[math-fun] rounded tetrahedron
MLB> And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron? ------ I nearly slid through a stopsign recently. It was on a sharp downgrade, preceded by gravel washed off a hillside. Had the gravel been perfectly spherical, I wouldn't even have slowed down. Except spherical gravel would have rolled away. What about Reuleaux tetrahedra (http://www.youtube.com/watch?v=jK7xPo1YXzY 3:18)? Neil 3D printed a lovely one, ~1.5", and then split for CES w/o photographing it. It's eerie to roll a book over. Why aren't they sold at science museum gift shops? But would one stay put on a hill? Neil claims, with sufficient friction, an amazing 60°! They're much less round than you might think. This has 007esque possibilities, not just for foot chases and booby-traps. Imagine pursuing Bond on a winding road where he releases a few pounds of Reuleaux gravel. You say http://www.zombiesurvivalwiki.com/thread/3341691/Liquid+Banana+Peel+-+Make+a... is cheaper? OK, but it leaves a residue. For the perfect crime, use dry ice Reuleaux gravel. --rwg http://gosper.org/reuleauxtet.png
I wonder if Meissner tetrahedra (true constant width) would work better or worse than Reuleaux tetrahedra (approximate constant width) for this purpose. Tom Bill Gosper writes:
MLB>
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron? ------ I nearly slid through a stopsign recently. It was on a sharp downgrade, preceded by gravel washed off a hillside. Had the gravel been perfectly spherical, I wouldn't even have slowed down. Except spherical gravel would have rolled away. What about Reuleaux tetrahedra (http://www.youtube.com/watch?v=jK7xPo1YXzY 3:18)? Neil 3D printed a lovely one, ~1.5", and then split for CES w/o photographing it. It's eerie to roll a book over. Why aren't they sold at science museum gift shops? But would one stay put on a hill? Neil claims, with sufficient friction, an amazing 60°! They're much less round than you might think. This has 007esque possibilities, not just for foot chases and booby-traps. Imagine pursuing Bond on a winding road where he releases a few pounds of Reuleaux gravel. You say http://www.zombiesurvivalwiki.com/thread/3341691/Liquid+Banana+Peel+-+Make+a... is cheaper? OK, but it leaves a residue. For the perfect crime, use dry ice Reuleaux gravel. --rwg http://gosper.org/reuleauxtet.png _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Does anyone know which sort of rollers the MoMath exhibit "Coaster Rollers" uses? Jim Propp On Tue, Jan 8, 2013 at 6:12 AM, Tom Karzes <karzes@sonic.net> wrote:
I wonder if Meissner tetrahedra (true constant width) would work better or worse than Reuleaux tetrahedra (approximate constant width) for this purpose.
Tom
Bill Gosper writes:
MLB>
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron? ------ I nearly slid through a stopsign recently. It was on a sharp downgrade, preceded by gravel washed off a hillside. Had the gravel been perfectly spherical, I wouldn't even have slowed down. Except spherical gravel would have rolled away. What about Reuleaux tetrahedra (http://www.youtube.com/watch?v=jK7xPo1YXzY 3:18)? Neil 3D printed a lovely one, ~1.5", and then split for CES w/o photographing it. It's eerie to roll a book over. Why aren't they sold at science museum gift shops? But would one stay put on a hill? Neil claims, with sufficient friction, an amazing 60°! They're much less round than you might think. This has 007esque possibilities, not just for foot chases and booby-traps. Imagine pursuing Bond on a winding road where he releases a few pounds of Reuleaux gravel. You say
http://www.zombiesurvivalwiki.com/thread/3341691/Liquid+Banana+Peel+-+Make+a...
is cheaper? OK, but it leaves a residue. For the perfect crime, use dry ice Reuleaux gravel. --rwg http://gosper.org/reuleauxtet.png _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Jim, There are four shapes of rollers in the MoMath ride-on exhibit. Three are acorn-like surfaces of revolution and the fourth is a Meissner tetrahedron. One of them is very similar to this very cool commercially available metal set: http://www.grand-illusions.com/acatalog/Solids_of_Constant_Width.html The second has a point at the apex but is differentiable everywhere else, and the third is based on a regular pentagon (extended with radii and rotated to a surface of revolution). (I also wanted to have one sphere in the mix, with a different color, to emphasize that all the others aren't spheres, but that was vetoed...) George http://georgehart.com/ On 1/8/2013 5:35 PM, James Propp wrote:
Does anyone know which sort of rollers the MoMath exhibit "Coaster Rollers" uses?
Jim Propp
On Tue, Jan 8, 2013 at 6:12 AM, Tom Karzes <karzes@sonic.net> wrote:
I wonder if Meissner tetrahedra (true constant width) would work better or worse than Reuleaux tetrahedra (approximate constant width) for this purpose.
Tom
Bill Gosper writes:
MLB>
And what exactly does "smooth" buy us? Let's just ask instead: how would we quadrangulate a rounded tetrahedron? ------ I nearly slid through a stopsign recently. It was on a sharp downgrade, preceded by gravel washed off a hillside. Had the gravel been perfectly spherical, I wouldn't even have slowed down. Except spherical gravel would have rolled away. What about Reuleaux tetrahedra (http://www.youtube.com/watch?v=jK7xPo1YXzY 3:18)? Neil 3D printed a lovely one, ~1.5", and then split for CES w/o photographing it. It's eerie to roll a book over. Why aren't they sold at science museum gift shops? But would one stay put on a hill? Neil claims, with sufficient friction, an amazing 60°! They're much less round than you might think. This has 007esque possibilities, not just for foot chases and booby-traps. Imagine pursuing Bond on a winding road where he releases a few pounds of Reuleaux gravel. You say
http://www.zombiesurvivalwiki.com/thread/3341691/Liquid+Banana+Peel+-+Make+a...
is cheaper? OK, but it leaves a residue. For the perfect crime, use dry ice Reuleaux gravel. --rwg http://gosper.org/reuleauxtet.png _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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[Thread mysteriously split into two ...] Consider the following construction of a continuously differentiable surface, with tetrahedral symmetry: (0) start from a regular tetrahedron of circumradius 1; (1) balloon each face out to a spherical cap of radius 1+r, bounded where it meets each of the other 3 faces extended; (2) balloon each vertex out to a cap of radius r, bounded similarly; (3) balloon each edge out the the unique Dupin cyclide tangent to all 4 adjacent spherical boundaries. Does this have constant width? Meissner's surface seems to be the case r = 0, which is not differentiable. A couple of recent references turned up via Google --- mathoverflow.net/.../are-there-smooth-bodies-of-constant-width www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf One response under the first thread suggests a method of constructing arbitrary _analytic_ surfaces of constant width, but is so cryptic that I can make neither head nor tail of it. Fred Lunnon On 1/8/13, George Hart <george@georgehart.com> wrote:
Jim,
There are four shapes of rollers in the MoMath ride-on exhibit. Three are acorn-like surfaces of revolution and the fourth is a Meissner tetrahedron.
One of them is very similar to this very cool commercially available metal set:
http://www.grand-illusions.com/acatalog/Solids_of_Constant_Width.html
The second has a point at the apex but is differentiable everywhere else, and the third is based on a regular pentagon (extended with radii and rotated to a surface of revolution).
(I also wanted to have one sphere in the mix, with a different color, to emphasize that all the others aren't spheres, but that was vetoed...)
George http://georgehart.com/
On 1/8/2013 5:35 PM, James Propp wrote:
Does anyone know which sort of rollers the MoMath exhibit "Coaster Rollers" uses?
Jim Propp
On Tue, Jan 8, 2013 at 6:12 AM, Tom Karzes <karzes@sonic.net> wrote:
I wonder if Meissner tetrahedra (true constant width) would work better or worse than Reuleaux tetrahedra (approximate constant width) for this purpose.
Tom
Thanks, Fred! There's a paper by Jay Fillmore linked from the Math Overflow post you linked to, and the corollary to Fillmore's Theorem 2 is this: "There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex.
Which is just what I was wondering (though the real analytic differentiability is a bonus). --Dan I wrote: << QUESTION: Has anyone proved that a 3D shape of constant width *cannot* have the symmetry of the tetrahedron? (Meaning no additional symmetries, either.)
On 2013-01-08, at 7:16 PM, Fred lunnon wrote:
[Thread mysteriously split into two ...]
Consider the following construction of a continuously differentiable surface, with tetrahedral symmetry: (0) start from a regular tetrahedron of circumradius 1; (1) balloon each face out to a spherical cap of radius 1+r, bounded where it meets each of the other 3 faces extended; (2) balloon each vertex out to a cap of radius r, bounded similarly; (3) balloon each edge out the the unique Dupin cyclide tangent to all 4 adjacent spherical boundaries.
Does this have constant width? Meissner's surface seems to be the case r = 0, which is not differentiable.
A couple of recent references turned up via Google ---
mathoverflow.net/.../are-there-smooth-bodies-of-constant-width
www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf
One response under the first thread suggests a method of constructing arbitrary _analytic_ surfaces of constant width, but is so cryptic that I can make neither head nor tail of it.
Fred Lunnon
The Fillmore paper (1968!) looks fascinating, but hard work for those not conversant with the jargon of spherical harmonic functions. The reference below does at least explain (pp. 8--9) what a "(Minkowski) support function" means. Referring back to Dan's original enquiry: according to p. 260 of Appendix I in V.V. Buldygin, A.B. Kharazishvili "Geometric Aspects of Probability Theory and Mathematical Statistics" --- for _any_ subset of Euclidean n-space, there exists a constant width set containing it, and having the same diameter (ie. width). Evidently from Fillmore's paper, the construction may fail to preserve the symmetry group --- of course, for any regular polytope, a sphere suffices. In fact, just at the moment, I am having difficulty imagining a set which is _not_ contained in a sphere of the same diameter ... Fred Lunnon On 1/9/13, Dan Asimov <dasimov@earthlink.net> wrote:
Thanks, Fred! There's a paper by Jay Fillmore linked from the Math Overflow post you linked to, and the corollary to Fillmore's Theorem 2 is this:
"There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex.
Which is just what I was wondering (though the real analytic differentiability is a bonus).
--Dan
I wrote:
<< QUESTION: Has anyone proved that a 3D shape of constant width *cannot* have the symmetry of the tetrahedron? (Meaning no additional symmetries, either.)
On 2013-01-08, at 7:16 PM, Fred lunnon wrote:
[Thread mysteriously split into two ...]
Consider the following construction of a continuously differentiable surface, with tetrahedral symmetry: (0) start from a regular tetrahedron of circumradius 1; (1) balloon each face out to a spherical cap of radius 1+r, bounded where it meets each of the other 3 faces extended; (2) balloon each vertex out to a cap of radius r, bounded similarly; (3) balloon each edge out the the unique Dupin cyclide tangent to all 4 adjacent spherical boundaries.
Does this have constant width? Meissner's surface seems to be the case r = 0, which is not differentiable.
A couple of recent references turned up via Google ---
mathoverflow.net/.../are-there-smooth-bodies-of-constant-width
www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf
One response under the first thread suggests a method of constructing arbitrary _analytic_ surfaces of constant width, but is so cryptic that I can make neither head nor tail of it.
Fred Lunnon
Come to think of it, Fillmore's Corollary quoted below is preceded by this remark (for readability I'll requote the Corollary, too): ----- If we imitate the construction of a Reuleux triangle in n-dimension, a parallel hypersurface at constant distance from this hypersurface is of class C1 and has the symmetry group (a finite group) of the regular n-simplex. Thus: Corollary. There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex. ----- OOPS. The Reuleaux tetrahedron, for instance, is *not* of constant width. (It's also spelled with an "a".) So Fillmore's Corollary is in serious doubt!!! In any case, the argument in his paper includes a crucial mistake. In fact, according to Wikipedia: ----- The Reuleaux tetrahedron is the intersection of four spheres of radius s centered at the vertices of a regular tetrahedron with side length s. The sphere through each vertex passes through the other three vertices, which also form vertices of the Reuleaux tetrahedron. The Reuleaux tetrahedron has the same face structure as a regular tetrahedron, but with curved faces: four vertices, and four curved faces, connected by six circular-arc edges. This shape is defined and named by analogy to the Reuleaux triangle, a two-dimensional curve of constant width. One can find repeated claims in the mathematical literature that the Reuleaux tetrahedron is analogously a surface of constant width, but it is not true: the two midpoints of opposite edge arcs are separated by a larger distance, (sqrt(3) - sqrt(2)/2)s. ----- So: I still don't know if there's a surface of constant width in 3-space having the symmetry group of the tetrahedron. --Dan On 2013-01-09, at 1:51 PM, Fred lunnon wrote:
The Fillmore paper (1968!) looks fascinating, but hard work for those not conversant with the jargon of spherical harmonic functions. The reference below does at least explain (pp. 8--9) what a "(Minkowski) support function" means.
Referring back to Dan's original enquiry: according to p. 260 of Appendix I in V.V. Buldygin, A.B. Kharazishvili "Geometric Aspects of Probability Theory and Mathematical Statistics" --- for _any_ subset of Euclidean n-space, there exists a constant width set containing it, and having the same diameter (ie. width).
Evidently from Fillmore's paper, the construction may fail to preserve the symmetry group --- of course, for any regular polytope, a sphere suffices. In fact, just at the moment, I am having difficulty imagining a set which is _not_ contained in a sphere of the same diameter ...
Fred Lunnon
On 1/9/13, Dan Asimov <dasimov@earthlink.net> wrote:
Thanks, Fred! There's a paper by Jay Fillmore linked from the Math Overflow post you linked to, and the corollary to Fillmore's Theorem 2 is this:
"There exists an analytic hypersurface of constant width in E^n having the same group of symmetries as a regular n-simplex.
Which is just what I was wondering (though the real analytic differentiability is a bonus).
--Dan
I wrote:
<< QUESTION: Has anyone proved that a 3D shape of constant width *cannot* have the symmetry of the tetrahedron? (Meaning no additional symmetries, either.)
On 2013-01-08, at 7:16 PM, Fred lunnon wrote:
[Thread mysteriously split into two ...]
Consider the following construction of a continuously differentiable surface, with tetrahedral symmetry: (0) start from a regular tetrahedron of circumradius 1; (1) balloon each face out to a spherical cap of radius 1+r, bounded where it meets each of the other 3 faces extended; (2) balloon each vertex out to a cap of radius r, bounded similarly; (3) balloon each edge out the the unique Dupin cyclide tangent to all 4 adjacent spherical boundaries.
Does this have constant width? Meissner's surface seems to be the case r = 0, which is not differentiable.
A couple of recent references turned up via Google ---
mathoverflow.net/.../are-there-smooth-bodies-of-constant-width
www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf
One response under the first thread suggests a method of constructing arbitrary _analytic_ surfaces of constant width, but is so cryptic that I can make neither head nor tail of it.
Fred Lunnon
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On 1/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
Consider an equilateral triangle.
--Dan
On 2013-01-09, at 1:51 PM, Fred lunnon wrote:
In fact, just at the moment, I am having difficulty imagining a set which is _not_ contained in a sphere of the same diameter ...
Groan --- or indeed a regular tetrahedron, which is where we came in ... Or indeed _any_ triangle or tetrahedron, according to the theorem I quoted earlier. On 1/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
...
OOPS. The Reuleaux tetrahedron, for instance, is *not* of constant width. (It's also spelled with an "a".)
So Fillmore's Corollary is in serious doubt!!! In any case, the argument in his paper includes a crucial mistake.
...
So: I still don't know if there's a surface of constant width in 3-space having the symmetry group of the tetrahedron.
A positive answer is guaranteed by that theorem (to which I can't discover the reference, discourtesy of Googlebooks). Part of the problem is the misnomer "Reuleaux tetrahedron" attached to the figure you described, which is as you observe irrelevant to the discussion. The "Meissner" tetrahedron --- a special case of the construction I posed earlier --- may well have constant width, but my own previous attempts to decide this question have been inconclusive. These fellows are presumably in possession of some high-powered construction. So what does the constant-width superset (with the same diameter) of a scalene triangle (including interior) look like, anyway? And is it unique, given the triangle? Fred Lunnon
I think the Meissner tetrahedron has constant width, but without full tetrahedral symmetry. Let L be the longest side length of a scalene triangle. Then isn't the intersection of the 3 disks of radius L, centered at the triangle's vertices, a shape of constant width? --Dan Fred wrote:
Part of the problem is the misnomer "Reuleaux tetrahedron" attached to the figure you described, which is as you observe irrelevant to the discussion. The "Meissner" tetrahedron --- a special case of the construction I posed earlier --- may well have constant width, but my own previous attempts to decide this question have been inconclusive.
|. . .
So what does the constant-width superset (with the same diameter) of a scalene triangle (including interior) look like, anyway? And is it unique, given the triangle?
Scratch that idea! Okay, second try: Build an equilateral triangle on the longest side of the scalene triangle, so that it contains the vertex opposite that side. Then the Reuleaux triangle in which this equilateral one is inscribed seems to be the natural shape of constant width containing the original scalene triangle and having the same diameter. I'm sure this is the least area of any such shape. I suspect it's a subset of any such shape, but am chastened by my earlier mistake, so will not try to assert this as more than a guess. --Dan On 2013-01-09, at 10:12 PM, Dan Asimov wrote:
I think the Meissner tetrahedron has constant width, but without full tetrahedral symmetry.
Let L be the longest side length of a scalene triangle.
Then isn't the intersection of the 3 disks of radius L, centered at the triangle's vertices, a shape of constant width?
--Dan
Fred wrote:
Part of the problem is the misnomer "Reuleaux tetrahedron" attached to the figure you described, which is as you observe irrelevant to the discussion. The "Meissner" tetrahedron --- a special case of the construction I posed earlier --- may well have constant width, but my own previous attempts to decide this question have been inconclusive.
|. . .
So what does the constant-width superset (with the same diameter) of a scalene triangle (including interior) look like, anyway? And is it unique, given the triangle?
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On 1/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
Scratch that idea!
Okay, second try: Build an equilateral triangle on the longest side of the scalene triangle, so that it contains the vertex opposite that side.
Then the Reuleaux triangle in which this equilateral one is inscribed seems to be the natural shape of constant width containing the original scalene triangle and having the same diameter.
Agreed --- once I had realised this, I felt much better about the theorem for arbitrary subsets of |E^n --- at any rate, for n = 2 ! Furthermore, in this case at least, the superset is fairly easily seen to be unique.
...
I think the Meissner tetrahedron has constant width, but without full tetrahedral symmetry.
Huh? It must have tetrahedral symmetry, by construction: 4 congruent spherical segments atop faces, 6 congruent cyclidal segments atop edges. A trivial computation --- can't understand why I balked at this before --- shows that, along a "bi-altitude" through the mid-points of the edges, the Meissner undershoots unit width by exactly the same amount that the "Reuleaux" overshoots it: sqrt(3) - 1/sqrt(2) - 1 = 0.0249 ; each attempt proves equally feeble (the same applies to my generalisation). Presumably a better approximation would be the mean of the two surfaces. Intuitively, I think it's pretty clear that the constant minimum width surface does exist in this case too (and is also unique). Simply place a hard Meissner inside a soft Reuleaux between parallel abrasive planes at unit distance, and randomly grind away ... Fred Lunnon
Well, here's the MathWorld description of the Meissner tetrahedron: ----- Meißner (1911) showed how to modify the Reuleaux tetrahedron (which is not a solid of constant width) to form a surface of constant width by replacing three of its edge arcs by curved patches formed as the surfaces of rotation of a circular arc. Depending on which three edge arcs are replaced (three that have a common vertex or three that form a triangle), one of two noncongruent shapes can be produced that are called Meissner tetrahedra or Meissner bodies (Lachand-Robert and Oudet 2007). ----- which seems to have less than full tetrahedral symmetry. --Dan On 2013-01-10, at 6:48 AM, Fred lunnon wrote:
On 1/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
Scratch that idea!
Okay, second try: Build an equilateral triangle on the longest side of the scalene triangle, so that it contains the vertex opposite that side.
Then the Reuleaux triangle in which this equilateral one is inscribed seems to be the natural shape of constant width containing the original scalene triangle and having the same diameter.
Agreed --- once I had realised this, I felt much better about the theorem for arbitrary subsets of |E^n --- at any rate, for n = 2 ! Furthermore, in this case at least, the superset is fairly easily seen to be unique.
...
I think the Meissner tetrahedron has constant width, but without full tetrahedral symmetry.
Huh? It must have tetrahedral symmetry, by construction: 4 congruent spherical segments atop faces, 6 congruent cyclidal segments atop edges.
A trivial computation --- can't understand why I balked at this before --- shows that, along a "bi-altitude" through the mid-points of the edges, the Meissner undershoots unit width by exactly the same amount that the "Reuleaux" overshoots it: sqrt(3) - 1/sqrt(2) - 1 = 0.0249 ; each attempt proves equally feeble (the same applies to my generalisation).
Presumably a better approximation would be the mean of the two surfaces. Intuitively, I think it's pretty clear that the constant minimum width surface does exist in this case too (and is also unique). Simply place a hard Meissner inside a soft Reuleaux between parallel abrasive planes at unit distance, and randomly grind away ...
Fred Lunnon
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P.S. Here's a nice write-up about the Meissner tetrahedron: (published in Math Intelligencer v.33 no.3 2011: <http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf> --Dan
Apologies: I had completely forgotten that the Meissner construction replaces only _three_ of the six edges. So yes, you (and he) are correct: it does have constant width, but not tetrahedral symmetry. The sectional curve along each edge which would achieve full symmetry must lie somewhere between the single circular arc used by Meissner, and the pair of arcs resulting from continuation of the spherical caps over the faces ("Releaux"). It's surprising that nobody seems to have explicitly given an analytical form for either implicitly or parametrically. Fred Lunnon On 1/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
P.S. Here's a nice write-up about the Meissner tetrahedron: (published in Math Intelligencer v.33 no.3 2011:
<http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf>
--Dan
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-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun- bounces@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Thursday, January 10, 2013 2:17 PM To: math-fun Subject: Re: [math-fun] rounded tetrahedron
Apologies: I had completely forgotten that the Meissner construction replaces only _three_ of the six edges. So yes, you (and he) are correct: it does have constant width, but not tetrahedral symmetry.
The sectional curve along each edge which would achieve full symmetry must lie somewhere between the single circular arc used by Meissner, and the pair of arcs resulting from continuation of the spherical caps over the faces ("Releaux"). It's surprising that nobody seems to have explicitly given an analytical form for either implicitly or parametrically.
Fred Lunnon
See http://www.xtalgrafix.com/Spheroform.htm and its links, including http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf Also, p.4 of http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf says that a Meissner-like tetrahedron with congruent "edge" parts is realizable as a Minkowski sum of the two Meissner tetrahedra.
On 1/10/13, Dan Asimov <dasimov@earthlink.net> wrote:
P.S. Here's a nice write-up about the Meissner tetrahedron: (published in Math Intelligencer v.33 no.3 2011:
<http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf>
--Dan
On 1/10/13, Huddleston, Scott <scott.huddleston@intel.com> wrote:
See http://www.xtalgrafix.com/Spheroform.htm and its links, including http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf
Looks like that probably wraps Dan's question up, albeit at some length!
Also, p.4 of http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf says that a Meissner-like tetrahedron with congruent "edge" parts is realizable as a Minkowski sum of the two Meissner tetrahedra.
That was obviously true --- but there seems no reason to expect that the width would be constant elsewhere. WFL
Right, assuming the proof holds up. But I'm more inclined to believe the validity of the claim in the Math Intelligencer article that you cite. In fact, they write: ----- Incidentally, the Minkowski sum (½)M_V ⊕ (½)M_F, which one obtains half way in the process of morphing M_V into M_F, would render a body with tetrahedal symmetry. It actually has the same constant width as MV and MF . Its volume, however, is larger than that of the Meissner bodies, due to the Brunn-Minkowski inequality. ----- Here M_V is the Meissner tetrahedron where starting from the Reuleaux tetrahedron, 3 edges sharing a vertex have been modified in the same way, to get a body of constant width. Likewise for M_F, but in this case the 3 edges bound one face of the Reuleaux tetrahedron. I believe the example in the example Scott Huddleston linked to below is identical to the example (½)M_V ⊕ (½)M_F mentioned in the Intelligencer article. --Dan P.S. Fred, what do you mean by "elsewhere" in your posted quoted below? On 2013-01-10, at 4:20 PM, Fred lunnon wrote:
On 1/10/13, Huddleston, Scott <scott.huddleston@intel.com> wrote:
See http://www.xtalgrafix.com/Spheroform.htm and its links, including http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf
Looks like that probably wraps Dan's question up, albeit at some length!
Also, p.4 of http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf says that a Meissner-like tetrahedron with congruent "edge" parts is realizable as a Minkowski sum of the two Meissner tetrahedra.
That was obviously true --- but there seems no reason to expect that the width would be constant elsewhere.
WFL
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On 1/11/13, Dan Asimov <dasimov@earthlink.net> wrote:
Right, assuming the proof holds up.
There does seem to be an lot of detail involved in Roberts' presentation. As far as I can tell however, it does at least attempt to cover all the options.
But I'm more inclined to believe the validity of the claim in the Math Intelligencer article that you cite. ... P.S. Fred, what do you mean by "elsewhere" in your posted quoted below?
Kawohl & Weber (Math. Intell., Scott's link) give no justification or reference for the constant-width claim, which I consider on the face of it very surprising and quite possibly erroneous [it takes one to know one ...]; "elsewhere" refers to width along diametral lines away from planes of symmetry. Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all. So the Minkowski yields a distinct surface. Fred Lunnon
In fact, they write:
----- Incidentally, the Minkowski sum (½)M_V ⊕ (½)M_F, which one obtains half way in the process of morphing M_V into M_F, would render a body with tetrahedal symmetry. It actually has the same constant width as MV and MF .
Its volume, however, is larger than that of the Meissner bodies, due to the Brunn-Minkowski inequality. -----
Here M_V is the Meissner tetrahedron where starting from the Reuleaux tetrahedron, 3 edges sharing a vertex have been modified in the same way, to get a body of constant width. Likewise for M_F, but in this case the 3 edges bound one face of the Reuleaux tetrahedron.
I believe the example in the example Scott Huddleston linked to below is identical to the example (½)M_V ⊕ (½)M_F mentioned in the Intelligencer article.
On 2013-01-10, at 4:20 PM, Fred lunnon wrote:
On 1/10/13, Huddleston, Scott <scott.huddleston@intel.com> wrote:
See http://www.xtalgrafix.com/Spheroform.htm and its links, including http://www.xtalgrafix.com/Reuleaux/Spheroform%20Tetrahedron.pdf
Looks like that probably wraps Dan's question up, albeit at some length!
Also, p.4 of http://www.mi.uni-koeln.de/mi/Forschung/Kawohl/kawohl/pub100.pdf says that a Meissner-like tetrahedron with congruent "edge" parts is realizable as a Minkowski sum of the two Meissner tetrahedra.
That was obviously true --- but there seems no reason to expect that the width would be constant elsewhere.
WFL
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But the Minkowski average in question does not involve the Reuleaux tetrahedron as summand. It is the sum of two types of Meissner tetrahedra: One (M_V) where 3 Reuleaux arcs that share a vertex are modified, and one (M_F) where 3 Reuleaux arcs that share a face are modified. --Dan On 2013-01-10, at 7:57 PM, Fred lunnon wrote:
Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all.
So the Minkowski yields a distinct surface.
My argument concerns just the cross-section of the region around one particular tetrahedral edge of the surface, and cut off by the extended neighbouring faces of the tetrahedron. The Minkowski average combines a Reuleaux "cusp" with a Meissner cyclide: its cross-section is the mean of a pair of adjacent arcs with a single arc, yielding some kind of non-circular curve with a rounded apex of greater curvature, and only its endpoints in common with either summand. The Roberts comprises an lower portion shared with the Releaux arcs, capped higher up by a circular arc. The two curves are evidently incongruent; therefore so are the completed surfaces. Finally, while I'm prepared to believe that there may be some obvious or well-known reason why this global Minkowski surface should retain the constant curvature of its summands, the matter remains currently beyond my comprehension --- can anybody cast any light here? Fred Lunnon On 1/11/13, Dan Asimov <dasimov@earthlink.net> wrote:
But the Minkowski average in question does not involve the Reuleaux tetrahedron as summand.
It is the sum of two types of Meissner tetrahedra: One (M_V) where 3 Reuleaux arcs that share a vertex are modified, and one (M_F) where 3 Reuleaux arcs that share a face are modified.
--Dan
On 2013-01-10, at 7:57 PM, Fred lunnon wrote:
Notice that the relevant portion of the Reuleaux cross-section comprises a pair of unit radius circular arcs meeting at a singular point, whereas the Meisner has a single small-radius circular arc, tangent to the larger pair. The Minkowski sum (ie. mean) of these two cannot possibly be the Roberts, which for part of its length coincides (only) with the Reuleaux: indeed, it can't contain circular arcs at all.
So the Minkowski yields a distinct surface.
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Light duly cast; twaddle below withdrawn: see Low Rollers thread. WFL On 1/11/13, Fred lunnon <fred.lunnon@gmail.com> wrote:
My argument concerns just the cross-section of the region around one particular tetrahedral edge of the surface, and cut off by the extended neighbouring faces of the tetrahedron.
The Minkowski average combines a Reuleaux "cusp" with a Meissner cyclide: its cross-section is the mean of a pair of adjacent arcs with a single arc, yielding some kind of non-circular curve with a rounded apex of greater curvature, and only its endpoints in common with either summand.
The Roberts comprises an lower portion shared with the Releaux arcs, capped higher up by a circular arc.
The two curves are evidently incongruent; therefore so are the completed surfaces.
Finally, while I'm prepared to believe that there may be some obvious or well-known reason why this global Minkowski surface should retain the constant curvature of its summands, the matter remains currently beyond my comprehension --- can anybody cast any light here?
Fred Lunnon
I hadn't know about Meissner tetrahedra, but was vaguely aware that there were no known 3D analogues of Reuleaux triangles that have constant width. QUESTION: Has anyone proved that a 3D shape of constant width *cannot* have the symmetry of the tetrahedron? (Meaning no additional symmetries, either.) --Dan On 2013-01-08, at 3:12 AM, Tom Karzes wrote:
I wonder if Meissner tetrahedra (true constant width) would work better or worse than Reuleaux tetrahedra (approximate constant width) for this purpose.
participants (7)
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Bill Gosper -
Dan Asimov -
Fred lunnon -
George Hart -
Huddleston, Scott -
James Propp -
Tom Karzes