Oloid

Formulas for the Oloid

Number of corners	${\displaystyle 0}$
Number of edges	${\displaystyle 2}$
Edge length	${\displaystyle {\tfrac {4}{3}}\pi r}$
Number of faces	${\displaystyle 1}$
surface	${\displaystyle 4\pi r^{2}}$
volume	${\displaystyle \approx 3{,}05r^{3}}$
Radius of the generating circles	${\displaystyle r}$
Side length of the dismantled cube

Structure of the oloid

The oloid (also known as a polysomatoloid ) is a geometric body that was discovered in 1929 by the sculptor and mechanical engineer Paul Schatz along with the evertable cube . It can be defined as the convex hull of two circles of the same size, intersecting perpendicularly , whose centers are at a distance from one another that is equal to their radius. It has no corners, two edges, namely a 240 ° arc of the two intersecting circles, and is otherwise smooth . It has properties that clearly distinguish it from other bodies and is considered a plausibility indicator for the inversion kinematics established by Schatz.

context

Evertable cube
(6 tetrahedra with red outer sides)
The ends of the diagonal (white line of constant length) move back and forth on two crossed circular arcs (blue and red), the line itself moves on a ruled surface .

Oloid model: 2 circular discs made of cardboard and string (= set of straight lines on the regular surface )

Paul Schatz discovered in the 1920s, a decomposition of the cube into three parts, one of which is irregular six tetrahedra is. If you connect these in an articulated manner at their two adjacent edges in the cube, a chain that can be completely turned inside out is created.

The extended chain has three diagonals of equal length between opposite joints. These are the spatial diagonals of the original cube, which are retained during the eversion and thus have a constant length. Schatz observed the path that such a diagonal takes when the chain was turned inside out, and discovered the oloid in the process. If one fixes one of the tetrahedra and observes the path of the diagonal opposite it (illustration on the left), one recognizes that the surface it sweeps over is a ruled surface and the surface of a geometric body, which Schatz called the Oloid.

The first description of the mathematical properties from an analytical point of view took place in 1997.

The oloid is part of the oloid stirrer, which is used to circulate and aerate water, e.g. B. is used in wastewater treatment and water remediation. Another form of application as an alternative to the ship propeller has not yet been able to exceed the prototype and test stage.

properties

The trace that an oloid leaves when rolling is like its network

The oloid is one of the few known bodies that roll over its entire surface . Its surface as a whole is a developable surface . In contrast to the cone or cylinder , the entire surface of the oloid (and not just a lateral surface ) can be produced from a single piece of cardboard without any kinks .

If you put it on a slope , it rolls down in a tumbling motion without ever rumbling over its edges. It is noteworthy that the surface is exactly as large as that of a sphere , which has the same radius as the two circles that create the oloid.

The angle at the midpoints of the edges is 60 °. If you look at the oloid perpendicular to the two edges, the contours form an exact square in the cross-section , which makes it possible to assess the quality of handcrafted olloids, since slight asymmetries are quickly recognized.

mathematics

The color gradient illustrates the position of the connecting lines between the edges for the entire parameter range of ${\displaystyle t}$

In addition, let the radius of the generating circles be. The two edges each have a length of . The surface is a ruled surface: for each point there is exactly one point (apart from reflection) on the other edge, so that the connecting line lies completely on the surface of the oloid. The length of this line is for all points , just the length of the three space diagonals of the tetrahedron chain and the decomposed cube, which thus has a side length of . ${\displaystyle r}$ ${\displaystyle {\tfrac {4}{3}}\pi r}$ ${\displaystyle x}$ ${\displaystyle y}$ ${\displaystyle {\sqrt {3}}r}$ ${\displaystyle r}$

The side length of the above-mentioned square, which the contours form in a certain viewing angle, is what the dimensions of the minimal cuboid that includes the oloid . ${\displaystyle {\sqrt {2}}r}$ ${\displaystyle 3r\cdot {\sqrt {2}}r\cdot {\sqrt {2}}r}$

construction

For embedding in three-dimensional Euclidean space, place the center of the lying circle on the origin , that of the standing circle . Thus for the point on the lying edge is given by and . The Pythagorean theorem then delivers the two points on the standing edge that are at a distance of : with and . Depending on the sign , this is a point on the upper or lower half of the oloid. For theoretical considerations, due to the symmetries in the oloid, it is possible to restrict the parameter range from to, for example (that is to say to a quarter of the surface and further to an eighth by defining the sign in ). This can also be useful for visualization. This bypasses the singular behavior of some of the relevant functions at the interval limits, i.e. the end points of the lying edge. ${\displaystyle (r,0,0)}$ ${\displaystyle t\in \left[{\tfrac {\pi }{3}},{\tfrac {5\pi }{3}}\right]}$ ${\displaystyle x=(x_{1},x_{2},0)}$ ${\displaystyle x_{1}=r\cos {t}}$ ${\displaystyle x_{2}=r\sin {t}}$ ${\displaystyle x}$ ${\displaystyle {\sqrt {3}}r}$ ${\displaystyle y=(y_{1},0,\pm y_{3})}$ ${\displaystyle y_{1}={\tfrac {r}{1-\cos {t}}}}$ ${\displaystyle y_{3}={\tfrac {r{\sqrt {1-2\cos {t}}}}{1-\cos {t}}}}$ ${\displaystyle t}$ ${\displaystyle \left[\pi ,{\tfrac {3\pi }{2}}\right]}$ ${\displaystyle y_{3}}$

Parameterization of the surface

With the help of the straight line equation , the following parameterization of the surface is reached: with ${\displaystyle x_{i}+s(y_{i}-x_{i})}$ ${\displaystyle \Phi \colon [0,1]\times \left[{\tfrac {\pi }{3}},{\tfrac {5\pi }{3}}\right]\to \mathbb {R} ^{3},}$ ${\displaystyle \Phi (s,t)=(\Phi _{1},\Phi _{2},\pm \Phi _{3})}$

${\displaystyle \Phi _{1}=r\left({\frac {s}{1-\cos {t}}}+(1-s)\cos {t}\right)}$

${\displaystyle \Phi _{2}=r(1-s)\sin {t}}$

${\displaystyle \Phi _{3}=r{\frac {s{\sqrt {1-2\cos {t}}}}{1-\cos {t}}}}$

For this is a point on the lying edge, for on the standing edge . A coordinate representation is given by the algebraic surface below . ${\displaystyle s=0}$ ${\displaystyle s=1}$

Parameterization of the volume

We obtain a parameterization for the full body of the Oberflächenparametrisierung by only a height parameter multiplied. With ${\displaystyle \Phi _{3}}$ ${\displaystyle h\in [0,1]}$ ${\displaystyle \Psi \colon [0,1]\times [0,1]\times \left[{\tfrac {\pi }{3}},{\tfrac {5\pi }{3}}\right]\to \mathbb {R} ^{3},}$ ${\displaystyle \Psi (h,s,t)=(\Psi _{1},\Psi _{2},\pm \Psi _{3})}$

${\displaystyle \Psi _{1}=\Phi _{1}}$

${\displaystyle \Psi _{2}=\Phi _{2}}$

${\displaystyle \Psi _{3}=h\,\Phi _{3}}$

For this gives the surface, for the horizontal section through the center of the oloid. It should be noted that part of the symmetries breaks, which is why the definition range can be restricted to only half (and no longer to a quarter). ${\displaystyle h=1}$ ${\displaystyle h=0}$ ${\displaystyle \Psi }$ ${\displaystyle t}$

Surface area

The size of the surface can be calculated exactly with the surface integral. To do this, one forms the Euclidean amount of the cross product of the six partial derivatives of the surface parameterization and integrates this according to and . It turns out that the surface is just the size of - just like a sphere of radius . ${\displaystyle s}$ ${\displaystyle t}$ ${\displaystyle 4\pi r^{2}}$ ${\displaystyle r}$

With the above parameterization of the surface and the restrictions mentioned, the following results for the surface content : ${\displaystyle F}$

${\displaystyle F=8\int _{\pi }^{\frac {3\pi }{2}}\int _{0}^{1}\left|{\frac {\partial \Phi }{\partial s}}\times {\frac {\partial \Phi }{\partial t}}\right|\,ds\,dt=}$

${\displaystyle 8\int _{\pi }^{\frac {3\pi }{2}}\int _{0}^{1}{{\sqrt {2}}r^{2}{\frac {(3s-2)\cos {t}+1}{\sqrt {(1-2\cos {t})(1-\cos {t})}}}}\,ds\,dt=}$

${\displaystyle 8r^{2}\int _{\pi }^{\frac {3\pi }{2}}{\frac {\sqrt {2}}{\sqrt {(1-2\cos {t})(1-\cos {t})}}}\underbrace {\int _{0}^{1}(3s\cos {t}-(2\cos {t}-1)\,ds} _{={\tfrac {1}{2}}\left(2-\cos {t}\right)}\,dt=}$

${\displaystyle 8r^{2}\int _{\pi }^{\frac {3\pi }{2}}{\frac {2-\cos {t}}{\sqrt {2(1-2\cos {t})(1-\cos {t})}}}\,dt=}$

${\displaystyle 8r^{2}\int _{0}^{\frac {\pi }{2}}{\frac {2+\cos {t}}{\sqrt {2(1+2\cos {t})(1+\cos {t})}}}\,dt=}$

${\displaystyle 8r^{2}\left(\arctan {\sqrt {2}}+\arcsin {\tfrac {1}{\sqrt {3}}}\right)=}$

${\displaystyle 8r^{2}\left(\arctan {\sqrt {2}}+\arctan {\tfrac {1}{\sqrt {2}}}\right)=8r^{2}\,{\frac {\pi }{2}}=4\pi r^{2}.}$

The integral transformation is based on which one obtains an antiderivative in which only two terms remain with the corresponding limits. The following applies to arcsine : (da ) and the last step is the functional equation of the arctangent . ${\displaystyle \cos {x}=-\cos {\left(x\pm \pi \right)}}$ ${\displaystyle \arcsin {x}=\arctan {\tfrac {x}{\sqrt {1-x^{2}}}}}$ ${\displaystyle |x|<1}$

Volume content

In contrast to this, every previously known volume formula for the oloid contains several elliptical integrals that can only be evaluated numerically . When analytical approach with the volume integral of the amount of the Jacobian of the Volumenparametrisierung the choice of providing for a simplification in the first steps: Since only by dependent, two of the partial derivatives equal to zero. This means that two thirds of the terms in the determinant calculation are omitted , and in particular none appear. The determinant is always positive within the limits and therefore equal to its amount. ${\displaystyle \Psi }$ ${\displaystyle \Psi _{3}}$ ${\displaystyle h}$ ${\displaystyle h}$

${\displaystyle V=4\int _{\pi }^{\frac {5\pi }{3}}\int _{0}^{1}\int _{0}^{1}\left|\det \,D\Psi \right|\,dh\,ds\,dt=}$

${\displaystyle 4r^{3}\int _{\pi }^{\frac {5\pi }{3}}{\frac {\sqrt {1-2\cos {t}}}{\left(1-\cos {t}\right)^{2}}}\underbrace {\int _{0}^{1}s\left(\left(3s-2\right)\cos {t}+1\right)\overbrace {\int _{0}^{1}1\,dh} ^{=1}\,ds} _{={\tfrac {1}{2}}}\,dt=}$

${\displaystyle 2r^{3}\int _{\pi }^{\frac {5\pi }{3}}{\frac {\sqrt {1-2\cos {t}}}{\left(1-\cos {t}\right)^{2}}}\,dt=}$ ${\displaystyle 2r^{3}\int _{0}^{\frac {2\pi }{3}}{\frac {\sqrt {1+2\cos {t}}}{\left(1+\cos {t}\right)^{2}}}\,dt=}$

${\displaystyle 2r^{3}\,\cdot {\tfrac {1}{\sqrt {3}}}\,\left(F({\tfrac {\pi }{3}}\,|\,{\tfrac {4}{3}})+E({\tfrac {\pi }{3}}\,|\,{\tfrac {4}{3}})\right)=}$

${\displaystyle 2r^{3}\,\left({\tfrac {1}{2}}K({\tfrac {3}{4}})+{\tfrac {2}{3}}E({\tfrac {3}{4}})-{\tfrac {1}{6}}K({\tfrac {3}{4}})\right)=}$

${\displaystyle {\tfrac {2}{3}}r^{3}\left(K({\tfrac {3}{4}})+2E({\tfrac {3}{4}})\right)}$

${\displaystyle \approx 3{,}0524\;r^{3}}$

The incomplete elliptic integrals of the first and second kind ( and ) can be expressed by the corresponding complete elliptic integrals ( and ), because the arguments are connected via the arccosecans . ${\displaystyle F}$ ${\displaystyle E}$ ${\displaystyle K}$ ${\displaystyle E}$

The irrational constant 3.052418468… can be calculated as precisely as you like, but no algebraic connections to other constants are known, and neither is it known whether it is transcendent .

The oloid surface

The oloid can be seen as part of an algebraic surface of degree 8 ( i.e. an octic ). The solution set of the defining polynomial equation provides the surface of an oloid with a radius , embedded in the three-dimensional space with the coordinate axes and , the center of the surface is enclosed . However, the restrictive constraints in order to obtain the oloid exclusively are not trivial. The polynomial equation consists of 48 terms with exclusively integer coefficients , the maximum of the exponent sums of the monomials is 8 and there is no constant term. If you replace with , the area on the -axis is shifted so that the center is at the zero point. ${\displaystyle O}$ ${\displaystyle r=1}$ ${\displaystyle x,y}$ ${\displaystyle z}$ ${\displaystyle ({\tfrac {1}{2}},0,0)}$ ${\displaystyle x}$ ${\displaystyle (x-{\tfrac {1}{2}})}$ ${\displaystyle x}$

${\displaystyle O\colon x^{8}-3y^{8}-3z^{8}-6x^{4}y^{4}}$

${\displaystyle -8x^{2}y^{6}-6x^{4}z^{4}-8x^{2}z^{6}+6y^{2}z^{6}}$

${\displaystyle +12x^{2}y^{2}z^{4}-9y^{4}z^{4}+6y^{6}z^{2}+12x^{2}y^{4}z^{2}}$

${\displaystyle +6x^{4}y^{2}z^{2}+4x^{7}+12x^{3}y^{4}+4xy^{6}}$

${\displaystyle -20xz^{6}-36x^{3}z^{4}+12x^{3}y^{2}z^{2}+24xy^{4}z^{2}}$

${\displaystyle -12x^{5}z^{2}+12x^{5}y^{2}+2x^{6}+10y^{6}}$

${\displaystyle -2z^{6}+22x^{2}y^{4}-46x^{2}y^{2}z^{2}-50x^{2}z^{4}}$

${\displaystyle -12y^{2}z^{4}-46x^{4}z^{2}+14x^{4}y^{2}-8x^{5}}$

${\displaystyle -8xy^{4}-8xz^{4}-52xy^{2}z^{2}-48x^{3}z^{2}}$

${\displaystyle -16x^{3}y^{2}-7x^{4}-11y^{4}+z^{4}}$

${\displaystyle -18x^{2}y^{2}-6x^{2}z^{2}-10y^{2}z^{2}+4x^{3}}$

${\displaystyle +4xy^{2}+4xz^{2}+4x^{2}+4y^{2}=0}$

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  The oloid as part of an algebraic surface of the 8th degree

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  ... zoomed out ...

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  ... zoomed out even further

literature

• Paul Schatz: Rhythm research and technology 3rd edition under the title: The world can be turned inside out: Rhythm research and technology. niggli Verlag 2008
• Spectrum of Science: Mathematical Conversations III, Article: A Journey into the Realm of the Cube , pages 12-17, Dossier 2/2004
• brand eins: Issue 12, 2017, pages 120–127 The secret of the inverted cube

Web links

Commons : Oloid  - collection of images, videos and audio files

• Franz Zahaurek: The Oloid according to Paul Schatz
• Jürgen Meier: further details and visualizations
• Mathematical context for the Oloid with illustrations (French)

Individual evidence

1. ↑ Hellmuth Stachel and Hans Dirnböck: The Development of the Oloid , Heldermann-Verlag, 1997 ( PDF )
2. ↑ Krajewski-Pumpentechnik, 2. Pond aeration / circulation (Oloid stirrer) Archived copy ( memento of the original from June 11, 2011 in the Internet Archive ) Info: The archive link was inserted automatically and not yet checked. Please check the original and archive link according to the instructions and then remove this notice. 
3. ↑ Unrolling an Oloid , YouTube Video
4. ↑ Calculation of the antiderivative to the surface with Wolfram's online integral calculator
5. ↑ Calculation of the antiderivative to the volume with Wolfram's online integral calculator
6. ↑ The identities used for F and E on wolfram.com (English)
7. ↑ Oloid and context (French)