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Permalink Reply by omer simsek on January 20, 2011 at 7:18am I am sorry about that but i could not get the file as well so i decided to post codes directly. thanks zigmond...
clc;
clear all;
close all;
global kp_x ki_x kd_x Intgr_x G perror_x distance_x ;
time=input('time=');
distance_x=input('distance_x=');
kp_x=input('kp_x=');
kd_x=input('kd_x=');
ki_x=input('ki_x=');
Intgr_x=0;
syms A1 A2 A3 A4 U1 U2 U3 U4
baslangic=[0 0 0 0 0 0 0 0 0 0 0 0];
for i=1:((time/0.2)+1)
z(i,:)=baslangic;
[t,x]=ode23('position_qdrt2_x',[0,0.2],[baslangic(1),baslangic(2),baslangic(3),baslangic(4),baslangic(5),baslangic(6),baslangic(7),baslangic(8),baslangic(9),baslangic(10),baslangic(11),baslangic(12)]);
a=size(x);
baslangic=[x(a(1),1),x(a(1),2),x(a(1),3),x(a(1),4),x(a(1),5),x(a(1),6),x(a(1),7),x(a(1),8),x(a(1),9),x(a(1),10),x(a(1),11),x(a(1),12)];
end
function xdot=position_qdrt2_x(t,x)
global G perror_x kp_x kd_x ki_x Intgr_x distance_x ;
perror_x=(distance_x-x(1))/0.2; %x(1)-displacement on x-axis%
Intgr_x=Intgr_x+0.2*perror_x;
G=(kp_x*perror_x)+(kd_x*(x(2)))+ki_x*Intgr_x; %x(2)-derivative of the x(1)%
%rotor speeds A1 A2 A3 A4%
A1=212.61-0.001*G;
A2=212.61;
A3=212.61+0.001*G;
A4=212.61;
b=54.2*10^-6;
d=1.1*10^-6;
U1=b*(A1^2+A2^2+A3^2+A4^2);
U2=b*(A4^2-A2^2);
U3=b*(A3^2-A1^2);
U4=d*(A2^2+A4^2-A1^2-A3^2);
B=A2+A4-A1-A3;
m=1;
g=9.8;
l=0.48;
Ix=8.1*10^-3;
Iy=8.1*10^-3;
Iz=14.2*10^-3;
Jr=71.88*10^-6;
y1=((cos(x(7))*sin(x(9))*cos(x(11))+sin(x(7))*sin(x(11)))*(U1/m));
y2=((cos(x(7))*sin(x(9))*sin(x(11))-sin(x(7))*cos(x(11)))*(U1/m));
y3=-g+(cos(x(7))*cos(x(9)))*(U1/m);
y4=x(12)*x(10)*((Iy-Iz)/Ix)-(Jr/Ix)*x(10)*B+((l*U2)/Ix);
y5=x(12)*x(8)*((Iz-Ix)/Iy)+(Jr/Iy)*x(8)*B+((l*U3)/Iy);
y6=x(10)*x(8)*((Ix-Iy)/Iz)+((l*U4)/Iz);
xdot=[x(2);y1;x(4);y2;x(6);y3;x(8);y4;x(10);y5;x(12);y6];
Season Two of the Trust Time Trial (T3) Contest has now begun. The fourth round is an accuracy round for multicopters, which requires contestants to fly a cube. The deadline is April 14th.51 members
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