机器人逆运动学求解（三轴相交）

腕部三根轴线相交的机器人，其末端执行器的位置和姿态可转换为腕部关节中心点的位置和末端执行器的姿态分别求解，其运行结果如下;

[J,T] = TX90_jacobian([90 0 90 45 30 60])

T =

   -0.9186    0.1768   -0.3536  -85.3553
   -0.2500    0.4330    0.8660  561.6025
    0.3062    0.8839   -0.3536  389.6447
         0         0         0    1.0000


theta_1 =

[ 90.0, 90.0, -90.0, 22.21, 110.7, 109.1]

T =

   -0.9186    0.1767   -0.3536  -85.3600
   -0.2501    0.4329    0.8660  561.6038
    0.3061    0.8839   -0.3535  389.6525
         0         0         0    1.0000


theta_2 =

[ 90.0, 90.0, -90.0, -157.8, -110.7, -70.89]

T =

   -0.9186    0.1769   -0.3534  -85.3449
   -0.2499    0.4329    0.8661  561.6100
    0.3062    0.8839   -0.3535  389.6525
         0         0         0    1.0000


theta_3 =

[ 90.0, 0.0002703, 90.0, 45.0, 30.0, 60.0]


theta_4 =

[ 90.0, 0.0002703, 90.0, -135.0, -30.0, -120.0]


theta_5 =

[ -77.98, -13.63, -74.75, -156.4, 24.45, 83.81]

T =

   -0.9184    0.1772   -0.3538  -85.3933
   -0.2501    0.4331    0.8660  561.5792
    0.3067    0.8838   -0.3534  389.7066
         0         0         0    1.0000

theta_6 =

[ -77.98, -13.63, -74.75, 23.57, -24.45, -96.19]

T =

   -0.9185    0.1768   -0.3536  -85.3739
   -0.2501    0.4330    0.8660  561.5836
    0.3062    0.8839   -0.3535  389.6979
         0         0         0    1.0000

theta_7 =

[ -77.98, -88.38, 74.75, -170.4, 97.28, 106.7]

T =

   -0.9186    0.1768   -0.3535  -85.3651
   -0.2499    0.4331    0.8660  561.5812
    0.3062    0.8838   -0.3536  389.6833
         0         0         0    1.0000

theta_8 =

[ -77.98, -88.38, 74.75, 9.607, -97.28, -73.3]

T =

   -0.9185    0.1768   -0.3536  -85.3764
   -0.2501    0.4331    0.8660  561.5768
    0.3062    0.8838   -0.3536  389.6838
         0         0         0    1.0000

上述结果表明，用逆运动学可得出8组逆解，逆解算出关节角求出的的末端姿态与正运动学求出的末端姿态基本一致。注：theta5为零时，机器人处于奇异位型，此时theta4与theta6有无数组解，尽量避免theta5为零
推导过程如下：




注：此处theta为与初始角的夹角，具体建模和正运动学过程