filling a matrix using for loops
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clc
clear all
Rmax=10;
Rcyl=.5;
Uinf=1;
Nr=50;
Ntheta=50;
theta=linspace(0,2*pi,Ntheta)
%discretize
dr=(Rmax-Rcyl)/(Nr-1);
dtheta= 2*pi/(Ntheta-1);
rj=zeros(Nr,1); %allocating rj vector
rtheta=zeros(Ntheta,1); %allocating rtheta vector
for jj=1:Nr
rj(jj)=Rcyl+(jj-1)*dr;
end
for kk=1:Ntheta
rtheta(kk)=(kk-1)*dtheta;
end
w=2/(1+sqrt(1-cos(pi/(Nr+1))*cos(pi/(Ntheta+1))));
term=((2./dr.^2)+(2./(rj.^2).*dtheta.^2)); %the term with psi on left side in eqn8
psi=zeros(Ntheta,Nr); %old psi
psi(1:Ntheta,1)=0; %BC's on wall side
psi(1:Ntheta,Nr)=Uinf*Rmax*sin(theta); %BC's on far field
psi(Ntheta,1:Nr)=0 ;%BC's on top
psi(1,1:Nr)= 0 ;%BC's on bottom
%SOR formula for new psi
psinew=zeros(Ntheta,Nr); %allocating new psi values
bracket=psi
for j=2:Nr-1
for k=2:Ntheta-1
psi(j,k)=((1./rj(j)).*((psi(j+1,k)-psi(j-1,k))./(2.*dr))+((psi(j+1,k)-psi(j-1,k))./(dr.^2))+(1./(rj(j).^2)).*((psi(j,k+1)+psi(j,k+1)))./(dtheta.^2)).*term;
deltapsi=bracket/((2/(dr^2))+(2./((rj.^2).*dtheta^2)))-psi(j,k)
psinew(j,k)=psi(j,k)+w.*deltapsi
end
end
im trying build a psinew matrix, based off psi(j,k) where psi(j,k) becvomes the bracket term. i get the error that "Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 50-by-1." psi matrix should be a 50 by 50
any help is appreciated
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답변 (1개)
Walter Roberson
2023년 2월 24일
Rmax=10;
Rcyl=.5;
Uinf=1;
Nr=50;
Ntheta=50;
theta=linspace(0,2*pi,Ntheta)
%discretize
dr=(Rmax-Rcyl)/(Nr-1);
dtheta= 2*pi/(Ntheta-1);
rj=zeros(Nr,1); %allocating rj vector
That is a vector
rtheta=zeros(Ntheta,1); %allocating rtheta vector
for jj=1:Nr
rj(jj)=Rcyl+(jj-1)*dr;
end
for kk=1:Ntheta
rtheta(kk)=(kk-1)*dtheta;
end
w=2/(1+sqrt(1-cos(pi/(Nr+1))*cos(pi/(Ntheta+1))));
term=((2./dr.^2)+(2./(rj.^2).*dtheta.^2)); %the term with psi on left side in eqn8
That uses all of rj, and rj is a vector, so term is non-scalar
psi=zeros(Ntheta,Nr); %old psi
psi(1:Ntheta,1)=0; %BC's on wall side
psi(1:Ntheta,Nr)=Uinf*Rmax*sin(theta); %BC's on far field
psi(Ntheta,1:Nr)=0 ;%BC's on top
psi(1,1:Nr)= 0 ;%BC's on bottom
%SOR formula for new psi
psinew=zeros(Ntheta,Nr); %allocating new psi values
bracket=psi
whos
for j=2:Nr-1
for k=2:Ntheta-1
psi(j,k)=((1./rj(j)).*((psi(j+1,k)-psi(j-1,k))./(2.*dr))+((psi(j+1,k)-psi(j-1,k))./(dr.^2))+(1./(rj(j).^2)).*((psi(j,k+1)+psi(j,k+1)))./(dtheta.^2)).*term;
That uses all of term on the right-hand side, and term is a vector, so the right hand side is a vector, but the left-hand side names a scalar destination.
deltapsi=bracket/((2/(dr^2))+(2./((rj.^2).*dtheta^2)))-psi(j,k)
psinew(j,k)=psi(j,k)+w.*deltapsi
end
end
댓글 수: 3
Walter Roberson
2023년 2월 24일
psi(j,k)=((1./rj(j)).*((psi(j+1,k)-psi(j-1,k))./(2.*dr))+((psi(j+1,k)-psi(j-1,k))./(dr.^2))+(1./(rj(j).^2)).*((psi(j,k+1)+psi(j,k+1)))./(dtheta.^2)).*term;
j and k are scalars
rj(j) is a scalar so 1./rj(j) is a scalar.
psi(j+1,k) is a scalar, and so is psi(j-1,k) so the subtraction is a scalar.
dr is a scalar.
So ((1./rj(j)).*((psi(j+1,k)-psi(j-1,k))./(2.*dr)) is operations only on scalars, giving a scalar result.
psi(j+1,k) is a scalar, as is psi(j-1,k) so the subtraction is a scalar. dr is a scalar. rj(j) is a scalar. So everything in ((psi(j+1,k)-psi(j-1,k))./(dr.^2))+(1./(rj(j).^2)) is operations on scalars, so that part is a scalar.
Likewise everything in (1./(rj(j).^2)).*((psi(j,k+1)+psi(j,k+1)))./(dtheta.^2)) is operations on scalars.
So every calculation on the line is a scalar. Until you get to the .*term part. term is a 50x1 vector. So you are multiplying some calculated scalar by a vector, which is going to give you a vector result. And you are trying to store that vector result into a scalar output location.
Perhaps you should be indexing term by something in the calculation, so that you get a scalar for that part?
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