Superellipse and a good coordinate system


by traianus
Tags: coordinate, superellipse
traianus
traianus is offline
#1
Mar28-06, 02:30 PM
P: 80
Hello guys, at the link

http://mathworld.wolfram.com/Superellipse.html

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
kryptyk is offline
#2
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
traianus is offline
#3
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
kryptyk is offline
#4
Mar30-06, 04:14 PM
P: 38

Superellipse and a good coordinate system


Ok...so 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
traianus is offline
#5
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

http://mathworld.wolfram.com/Ellipti...ordinates.html

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
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#6
Apr4-06, 01:00 PM
P: 80
Is this topic so difficult? Nobody knows? Please help me if you can!
traianus
traianus is offline
#7
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
traianus is offline
#8
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
traianus is offline
#9
Apr23-06, 12:51 PM
P: 80
Come on, please!


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