Superellipse and a good coordinate system

by traianus
Tags: coordinate, superellipse
traianus is offline
Mar28-06, 02:30 PM
P: 80
Hello guys, at the link

you can find the definition of superellipse. Now consider the particular super ellipse

[tex]\frac{x^{2n}}{A^{2n}} +\frac{y^{2n}}{B^{2n}} = 1 [/tex]

In which A,B, are constant and n is a positive integer.
What is the coordinate system that has two families of curves in which one represents the superellipse and the other one is perpendicular to it? In the particular case of A = B, n = 1, the coordinate system is

[tex]x = r\cos{\varphi}[/tex]
[tex]y = r\sin{\varphi}[/tex]

If A is different than B but still n =1, the coordinate system is similar but it involves also hyperbolic sine and cosine and one family of curves is the generic ellipse and the other family is the generic hyperbola and they are perpendicular to each other, like it happens in the polar coordinates. So, is there a similar curvilinear coordinate system for the particular superellipse I described?
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kryptyk is offline
Mar30-06, 01:40 AM
P: 38
What about:

[tex]x = A (\cos \varphi)^{\frac{1}{n}}[/tex]
[tex]y = B (\sin \varphi)^{\frac{1}{n}}[/tex]

traianus is offline
Mar30-06, 01:05 PM
P: 80
This is a parametric representation of the superellipse, not a coordinate system. In fact, you have only a "free" parameter [tex]\varphi[/tex] and you must have two parameters, like for polar coordinates, where you have [tex]\varphi[/tex] and [tex]r[/tex].

kryptyk is offline
Mar30-06, 04:14 PM
P: 38

Superellipse and a good coordinate system consider then:

[tex]x = A\, r [\cos \varphi]^{\frac{1}{n}}[/tex]
[tex]y = B\, r [\sin \varphi]^{\frac{1}{n}}[/tex]

from which we then get:

[tex]\frac{x^{2n}}{A^{2n}} +\frac{y^{2n}}{B^{2n}} = r^{2n}[/tex]

Let [itex]\{\mathbf{e}_1, \mathbf{e}_2} \}[/itex] form an orthonormal basis for the rectangular coordinate system and let [itex]\mathbf{w} = x\, \mathbf{e}_1 + y\, \mathbf{e}_2\;[/itex].

So then,

[tex]\mathbf{w} = r (A [\cos \varphi]^{\frac{1}{n}} \mathbf{e}_1 + B [\sin \varphi]^{\frac{1}{n}} \mathbf{e}_2)[/tex]

To find a tangent vector to some superellipse given by fixed [itex]r\,[/itex], we find:

[tex]\frac{\partial \mathbf{w}}{\partial \varphi} = \frac{r}{\sin \varphi \cos \varphi} (-\frac{A\, \sin^2 \varphi}{n} [\cos \varphi]^{\frac{1}{n}} \mathbf{e}_1 + \frac{B\, \cos^2 \varphi}{n} [\sin \varphi]^{\frac{1}{n}} \mathbf{e}_2)[/tex]

And I have no idea where I'm going with this but I'm having fun with the TeX stuff :p
traianus is offline
Mar30-06, 06:17 PM
P: 80
It is not difficult to find a curvilinear coordinate system, the difficulty is to find one in which the two curves that you obtain when you set one parameter constant are perpendicular to each other. See for example the link

From which it appears clear that the two families of curves (ellipses and hyperbolas) are perpendicular in each point they intercept. My desire is to find the analytical expressions for the curve similar to the link I posted. But instead of ellipses I have a particular kind of superellipses in which n is positive integer (see previous equation).
Notice that when you change superellipse, A,B change as well, like in polar coordinate system, when you move from a circle to another one r changes.
traianus is offline
Apr4-06, 01:00 PM
P: 80
Is this topic so difficult? Nobody knows? Please help me if you can!
traianus is offline
Apr6-06, 12:46 AM
P: 80
So far, we obtained the parametric representaion. Nobody knows better? Should I give up? Is not here any expert?
traianus is offline
Apr14-06, 07:19 PM
P: 80
Nobody in this forum is able to help me? Can you please tell me another forum where I can ask the same question?
traianus is offline
Apr23-06, 12:51 PM
P: 80
Come on, please!

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