Array indices must be positive integers or logical values.

2 vues (au cours des 30 derniers jours)
Ali Deniz
Ali Deniz le 16 Oct 2021
I am trying to solve an equation by using Runge-Kutta Euler Method. Why do I get "Array indices must be positive integers or logical
values." error?
%Euler Method
%parameters
g=9.81;
rho=1.2;
s=0.00011;
m=0.023;
Cd=0.9;
%Initial Condition
V0=0;
V=0;
%I choose dt as
dt=2;
%time interval
t0=0;
ts=10;
t=0;
i=0;
while dt<ts;
d1=g-0.5.*rho.*V0.^2*(s/m).*Cd;
phiAvg=d1;
V(i)=V0+dt.*phiAvg;
V0=V;
t(i)=t;
i=i+1;
t=t+dt;
end
plot(V,t)

Réponse acceptée

KSSV
KSSV le 16 Oct 2021
%Euler Method
%parameters
g=9.81;
rho=1.2;
s=0.00011;
m=0.023;
Cd=0.9;
%Initial Condition
V0=0;
V=zeros([],1);
%I choose dt as
dt=0.2;
%time interval
t0=0;
ts=10;
t=0;
i=0;
while t<ts
i=i+1;
d1=g-0.5.*rho.*V0.^2*(s/m).*Cd;
phiAvg=d1;
V(i)=V0+dt.*phiAvg ;
V0=V(i);
% t(i)=t;
t=t+dt;
end
plot(V)
  3 commentaires
dpb
dpb le 16 Oct 2021
" I must get V(0)=9.81..."
We can't always get what we want. MATLAB arrays are unalterably and inviolatably 1-based, NOT zero-based.
You must create a secondary variable to plot against to have a zero origin graph.
Ali Deniz
Ali Deniz le 16 Oct 2021
Okay, thank you.

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Plus de réponses (3)

dpb
dpb le 16 Oct 2021
In
...
i=0;
while dt<ts;
d1=g-0.5.*rho.*V0.^2*(s/m).*Cd;
phiAvg=d1;
V(i)=V0+dt.*phiAvg;
V0=V;
t(i)=t;
...
what is i first time through the loop?
Using the debugger is very helpful in such cases...
  1 commentaire
Ali Deniz
Ali Deniz le 16 Oct 2021
I want to get a graph so I wanted to store values in V(i). For example V(0)=9.81, V(2)=19.62 ... when t=0, t=2 ....

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John D'Errico
John D'Errico le 16 Oct 2021
Modifié(e) : John D'Errico le 16 Oct 2021
It does not matter that you WANT V(0) to be something. MATLAB does not support zero based indexing. PERIOD.
However, nothing stops you from starting the vector at V(1). And then when you index into the vector, just use V(ind + 1). Now when ind == 0, there is no problem.
If you want to do a plot?
t = 0:1:10; % or whatever it should be
plot(t,V(t+1))

Image Analyst
Image Analyst le 18 Oct 2021

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