clear; clc; du = pi/180; L1(1)=Link('theta',-123*du,'a', 0, 'alpha',0,'qlim',[180,365],'modified'); L1(2)=Link('d',0,'a',185, 'alpha',0,'qlim',[3*du,63*du],'modified'); L1(3)=Link('d',90,'a', 0, 'alpha',pi/2,'qlim',[60*du,120*du],'modified'); L1(4)=Link('theta',0,'a', 120, 'alpha',pi/2,'qlim',[230,326],'modified'); % L1(4)=Link('offset',205, 'theta',pi/2,'a',115, 'alpha',pi/2,'qlim',[0,0],'modified'); Needle=SerialLink(L1,'name','Needle'); Needle.display % Needle.plot([100,pi/4,pi/2,0]) figure Needle.teach T=Needle.fkine([180,pi/4,pi/2,0]) T1=myfkine_B([0,pi/4,pi/2,0])这段程序如何求取逆运动学
时间: 2023-08-06 11:18:33 浏览: 54
这段程序可以通过以下步骤求解机械臂的逆运动学:
1. 设定期望的末端执行器位姿,这里假设期望的位姿为 `T_desired`。
```matlab
T_desired = transl(x_desired, y_desired, z_desired) * trotx(alpha_desired) * troty(beta_desired) * trotz(gamma_desired);
```
2. 调用机械臂工具箱中的 `ikine` 函数,求解关节角度。
```matlab
q_desired = Needle.ikine(T_desired, 'mask', [1 1 1 0 0 0]);
```
其中,`'mask', [1 1 1 0 0 0]` 表示限制关节 1、2、3 的运动范围,而关节 4 不限制。函数返回的 `q_desired` 即为所求的关节角度。
完整的代码如下所示:
```matlab
clear; clc;
du = pi/180;
L1(1)=Link('theta',-123*du,'a', 0, 'alpha',0,'qlim',[180,365],'modified');
L1(2)=Link('d',0,'a',185, 'alpha',0,'qlim',[3*du,63*du],'modified');
L1(3)=Link('d',90,'a', 0, 'alpha',pi/2,'qlim',[60*du,120*du],'modified');
L1(4)=Link('theta',0,'a', 120, 'alpha',pi/2,'qlim',[230,326],'modified');
Needle=SerialLink(L1,'name','Needle');
Needle.display
figure
Needle.teach
% 正运动学
T = Needle.fkine([180,pi/4,pi/2,0])
% 逆运动学
T_desired = transl(x_desired, y_desired, z_desired) * trotx(alpha_desired) * troty(beta_desired) * trotz(gamma_desired);
q_desired = Needle.ikine(T_desired, 'mask', [1 1 1 0 0 0])
```
其中,`x_desired`、`y_desired`、`z_desired`、`alpha_desired`、`beta_desired`、`gamma_desired` 为期望的末端执行器位姿。
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