function call and getting array error
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Thin Rupar Win
am 9 Nov. 2021
Kommentiert: Steven Lord
am 10 Jan. 2024
Dear Sir or Madam,
I recently learn matlab for my education purpose. I would like to know how to solve the Index error in my code as I call function in for loop and getting error. I have already check array boundary and can't find the answer for the case. I deeply request you how to solve this case.
clear
% nx and ny must be the same size with t, n_part of the DEM loop
nx=4;ny=9;
% f_ equation for collisions
f=zeros(nx,ny,9);
feq=zeros(nx,ny,9);feq_f=zeros(nx,ny,9);
% velocities for x and y direction
u=zeros(nx,ny);v=zeros(nx,ny);
% velocities on solid particle
% for x direction
U_s_x=zeros(nx,ny,9);U_px=[0,0];Omega_px=[0,0];Xpx=[0,0];
% for y direction
U_s_y=zeros(nx,ny,9);U_py=[0,0];Omega_py=[0,0];Xpy=[0,0];
% position of particle and omega_s
omega_s=zeros(nx,ny,9);
% force term Ff,Tf on solid particle
Ffx=zeros(nx,ny,9);Ffy=zeros(nx,ny,9);
Tfx=zeros(nx,ny,9);Tfy=zeros(nx,ny,9);
rho=ones(nx,ny);x=zeros(nx);y=zeros(ny);
w(9)=zeros;
% initialization of the system
w=[1/9 1/9 1/9 1/9 1/36 1/36 1/36 1/36 4/9];
cx=[1 0 -1 0 1 -1 -1 1 0];
cy=[0 1 0 -1 1 1 -1 -1 0];
c2=1./3.;
% derivative rate
dx=1.0;dy=1.0;
x1=(nx-1)/(ny-1);y1=1.0;
dx=x1/(nx-1);
dy=y1/(ny-1);
x=(0:dx:x1);
y=(0:dy:y1);
u0=0.2;
% relaxiation time
alpha=0.02;
Re=u0*(ny-1)/alpha;
% omega = 1/tau
omega=1./(3.*alpha+0.5);
% toleriant and error
%count=0;tol=1.0e-4;error=10.;erso=0.0;
% setting velocity
for j=2:ny-1
u(1,j)=u0;
end
t_lbm=50;
for t=1:t_lbm
[f,Ffx,Ffy,Tfx,Tfy]=up_collition(nx,ny,u,v,cx,cy,U_s_x,U_px,Omega_px,Xpx,U_s_y,U_py,Omega_py,Xpy,feq,rho,w,feq_f,f,omega_s,Ffx,Ffy,Tfx,Tfy);
end
function[f,Ffx,Ffy,Tfx,Tfy]=up_collition(nx,ny,u,v,cx,cy,U_s_x,U_px,Omega_px,Xpx,U_s_y,U_py,Omega_py,Xpy,feq,rho,w,feq_f,f,omega_s,Ffx,Ffy,Tfx,Tfy)
% the weighting function of solid particles k Bk,
% the local solid ratio e_k divided
e_k=[1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8 0];
e_total=1;
for j=1:ny
for i=1:nx
t1=u(i,j)*u(i,j)+v(i,j)*v(i,j);
for k=1:9
x=zeros(length(i));% later add with dx term, let dt =1;
t2=u(i,j)*cx(k)+v(i,j)*cy(k);
Bk(k)=e_k(k).*(0.56-0.5)/(1-e_total+(0.56-0.5));
% velocity on solid particle
% the equaiton is Us=UP+omega_px *(x +0.5 *c(k)*dt)-Xp
U_s_x(i,j,k)=U_px(i,j)+Omega_px(i,j)*(x(i)+0.5*(cx(k)))-Xpx(i,j);
U_s_y(i,j,k)=U_py(i,j)+Omega_py(i,j)*(x(i)+0.5*(cy(k)))-Xpy(i,j);
t22=U_s_x(i,j,k)*cx(k)+U_s_y(i,j,k)*cy(k);
t11=U_s_x(i,j,k)*U_s_x(i,j,k)+U_s_y(i,j,k)*U_s_y(i,j,k);
feq(i,j,k)=rho(i,j)*w(k)*(1.0+3.0*t2+4.5*t2*t2-1.5*t1);
feq_f(i,j,k)=rho(i,j)*w(k)*(1.0+3.0*t22+4.5*t22*t22-1.5*t11);
f(i,j,k)=f(i,j,k)-(0.9091*(f(i,j,k)-feq(i,j,k)))+(Bk(k)*(feq_f(i,j,k)-feq(i,j,k)));
% additional collision terms
omega_s(i,j,k)=feq_f(i,j,k)-f(i,j,k)+((f(i,j,k)-feq(i,j,k)).*0.42);
% force term and torque term for DEM
Ffx(i,j,k)=sum(Bk(k)).*sum(omega_s(i,j,k))*cx(k);
Ffy(i,j,k)=sum(Bk(k)).*sum(omega_s(i,j,k))*cy(k);
Tfx(i,j,k)=sum((x(i)-Xpx)*Bk(k)).*sum(omega_s(i,j,k)).*cx(k);
Tfy(i,j,k)=sum((x(i)-Xpy)*Bk(k)).*sum(omega_s(i,j,k)).*cy(k);
end
end
end
end
4 Kommentare
Rik
am 9 Nov. 2021
You're indexing several variables. Did you check each one? One of them is resulting in an error. You best chance to solve the problem is to figure out which variable is causing the problem.
Akzeptierte Antwort
Sudharsana Iyengar
am 9 Nov. 2021
U_s_x is a 4x9x9 matrix but U_px is 1x2 matrix. So there is no U_px(3,4) did you check on that.
U_s_x(i,j,k)=U_px(i,j)+Omega_px(i,j)*(x(i)+0.5*(cx(k)))-Xpx(i,j);
U_s_y(i,j,k)=U_py(i,j)+Omega_py(i,j)*(x(i)+0.5*(cy(k)))-Xpy(i,j);
3 Kommentare
Steven Lord
am 10 Jan. 2024
FYI, one way to help debug this type of issue is to set an error breakpoint, run the code to reproduce the error, and examine the variables used on the line where MATLAB enters debug mode to make sure you're not trying to retrieve an element that doesn't exist of one of those variables.
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