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November 8, 2000, 12:23 |
conformal transformations
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#1 |
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Trying to transform an airfoil into a near circle using the inverse of the joukowski formula x = z + a^2/z where x,z complex , "a" effectively constant and getting stupid results. Any suggestions?
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November 8, 2000, 12:46 |
Re: conformal transformations
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#2 |
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November 10, 2000, 09:37 |
Re: conformal transformations
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#3 |
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Thanks Andy, you seem to have a certain knowledge about this, have you ever come across a reverse solution for the specific problem of the inverse of x= 1/2(z+1/z)?
The site was helpful regardless. Alex |
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November 10, 2000, 12:28 |
Re: conformal transformations
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#4 |
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I have some experience with the inverse transformation of Karman-Treffz airfoils. I encountered a difficulty in getting a circle for highly cambered airfoil. I found that the mapping is not one-to-one inside a certain lense shape. So, if the airfoil is so cambered that a part of it (typically, lower surface, near trailing edge) will be contained in this lense shape, then the points in there will not be mapped back onto the circle from which the airfoil was generated. So, going from circle to airfoil is fine, but its inverse is not necessarily fine.
I don't know how to get around this yet. Do you have a problem for cambered airfoil? |
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November 10, 2000, 12:36 |
Re: conformal transformations
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#5 |
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Sorry. I last looked at conformal transformation nearly 20 years ago and most things are pretty vague now - if not completely forgotten!,
Andy. |
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November 10, 2000, 13:58 |
Re: conformal transformations
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#6 |
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Thanks Nishikawa, I haven't come across Karman-Treffz airfoils but I assume the transform for them is similar that of Joukowsky's. What method did you use to transform it to the circle? I'll look into the discontinuities to see if I can give you any input,
Alex |
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November 10, 2000, 20:51 |
Re: conformal transformations
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#7 |
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Karman-Treffz airfoil is more general than Joukowsky's, it generates airfoils with trailing edge of finite angle. In both cases, a circle is mapped to an airfoil. Inverse transformation will map the airfoil to a circle.
So, in the case of Joukowsky, x = z + a^2/z which transforms a circle in z-plane to an airfoil in x-plane, if we have a solution for a circle, we get a solution for an airfoil in x. This transformation can be written also as (x-2a)/(x+2a)=(z-a)^2/(z+a)^2. You can solve this for z easily, and get the inverse transformation: z = a [sqrt(x+2a)+sqrt(x-2a)]/[sqrt(x+2a)-sqrt(x-2a)]. This is what I use to transform an airfoil into a circle. It sounds like what you're saying is something different... |
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November 14, 2000, 14:59 |
Re: conformal transformations
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#8 |
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Nishikawa, I think what you have given me is the solution to what I'm looking for but in practice I cannot get it to produce a circle in the z plane.
Do you have any suggestions as to books I could find this reverse transfrom in to be used in practicle applications? |
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November 14, 2000, 19:54 |
Re: conformal transformations
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#9 |
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I don't know any books that describe the inverse transformation.
I wonder why you can't produce a circle. What is the problem? What do you get? As I mentioned, it has a problem when the sirfoil has large camber, but it works for others. But of course it does not transform an arbitrary airfoil into a perfect circle. |
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November 15, 2000, 04:59 |
Re: conformal transformations
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#10 |
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I can produce an airfoil fine but when I reverse transform it it comes out as a smaller shifted version of the original airfoil. To get this I got z in terms of x using a quadratic to solve for z. Is this an incorrect procedure in general?
With your solution I am getting similar results. should I be applying it by replacing x and z with complex numbers in the form of e.g. u+iv? Thanks for you help in the matter |
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November 16, 2000, 03:57 |
Re: conformal transformations
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#11 |
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Well, I never had such a problem. The inverse transformation formula I wrote down worked for me. I wrote a code to do the transformation in FORTRAN 77 and 90 which handle complex numbers. I don't have to break z=u+iv into u and v. I use the formula as it is.
Good Luck! |
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