: From kinematic simulation and program, you can see that I set certain value not arbitrary when I call link. That is because I want the simulation can be more close to the real situation .So; I adopt the parameter of puma560 (you can see it from the program) and there is
31、no difference between my robot and puma560 radically.The parameter of: When I choose the parameter of, I just want to test something.For example, when you denote the parameter of like this , you want to use the function fkine(p560, ) to obtain the homogenous function T, then, you want to use ikine(p
32、560,T) to test whether the is what you have settled before.The result is as follows:=0 -pi/4 -pi/4 0 pi/8 0;T=fkine(p560, );=ikine(p560,T)=0 -pi/4 -pi/4 0 pi/8 0 Actually, not all of the parameter can do like this. For example, when you try =pi/2,pi/2,pi/2,pi/2,pi/2,pi/2 , the answer is not itself.
33、VI. References1 http:/en.wikipedia.org/wiki/Robot_ online, 7-ferbury-20152 http:/ online, 7-ferbury-2015 3 http:/en.wikipedia.org/wiki/Jacobian_algorithm accessed 8-February-20154 Youlun Xiong, Han Ding, Encang Liu, RobotM, Tinghua university press ,1993VIIAppendixclc; clear;%modified 改进的D-H法L1=link
34、(0 0 pi 0 0,modified);L2=link(-pi/2 0 0 0.1491 0,modified);L3=link(0 0.4318 -pi/2 0 0,modified);L4=link(-pi/2 0.0203 0 0.4318 0,modified);L5=link(pi/2 0 0 0 0,modified);L6=link(-pi/2 0 0 0 0,modified);r=robot(L1 L2 L3 L4 L5 L6);r.name=zhaoshibo;%模型的名称drivebot(r)track.m%前 3 个关节对机械手位置的影响qA=0,0,0,0,0,0
35、; %起始点关节空间矢量qB= 0 -pi/4 -pi/4 0 pi/8 0; %终止点关节空间矢量t=0:0.1:10; %仿真时间q,qd,qdd=jtraj(qA,qB,t); %关节空间规划plot(r,q)%关节 3 的角速度、角速度和角加速度曲线figuresubplot(1,3,1)plot(t,q(:,3)%关节 3 的位移曲线xlabel(时间 t/s);ylabel(关节的角位移/rad);grid onsubplot(1,3,2)plot(t,qd(:,3)%关节 3 的位移曲线xlabel(时间 t/s);ylabel(关节的角速度/（rad/s);grid onsubplot(1,3,3)plot(t,qdd(:,3)%关节 3 的位移曲线xlabel(时间 t/s);ylabel(关节的角加速度/（rad/s2);grid on11软硬件