puma560的运动学及matlab实现（正解+逆解）

正解源码

DEG = pi/180;

cta1=-70.4385

cta2=182.6918

cta3=-90.0000

cta4=-82.4708

cta5=-19.7387

cta6=-97.9933

T01=[cosd(cta1),-sind(cta1),0,0;

sind(cta1), cosd(cta1),0,0;
0,0,1,0;
0,0,0,1];
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T02=T01*[cosd(cta2),-sind(cta2),0,0;

0,0,1, 149.09;
-sind(cta2),-cosd(cta2),0,0;
0,0,0,1] ;
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T03=T02*[cosd(cta3),-sind(cta3),0,431.8;

sind(cta3), cosd(cta3),0,0;
0,0,1,0;
0,0,0,1];
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T04=T03*[cosd(cta4),-sind(cta4),0,20.32;

0,0,1,433.07;
-sind(cta4),-cosd(cta4),0,0;
0,0,0,1];
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T05=T04*[cosd(cta5),-sind(cta5),0,0;

0,0,-1,0;
sind(cta5), cosd(cta5), 0,0;
0,0,0,1];
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T06=T05*[cosd(cta6),-sind(cta6),0,0;

0,0,1,0;
-sind(cta6),-cosd(cta6),0,0;
0,0,0,1];
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O=T06*[0;0;0;1];

=====================================================

逆解源码

fid = fopen('inverseout.txt','w');%逆解的保存文件 %赋初值 T06 =[0.0000    1.0000    0.0000 -149.0900; 0.0000   -0.0000    1.0000  864.8700; 1.0000         0   -0.0000   20.3200; 0         0         0    1.0000] ; a0=0;  a1=0;  a2=431.8;  a3=20.32;  a4=0;  a5=0; d1=0;  d2=149.09;  d3=0;  d4=433.07;  d5=0; d6=0; n_x=T06(1);   n_y=T06(2);  n_z=T06(3); o_x=T06(5);   o_y=T06(6);  o_z=T06(7); a_x=T06(9);   a_y=T06(10);  a_z=T06(11); p_x=T06(13);  p_y=T06(14);   p_z=T06(15); disp(['八组解分别是：']); for i=1:2 for j=1:2 for k=1:2 %求解theta1(为弧度) sqr1=[sqrt(p_x^2+p_y^2-d2^2),-sqrt(p_x^2+p_y^2-d2^2)]; ta1=atan2(p_y,p_x)-atan2(d2,sqr1(i)); %求解theta3（弧度表示） k1=(p_x^2+p_y^2+p_z^2-a2^2-a3^2-d2^2-d4^2)/(2a2); sqr3=[sqrt(a3^2+d4^2-k1^2),-sqrt(a3^2+d4^2-k1^2) ]; ta3=atan2(a3,d4)-atan2(k1,sqr3(j)); fs23=-((a3+a2cos(ta3))p_z)+(cos(ta1)p_x+sin(ta1)p_y)(a2sin(ta3)-d4); sc23=(-d4+a2sin(ta3))p_z+(cos(ta1)p_x+sin(ta1)p_y)(a2cos(ta3)+a3); ta23=atan2( fs23,sc23); %求解theta2 （弧度表示） ta2=ta23-ta3; %求解theta4 （弧度表示） fs4=[ -a_xsin(ta1)+a_ycos(ta1),a_xsin(ta1)-a_ycos(ta1)]; sc4=[ -a_xcos(ta1)cos(ta23)-a_ysin(ta1)cos(ta23)+a_zsin(ta23), a_xcos(ta1)cos(ta23)+a_ysin(ta1)cos(ta23)-a_zsin(ta23)]; fprintf(fid,'%d,',sc4(1,1)); fprintf(fid,'\t'); fprintf(fid,'%d,',sc4(2,1)); fprintf(fid,'\t'); fprintf(fid,'%d,',fs4(1,1)); fprintf(fid,'\t'); fprintf(fid,'%d,',fs4(1,2)); fprintf(fid,'\t'); fprintf(fid,'\n'); ta4=atan2(fs4(k),sc4(k)); %求解theta5 （弧度表示） fs5=-a_x(cos(ta1)*cos(ta23)*cos(ta4)+sin(ta1)sin(ta4))... -a_y(sin(ta1)*cos(ta23)cos(ta4)-cos(ta1)sin(ta4))... +a_z(sin(ta23)cos(ta4)); sc5=a_x(-cos(ta1)sin(ta23))+a_y(-sin(ta1)sin(ta23))+a_z(-cos(ta23)); ta5=atan2(fs5,sc5); %求解theta6 （弧度表示） fs6=-n_x(cos(ta1)*cos(ta23)*sin(ta4)-sin(ta1)cos(ta4))... -n_y(sin(ta1)*cos(ta23)*sin(ta4)+cos(ta1)cos(ta4))... +n_z(sin(ta23)*sin(ta4));

sc6= n_x*(cos(ta1)*cos(ta23)*cos(ta4)+sin(ta1)*sin(ta4))*cos(ta5)...
              -n_x*cos(ta1)*sin(ta23)*sin(ta5)...
              +n_y*(sin(ta1)*cos(ta23)*cos(ta4)+cos(ta1)*sin(ta4))*cos(ta5)...
              -n_y*sin(ta1)*sin(ta23)*sin(ta5)...
              -n_z*(sin(ta23)*cos(ta4)*cos(ta5)+cos(ta23)*sin(ta5));
            ta6=atan2(fs6,sc6);
            %save 
            %将其化为角度
            Theta=[ta1 ta2 ta3 ta4 ta5 ta6]./pi*180
      end
  end   
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end