# Index in position 1 exceeds array bounds (must not exceed 1)?

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Abdirahman Mohamed on 10 Oct 2020
Answered: per isakson on 10 Oct 2020
When i run my program i get the error "Index in position 1 exceeds array bounds (must not exceed 1)." the error points to the line where i have put the arrow.
I assume the error is related to my v but i dont seem to get the error fixed. Any help would be appreciated
Thanks!
v=[];
v(1:4,1)=0;
for i=1:length(alpha)
steg=alpha(i)*hmax;
t=0:steg:T;
for n=1:(T/steg)
end
end

per isakson on 10 Oct 2020
It's easier to debug code like yours if one
• use intermediary variables to breaks the long expression into smaller expressions.
• can run the code, which requires a realistic set of variable values ( alpha, hmax, etc.)
Abdirahman Mohamed on 10 Oct 2020
Sorry here is the complete version of my code:
I have done my best to condense the expression in the second for loop
clear
clc
close all
theta=0.5;
alpha=[];
alpha=[1, 10, 100];
v=[];
v(1:4,1)=0; %initial villkor
steg=0;
T=0.5;
t=[];
m1=460; m2=60; k1=5500;k2=130000; c1=300; c2=1300; v=60/3.6;
H=0.2; L=1;
nyak2=100*k2;
A = [0 0 1 0;0 0 0 1;-k1/m1 k1/m1 -c1/m1 c1/m1;k1/m2 -(k1+nyak2)/m2 c1/m2 -(c1+c2)/m2];
eigen=eig(A); %egenvärde för matrisen B
for i=1:4
emax(i)=-2*real(eigen(i))/(norm(eigen(i))^2);
end
hmax=min(emax); %nya maximala tidssteget
options=odeset('RelTol',1e-9,'AbsTol',1e-9,'Refine',1);
idenmat=[1 0 0 0;0 1 0 0; 0 0 1 0; 0 0 0 1];
for i=1:length(alpha)
steg=alpha(i)*hmax;
t=0:steg:T;
for n=1:(T/steg)
end
end

per isakson on 10 Oct 2020
I replaced the inner for-loop by
for n=1:(T/steg)
f01 = idenmat - steg*theta*A;
f22 = quartercar( t(n), v(1:4,n), A, nyak2, c2, m2, H, L, v );
f23 = roadprofile( H, L, v, t(n+1), nyak2 );
tmp = f01 \ (v(1:4,n)+(steg*theta*( f22 + f23 ) ) );
%
end
to make it easier to spot the error and inspect variable values. (My variable names indicate lack of fantacy. )
Next, I turned your script into a function
function cssm
end
to avoid litter my base workspace (and because I prefer functions)
Next, I enabled Pause on Errors (in the Run menu)
And clicked Run. The execution halted on the line
f22 = quartercar( t(n), v(1:4,n), A, nyak2, c2, m2, H, L, v );
Hovering with the cursor over v(1:4,n) showed that the value of v is a scalar. v is assigned the scalar value at the line
m1=460; m2=60; k1=5500;k2=130000; c1=300; c2=1300; v=60/3.6;