|
|
EDA365欢迎您登录！
您需要 登录 才可以下载或查看，没有帐号？注册
x
matlab robotics toolbox的逆运动学求解ikine为题。! m. y7 r! b# B+ s7 Z9 Y
代码如下：
% u! b. ]. s1 \; Y0 z* L$ [clear all;
. |" W2 L# x$ D1 k& @; rclc;
3 B3 F2 q7 u4 \, E9 j: h% e' nstartup_rvc;4 t: R' W! b5 E/ w
L(1) = Link([0,0.194,0.45,0,0]);% p+ M; ?* J6 ^ b
L(1).qlim=([-130*pi/180,130*pi/180]);
$ k w+ g$ F( hL(2) = Link([0,0.072,0.35,0,0]);+ c+ V% X+ a& V* L8 b0 T
L(2).qlim=([-5*pi/6,5*pi/6]);
& x! G+ E# M+ G- mL(3) = Link([0,0,0,0,1]);- M% M% v* y0 g3 @+ V: n
L(3).qlim=[0,0.5];
$ @; x7 E4 r; t: u; bL(4) = Link([0,0,0,0,0]);* |% X; c& i8 z% P l
L(4).qlim=([-pi,pi]);
4 }0 { j1 g/ V% P. f' n1 z mbot = SerialLink(L); %连接连杆
5 J/ D5 j, x0 o& d" `5 |, Bbot.name='SCARA';
& \( |" W4 m4 O' q3 [bot.display() %显示D-H参数表
; T C6 R0 H) O4 n5 n! ~bot.teach()
& |# _; ], e2 ~ A: z7 P' Z
/ P& p9 F5 {3 S. R- @9 v' xT=[-0.2588 0.9659 0 0.1344;-0.9659 -0.2588 0 0.7278;0 0 1 0.366;0 0 0 1]& I. A0 }4 y# X# L( w
inverse_kinematics=bot.ikine(T) %逆向运动学. 这个逆解为啥就是不能求？？？？？1 j) b% ^. f0 S! ]8 W w R
- s3 S) t7 g/ y# U. x; D; z8 r' p8 e4 i9 B6 v
theta1=inverse_kinematics(1);' q4 n/ c- a" t, e: j$ a
theta2=inverse_kinematics(2);
! _# l5 g2 [! V' C# r- jd=inverse_kinematics(3);% C7 X' [, [5 k4 }0 l5 ~5 c& @
theta4=inverse_kinematics(4);
* v1 @ G9 A2 C# Y1 h# d% R; Yforward_kinematics=bot.fkine([theta1 theta2 d theta4])# S7 t, m# j& d4 Y
错误信息：
5 _3 s! B3 U) r7 v& OError using SerialLink/ikine (line 164) B3 k5 g9 {% @% G5 ]2 I
Number of robot DOF must be >= the same number of 1s in the mask matrix
/ h* J# V) A) R5 R/ ?% L0 Q( [$ U' z6 e d7 T- m2 z
Error in ikine (line 19). e+ c' F- J" K' {' @- Q
inverse_kinematics=bot.ikine(T). h6 _& G; {; c/ z
' F: R1 z: }' Q* p( O
; Y5 r, n8 f, a c, O8 O% g; H
6 z# c% u* o/ q3 r6 X5 l6 i4 O
求助大神！
5 L8 ?; a! ]! q9 |. y* x |
|
/1 
发表于 2020-9-15 13:33
收藏
淘帖
支持！
反对！