Why the shape of Pringles potato chip is hyperbolic paraboloid?
I found several articles that say the shape is hyperbolic paraboloid, but cannot find out why it is so. Does anyone have reasonable (and/or mathematical) answers?
[Math] The shape of Pringles potato chip
geometryhyperbolic-geometry
Related Solutions
You've written $$r(u,v)=\langle a(u+v), b(u-v), uv\rangle$$ and that's the same thing as saying $\langle x,y,z\rangle = \langle a(u+v), b(u-v), uv\rangle$, so that $$ \begin{align} x & = a(u+v), \\ y & = b(u-v), \\ z & = uv. \end{align} $$ So $$ \frac{y^2}{b^2} - \frac{x^2}{a^2} = \frac{(b(u-v))^2}{b^2} - \frac{(a(u+v))^2}{a^2}. $$ You can cancel $a$ and $b$ and then do routine simplifications and see if it turns into $-4uv$, which then turns into $-4z$, so that $\frac{y^2}{b^2} - \frac{x^2}{a^2} = \frac{z}{c}$ with $c = -\frac14$.
That shows that the parametrized surface you've got is a subset of the surface defined by the equations involving $x$, $y$, and $z$. Then you need to show that it's the whole set. One way to do that would be by solving for $u$ and $v$ in terms of $x$ and $y$, thereby showing that all $(x,y,z)$ points on the surface actually appear within the surface parametrized by $u$ and $v$.
Lets start at the beginning
There is no 3 dimensional Euclidean surface that represents the complete hyperbolic plane (Hilberts theorem )
There are Surfaces of revolution that have (outside some boundaries ) a constant negative curvature $K$ .
All surfacesof revolution revolving around the $x_3 $ axis can be formulated as:
$ x_1 = \phi(v) \cos u $ , $ x_2 = \phi(v) \sin u $ and $ x_3 = \psi(v) $
For a revolution surfaces that have (outside some boundaries ) a constant negative curvature of $K = -1 $ also needs:
- $(\phi')^2+(\psi')^2 = 1 $ and
- $\phi'' = \phi $
The meridian curve (generatrix ) $ x_1 = \phi(v) , x_2 = 0 $ and $ x_3 = \psi(v) $ is a geodesic on the surface.
combining this with the hyperbolic plane the meridian curve is an hyperbolic line
There are 3 types of these surfaces
- the "pseudo spherical surfaces of the elliptic type":
also called the "conic type" $\phi(v)=C\sinh v$ and $\psi(v)=\int_0^v \sqrt{1-C^2\cosh^2v} dv$
The meridians are hyperbolic lines trough the vertex and the parallels are arcs of hyperbolic circles with the vertex as center.
- the "pseudo spherical surface of the parabolic type":
This is the one that is also known as the pseudosphere or the tractoid
$\phi(v)=e^v$ and $\psi(v)=\int_0^v \sqrt{1-e^{2v}} dv$ or
$\phi(v)= \frac{1}{\cosh v} $ and $\psi(v) = v -\tanh v $
(there are also other formula's)
The meridians are hyperbolic lines asymtopic to each other horoparallel) and the parallels are arcs of horocycles.
- the "pseudo spherical surface of the hyperbolic type":
$\phi(v)=C\cosh v$ and $\psi(v)=\int_0^v \sqrt{1-C^2\sinh^2v} dv$
PS this is not an one sheeted hyperboliod (an hyperboloid doesn't have an constant curvature.)
The meridians are hyperbolic lines athat share a common perpendicular and the parallels are the common perpendicular and arcs of hypercycles equidistant to this common perpendicular) .
I think this is the surface than Riemann describes in his famous Habilationsvortrag, (section II $ 5):
The theory of surfaces of constant curvature will serve for a geometric illustration. It is easy to see that surface whose curvature is positive may always be rolled on a sphere whose radius is unity divided by the square root of the curvature; but to review the entire manifoldness of these surfaces, let one of them have the form of a sphere and the rest the form of surfaces of revolution touching it at the equator. [...] The surface with curvature zero will be a cylinder standing on the equator; the surfaces with negative curvature will touch the cylinder externally and be formed like the inner portion (towards the axis) of the surface of a ring.
pictures of all three types of "pseudosperical surfaces" can be seen at:
virtualmathmuseum.org/Surface/gallery_o.html#PseudosphericalSurfaces
and
demonstrations.wolfram.com/SurfacesOfRevolutionWithConstantGaussianCurvature/
So now back to the question which did Beltrami use?
First of all does it matter?
Surfaces that have the same constant curvature are isometric, and you can take this very literal:
a triangle were the sides have lengths a,b, and c on one surface is congruent to an triangle on one another surface where the sides have lengths. (theorem of Minding 1938))
the different surfaces are just different "cut-outs" of the hyperbolic plane
But besides that I am not sure about it.
Reading Stillwell's translation of Beltrami's "Saggio" in "Sources of hyperbolic geometry" it first reads like he is using the hyperbolic type (this is the only surface where there are easy identifiable orthogonal hyperbolic lines) but later he changes to the tractoid.
Maybe he changed to the tractoid because this is the only one with a easy formulation.
But again does it really matter?
- About the pictures:
The rolled out version to me looks like a small part of the the hyperbolic type if only it was rolled out over an curved surface (a cylinder orthogonal to the lenght of the surface bump up) so that the frills disappear. and become a side of a ring or indeed a big pringle.
Maybe best compare it to a small part of the surface generated in the wolfram demonstration "negative curvature , first type" with the slider at the right and then the yellow side up and taking only a small part of it (around $30^o $ of the big circle)
Hopes this helps
Best Answer
But these are all true in hindsight; the real procedure is to do ridiculous amount of user study and then figure out why it works.
For example why washing machine doors are circular? Why toothbrush manufacturers want to have angles in fact you can easily rely on the technology known as your elbow etc.