Path planning for autonomous empty container handler based on A* algorithm and spiral curve optimization strategy
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摘要: 集装箱堆场作业环境复杂,实现无人堆高机精准停位是亟须解决的关键问题。对此,设计了基于改进A*算法与多项式螺旋曲线结合的混合策略。首先,基于堆高机作业工况及抓取作业流程,设计多目标点改进A*全局规划算法,通过单次规划依次通过多个给定的目标点;其次,根据非完整约束条件,在靠近最终目标点处设计基于多项式螺旋线的局部规划。选取多种典型作业场景,对提出的算法进行仿真实验及分析。研究结果表明,该策略能使堆高机在全局路径中依次通过多个目标点,并经局部规划调整后满足作业目标点的约束,符合作业要求。Abstract: The operational environment of container yards is complex, and achieving precise parking for autonomous empty container handler is a critical problem that need to be solved. To address this, a hybrid strategy combining an improved A* algorithm with polynomial spiral curves was designed. Considering the working conditions and grasping workflow of the empty container handler, a multi-target improved A* global planning algorithm was developed, enabling the sequential traversal of multiple given target points in a single planning process. Furthermore, based on non-holonomic constraints, a local planning strategy utilizing polynomial spiral curves was designed near the final target point. Simulations and analyses were conducted on selected typical operational scenarios to validate the proposed algorithm. The results indicate that the strategy enables the stacker crane to sequentially pass through multiple target points along the global path. After local planning adjustments, it can meet the constraints of the operational target points, thereby fulfilling operational requirements.
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表 1 空箱堆高机技术参数
Table 1. Technical parameters of ECH
单位:m 车身长度 轮轴距 最小转弯半径 吊具尺寸 6.9 4.5 6 6.11 ~ 12.17 表 2 不同场景下局部规划算法的参数结果
Table 2. Solution results of local planning parameters in different scenarios
场景 起点曲率/m-1 目标点坐标 实到点坐标 起始航向角/rad 目标航向角/rad b c sfinal 1 0.1 (10, 1) (10.01, 0.99) 0 0 0.695 −0.402 10.067 2 0.1 (10, 1) (10.02, 1.00) 0 π/2 −0.550 0.383 12.784 3 0.1 (10, 1) (10.00, 1.02) π/2 π/2 −0.380 0.384 16.694 4 0.05 (10, 10) (10.02, 9.99) 0 0 0.180 −0.213 15.969 5 0.05 (10, 10) (10.00, 10.02) 0 π/2 0.094 0.133 16.080 6 0.05 (10, 10) (10.03, 9.99) π/2 π/2 −0.250 0.214 16.507 7 0 (10, 10) (10.04, 10.00) 0 −π/2 0.159 −0.319 26.124 -
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