Having errors in a multivariable for loop.

1 次查看（过去 30 天）
Gabriel Quattromani
编辑： VBBV 2022-3-30
I have this following code in order to find the values of vmax with varying h and x. The values are to be inputted into a single matrix. I am receiving the error message: Array indices must be positive integers or logical values.
function [vmax] = v1(temp_diff,h)
vmax=0.083*sqrt(temp_diff*h);
end
function [vmax] = v2(x,temp_diff,h)
vmax=[0.143/(x+1.32)]*sqrt(temp_diff*h);
end
function [vmax] = v3(temp_diff,h)
vmax=0.043*sqrt(temp_diff*h);
end
temp_int=24;
temp_ext=10;
U=5;
hi=4;
h=1:5;
temp_diff=(U/hi)*(temp_int-temp_ext);
x=0:0.1:2.5;
X=repmat(x,size(h,2),1);
H=repmat(x,size(x,2),2);
n=length(x);
m=length(h);
V=zeros(n,m);
for i=0:0.1:2.5
for h=1:5
if i<0.4
vmax(i)=v1(temp_diff(i),h(i));
elseif i<=2
vmax(i)=v2(temp_diff(i),h(i),x(i));
else
vmax(i)=v3(temp_diff(i),h(i));
end
end
end
Hoping someone with more experience can help me out!

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采纳的回答

VBBV
VBBV 2022-3-30
编辑：VBBV 2022-3-30
Ran in:
temp_int=24;
temp_ext=10;
U=5;
hi=4;
h=1:5;
temp_diff=(U/hi)*(temp_int-temp_ext);
I = 0:0.1:2.5;
x = 0:0.1:2.5;
n=length(x);
m=length(h);
V=zeros(n,m);
X=repmat(x,size(h,2),1);
H=repmat(x,size(x,2),2);
for ii = 1:length(I) % chnage the loop index // variable //from 0 to 1
for h=1:5
if I(ii)<0.4
vmax(ii)=v1(temp_diff,h);
elseif I(ii)<=2
vmax(ii)=v2(temp_diff,h,x(ii));
else
vmax(ii)=v3(temp_diff,h);
end
end
end
plot(I,vmax)
function [vmax] = v1(temp_diff,h)
vmax=0.083*sqrt(temp_diff*h);
end
function [vmax] = v2(x,temp_diff,h)
vmax=[0.143/(x+1.32)]*sqrt(temp_diff*h);
end
function [vmax] = v3(temp_diff,h)
vmax=0.043*sqrt(temp_diff*h);
end
Modfy the loop so that array indices are positive integers and not zeros

更多回答（1 个）

Arif Hoq
Arif Hoq 2022-3-30
try this:
temp_int=24;
temp_ext=10;
U=5;
hi=4;
h=1:5;
temp_diff=(U/hi)*(temp_int-temp_ext);
x=0:0.1:2.5;
X=repmat(x,size(h,2),1); % you are not using this parameter in your code
H=repmat(x,size(x,2),2); % you are not using this parameter in your code
n=length(x);
m=length(h);
V=zeros(n,m);
for i=1:5 % for loop for value of h
for j=1:length(x) % for loop for value of x
if i<0.4
vmax(i)=v1(temp_diff,h(i)); % as temp_diff is a scalar so you don't need indexing
elseif i<=2
vmax(i)=v2(temp_diff,h(i),x(j));
else
vmax(i)=v3(temp_diff,h(i));
end
end
end

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