Array indices must be positive integers or logical values.

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Ali Deniz
Ali Deniz el 16 de Oct. de 2021
Respondida: Image Analyst el 18 de Oct. de 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)

Respuesta aceptada

KSSV
KSSV el 16 de Oct. de 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 comentarios
dpb
dpb el 16 de Oct. de 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.

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Más respuestas (3)

dpb
dpb el 16 de Oct. de 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 comentario
Ali Deniz
Ali Deniz el 16 de Oct. de 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 el 16 de Oct. de 2021
Editada: John D'Errico el 16 de Oct. de 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 el 18 de Oct. de 2021

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