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?c123?s0T3=0T11T22T3=?123?0??0?s123c123000L1c1?L2c12?0L1s1?L2s12??
?10?01?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???'????
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?'?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????L0?2??y??0?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??因为,??Jv,且v=[1, 0]T,即vX=1m/s,vY=0,因此
?1??1?l2c12l2s12??1?1??????lc??lc??0??ls??lslls?????1121211212???122?2?c1?1?12?rad/s=-2rad/s
l1s?20.5?2??c?1c?12?4rad/sl2s?2l1s?2(算出最后结果2分)
?2=4 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,rad/s;
手部瞬时速度为1m/s。
sθ1=sinθ1式中:s12=sin(θ1+θ2)
??????6.解:建立如图的坐标系,则各连杆的DH参数为:
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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?1?0?0有
0?s?i?1c?i?100??dis?i?1?? dic?i?1??1??s?3c?3000?00?? 1L2??01?0ai?1?c?1?s?0T??11?0??0故
?s?10c?1000??1?000?1?,2T???01L1???01??0?c?1s?3?s?1s?3c?300??c?3?s?0?1?d2?2?,3T??3?0100???001??00s?1?c?100?c?1c?3?s?c?0012T2T3T??133T?1?s?3??0s?1L2?s?1d2??c?1L2?c?1d2??(写出最后结果2分)
?L1?1?s?1?sin?1式中:c?1?cos?1
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