机器人学基础机器人运动学 蔡自兴PPT学习教案

1、会计学1机器人学基础机器人学基础 机器人运动学机器人运动学 蔡自兴蔡自兴 从几何学几何学的观点来处理手指位置手指位置P与关节变量关节变量L1, L2, 和 的关系称为运动学运动学(Kinematics)。122 3.0 Introduction to Robot Kinematics第1页/共68页joint rotations and orientations to achieve this?3 3.0 Introduction to Robot Kinematics第2页/共68页xry 1211212coscos()xLL11212sinsin()yLLThe general vecto

2、r form( )rf4 3.0 Introduction to Robot Kinematics第3页/共68页2221122sinarctan( ) arctan()cosLyxLL2222121 2()arccos2xyLLLL式中同样，如果用向量表示上述关系式，其一般可表示为1( )fr5 3.0 Introduction to Robot Kinematics第4页/共68页1机器人到达给定的手爪位置 P有两个姿态满足要求，即图中的 也是其解。此时 和 变成为另外的值，即逆逆运动学的解不是惟一的运动学的解不是惟一的。2 将运动学公式 两边微分即可得到机器人手爪的速度和关节速度的关系，

3、再进一步进行微分将得到加速度之间的关系，处理这些关系也是机器人的运动学问题。 ( )rf6 3.0 Introduction to Robot Kinematics第5页/共68页73.1 Representation of Kinematics Equation of Robot Manipulator3.1 Representation of Kinematics Equation of Manipulator第6页/共68页83.1 Representation of Kinematics Equation of Robot Manipulator3.1 Representation o

4、f Kinematics Equation of Manipulator第7页/共68页9(3.1)6123456TA A A A A A3.1 Representation of Kinematics Equation of Manipulator3.1 Representation of Kinematics Equation of Robot Manipulator第8页/共68页10Forming a right-hand frame： n = o a or a= n o3.1n,o,ap图 矢量和3.1 Representation of Kinematics Equation of

5、 Manipulator第9页/共68页1160001xxxxyyyyzzzznoapnoapTnoap(3.2)3.1 Representation of Kinematics Equation of Manipulator第10页/共68页12n欧拉变换Euler可由连乘三个旋转矩阵来求得，即（3.3）3.2图 欧拉角的定义),(),(),(),(zRotyRotzRotEuler3.1 Representation of Kinematics Equation of Manipulator第11页/共68页133.3图 用横滚、俯仰和偏转表示机械手运动姿态3.1 Representati

6、on of Kinematics Equation of Manipulator第12页/共68页14(3.4)( , ,)( , )( , )( ,)RPYRot zRot yRot x 3.1 Representation of Kinematics Equation of Manipulator第13页/共68页1561000100010001xyzppTp某姿态变换(3.6)3.1 Representation of Kinematics Equation of Manipulator第14页/共68页163.4图 用柱面坐标和球面坐标表示位置)0 , 0 ,(),(), 0 , 0(

7、),(rTranszRotzTransrzCyl3.1 Representation of Kinematics Equation of Manipulator第15页/共68页17( , , )( , )( ,)(0,0, )SphrRot zRot yTransr (3.9)3.1 Representation of Kinematics Equation of Manipulator第16页/共68页183.1 Representation of Kinematics Equation of Manipulator第17页/共68页193.5图 转动关节连杆四参数示意图iiaa 和垂直于

8、iidi3.1 Representation of Kinematics Equation of Manipulator第18页/共68页20ii3.1 Representation of Kinematics Equation of Manipulatorai-1: Link Length - mutual perpendicular unique except for parallel axis : Link Twist - measured in the right-hand sense about1ia1i第19页/共68页213.1 Representation of Kinemat

9、ics Equation of Manipulatordi: Link Offset - variable if joint i is prismatic (平动关节) : Joint Angle - variable if joint i is revolute (转动关节)i第20页/共68页223.1 Representation of Kinematics Equation of Manipulator,iiiia drevolute jointprismatic jointiid第21页/共68页233.1 Representation of Kinematics Equation

10、of Manipulatory-vectors: complete right-hand frames第22页/共68页243.1 Representation of Kinematics Equation of Manipulator1. Normals 2. Origins3. Z-axes4. X-axes第23页/共68页253.1 Representation of Kinematics Equation of Manipulator第24页/共68页263.1 Representation of Kinematics Equation of ManipulatorExample-R

11、RR Arm第25页/共68页273.1 Representation of Kinematics Equation of ManipulatorExample-RRR Armlinkdi1231i1iaiiidiia= distance from zi to zi+1 along xi= distance from xi-1 to xi along zi= angle from zi to zi +1 about xi= angle from xi-1 to xi about zi第26页/共68页283.1 Representation of Kinematics Equation of

12、ManipulatorExample-RRR Armlinkdi1000120L10230L2031i1iaiiidiia= distance from zi to zi+1 along xi= distance from xi-1 to xi along zi= angle from zi to zi +1 about xi= angle from xi-1 to xi about zi第27页/共68页293.5图 转动关节连杆四参数示意图iidiiaDenavit-Hartenberg notation= distance from zi to zi+1 along xi= distan

13、ce from xi-1 to xi along zi= angle from zi to zi +1 about xi= angle from xi-1 to xi about zi3.1 Representation of Kinematics Equation of Manipulator第28页/共68页30(1) 绕 轴旋转 角，使 轴转到与 同一平面内。(2) 沿 轴平移一距离 ，把 移到与 同一直线上。(3) 沿 xi 轴平移一距离 ai-1 ，使连杆 坐标系的原点与 连杆 i 的坐标系原点重合。(4) 绕 xi 轴旋转 角，使 zi1 转到与 zi 同一直线上。1izi1ixi

14、x1izid1ixix1i1i3.1 Representation of Kinematics Equation of Manipulator第29页/共68页（3.13）11( ,)(0,0,)(,0,0)( ,)iiiiiARot zTransd Trans aRot x1000011111111iiiiiiiiiiiiiiiiiidcssascccscasscscAiA313.1 Representation of Kinematics Equation of Manipulator第30页/共68页321i61Ti6161AAATiii(3.15)111111111100001iiii

15、iiiiiiiiiiiiiiicsas cc csd sTs sc scd c (3.16)3.1 Representation of Kinematics Equation of Manipulator第31页/共68页33 3.2 Solving Kinematics Equation6T6T6T1111A6T61611TTA16T2第32页/共68页上列各方程的左式为和前个关节变量的函数。可用这些方程来确定各关节的位置。6261112TTAA636111213TTAAA64611121314TTAAAA6561112131415TTAAAAA6T) 1( i34 3.2 Solving

16、Kinematics Equation第33页/共68页35TEuler),(3.23)10000001000csscsssccscsssccssccssccsscccpaonpaonpaonzzzzyyyyxxxx(3.24) 3.2 Solving Kinematics Equation3.2 Solving Kinematics Equation第34页/共68页36xyzxyzxyznc c cs sns c cc sns coc c ss cos c sc cos sac sas sac (3.25)(3.26)(3.27)(3.28)(3.29)(3.30)(3.31)(3.32)

17、(3.33)1zca1xcas1zcns 3.2 Solving Kinematics Equation第35页/共68页371zca1xcas1zcnssinsin0180 3.2 Solving Kinematics Equation第36页/共68页383.8图 反正切函数atan2 3.2 Solving Kinematics Equation用双变量反正切函数用双变量反正切函数(two-argument arc tangent function) 确定角度确定角度第37页/共68页39),(),(),(1zRotyRotTzRot),(),(),(11zRotTzRotyRot(3.

18、37)(3.38) 3.2 Solving Kinematics Equation第38页/共68页40),(2atan),(2atan180),(2atanyxyxzyxxyocosncnsaasacaa(3.46)3.2.1 Solution of the Euler Transformation 3.2 Solving Kinematics Equation第39页/共68页41),(2atan),(2atan180),(2atanyxyxyxzxyocosacasnsncnnn(3.52) 3.2 Solving Kinematics Equation第40页/共68页42zyxzyx

19、xypcpspcsrppspcpp)(),(2atan180),(2atan(3.58) 3.2 Solving Kinematics Equation第41页/共68页43。 3.3 Kinematics Equation of PUMA 560第42页/共68页44 3.3 Kinematics Equation of PUMA 560第43页/共68页453.9 PUMA 560图 机器人的连杆坐标系( )a 结构图 3.3 Kinematics Equation of PUMA 560第44页/共68页463.9 PUMA 560图 机器人的连杆坐标系( )a 结构图 3.3 Kine

20、matics Equation of PUMA 560第45页/共68页473.9 PUMA 560图 机器人的连杆坐标系( )a 结构图 3.3 Kinematics Equation of PUMA 560第46页/共68页483.9 PUMA 560图 机器人的连杆坐标系( )a 结构图 3.3 Kinematics Equation of PUMA 560第47页/共68页493.9 PUMA 560图 机器人的连杆坐标系( )a 结构图( )b 坐标图 3.3 Kinematics Equation of PUMA 560第48页/共68页501i61Ti6161AAATiii(3.

21、15)111111111100001iiiiiiiiiiiiiiiiiiicsas cc csd sTs sc scd c (3.16)3.3.1 Motion Analysis of PUMA 560 3.3 Kinematics Equation of PUMA 560第49页/共68页51100001000000111110csscT100000100002222221csdscT100001000003323332csascT100000100044434443csdascT 3.3 Kinematics Equation of PUMA 560第50页/共68页52100000010

22、000555554csscT100000010000666665csscT)()()()()()(66555444333222111060TTTTTTT 621,(3.59) 3.3 Kinematics Equation of PUMA 560第51页/共68页531000611060zzzzyyyyxxxxpaonpaonpaonTTT 3.3 Kinematics Equation of PUMA 560第52页/共68页54 3.3 Kinematics Equation of PUMA 560.,;,)(,)(,)(),()(),()(;)(),()(),()(23422233122

23、3423322112234233221523542354152354231541523542316523646542365464165236465423165464165236465423165236465423646541652364654231646541652364654231cdsasapcdsdcacaspsdsdcacacpccscsassccssccsassscsscccassccssccsoccscccssscsscccsoscsccsssscssccccocscsscccsnscccsccssssccccsnscccsscsssscccccnzyxzyxzyxzyx第53页/

24、共68页55)()()()()()(100066555444333222111060TTTTTTpaonpaonpaonTzzzzyyyyxxxx(3.65)pao,n,和621,621, 3.3 Kinematics Equation of PUMA 560第54页/共68页561)()()()()()(100066555444333222111060TTTTTTpaonpaonpaonTzzzzyyyyxxxx(3.65)010123451162233445566()()()()()TTTTTTT0111T1111111111111611110 00 0001 00001000 10001

25、xxxxxxxxyyyyyyyyzzzzzzzzcsnoapnoapscnoapnoapTnoapnoap 3.3 Kinematics Equation of PUMA 560第55页/共68页571010123451162233445566()()()()()TTTTTTT1111111111111611110 00 0001 00001000 10001xxxxxxxxyyyyyyyyzzzzzzzzcsnoapnoapscnoapnoapTnoapnoap1112xyys pc ppd利用三角代换：cos ;sinxypp其中22;atan2,xyyxpppp(3.67)两边(2,4)

26、项元素对应相等：(3.68) 3.3 Kinematics Equation of PUMA 560第56页/共68页58212122221222122sin()/;cos()1 (/)atan2,1atan2(,)atan2(,)yxxyddddppdppd 1(3.70)11112xyys pc ppdcos ;sinxypp22;atan2,xyyxpppp 3.3 Kinematics Equation of PUMA 560第57页/共68页5931111111111111611110 00 0001 00001000 10001xxxxxxxxyyyyyyyyzzzzzzzzcsn

27、oapnoapscnoapnoapTnoapnoap(3.67)1113 234 232213 234232 2xyxzzc ps ppa cd sa cppa sd ca s 两边(1,4)项和(3,4)项元素对应相等：3 34 3a cd sk其中2222222232422xyzpppaaddka 3.3 Kinematics Equation of PUMA 560第58页/共68页601),(2atan),(2atan22423433kdakda3(3.73)正、负号对应 的两种可能解。3 34 3a cd sk1112222122atan2(,)atan2(,)xyyyxxys pc

28、 ppdppdppd求 3 式中，2222222232422xyzpppaaddka 3.3 Kinematics Equation of PUMA 560第59页/共68页61201034531236445566,()()()TTTTT 1 231 23232 31 231 23232 336112000010001xxxxyyyyzzzzc cs csa cnoapc ss sca snoapTnoapscd1 231 23232 331 231 23232 34xyzxyzc c ps c ps pa cac s ps s pc pa sd(3.75)两边(1,4)项和(2,4)项元素对

29、应相等： 3.3 Kinematics Equation of PUMA 560第60页/共68页62231和2323232(3.78)2332 3112 3423221142 3112 33232211232332 3112 3442 3112 33atan2,zxyzxyzxyzxyzxyzxyaa cpc ps pa sdspc ps pda cpc ps pa cacpc ps paa cpc ps pa sdda cpc ps pa ca(3.77) 3.3 Kinematics Equation of PUMA 560第61页/共68页63401034531236445566,()()()TTTTT 1 231 23232 31 231 23232 336112000010001xxxxyyyyzzzzc cs csa cnoapc ss sca snoapTnoapscd1 231 23234 5114 5xyzxyc c as c as ac ss ac as s (3.75)两边(1,3)项和(3,3)项元素对应相等：4111 231 2323atan2(,)xyxyza sa ca c ca s ca s(3.79) 3.3 Kinematics Equation of PUMA 560第62页/共68页645010454123465566,()()TT