# Subscripted assignment dimension mismatch error in kinematic analysis of a 1 DOF mechanism

1 view (last 30 days)
Zafer Duyenli on 1 Dec 2019
Commented: Zafer Duyenli on 15 Dec 2019
Greetings.
I have been trying to analyze a 1DOF mechanism by Newton Raphson algorythm but I always get the "Subscripted assignment dimension mismatch." error in line 59. How can I solve this? Thank you for your answers. The whole code is below:
clc,
clear,
% Input - defining physical parameters of mechanism
a2=31;a3=64;a4=35;
% Maximum Iteration
Nmax=100;
% Input - defining initial guess values for th3,th4,tau,s
x=[205*pi/180,66*pi/180,76,57];
% Error Tolerance
xe=0.001*abs(x);
% Input - System inputs (th2, w2, al2)
dth=5*pi/180;
th2=0*pi/180:dth:180*pi/180;
w2=10*ones(1,length(th2));
al2=0*ones(1,length(th2));
%--------------
xe=transpose(abs(xe));
kerr=1; %if kerr=1, results won't be converged
for k=1:1:length(th2);
for n=1:Nmax;
th3(k)=x(1);th4(k)=x(2);tau(k)=x(3);s(k)=x(4);
% Input - Jacobian Matrix
J=zeros(4,4);
J(1,2)=-a4*sin(th4(k));
J(2,2)=a4*cos(th4(k));J(2,3)=-1;
J(3,1)=-a3*sin(th3(k));J(3,4)=1;
J(4,1)=a3*cos(th3(k));
% Input - f vectors
f=zeros(4,1);
f(1,1)=-(a2*cos(th2(k))+a4*cos(th4(k))+4);
f(2,1)=-(a2*sin(th2(k))+a4*sin(th4(k))-s);
f(3,1)=-(a2*cos(th2(k))+a3*cos(th3(k))+tau);
f(4,1)=-(a2*sin(th2(k))+a3*sin(th3(k)));
%-----------------
eps=inv(J)*f;x=x+transpose(eps);
if abs(eps)<xe;
kerr=0;break
end
end
if kerr==1;
'Error nr';
end
th3(k)=x(1);th4(k)=x(2);
tau5(k)=x(3);tau6(k)=x(4);
% Velocity analysis
fv(1,1)=w2(k)*a2*sin(th2(k));
fv(2,1)=-w2(k)*a2*cos(th2(k));
fv(3,1)=w2(k)*a2*sin(th2(k));
fv(4,1)=-w2(k)*a2*cos(th2(k));
vel=inv(J)*fv;
w3(k)=vel(1);w4(k)=vel(2);
Vtau(k)=vel(3);Vs(k)=vel(4);
%Acceleration analysis
fa(1,1)=al2*a2*sin(th2(k))+w2.^2+a2*cos(th2(k))+w4.^2*a4*cos(th4(k));
fa(2,1)=-al2*a2*cos(th2(k))+w2.^2*a2*sin(th2(k))+w4.^2*a4*sin(th4(k));
fa(3,1)=al2*a2*sin(th2(k))+w2.^2*a2*cos(th2(k))+w3.^2*a3*cos(th3(k));
fa(4,1)=-al2*a2*cos(th2(k))+w2.^2*a2*sin(th2(k))+w3.^2*a3*sin(th3(k));
acc=inv(J)*fa;
al3=acc(1);al4=acc(2);atau=acc(3);as=acc(4);
end
th2d=th2*180/pi;
th3d=th3*180/pi;
th4d=th4*180/pi;
% Plots
figure(1),
subplot(4,3,1),plot(th2d,th3d,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\theta_3 (^o)'),xlim([0 180])
subplot(4,3,2),plot(th2d,w3,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\omega_3 (r/s)'),xlim([0 180])
subplot(4,3,3),plot(th2d,al3,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\alpha_3 (r/s^2)'),xlim([0 180])
subplot(4,3,4),plot(th2d,th4d,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\theta_4 (^o)'),xlim([0 180])
subplot(4,3,5),plot(th2d,w4,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\omega_4 (r/s)'),xlim([0 180])
subplot(4,3,6),plot(th2d,al4,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\alpha_4 (r/s^2)'),xlim([0 180])
subplot(4,3,7),plot(th2d,tau,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\tau_5 (cm)'),xlim([0 180])
subplot(4,3,8),plot(th2d,Vtau,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\v tau_5 (cm/s)'),xlim([0 180])
subplot(4,3,9),plot(th2d,atau,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\a tau_5 (cm/s^2)'),xlim([0 180])
subplot(4,3,10),plot(th2d,s,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\tau_6 (cm)'),xlim([0 180])
subplot(4,3,11),plot(th2d,Vs,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\v tau_6 (cm/s)'),xlim([0 180])
subplot(4,3,12),plot(th2d,as,'r','linewidth',2),xlabel('\theta_2 (^o)'),ylabel('\a tau_6 (cm/s^2)'),xlim([0 180])

Vimal Rathod on 5 Dec 2019
You could try initializing fa varible with zeros using "zeros function" before assigning it to the above expression in your code. I think that would remove the dimension mismatch error you are getting.
Try debugging your code and find out the size of Right hand side expression. Then initializing variable fa with proper size would solve your error.

#### 1 Comment

Zafer Duyenli on 15 Dec 2019
Okay I found out that I didn't initialize the variable such as w2, al2, etc in iterations. I should have inputted them in form of w2(k), al2(k), etc.
Thank you so much!

R2018a