DH参数法 例题 机器人学
四、已知操作臂各连杆的连接方式，计算末端执行器的位姿，建议遵循DH方法的基本原则Exp.1:如图1所示为3-自由度的机械手，这三个关节都是转动的。关节轴3垂直于关节轴1和关节轴2形成的平面。给出连杆坐标系的D-H参数，并推导出坐标系{3}到坐标系{1}的变换矩阵。D-H 参数αi−1\alpha_{i-1}αi−1ai−1a_{i-1}ai−1did_{i}diθi\theta_{i}θi
四、已知操作臂各连杆的连接方式，计算末端执行器的位姿，建议遵循DH方法的基本原则
Exp.1:如图1所示为3-自由度的机械手，这三个关节都是转动的。关节轴3垂直于关节轴1和关节轴2形成的平面。给出连杆坐标系的D-H参数，并推导出坐标系{3}到坐标系{1}的变换矩阵。
| D-H 参数 | α i − 1 \alpha_{i-1} αi−1 | a i − 1 a_{i-1} ai−1 | d i d_{i} di | θ i \theta_{i} θi |
|---|---|---|---|---|
| 1 | X | X | X | θ 1 \theta_{1} θ1 |
| 2 | 0 | a 0 a_{0} a0 | 0 | θ 2 \theta_{2} θ2 |
| 3 | − 9 0 ∘ -90^{\circ} −90∘ | a 1 a_{1} a1 | 0 | θ 3 \theta_{3} θ3 |
DH参数法的一般式：
2
1
T
=
[
1
0
0
0
0
cos
α
1
−
sin
α
1
0
0
sin
α
1
cos
x
1
0
0
0
0
1
]
[
1
0
0
a
1
0
1
0
0
0
0
1
0
0
0
0
1
]
[
cos
θ
2
−
sin
θ
2
0
0
sin
θ
2
cos
θ
2
0
0
0
0
1
0
0
0
0
1
]
[
1
0
0
0
0
1
0
0
0
0
1
d
2
0
0
0
1
]
{ }^{1}_{2} T=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \cos \alpha_{1} & -\sin \alpha_{1} & 0 \\ 0 & \sin \alpha_{1} & \cos x_{1} & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\left[\begin{array}{llll} 1 & 0 & 0 & a_{1} \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} \cos \theta_{2} & -\sin \theta_{2} & 0 & 0 \\ \sin \theta_{2} & \cos \theta_{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & d_{2} \\ 0 & 0 & 0 & 1 \end{array}\right]
21T=⎣⎢⎢⎡10000cosα1sinα100−sinα1cosx100001⎦⎥⎥⎤⎣⎢⎢⎡100001000010a1001⎦⎥⎥⎤⎣⎢⎢⎡cosθ2sinθ200−sinθ2cosθ20000100001⎦⎥⎥⎤⎣⎢⎢⎡10000100001000d21⎦⎥⎥⎤
= [ cos θ 2 − sin θ 2 0 a 1 cos α 1 sin θ 2 cos α 1 cos θ 2 − sin α 1 − d 2 sin α 1 sin α 1 sin θ 2 sin α 1 cos θ 2 cos α 1 d 2 cos α 1 0 0 0 1 ] =\left[\begin{array}{cccc} \cos \theta_{2} & -\sin \theta_{2} & 0 & a_{1} \\ \cos \alpha_{1} \sin \theta_{2} & \cos \alpha_{1} \cos \theta_{2} & -\sin \alpha_{1} & -d_{2} \sin \alpha_{1} \\ \sin \alpha_{1} \sin \theta_{2} & \sin \alpha_{1} \cos \theta_{2} & \cos \alpha_{1} & d_{2} \cos \alpha_{1} \\ 0 & 0 & 0 & 1 \end{array}\right] =⎣⎢⎢⎡cosθ2cosα1sinθ2sinα1sinθ20−sinθ2cosα1cosθ2sinα1cosθ200−sinα1cosα10a1−d2sinα1d2cosα11⎦⎥⎥⎤
代入参数
α
1
=
0
,
a
1
=
a
0
,
d
2
=
0
\alpha_{1}=0, a_{1}=a_{0}, d_{2}=0
α1=0,a1=a0,d2=0
2
1
T
=
[
cos
θ
2
−
sin
θ
2
0
a
0
sin
θ
2
cos
θ
2
0
0
0
0
1
0
0
0
0
1
]
{ }_{2}^{1} T=\left[\begin{array}{cccc} \cos \theta_{2} & -\sin \theta_{2} & 0 & a_{0} \\ \sin \theta_{2} & \cos \theta_{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]
21T=⎣⎢⎢⎡cosθ2sinθ200−sinθ2cosθ2000010a0001⎦⎥⎥⎤
同理
3
2
T
=
[
cos
θ
3
−
sin
θ
3
0
a
2
cos
α
2
sin
θ
3
cos
α
2
cos
θ
3
−
sin
α
2
−
d
3
sin
α
2
sin
α
2
sin
θ
3
sin
α
2
cos
θ
3
cos
α
2
d
3
cos
α
2
0
0
0
1
]
{}^{2}_{3} T=\left[\begin{array}{cccc} \cos \theta_{3} & -\sin \theta_{3} & 0 & a_{2} \\ \cos \alpha_{2} \sin \theta_{3} & \cos \alpha_{2} \cos \theta_{3} & -\sin \alpha_{2} & -d_{3} \sin \alpha_{2} \\ \sin \alpha_{2} \sin \theta_{3} & \sin \alpha_{2} \cos \theta_{3} & \cos \alpha_{2} & d_{3} \cos \alpha_{2} \\ 0 & 0 & 0 & 1 \end{array}\right]
32T=⎣⎢⎢⎡cosθ3cosα2sinθ3sinα2sinθ30−sinθ3cosα2cosθ3sinα2cosθ300−sinα2cosα20a2−d3sinα2d3cosα21⎦⎥⎥⎤
α 2 = − 9 0 ∘ , a 2 = a 1 , d 3 = 0 \alpha_{2}=-90^{\circ}, \quad a_{2}=a_{1}, \quad d_{3}=0 α2=−90∘,a2=a1,d3=0
3 2 T = [ cos θ 3 − sin θ 3 0 a 1 0 0 1 0 − sin θ 3 − cos θ 3 0 0 0 0 0 1 ] ^{2}_{3} T=\left[\begin{array}{cccc} \cos \theta_{3} & -\sin \theta_{3} & 0 & a_{1} \\ 0 & 0 & 1 & 0 \\ -\sin \theta_{3} & -\cos \theta_{3} & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] 32T=⎣⎢⎢⎡cosθ30−sinθ30−sinθ30−cosθ300100a1001⎦⎥⎥⎤
3 1 T = 2 1 T 3 2 T = [ cos θ 2 − sin θ 2 0 a 0 sin θ 2 cos θ 2 0 0 0 0 1 0 0 0 0 1 ] [ cos θ 3 − sin θ 3 0 a 1 0 0 1 0 − sin θ 3 − cos θ 3 0 0 0 0 0 1 ] ^{1}_{3} T={}^{1}_{2}T_{3}^{2} T=\left[\begin{array}{cccc} \cos \theta_{2} & -\sin \theta_{2} & 0 & a_{0} \\ \sin \theta_{2} & \cos \theta_{2} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\left[\begin{array}{cccc} \cos \theta_{3} & -\sin \theta_{3} & 0 & a_{1} \\ 0 & 0 & 1 & 0 \\ -\sin \theta_{3} & -\cos \theta_{3} & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] 31T=21T32T=⎣⎢⎢⎡cosθ2sinθ200−sinθ2cosθ2000010a0001⎦⎥⎥⎤⎣⎢⎢⎡cosθ30−sinθ30−sinθ30−cosθ300100a1001⎦⎥⎥⎤
= [ cos θ 2 cos θ 3 − cos θ 2 sin θ 3 − sin θ 2 a 1 cos θ 2 + a 0 sin θ 2 cos θ 3 − sin θ 2 sin θ 3 cos θ 2 a 1 sin θ 2 − sin θ 3 − cos θ 3 0 0 0 0 0 1 ] =\left[\begin{array}{cccc} \cos \theta_{2} \cos \theta_{3} & -\cos \theta_{2} \sin \theta_{3} & -\sin \theta_{2} & a_{1} \cos \theta_{2}+a_{0} \\ \sin \theta_{2} \cos \theta_{3} & -\sin \theta_{2} \sin \theta_{3} & \cos \theta_{2} & a_{1} \sin \theta_{2} \\ -\sin \theta_{3} & -\cos \theta_{3} & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] =⎣⎢⎢⎡cosθ2cosθ3sinθ2cosθ3−sinθ30−cosθ2sinθ3−sinθ2sinθ3−cosθ30−sinθ2cosθ200a1cosθ2+a0a1sinθ201⎦⎥⎥⎤
出自：机器人学入门必看
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