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基于蒙特卡洛法的機器人工作空間仿真

matlab代碼

clear,clc,close all format compact alpha=[0,90,0,0,90,-90]; a=[0,0,-0.3,-0.276,0,0]; d=[0.1215,0.1225,-0.102,0.09,0.09,0.082]; offset=[0,-90,0,-90,0,0]; DH=[alpha',a',d',offset']; qlim=[-170,170;-84,130;-188,50;-170,170;-117,117;-360,360]; for i=1:length(alpha)L(i)=RevoluteMDH('d',d(i),'a',a(i),'alpha',deg2rad(alpha(i)),'offset', ...deg2rad(offset(i)),'qlim',deg2rad([qlim(i,1),qlim(i,2)])); end robot=SerialLink(L,'name','Arm6');base=transl(0,0,0)*rpy2tr(0,0,0,'deg'); robot.base=base; tool=transl(0,0,0)*rpy2tr(0,0,0,'deg','xyz'); %末端工具 robot.tool=tool; robot.display()xita = [0 0 0 0 0 0];for i = 1:10000for m = 1:6xita(m) = qlim(m,1) + (qlim(m,2)-qlim(m,1))*rand;endpx = d(6)*cos(xita(5))*sin(xita(1))...- d(6)*cos(xita(2)+xita(3)+xita(4))*cos(xita(1))*sin(xita(5))...+ cos(xita(1))*a(3)*cos(xita(2)+xita(3))...+ cos(xita(1))*a(2)*cos(xita(2))...+ d(4)*sin(xita(1))...+ d(5)*sin(xita(2)+xita(3)+xita(4))*cos(xita(1));py = sin(xita(1)) * (a(3) * cos(xita(2) + xita(3)) + a(2)*cos(xita(2))) ...-d(4)*cos(xita(1))...-d(6)*(cos(xita(1)*cos(xita(5)) + cos(xita(2)+xita(3)+xita(4))*sin(xita(1))*sin(xita(5)))) ...+d(5)*sin(xita(2)+xita(3)+xita(4))*sin(xita(1));pz = d(1)+a(3)+ sin(xita(2)+xita(3)) + a(2)*sin(xita(2))...-d(5)*cos(xita(2)+xita(3)+xita(4))...-d(6)*sin(xita(2)+xita(3)+xita(4))*sin(xita(5));%旋轉矩陣中移動部分向量p = [px py pz]';plot3(p(1),p(2),p(3),'r*')hold on end

取100000個采樣得到的三維圖如下所示：

總結

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