機器人逆運動學求解（三軸相交）

腕部三根軸線相交的機器人，其末端執行器的位置和姿態可轉換爲腕部關節中心點的位置和末端執行器的姿態分別求解，其運行結果如下;

[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

上述結果表明，用逆運動學可得出8組逆解，逆解算出關節角求出的的末端姿態與正運動學求出的末端姿態基本一致。注：theta5爲零時，機器人處於奇異位型，此時theta4與theta6有無數組解，儘量避免theta5爲零
推導過程如下：




注：此處theta爲與初始角的夾角，具體建模和正運動學過程