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  • 大小: 3KB
    文件類型: .m
    金幣: 1
    下載: 0 次
    發布日期: 2021-05-14
  • 語言: Matlab
  • 標簽: 數據文件??

資源簡介

本程序在MATLAB環境下編寫PUMA機器人的運動學正解算法,根據輸入的機器人各關節角度得出機器人末端執行器的空間位姿,并且通過仿真驗證。最終所有的關節角度數據和末端空間位姿均存儲在相應的數據文件中,便于后續程序的引用和處理。

資源截圖

代碼片段和文件信息

%this?program?is?to?calculate?the?transformation?matrix?from?joint?1?to
%joint?6;?and?use?the?cycling?program?to?generate?the?position?data?and
%joint?data;


%this?parameter?is?for?PUMA?560
alf=[pi/2?0?-pi/2?pi/2?-pi/2?0];
a=[0?0.4318?0.0203?0?0?0];
d=[0?0?0.15?0.4318?0?0];
%?this?is?the?Home_Position
%thta=[0?0?0?0?0?0];
%?qz?position?
%thta=[0?0+pi/2?0?0?0+pi?0+pi];
%?qr?position?joint?is?exceed?the?joint?value
%thta=[0?pi/2+pi/2?-pi/2?0?0+pi?0+pi];
%?qs?position?
%thta=[0?0+pi/2?-pi/2?0?0+pi?0+pi];
%?qn?position?
%thta=[0?pi/6+pi/2?-2*pi/3?0?pi/4+pi?0+pi];

%thta=[-pi/2?pi/2-pi/4?-pi/3?-2*pi/3?pi-pi/2?pi-2*pi/3];
thta1=[-pi/2?-pi/3?0?pi/3?pi/2];
????thta2=[-pi?-pi/2?-pi/6?0?pi/6];
???????thta3=[-pi/6?0?pi/4?pi/2?2*pi/3];
????????thta4=[-pi/2?0?pi/4?pi/3?pi/2?2*pi/3];
????????thta5=[-pi/2?-pi/3??pi/6?pi/4?pi/2];
????????thta6=[-pi?-2*pi/3?-pi/2?0?pi/4?pi?4*pi/3];
????????pjoints=[];
????????ppos=[];
????????pjoint1=[];
????????save?‘pumajoints.mat‘?pjoints;
????????save?‘pumaposition.mat‘?ppos;
????????save?‘pumafirst.mat‘?pjoint1;
x=sin(alf);
x(abs(x)<1e-16)=0;
y=cos(alf);
y(abs(y)<1e-16)=0;

for?j=2:6
for?k=2:4
????for?l=1:5
????????for?m=2:5
????????????for?n=1:4
????????????????for?s=3:4
????????????????????total=[thta1(s)-0.001?thta2(n)?thta3(l)?thta4(m)?thta5(k)?thta6(j)];
????????????????????%actjoin=total-thta;
%Robotics?Vision?&?Control:?(c)?Peter?Corke?result?后置坐標法建模(標準建模方法std-DH)
%矩陣變換順序為Trans(00di)Rot(zthta)trans(ai00)Rot(xalf)
%運算結果與p560.fkine(q

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