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sig = 1; % Source strength % GRID: x = -150:.02:150; y = 0:.02:100; for m = 1:length(x) for n = 1:length(y) xx(m,n) = x(m); yy(m,n) = y(n); % Velocity potential function: phi_Source(m,n) = (sig/4/pi) * log(x(m)^2+(y(n)+.01)^2); % Stream function: psi_Source(m,n) = (sig/2/pi) * atan2(y(n),x(m)); end end % Plots % Source at origin of coordinat systex: contour(xx,yy,psi_Source,(-0.5:.5:5),'k') hold on contour(xx,yy,phi_Source,10,'r') legend('streamlines' ,'potential') title(' Source at origin') axis image hold off

1 个评论

Here is the code properly formatted (you're welcome, Ricardeo):
sig = 1;
% Source strength
% GRID:
x = -150:.02:150;
y = 0:.02:100;
for m = 1:length(x)
for n = 1:length(y)
xx(m,n) = x(m);
yy(m,n) = y(n);
% Velocity potential function:
phi_Source(m,n) = (sig/4/pi) * log(x(m)^2+(y(n)+.01)^2);
% Stream function:
psi_Source(m,n) = (sig/2/pi) * atan2(y(n),x(m));
end
end
% Plots
% Source at origin of coordinate system:
contour(xx,yy,psi_Source,(-0.5:.5:5),'k')
hold on
contour(xx,yy,phi_Source,10,'r')
legend('streamlines' ,'potential')
title(' Source at origin')
axis image
hold off
ProTip: Check out meshgrid to transform it into:
[yy xx] = meshgrid(y,x);
phi_Source = (sig/4/pi) .* log(xx.^2+(yy+0.01).^2);
psi_Source = (sig/2/pi) .* atan2(yy,xx);
and avoid for loops.

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回答(2 个)

Image Analyst
Image Analyst 2017-11-6

0 个投票

It runs for me. Takes like 25 minutes though. What is your question?
Eric
Eric 2017-11-6

0 个投票

Be patient, it just takes a long time to display the plots. For me it took about 2 seconds for the contour functions to return and about 86 seconds for the figure to display. You may want to look into alternative ways to plot, or ways to reduce the number of elements plotted, if you desire speed.

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2017-11-6

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