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1、大作业：PUMA机器人1 .坐标系建立利用D-H参数法建立坐标系:2 . D-H参数表关节i()i()lidi运动范围190-9000-160 160200a2d2-225 4539090-a30-45 22540-900d4-110 170509000-100 1006000de-2660 2663 .正运动学推导变换矩阵求取i1AiRot(zi-1, i ) Trans(0,0,di ) Trans(li ,0,0) Rot(xi, i) =cic00s ic ic ic isi10s is ic isci10lic lis di1列各 D-H 变换矩阵如下：0A1c1s1000010s1
2、c10000011A2c2s200s2c2000010a2c2a2s2d212A3c3s3000010s3c300a3c3 a3s3 0 1c4s4000010s4c40000d414A5c5s5000010s5c50000015A6c6s600-s6c600001000d61注：为书写方面，本文中本文中cicosi,sisin又由 0T60A1 1A2 2A33A44A55A6 ，利用Matlab进行符号运算，运行程序运行程序PUMAzhengyundongxue （程序详见附录）得:nxsxaxpxnysyaypynzszazpz00010T6其中nx - c6*(s5*(c1*c2*s3
3、 + c1*c3*s2) + c5*(s1*s4 - c4*(c1*c2*c3 - c1*s2*s3) - s6*(c4*s1 + s4*(c1*c2*c3 - c1*s2*s3)ny s6*(c1*c4 - s4*(c2*c3*s1 - s1*s2*s3) - c6*(s5*(c2*s1*s3 + c3*s1*s2)- c5*(c1*s4 + c4*(c2*c3*s1 - s1*s2*s3)nz s4*s6*(c2*s3 + c3*s2) - c6*(s5*(c2*c3 - s2*s3) + c4*c5*(c2*s3 + c3*s2)sx s6*(s5*(c1*c2*s3 + c1*c3*s
4、2) + c5*(s1*s4 - c4*(c1*c2*c3 - c1*s2*s3)- c6*(c4*s1 + s4*(c1*c2*c3 - c1*s2*s3)sy s6*(s5*(c2*s1*s3 + c3*s1*s2) - c5*(c1*s4 + c4*(c2*c3*s1 - s1*s2*s3)+ c6*(c1*c4 - s4*(c2*c3*s1 - s1*s2*s3)sz s6*(s5*(c2*c3 - s2*s3) + c4*c5*(c2*s3 + c3*s2) + c6*s4*(c2*s3 + c3*s2)ax c5*(c1*c2*s3 + c1*c3*s2) - s5*(s1*s4
5、- c4*(c1*c2*c3 - c1*s2*s3)ay c5*(c2*s1*s3 + c3*s1*s2) + s5*(c1*s4 + c4*(c2*c3*s1 - s1*s2*s3)az c5*(c2*c3 - s2*s3) - c4*s5*(c2*s3 + c3*s2)px - d2*s1+d*(c1*c2*s3 + c1*c3*s2) 6*d5*(s1*s4 - c4*(c1*c2*c3 - c1*s2*s3)+ 2*c1*c2 + c5*d6*(c1*c2*s3 + c1*c3*sZ*c+甘s2*s3 -0rc1*c2*c3py c1*d2 + d*(c2*s1*s3 + c3*s1*
6、s2) 6*s5*(c1*s4 + c4*(c2*c3*s1 - s1*s2*s3) + 2*c2*s1 + c5*d*(c2*s1*s3 + c3*s1*s2)3*s1*s2*s3 - 3*c2*c3*s1pz d4*(c2*c3 - s2*s3) -, *s2 + d*(c5*(c2*c3 - s2*s3) - c4*s5*(c2*s3 + c3*s2) + 3*c2*s3 +a*c3*s2变换矩阵验证初步验证根据DH参数表中8 i的初始值，将其带入0T6后得到末端执行器的坐标为px= -C2py=a2+d4+d6pz=as与初始设定一致，因此初步判断变换矩阵正确。使用matlab机器人工
7、具箱详细验证关节i()i()li (m)di (m)运动范围190-9000-160。160。200a2=d2=-225 4539090-a3=0-450 22540-900d4=-110 170509000-100 1006000d6=-2660 266利用matlab机器人工具箱按照以上 DH参数，编写并运行程序 PUMAyanzheng （程序详 见附录）得到如下图，与 PUMA560机器人一致，表明 DH参数法建模正确。口丁4.逆运动学推导反变换法求逆0T6 0A 1A22A 3A4 4A5 5A6 ,且有nx sx ax px o ny sy ay pyI 6 nz sz az pz
8、 0001(D求i由 Bi A 1 0T61A2 2A3 3A 4A 5A6 T61 得nx* c1 ny* s1sx* c1 sy* s1 ax* c1 ay* s1px* c1 py* s1-nzny* c1 - nx* s10-szsy* c1 - sx* s10-azay* c1 - ax* s10-pzpy* c1 - px* s1nx sx ny sy nz sz00ax px ay py az pz 01nx s4* s6*( c2* c3 s2* s3) c6*( s5*( c2* s3 c3* s2) c4* c5*( c2* c3 s2* s3) ny c6*( s5*(
9、c2* c3 s2* s3) c4* c5*( c2* s3 c3* s2) s4* s6*( c2* s3 c3* s2) nz c4*s6 + c5*c6*s4sxs6*(s5*(c2*s3 + c3*s2) - c4*c5*(c2*c3 - s2*s3) - c6*s4*(c2*c3 - s2*s3)sy s6*(s5*(c2*c3 - s2*s3) + c4*c5*(c2*s3 + c3*s2) - c6*s4*(c2*s3 + c3*s2)sz c4*c6 + c5*s4*s6ax c5*(c2*s3 + c3*s2) + c4*s5*(c2*c3 - s2*s3)ay c4*s5*
10、(c2*s3 + c3*s2) - c5*(c2*c3 - s2*s3)az s4*s5px=a2*c2 + d4*(c2*s3 + c3*s2) - d 6*(c5*(c2*s3 + c3*s2) + c4*s5*(c2*c3 - s2*s3)-a3*c2*c3 + a 3*s2*s3py a2*s2 - d4*(c2*c3 - s2*s3) + d 6 *(c5*(c2*c3 - s2*s3) - c4*s5*(c2*s3 + c3*s2)-a3*c2*s3 - a 3*c3*s2pz d2 - d6*s4*s5由矩阵对应元素均相等，B1(3,3) T61(3,3) , B1(3,4) T
11、61(3,4)得ay* cl - ax* si s4*s5 py* ci - px* si d2 - d6*s4*s5得到：py* ci px* si d2 d6* ay* ci - ax*s1整理有ci* (py d6* ay) (px d6* ax)* si d2三角变换px d6* ax r sinpy d6* ay r cos其中r . (d6*ax px)2 (py d6* ay)2px d6 * axarctan()py d6 * ax得到qarccos2)r(2)求 32 i i i 0 i 0.345由 B3= A3 g A2 g Ai g T6A4g A6gA6 T 63 付
12、b11b12b13b14c4*c5*c6 + s4*s6- c6*s4 + c4*c5*s6c4*s5-d 6*c4*s5b21b22b23b24-c4*s6 + c5*c6*s4c4*c6 + c5*s4*s6s4*s5-d 6*s4*s5b31b32b33b34-c6*s5-s5*s6c5d4 d6 *c5b41b42b43b440001其中 B3 个元素如下：b11 ny* s1* c23 - nz* s23 + c1* nx* c23b12 sx* c1* c23 - sz* s23+ sy* s1* c23b13 ax* c1* c23 - az* s23 + ay* s1* c23
13、b14 py* s1* c23 - pz* s23 + c1* px* c23 a3 a2* c3b21= ny*c1 - nx*s1b22= sy*c1 -sx *s1b23= ay*c1 - ax*s1b24= py*c1 - px*s1-d 2b31 nz* c23 + ny* s1* s23 + nx* c1* s23b32 sz* c23 + sx* c1* s23+ sy* s1* s23b33 az* c23 + ax* c1* s23+ ay* s1* s23b34 pz* c23 + py* s1* s23 + px* c1* s23 a2 * s3b41 0b42 0b43 0b44 1注：为书写方便，本文中c23 cos( 23), s23 sin( 23) 。由矩阵元素对应相等， B3(1,4) d6* B3(1,3) ， B3(3,4) d4 d6* B3(3,3) 得a3 (pz + az*d 6)*s23 + d6*(ax*c1 + ay*s1) + px*c1 + py*s1 *c23 = a 2*c3-d4 + d6*(ax*c1 + ay*s1) + px*c1 + py*s1 *s23+(pz + az*d 6)*c23 = a 2 *s3两式平方求和，整
