Finding the streamlines of a complex potential for a Joukowsky transform from a unit circle to a flat plate.

In page 36 of the paper ''On dynamic interactions between body motion and fluid motion by Frank T. Smith, Samire Balta, Kevin Liu and Edward R. Johnson'', there is the equation (65) giving the complex velocity potential. I am taking the imaginary part of this equation to reproduce Figure 18 on the next page.

Here is my Matlab code:

x = -2:0.02:2;

y = -1:0.01:1;

[X, Y] = meshgrid(x, y);

z = X + i.*Y;

zc = X - i.*Y;

a = sqrt(z.^2 - 1);

w = -i.*(z + a + log(1./(z + a)));

f = imag(w)

hold on

[C, h] = contour(X, Y, f, 'k');

y = 0;

line([-1,1],[y,y], 'linewidth', 2, 'color', 'k');

axis equal

hold off

For some reason starting off at x = 0.22 and carrying on the negative x values, I can’t get the same plot as in Figure 18.Can you see where have I gone wrong in my code?

Best regards,

Marialis Simoni

[Stephen23]

https://web.archive.org/web/20201023060240/https://webcache.googleusercontent.com/search?q=cache%3A9uWnmVZWBdAJ%3Ahttps%3A%2F%2Fwww.mathworks.com%2Fmatlabcentral%2Fanswers%2F622318-finding-the-streamlines-of-a-complex-potential-for-a-joukowsky-transform-from-a-unit-circle-to-a-fla

https://web.archive.org/web/20201023071347/https://www.mathworks.com/matlabcentral/answers/622318-got-it-right

Answer 1

[Alan Stevens]

There are some typos in the paper by Smith et al, and some steps left out. I think the following generates figure 18:

x = [-2:0.01:-0.01 0.01:0.01:2];
y = -1:0.01:1;
[X, Y] = meshgrid(x, y);
z = X + 1i.*Y;
zeta = -z + sqrt(z.^2-1);   % Equation (59)
idx = X<0;
zeta(idx) = 1./zeta(idx);
w = 1i.*(1./zeta - log(zeta));  % First line of equation (65)
f = imag(w);
hold on
[C, h] = contour(X, Y, f, 'k');
line([-1,1],[0,0], 'linewidth', 2, 'color', 'k');
axis equal
hold off 

This results in

[Marialis Simoni]

Hi Alan,

Thank you very much for your answer. It really helped a lot. There are some typos I think in the notes. However, if you want to obtain the second line of equation (65), you multiply by sqrt(z^2-1) +1 the expression for \zeta. Also the expression inside the log is log(1/zeta), because there is a minus in front of the zeta.

[Marialis Simoni]

Hi Alan,

I have one more question. Why do you leave out the x = 0 value, when defining vector x? Also why is zeta = 1/zeta for the negative values of x?

Please let me know.

[Rik]

Why should he bother answer your question? You removed your question body, so it is clear you have no respect for him or this forum.

[Marialis Simoni]

Dear Rik,

Thank you for sharing your concern. It would be good for you to identify yourself as well. You do have a surname, don't you?

First things first, it's the first time I am using this forum and I'm not that familiar with the rules. I did remove the question, because I'm writing a thesis and I would not want my work to be stolen. I do need help with the coding time after time, as I get stuck sometimes when debugging the code.

I thought about it, and maybe when asking questions with regards to code debugging I should spare some details.

Also, I thought when I write my thesis, all of this will be on the web, therefore plagiarism softwares will detect it as plagiarism, when it is my work.

Do you think my reasoning counts?

Please let me know about your thoughts.

Best regards,

Marialis Simoni

[Stephen23]

" I did remove the question, because I'm writing a thesis and I would not want my work to be stolen"

By using MATLAB Answers you agreed to the terms and conditions given here:

https://www.mathworks.com/matlabcentral/termsofuse.html#answers

where it is clearly stated that "You agree that all Content that you contribute to the MATLAB Answers portion of MATLAB Central will be licensed under the Creative Commons Attribution Share Alike 3.0 license", which means that already given everyone on the entire planet permission to copy, distribute, and use your code (with the appropriate attribution of course).

[Rik]

I don't think you require my surname to have a productive conversation. You can find or deduce it if you look around a bit. It is not secret, but I do keep it slightly more private than I used to.

If you don't want your work public, you shouldn't be using a public forum. When you posted your question you agreed to the terms of use, which state that you publish everything on Matlab Answers under CC-by-sa 3.0. That means you already gave permission to everybody interested to use your work. If you need private consultation you should hire a private consultant.

I don't think plagiarism detectors will pose a problem, as you seem to be using your actual name. From what I have heard, most institutions will have a manual look if their software finds something. If they don't you can fight any accusation of plagiarism with great ease when it comes to this.

So, no, I don't think your reasons are enough to warrant deletion.

Also, the site admins are generally able to restore the original post, and we can also re-post it from the Google cache capture that Stephen posted.

[Alan Stevens]

@Rik In fact I had already answered. However, because I did so on my phone with its limited view, I inadvertently replied as a second answer rather than as a comment!

@Marialis Your question could not possiby be considered to be original research as it simply asked to reproduce a previously published result.

[Marialis Simoni]

Hi everyone,

I reposted the question. Hope everything is fine now.

Also, Alan I meant what I wrote in a future context, when I produce my own original figures.

Thanks everyone for your constructive criticism.

Answer 2

[Alan Stevens]

x=0:. There was a strange vertical line that O thought was related to this. However, you could put it back and see if it makes any difference.

It was clear from the plot without inverting zeta for negative X that the contours were inverted. I don't have a technical justification for it!

[Marialis Simoni]

Thank you very much.