[math-fun] Low Rollers: Bodies with Minimal Constant Width and Tetrahedral Symmetry
Bodies with Minimal Constant Width and Tetrahedral Symmetry ___________________________________________________________ Fred Lunnon, Jan 2013 Ernst Meissner was an eminent applied mathematician in his day (1912). On first encountering this topic --- some 30 years ago I suppose, perhaps excusing numerous recent lapses of memory --- I wondered why a good engineer would be satisfied with something as ugly as the asymmetrical solutions now given his name. Failing to to appreciate the part played by minimal surface area and volume in his investigations --- discussed by Kawohl & Weber --- and preferring full tetrahedral symmetry, in I had waded; just as Patrick Roberts has, and no doubt a few more folk besides. Failing somehow to persuade my construction to hang together, I eventually was obliged to put it aside. This time round, Roberts has proved doughtier; sadly however, despite provision of a plethora of painstakingly prepared pictures, his report manages to follow such an circuitous logical path, that just having a another shot myself looked easier than attempting to check his. The process proves so gratifyingly straightforward that I can no longer imagine what my earlier difficulty could have been --- which must after all be preferable to the alternative! Stage (O): inflate the 4 faces of the unit-edge tetrahedron out to congruent caps of unit-radius spheres, meeting the neighbouring vertices, but with great circle boundaries to be specified later. The gaps over the 6 edges are to be filled by congruent lunes of cyclides: envelopes of pencils of spheres in two orthogonal ways, having circles of principal curvature, and 9 degrees of freedom. Stage (A): nail down the cyclide. Take a section through a plane of symmetry of the tetrahedron, and inflate the isosceles triangle out to constant width. Attach arcs of radius 1 over two sides, centred on triangle corners; and arcs of radius b over the base, c over the opposite corner, both with centre P at distance d along the altitude and outside the corner. [The disarmingly detailed diagram on page 2 of Roberts assumes without justification that d = 0.] Inspection of the lines joining P to the mid-point and corners of the base, together with Pythagoras and constant unit width, establishes b + c = 1 ; 2(c + d) + rt2 / 2 = 1 ; b^2 = 1/4 + (d + rt2 / 2)^2 ; which solve easily yielding c = d = (2 - rt2) / 8 ; b = 1 - d = (6 + rt2) / 8 . In particular, the completed small circle actually passes through the nearby triangle corner. It is a little harder to justify an assumption that the cyclide has pinch points at tetrahedron vertices, but we're going to assume that anyway. Now the cyclide is determined, and furthermore the tetrahedron edge lies on its surface --- a bonus feature suggesting that we're on the right track. Stage (B): take x-axis along the base, z-axis along the altitude, y-axis along the edge through the opposite corner. Using Pythagoras, the pencil sphere with centre (x, 0, z) and radius f(x) satisfies f(x) = z ; x^2 + (z + rt2 / 2 + d)^2 = (b - z)^2 ; whence f(x) = (1 - 4 x^2) d . Stage (C): consider a general diameter joining points on cyclides at opposite tetrahedral edges. This passes through the centres (x, 0, f(x)), (0, y, -f(y) - rt2 / 2) of pencil spheres tangent to the envelope at those points. By Pythagoras, the distance between their centres equals the square root of x^2 + y^2 + (f(x) + f(y) + rt2 / 2)^2 = (1 - f(x) - f(y))^2 , using f(x) definition, with (f(x) + f(y))^2 conveniently cancelling. Hence the width along that diameter --- the distance between cyclide surface points on the spheres --- equals constant 1 . Stage (D): inspect the boundary between cyclide and unit radius spherical cap over a tetrahedral face. This is a longtitudinal circle on the cyclide at x = +/- 1/2 with radius 1 - f(1/2) = 1 , coincident with a great circle of the sphere passing through the tetrahedral vertices. Furthermore cyclide and sphere meet tangentially, since both are perpendicular to radii from the vertex; so the composite surface remains differentiable, away from vertices. Bingo! Now what about the Minkowski mean of a pair of non-congruent Meissner bodies, mentioned by Kawohl & Weber? Elementary properties of Minkowski sums suggest --- now WFL has at last got his head around some, courtesy of strenuous combined efforts by DA and WDS --- that this also has minimal constant width, and composes spheres and cyclides as before. Any local alteration in curvature of such a surface would entail some counter-balancing alteration at its opposite side, in order to maintain the width at the expense of destroying the symmetry. Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist? The singularities at the vertices may be avoided via expansion to a parallel surface at the cost of sacrificing minimal width, the larger surface composed of segments from 8 spheres and 6 cycloids. Though differentiable, this solution is less elegant than the analytical surface proposed by Fillmore. It would be nice to see analytical parametric forms or implicit equations for the latter, provided some heroic geek is prepared to decipher the spherical harmonic technicalities involved! [The Asimov discussion thread critique of the Fillmore n-space simplex corollary is inapplicable here, since we already are in possession of a continuous body for submission to Fillmore's finite analytic smoothing procedure.] References contributed by various math-fun sources: General page: en.wikipedia.org/wiki/Reuleaux_tetrahedron Discussion thread: http://mathoverflow.net/questions/54252/are-there-smooth-bodies-of-constant-... Newslist thread: [math-fun] rounded tetrahedron Demo video: http://www.youtube.com/watch?v=jYf3nOYM_mQ Patrick Roberts "Proof of Constant Width of Spheroform with Tetrahedral Symmetry" Corvallis Oregon, (August 2012) http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf Bernd Kawohl & Christof Weber "Meissner's Mysterious Bodies" Mathematical Intelligencer vol 33 (2011) 94--101 http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf Jay P. Fillmore "Symmetries of surfaces of constant width" J. Differential Geom. vol 3 (1969) 103--110. intlpress.com/JDG/archive/1969/3-1&2-103.pdf Thomas Lachand-Robert & Edouard Oudet "Bodies of constant width in arbitrary dimension" Mathematische Nachrichten vol. 7 (2005) 740--750 http://www.lama.univ-savoie.fr/~lachand/pdfs/spheroforms.pdf ___________________________________________________________
Despite initially having been hazy about the details of Minkowski sums, it turns out that my intuition was not quite so wonky as earlier appeared! Lemma: The Minkowski sum of a surface and a sphere of radius r is (congruent to) the parallel surface at distance r ; proof elementary. Consider the mean M_T = (M_V (+) M_F)/2 , the summands positioned so that all three coincide in the spherical caps over each tetrahedral face, and the mean has the symmetry of a regular tetrahedron. Section it along a plane of symmetry: over each edge, the mean composes a circular arc of small radius e from one summand, with a pair of arcs of unit radius from the other meeting in nodal line; at their free ends, the small and unit arcs touch tangentially. * * * * * * * * * * * * * * * * * * * * * * * * * By the lemma, the sum is parallel to the unit arcs at distance e : rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the corresponding portion of the Roberts section is a single arc of radius d = (2 - rt2) / 8 . Finally, the Minkowski and Roberts bodies are both minimal constant-width, with tetrahedral symmetry, but incongruent. The screed is already in need of updating ... comments are invited. While it suffers from an absence of diagrams, that on page 2 of Roberts report provides a partial substitute. In that connection I have already spotted one misprint: Stage (A): for "d = 0" read "d = c" Fred Lunnon On 1/12/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist?
Groan ... Below should have read: " whereas the corresponding portion of the Roberts section initially continues from the faces as unit-radius arcs, capped by a third arc of radius d = (2 - rt2) / 8 . " WFL On 1/14/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
Despite initially having been hazy about the details of Minkowski sums, it turns out that my intuition was not quite so wonky as earlier appeared!
Lemma: The Minkowski sum of a surface and a sphere of radius r is (congruent to) the parallel surface at distance r ; proof elementary.
Consider the mean M_T = (M_V (+) M_F)/2 , the summands positioned so that all three coincide in the spherical caps over each tetrahedral face, and the mean has the symmetry of a regular tetrahedron. Section it along a plane of symmetry: over each edge, the mean composes a circular arc of small radius e from one summand, with a pair of arcs of unit radius from the other meeting in nodal line; at their free ends, the small and unit arcs touch tangentially.
* * * * * * * * * * * * * * * * * * * * * * * * *
By the lemma, the sum is parallel to the unit arcs at distance e : rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the corresponding portion of the Roberts section is a single arc of radius d = (2 - rt2) / 8 .
Finally, the Minkowski and Roberts bodies are both minimal constant-width, with tetrahedral symmetry, but incongruent.
The screed is already in need of updating ... comments are invited. While it suffers from an absence of diagrams, that on page 2 of Roberts report provides a partial substitute. In that connection I have already spotted one misprint:
Stage (A): for "d = 0" read "d = c"
Fred Lunnon
On 1/12/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist?
Hi Fred, I suspect the Minkowski mean of Meissner tetrahedra is the same constant width solid Roberts describes. If so, your argument below must have a flaw. When you say
rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the
you seem to assume that the "arc" length of radius (1 + e)/2 in your cross section is nonzero. I believe it must be zero. If you think it's nonzero, look carefully at the diameters normal to that arc segment, and where the trajectory of the opposite end of those diameters would have to lie. Best, - Scott
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun- bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Monday, January 14, 2013 9:05 AM To: math-fun Subject: Re: [math-fun] Low Rollers: Bodies with Minimal Constant Width and Tetrahedral Symmetry
Despite initially having been hazy about the details of Minkowski sums, it turns out that my intuition was not quite so wonky as earlier appeared!
Lemma: The Minkowski sum of a surface and a sphere of radius r is (congruent to) the parallel surface at distance r ; proof elementary.
Consider the mean M_T = (M_V (+) M_F)/2 , the summands positioned so that all three coincide in the spherical caps over each tetrahedral face, and the mean has the symmetry of a regular tetrahedron. Section it along a plane of symmetry: over each edge, the mean composes a circular arc of small radius e from one summand, with a pair of arcs of unit radius from the other meeting in nodal line; at their free ends, the small and unit arcs touch tangentially.
* * * * * * * * * * * * * * * * * * * * * * * * *
By the lemma, the sum is parallel to the unit arcs at distance e : rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the corresponding portion of the Roberts section is a single arc of radius d = (2 - rt2) / 8 .
Finally, the Minkowski and Roberts bodies are both minimal constant-width, with tetrahedral symmetry, but incongruent.
The screed is already in need of updating ... comments are invited. While it suffers from an absence of diagrams, that on page 2 of Roberts report provides a partial substitute. In that connection I have already spotted one misprint:
Stage (A): for "d = 0" read "d = c"
Fred Lunnon
On 1/12/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist?
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Scott, I think you may have intended your message for my eyes only --- but you posted it to math-fun, and in any case your point is worth public discussion. I hadn't just assumed that a radius (1 + e)/2 arc has nonzero length, although I did omit the justification for the sake of brevity: it starts (on an extended tetrahedral face plane) where arcs of radius 1 and e touch, and finishes where it becomes parallel to the other end of the radius 1 arc (at an apical node). However you may be assuming that the opposite ends of those diameters trace a circular arc. They don't: their paths are similarly composed of 3 arcs, albeit now longtitudinal rather than transverse. I don't know how much of this might usefully be put into the screed ... [A flaw in my argument, indeed --- has this young man no fear?] Regards, Fred On 1/15/13, Huddleston, Scott <scott.huddleston@intel.com> wrote:
Hi Fred,
I suspect the Minkowski mean of Meissner tetrahedra is the same constant width solid Roberts describes. If so, your argument below must have a flaw. When you say
rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the
you seem to assume that the "arc" length of radius (1 + e)/2 in your cross section is nonzero. I believe it must be zero. If you think it's nonzero, look carefully at the diameters normal to that arc segment, and where the trajectory of the opposite end of those diameters would have to lie.
Best, - Scott
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun- bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Monday, January 14, 2013 9:05 AM To: math-fun Subject: Re: [math-fun] Low Rollers: Bodies with Minimal Constant Width and Tetrahedral Symmetry
Despite initially having been hazy about the details of Minkowski sums, it turns out that my intuition was not quite so wonky as earlier appeared!
Lemma: The Minkowski sum of a surface and a sphere of radius r is (congruent to) the parallel surface at distance r ; proof elementary.
Consider the mean M_T = (M_V (+) M_F)/2 , the summands positioned so that all three coincide in the spherical caps over each tetrahedral face, and the mean has the symmetry of a regular tetrahedron. Section it along a plane of symmetry: over each edge, the mean composes a circular arc of small radius e from one summand, with a pair of arcs of unit radius from the other meeting in nodal line; at their free ends, the small and unit arcs touch tangentially.
* * * * * * * * * * * * * * * * * * * * * * * * *
By the lemma, the sum is parallel to the unit arcs at distance e : rescaling, the mean comprises two arcs of radius (1 + e)/2 , touching the summands where they meet, and capped by a third arc of radius e/2 . Since e = 1 - rt3 / 2 is nonzero, the mean section is non-circular; whereas the corresponding portion of the Roberts section is a single arc of radius d = (2 - rt2) / 8 .
Finally, the Minkowski and Roberts bodies are both minimal constant-width, with tetrahedral symmetry, but incongruent.
The screed is already in need of updating ... comments are invited. While it suffers from an absence of diagrams, that on page 2 of Roberts report provides a partial substitute. In that connection I have already spotted one misprint:
Stage (A): for "d = 0" read "d = c"
Fred Lunnon
On 1/12/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Hence it looks a safe bet that Roberts = Minkowski holds, as conjectured by Dan Asimov, and earlier dismissed by myself with gratuitous contempt. A more detailed investigation along the lines above would not come amiss ... But now more generally, does minimal constant-width with tetrahedral symmetry uniquely determine the body modulo similarity --- why should no others exist?
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Huddleston, Scott