机器人学基础机器人轨迹规划.pptx
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1、Ch.7 Trajectory Planning of Robots 17.1General Considerations in Robot Trajectory Planning7.2Interpolated Calculation of Joint Trajectories 7.3Planning of Cartesian Path Trajectories7.4Real Time Generation of Planning Trajectories7.5SummaryCh.7 Trajectory Planning of Robots第1页/共52页7.1 General Consid
2、erations in Trajectory Planning 轨迹规划应考虑的问题Basic Problem:Move the manipulator arm from some initial position to some desired final position(May be going through some via points).27.1 General considerations第2页/共52页7.1 General Considerations in Trajectory PlanningTrajectory:Time history of position,vel
3、ocity and acceleration for each DOFPath points:Initial,final and via pointsConstraints:Spatial,time,smoothness37.1 General considerations第3页/共52页Joint spaceEasy to go through via points(Solve inverse kinematics at all path points)No problems with singularitiesLess calculationsCan not follow straight
4、 lineCartesian spaceWe can track a shape(for orientation:equivalent axes,Euler angles,)More expensive at run time(after the path is calculated need joint angles in a lot of points)Discontinuity problems4General Considerations-Solution Space7.1 General considerations第4页/共52页Cartesian planning difficu
5、lties:5General Considerations-Solution Space7.1 General considerationsInitial(A)and Goal(B)Points are reachable,but intermediate points(C)unreachable.第5页/共52页Ch.7 Trajectory Planning of Robots 67.1General Considerations in Robot Trajectory Planning7.2Interpolated Calculation of Joint Trajectories 7.
6、3Planning of Cartesian Path Trajectories7.4Real Time Generation of Planning Trajectories7.5SummaryCh.7 Trajectory Planning of Robots第6页/共52页Joint-Space SchemesEach path point is converted into a set of desired joint angles by application of the inverse kinematics.A smooth function is found for each
7、of the n joints which pass through the via points and end at the goal point.Time required for each segment is the same for each joint.The determination of the desired joint angle function for a particular joint is independent with other joints.77.2 Interpolated Calculation of Joint Trajectories 关节轨迹
8、的插值计算7.2 JointSpace Schemes第7页/共52页Choice of interpolation function is not unique!8Joint-Space Schemes 7.2 JointSpace SchemesSeveral possible path shapes for a single joint.第8页/共52页Some possible interpolation functions:Cubic polynomials Cubic polynomials for a path with via pointsHigher-order polyno
9、mials Linear function with parabolic blendsLinear function with parabolic blends for a path with via points9Joint-Space Schemes 7.2 JointSpace Schemes第9页/共52页In making a single smooth motion,at least four constraints on are evident:107.2.1 Cubic Polynomials 三次多项式插值三次多项式插值7.2 JointSpace Schemes第10页/共
10、52页Combining the four constraints yields four equations with four unknowns:117.2.1 Cubic Polynomials7.2 JointSpace Schemes第11页/共52页These four constraints uniquely specify a particular cubic:127.2.1 Cubic PolynomialsThe joint velocity and acceleration along this path are:7.2 JointSpace Schemes第12页/共5
11、2页Eg.7.1 A single-link robot with a rotary joint is motionless at =15 degrees.It is desired to move the joint in a smooth manner to=75 degrees in 3 seconds.Find the coefficients of a cubic which accomplishes this motion and brings the manipulator to rest at the goal.Plot the position,velocity,and ac
12、celeration of the joint as a function of time.137.2.1 Cubic Polynomials7.2 JointSpace Schemes第13页/共52页Solution:Plugging 0=15,f=75,tf=3 into(7.6),we find147.2.1 Cubic Polynomials7.2 JointSpace Schemes第14页/共52页Solution:157.2.1 Cubic Polynomials7.2 JointSpace SchemesStarts at 15 degrees and ends at 75
13、degrees!第15页/共52页Solution:167.2.1 Cubic Polynomials7.2 JointSpace SchemesStarts and ends at rest!第16页/共52页Solution:177.2.1 Cubic Polynomials7.2 JointSpace SchemesAcceleration profile is linear!第17页/共52页If we come to rest at each pointuse formula from previous slideor continuous motion(no stops)need
14、velocities at intermediate points:Initial Conditions:187.2.2 Cubic polynomials with via points 过路径点的三次多项式插值过路径点的三次多项式插值7.2 JointSpace SchemesSolutions:第18页/共52页How to specify velocity at the via points:The user specifies the desired velocity at each via point in terms of a Cartesian linear and angul
15、ar velocity of the tool frame at that instant.The system automatically chooses the velocities at the via points by applying a suitable heuristic in either Cartesian space or joint space(average of 2 sides etc.).The system automatically chooses the velocities at the via points in such a way as to cau
16、se the acceleration at the via points to be continuous.197.2 JointSpace Schemes7.2.2 Cubic polynomials with via points第19页/共52页Higher order polynomials are sometimes used for path segments.For example,if we wish to be able to specify the position,velocity,and acceleration at the beginning and end of
17、 a path segment,a quintic polynomial is required:207.2.3 Higher-order polynomials高阶多项式插值高阶多项式插值7.2 JointSpace Schemes第20页/共52页Where the constraints are given as:217.2.3 Higher-order polynomials7.2 JointSpace Schemes第21页/共52页Solution to these equations:227.2.3 Higher-order polynomials7.2 JointSpace S
18、chemes第22页/共52页Linear interpolation(Straight line):Note:Although the motion of each joint in this scheme is linear,the end-effector in general does not move in a straight line in space.237.2.4 Linear function with parabolic blends 用抛物线过渡的线性插值用抛物线过渡的线性插值7.2 JointSpace SchemesDiscontinuous velocity-ca
19、n not be controlled!第23页/共52页To create a smooth path with continous position and velocity,we start with the linear function but add a parabolic blend region at each path point.Constant acceleration is used during the blend portion to change velocity smoothly.247.2.4 Linear function with parabolic bl
20、ends7.2 JointSpace Schemes第24页/共52页Assume that the parabolic blends both have the same duration,and therefore the same constant acceleration(modulo a sign).There are many solutions to the problem-but the answer is always symmetric about the halfway point.257.2.4 Linear function with parabolic blends
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