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−1	ai−1a_{i-1}ai−1	did_{i}di	θi\theta_{i}θi
1	X	X	X	θ1\theta_{1}θ1
2	0	a0a_{0}a0	0	θ2\theta_{2}θ2
3	−90∘-90^{\circ}−90∘	a1a_{1}a1	0	θ3\theta_{3}θ3

DH参数法的一般式：
21T=[10000cos⁡α1−sin⁡α100sin⁡α1cos⁡x100001][100a1010000100001][cos⁡θ2−sin⁡θ200sin⁡θ2cos⁡θ20000100001][10000100001d20001]{ }^{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⁡θ2−sin⁡θ20a1cos⁡α1sin⁡θ2cos⁡α1cos⁡θ2−sin⁡α1−d2sin⁡α1sin⁡α1sin⁡θ2sin⁡α1cos⁡θ2cos⁡α1d2cos⁡α10001]=\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α1=0,a1=a0,d2=0\alpha_{1}=0, a_{1}=a_{0}, d_{2}=0α1=0,a1=a0,d2=0
21T=[cos⁡θ2−sin⁡θ20a0sin⁡θ2cos⁡θ20000100001]{ }_{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
32T=[cos⁡θ3−sin⁡θ30a2cos⁡α2sin⁡θ3cos⁡α2cos⁡θ3−sin⁡α2−d3sin⁡α2sin⁡α2sin⁡θ3sin⁡α2cos⁡θ3cos⁡α2d3cos⁡α20001]{}^{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⎦⎥⎥⎤

α2=−90∘,a2=a1,d3=0\alpha_{2}=-90^{\circ}, \quad a_{2}=a_{1}, \quad d_{3}=0 α2=−90∘,a2=a1,d3=0

32T=[cos⁡θ3−sin⁡θ30a10010−sin⁡θ3−cos⁡θ3000001]^{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⎦⎥⎥⎤

31T=21T32T=[cos⁡θ2−sin⁡θ20a0sin⁡θ2cos⁡θ20000100001][cos⁡θ3−sin⁡θ30a10010−sin⁡θ3−cos⁡θ3000001]^{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⁡θ2cos⁡θ3−cos⁡θ2sin⁡θ3−sin⁡θ2a1cos⁡θ2+a0sin⁡θ2cos⁡θ3−sin⁡θ2sin⁡θ3cos⁡θ2a1sin⁡θ2−sin⁡θ3−cos⁡θ3000001]=\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