Index exceeds the number of array elements (1) error

1 Ansicht (letzte 30 Tage)
Zafer Duyenli
Zafer Duyenli am 18 Dez. 2018
Beantwortet: Aoi Midori am 19 Dez. 2018
greetings.
I have asked another question about that topic a time ago but I have the same issue again and I can't find the solution no matter how deep I look at into the program.
I am trying to analyze velocities, accelerations and positions of each link in a mechanism by using Newton Raphson method. I have no issue until the analysis of acceleration matrix and I have the issue stated above in line 65. How can I reconstruct that line for solution?
Thank you for your time.
clc, clear;
%constants
a2=29;a3=15;a4=29;a5=27;a6=25;
%max iteration nr
Nmax=100;
%input:initial guess values th3,th5,th6,to
x=[12*pi/180,61*pi/180,323*pi/180,31];
%Error tolerance
xe=0.001*abs(x);
%System inputs
dth=5*pi/180;
th2=290*pi/180:dth:368*pi/180;
w2=-5*ones(1,length(th2));
al2=0*ones(1,length(th2));
xe=transpose(abs(xe));
kerr=1;
for k=1:1:length(th2)
for n=1:Nmax
th3(k)=x(1);th5(k)=x(2);
th6(k)=x(3);to(k)=x(4);
%Jacobian Matrix
J=zeros(4,4);
J(1,1)=-a3*sin(th3(k));J(1,2)=-a4*sin(th5(k)+210*pi/180);
J(2,1)=a3*cos(th3(k));J(2,2)=a4*cos(th5(k)+210*pi/180);
J(3,2)=-a5*sin(th5(k));J(3,3)=-a6*sin(th6(k));J(3,4)=-1;
J(4,2)=a5*cos(th5(k));J(4,3)=a6*cos(th6(k))
%position matrix
f=zeros(4,1);
f(1,1)=-(a2*cos(th2(k))+a3*cos(th3(k))+a4*cos(th5(k)+210*pi/180)-43);
f(2,1)=-(a2*sin(th2(k))+a3*sin(th3(k))+a5*sin(th5(k)+210*pi/180)+33);
f(3,1)=-(a5*cos(th5(k))+a6*cos(th6(k))-to(k));
f(4,1)=-(a5*sin(th5(k))+a6*sin(th6(k))-10);
eps=inv(J)*f;
x=x+transpose(eps);
if abs(eps)<xe
kerr=0;break
end
if kerr==1
'Error nr'
end
th3(k)=x(1);th5(k)=x(2);th6(k)=x(3);to(k)=x(4);
%velocity analysis
fv=zeros(4,1);
fv(1,1)=w2(k)*a2.*sin(th2(k));
fv(2,1)=-w2(k)*a2*cos(th2(k));
vel=inv(J)*fv;
w3(k)=vel(1);w5(k)=vel(2);w6=vel(3);vt(k)=vel(4);
%acceleration analysis
fa=zeros(4,1);
fa(1,1)=al2(k)*a2*sin(th2(k))-w2(k)^2*a2*cos(th2(k))-w3(k)^2*a3*cos(th3(k))-w5(k)^2*a4*cos(th5(k)+210*pi/180);
fa(2,1)=-al2(k)*a2*cos(th2(k))+w2(k)^2*a2*sin(th2(k))+w3(k)^2*a3*sin(th3(k))+w5(k)^2*a4*sin(th5(k)+210*pi/180);
fa(3,1)=w5(k)^2*a5*cos(th5(k))+w6(k)^2*a6*cos(th6(k)); %LINE 65
fa(4,1)=w5(k)^2*a5*sin(th5(k))+w6(k)^2*a6*sin(th6(k));
acc=inv(J)*fa;
al3(k)=acc(1);al5(k)=acc(2);
al6(k)=acc(3);at(k)=acc(4);
end
end
th2d=th2*180/pi;
th3d=th3*180/pi;
th5d=th5*180/pi;
th6d=th6*180/pi;
%--------Plots---------------
figure(1),
subplot(4,3,1),plot(th2d,th4d,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\theta_3 (^o)'),xlim([300 368])
subplot(4,3,2),plot(th2d,w4,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\omega_3 (r/s)'),xlim([300 368])
subplot(4,3,3),plot(th2d,al4,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\alpha_3 (r/s^2)'),xlim([300 368])
subplot(4,3,4),plot(th2d,th5d,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\theta_5 (^o)'),xlim([300 368])
subplot(4,3,5),plot(th2d,w5,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\omega_5 (r/s)'),xlim([300 368])
subplot(4,3,6),plot(th2d,al5,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\alpha_5 (r/s^2)'),xlim([300 368])
subplot(4,3,7),plot(th2d,s,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\theta_6 (cm)'),xlim([300 368])
subplot(4,3,8),plot(th2d,Vs,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\omega_6 (cm/s)'),xlim([300 368])
subplot(4,3,9),plot(th2d,as,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\alpha_6 (cm/s^2)'),xlim([300 368])
subplot(4,3,10),plot(th2d,to,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('t (cm)'),xlim([300 368])
subplot(4,3,11),plot(th2d,Vt,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('V_t (cm/s)'),xlim([300 368])
subplot(4,3,12),plot(th2d,at,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('a_t (cm/s^2)'),xlim([300 368])

Akzeptierte Antwort

Aoi Midori
Aoi Midori am 19 Dez. 2018
I assumed that you wanted to write in LINE 59 as:
w3(k)=vel(1);w5(k)=vel(2);w6(k)=vel(3);vt(k)=vel(4);
where it's not w6=vel(3) but w6(k)=vel(3), otherwise w6 would be just a scalar and you wouldn't be able to access to w6(k) as you wrote in LINE 65.

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