其中c123?cos??1??2??3?,s123?sin??1??2??3?. 3.解:如图2所示,逆运动学有两组可能的解。
Pθ2'yα'θ2αθ1'θ1x
图2
第一组解:由几何关系得
x?L1cos?1?L2cos??1??2? (1) y?L1sin?1?L2sin??1??2? (2)
(1)式平方加(2)式平方得
2x2?y2?L1?L22?2L1L2cos?2
2?x2?y2?L1??L22???2?arccos??
2LL?12??L2sin?2??y??1?arctan???arctan??
xL?Lcos????122?2???x2?y2??L1??L22? 第二组解:由余弦定理, ??arccos?2LL??12???'????
2?'?1??????arctan?y?2??
?x?4.解:由关节角给出如下姿态:
??L1sin?1?L2sin??1??2?J????L1cos?1?L2cos??1??2?由静力学公式??JF
T?L2sin??1??2????L2?L2?????
L0L2cos??1??2?????1??L2L1??fx???L2fx??A?JFA????????Lf?
?L0?2??0??2x?T??L2L1??0??L1fy??B?JFB??? ??f????L00?2??y???T??l1s?1?l2s125.解:因为:J???l1c?1?l2c12因此,逆雅可比矩阵为:J?l2s12? l2c12???11?l1l2s?2l2c12???lc??lc?11212l2s12?
?l1s?1?l2s12??&?Jv,且v=[1, 0]T,即vX=1m/s,vY=0,因此 因为,??1&???l2c12l2s12??1?1?1??&???lc??lc??0??ls??lslls?????1121211212???122?2?c121?&??rad/s=-2rad/s 1l1s?20.5?&2??c?1c?12?4rad/sl2s?2l1s?2(算出最后结果2分)
&因此,在该瞬时两关节的位置分别为, θ1=30°,θ2=-60°;速度分别为?&1=-2rad/s,?2=4 rad/s;手
部瞬时速度为1m/s。
sθ=sinθ11+θ) 式中:s12=sin(θ12??????6.解:建立如图的坐标系,则各连杆的DH参数为:
z1y1z0x1x2z2y3x3z3y0L2x0
连杆 1 2 3 转角?n 偏距dn 扭角?i?1 0 90° 0 杆长ai?1 0 0 0 ?1 0 ?3 L1 d2 L2 由连杆齐次坐标变换递推公式
?s?i?c?i?s?c?c?ic?i?1i?1Ti??ii?1?s?is?i?1c?is?i?10?s?i?1c?i?1??dis?i?1?? dic?i?1?ai?1??00有
??c?1?s?100??000T??s?1c?100??1100?11??001L?,2T??1??010?0001????000故
??c?1c?3?c?1s?3s?10T?0T12s?c?3?s?1s?3?c?1312T3T??1??s?3c?30?000s?1?sin?1式中:c?1?cos?1
LL01??0??c?3?s?300??d?0??2?,2??s?3c?300?3T?001L? 2?1????0001??s?1L2?s?1d2??c??1L2?c?1d2L?(写出最后结果2分)1?1??