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。

DH参数法 例题 机器人学_第1张图片

D-H 参数 α i − 1 \alpha_{i-1} αi1 a i − 1 a_{i-1} ai1 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=10000cosα1sinα100sinα1cosx100001100001000010a1001cosθ2sinθ200sinθ2cosθ2000010000110000100001000d21

= [ 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] =cosθ2cosα1sinθ2sinα1sinθ20sinθ2cosα1cosθ2sinα1cosθ200sinα1cosα10a1d2sinα1d2cosα11

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=cosθ2sinθ200sinθ2cosθ2000010a0001
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=cosθ3cosα2sinθ3sinα2sinθ30sinθ3cosα2cosθ3sinα2cosθ300sinα2cosα20a2d3sinα2d3cosα21

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=cosθ30sinθ30sinθ30cosθ300100a1001

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=cosθ2sinθ200sinθ2cosθ2000010a0001cosθ30sinθ30sinθ30cosθ300100a1001

= [ 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] =cosθ2cosθ3sinθ2cosθ3sinθ30cosθ2sinθ3sinθ2sinθ3cosθ30sinθ2cosθ200a1cosθ2+a0a1sinθ201

出自:机器人学入门必看

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