Index exceeds array bounds

i have written a code which gets data at time t and location i in two for loops, and calculates parameters. i haven't assigned a size to the arrays.
for t=1:24
Tfodeg(1,1)=20;Tfo(1,1)=293.15;Tfodeg(2,1)=20;Tfo(2,1)=293.15;Tpmh(1,1)=293;
for i=1:16
rho_i(t,i)=999.9-((4.48e-3).*(Tfodeg(t,i-1).^2));
mu_i(t,i)=0.001.*exp(-1.6-(1150./Tfo(t,i-1))+((690/Tfo(t,i-1))^2));
first I have assigned the preliminiary conditions. when i run this, i get the error:
Index in position 2 is invalid. Array indices must be positive integers or logical values.
which was rational. so, i changed the code
for t=2:25
Tfodeg(1,1)=20;Tfo(1,1)=293.15;Tfodeg(2,1)=20;Tfo(2,1)=293.15;Tpmh(1,1)=293;
for i=2:17
rho_i(t,i)=999.9-((4.48e-3).*(Tfodeg(t,i).^2));
mu_i(t,i)=0.001.*exp(-1.6-(1150./Tfo(t,i))+((690/Tfo(t,i))^2));
and i get this one:
Index in position 2 exceeds array bounds (must not exceed 1).
I think my code is mistake-free. anyone can help?

3 commentaires

G A
G A le 27 Juin 2021
You have to define Tfodeg(t,i) before using it in expression for rho_i(t,i)
Hesam Jokar
Hesam Jokar le 28 Juin 2021
thanks. I have defined the Tfodeg(t,i) before the for loops. You think i have to define it inside the loop?
G A
G A le 30 Juin 2021
It is difficult to say anythig without seing the whole code.

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Image Analyst
Image Analyst le 27 Juin 2021

1 vote

2 commentaires

Hesam Jokar
Hesam Jokar le 28 Juin 2021
Thanks
You say "I have defined the Tfodeg(t,i) before the for loops." but what are its dimensions. Before the for loop, do this
sizeTfo = size(Tfodeg) % No semicolon
for t=2:25
Tfodeg(1,1)=20;Tfo(1,1)=293.15;Tfodeg(2,1)=20;Tfo(2,1)=293.15;Tpmh(1,1)=293;
for i=2:17
rho_i(t,i)=999.9-((4.48e-3).*(Tfodeg(t,i).^2));
mu_i(t,i)=0.001.*exp(-1.6-(1150./Tfo(t,i))+((690/Tfo(t,i))^2));
What does it say? My guess is that the second dimension is only 1m,, not 15. So to fix, either make it 15 or 16 or limit your inner for loop to the number of columns it has:
sizeTfo = size(Tfo) % No semicolon
sizeTfodeg = size(Tfodeg) % No semicolon
for t=2:25
Tfodeg(1,1)=20;
Tfo(1,1)=293.15;
Tfodeg(2,1)=20;
Tfo(2,1)=293.15;
Tpmh(1,1)=293;
lastIndex = min([sizeTfo(2), sizeTfodeg(2)])
for i=2 : lastIndex
rho_i(t,i)=999.9-((4.48e-3).*(Tfodeg(t,i).^2));
mu_i(t,i)=0.001.*exp(-1.6-(1150./Tfo(t,i))+((690/Tfo(t,i))^2));

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le 30 Juin 2021

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