Plotting Inclined Rankine Oval

Question

tag 2020-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/519951-plotting-inclined-rankine-oval

评论： Omer 2022-10-15

I'm trying to plot Inclined Rankine Oval in Matlab. I use quite simple code:

[x,y] = meshgrid(-25:0.1:25); 
m=5; 
a=2; 
U=1; 
figure (2) 
psi = U*(y*cos(pi/4)-x*sin(pi/4))-m*atan2(y,x-a)+m*atan2(y,x+a)-m*atan2(y+4,x-7)+m*atan2(y+8,x-3); 
contour(x,y,psi,100); 
grid on;

For comparison, I also plotted horizontal Rankine Oval. Please, see the picture. I don't understand what are these horizontal lines on inclined Rankine Oval. And how I get rid of them?

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tag 2020-4-22

I know, I can plot horizontal Oval with horizontal stream without a problem. But when I try to plot inclined Oval with inclined stream there is a problem.

Omer 2022-10-15

Can u send me the code for the horizontal one?

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Answer 1

David Goodmanson 2020-4-22

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https://ww2.mathworks.cn/matlabcentral/answers/519951-plotting-inclined-rankine-oval#answer_427740

编辑：David Goodmanson 2020-4-22

在 MATLAB Online 中打开

Hi tag,

Atan2 is restricted to -pi < atan <= pi, so when the angle increases past pi you get a jump down to -pi That occurs when x is negative and y passes through 0 on the negative x axis. If you consider x and y as z = x + iy in the complex plane, those lines are branch cuts and Matlab is doing a good job of showing them.

When both points have the same y value, the branch cuts cancel out to the left of the second point and you don't see any discontinuity there. (The branch cut remains in between the two points). With different y values you see both branch cuts.

A good way to get out of this is to go to complex notation. The psi2 calculation below puts the original code into the complex domain. The angle function does basically the same thing as atan2, so by using two angle functions psi2 comes out the same as psi1.

The psi3 version divides two complex functions to find the difference in angle between the the two of them, then finds that angle. This effectively allows the two branch cuts to cancel, and the resulting plot has no discontinuities, except on the line between the two points.

[x,y] = meshgrid(-25:0.1:25); 
m=5; 
%a=2; 
U=1; 
p1 = [3 -8];
p2 = [7 -4];
psi1 = U*(y*cos(pi/4)-x*sin(pi/4))-m*atan2(y-p2(2),x-p2(1))+m*atan2(y-p1(2),x-p1(1)); 
figure(1)
contour(x,y,psi1,100);    % original result 
colorbar
grid on;
z = x+i*y;
z1 = p1(1) + i*p1(2);
z2 = p2(1) + i*p2(2);
u = (z2-z1)/abs(z2-z1);   % direction vector on the unit circle
psi2 = imag(z/u) -m*angle(z-z2) + m*angle(z-z1); 
figure(2)
contour(x,y,psi2,100);    % same result
colorbar
grid on;
psi3 = imag(z/u) -m*angle((z-z2)./(z-z1)); 
figure(3)
contour(x,y,psi3,100);   % new result
colorbar
grid on;

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David Goodmanson 2020-4-23

编辑：David Goodmanson 2020-4-23

I don't see how that could make a difference. For me, plots 1 and 2 are identical because phi1 and phi2 are identical, within 1e-15 or values like that. So that is the first thing to check. Have you possibly, earlier on, assigned to 'i' a value different than sqrt(-1) ? Although I think it's kind of ugly notation, one way to guard against that is to replace i with 1i in the definition of z1 and z2.

tag 2020-4-23