Talk:Octahedron

Octahedron equation
¿Shouldnt be on the article the equation of the octahedron?
 * Which is:
 * for the volume $$|x|+|y|+|z| \le k$$
 * for the surface $$|x|+|y|+|z| = k$$
 * where $$k \in R $$

Andreini Tessellations
This article says there are 5 Andreini tessellations, but the article Andreini tessellation says there are 28. Which is correct? --Auximines 13:14, 21 Apr 2004 (UTC)


 * Maybe someone meant to say that five (would you believe six?) of the 28 contain octahedra. &mdash;Tamfang 06:53, 25 November 2006 (UTC)

New stat table
I replace stat table with template version, which uses tricky nested templates as a "database" which allows the same data to be reformatted into multiple locations and formats. See here for more details: User:Tomruen/polyhedron_db_testing
 * Tom Ruen 00:53, 4 March 2006 (UTC)

volume
i really need to know how i would get the formula that is given for the volume of the octahedron.... like.. the working out to get the formula...

PLEASE, THANKYOU

-Dave —Preceding unsigned comment added by 58.169.29.65 (talk • contribs) 17:39, 9 October 2006


 * Well, let's see. Starting with the octahedron whose vertices are (0,0,&plusmn;1) etc., consider the "first octant", that portion in which x,y,z are all positive.  That's a (skew) pyramid whose base is a right triangle.  A pyramid's volume is (height)&times;(area of base)/3; the base is half of a unit square, and the height is 1, so the pyramid's volume is 1/6.  There are eight such pyramids, so the total volume is 8/6 = 4/3.  But it's usual to express volume in terms of the edge length, and this octahedron has edge &radic;2; so we divide 4/3 by the cube of &radic;2 and get (&radic;2)/3 for the volume of an octahedron of unit edge.  I didn't look at the answer while deriving this, so now I will &mdash; ah, good. ;)  &mdash;Tamfang 04:52, 11 October 2006 (UTC)


 * I ought perhaps to have mentioned that I looked it up in Regular Polytopes. —Tamfang (talk) 05:43, 6 February 2009 (UTC)

Would you agree that the configuration of my models in Talk:Nuclear model is that of an octahedron? If so, does the method of construction interest you? How about that they can be built using Neodymium magnets?WFPM (talk) 20:30, 3 May 2010 (UTC)


 * Yes, no, no. —Tamfang (talk) 01:38, 4 May 2010 (UTC)

Gee! Thanks.WFPM (talk) 02:23, 4 May 2010 (UTC)

hexagonal prism
I changed the hexagonal prism from 6 triangle to 6 squares. To the best of my knowledge, that is how a hexagonal prism is constructed. Change it back if I was in error. Scott MacHardy 02:35, 23 October 2006 (UTC)


 * Everyone makes mistakes! &mdash;Tamfang 04:14, 28 October 2006 (UTC)

Alum
The substance Alum is generally octahedral i beleive (at least, mine always is). Perhapse it should be mentioned somewhere? 71.28.13.158 03:35, 28 November 2006 (UTC)

subgroups
It occurs to me that the subgroups of Oh include Th as well as Td. Is there a figure that we can associate with that subgroup? &mdash;Tamfang 22:56, 25 April 2007 (UTC)

octahedra in music
What does it mean "place notes at the vertex"? 205.155.71.58 (talk) 21:49, 31 July 2008 (UTC)


 * It means that each vertex represents a note in the hexany scheme. I'd take the section out. —Tamfang (talk) 07:38, 11 August 2008 (UTC)

Symmetry group
From the article: "The octahedron's symmetry group is Oh, of order 48, the three dimensional hyperoctahedral group." From Hyperoctahedral group: "In three dimensions, the hyperoctahedral group is known as the binary octahedral group." But the Binary octahedral group has only one involution, while the symmetry group of the octahedron clearly has more than one.

There is something wrong here, but I don't know the terminology well enough to say where. Maproom (talk) 15:18, 8 September 2009 (UTC)


 * Ok, this has been fixed, JackSchmidt has corrected the Hyperoctahedral group article. Maproom (talk) 22:21, 10 September 2009 (UTC)

Symmetry group (2)
It is unclear whether or not the triangular antiprism is an octahedron. If so, a non-technical explanation about the difference between the three members of the subgroup would be nice.--Wickey-nl (talk) 14:13, 14 January 2010 (UTC)


 * The triangular antiprism is an octahedron. Or to put it another way, an octahedron can be regarded as a triangular antiprism; and a regular octahedron as a regular triangular antiprism.


 * What three members, and what subgroup, do you mean?


 * I have reverted your more recent edit. It is clear what it means to say "This group's subgroups include [certain groups]". But "Group Oh includes the subgroups ..." is less clear. I guess "Group Oh has as subgroups ..." would be ok. Maproom (talk) 14:55, 14 January 2010 (UTC)


 * All pictures in the table seem to show about the same polyhedron, except that the triangular antiprism could have either 6 or 8 planes. Are rectified tetrahedrons, triangular antiprisms and square bipyramids different polyhedrons or the same with only different origin? "These symmetries can be emphasized by different decorations of the faces" is a fantastic explanation of these three subgroups. Or are they subsubgroups?


 * Symmetry group Oh obviously needs a separate article and the table could better be placed there. Symmetry group is a very, very bad article and the intro of Hyperoctahedral group is a nothing saying technical string of internal links.


 * Uniform colorings does not need a separate chapter in the article.--Wickey-nl (talk) 15:20, 15 January 2010 (UTC)


 * All the pictures in the table show the same polyhedron, differently coloured (though this is not at all clear) to show how they can be regarded as less regular polyhedra, of lower symmetry.
 * It's like saying that a square, with symmetry group D8, can be regarded as a rectangle which just happens to have all its sides the same length. The symmetry group of a rectangle is C2XC2, which is therefore a subgroup of the symmetry group of a square.
 * I have replaced the word "decorations" by "colorings" - I hope that helps a little. Maproom (talk) 16:53, 15 January 2010 (UTC)


 * A little bit. Is tetragonal bipyramid synonym with octahedron?--Wickey-nl (talk) 17:04, 15 January 2010 (UTC)


 * I wouldn't quite say it was a synonym - but they describe the same polyhedron. A tetragonal bipyramid, more simple called a square bipyramid, is two square pyramids, stuck together base to base. Maproom (talk) 18:39, 15 January 2010 (UTC)


 * Octahedron will usually imply regular octahedron with octahedral symmetry, while a (Th) tetragonal bipyramid (or especially square bipyramid), being in the bipyramid family, would more likely to be considered to have only dihedral symmetry. Tom Ruen (talk) 21:10, 15 January 2010 (UTC)

I would conclude: A tetragonal bipyramid and the special case of it, the square bipyramid are always octahedrals. Equilateral tetragonal bipyramids, which are a special case of square bipyramids, are identical with regular octahedrons.--Wickey-nl (talk) 15:21, 16 January 2010 (UTC)

Edge length of dual
Now that I think about it again, the whole thing about the edge length of the dual cube of an octahedron is completely dubious, because the result of the inversion operation depends on the radius of the sphere you use to invert it. There is no unique dual cube for which one may say decisively that if the octahedron has edge length A then the dual cube must have edge length B. So the whole idea is unfounded in the first place. It suffices to state that the cube is the dual of the octahedron. Edge length assertions are meaningless unless it is explicitly stated how the octahedron is to be inverted.&mdash;Tetracube (talk) 16:59, 29 April 2010 (UTC)
 * If the sentence is to mean anything, it must assume a method of inversion that, repeated, will get you back to where you started (this is obviously not the case for the diagram in the article). And the method needs specifying. You could try keeping the volume, or the surface area, or the diameter, fixed. But I think that the best way is to keep the mid-points of the edges fixed, as one way to form a dual is to rotate each edge through a right angle about its midpoint, using an axis through the centre of symmetry.
 * Either we should specify the way of forming the dual, or forget about comparing edge lengths. Maproom (talk) 18:29, 29 April 2010 (UTC)


 * I agree.&mdash;Tetracube (talk) 00:37, 30 April 2010 (UTC)
 * Me too. (This message will cost the Net hundreds if not thousands of dollars to send everywhere) —Tamfang (talk) 06:25, 1 May 2010 (UTC)

Duality should be a self-inverse operation, such that applying the duality operation twice to a given object restores it exactly, with the same size and orientation. One easy way of forming a dual such that this property is true is to have the midpoints of the edges of the octahedron and cube coinciding. The diagram showing the dual does not satisfy this property, which is where the confusion is arising. If we were to adopt this convention, the edge length of the octahedron would be exactly sqrt(2) times larger than that of its dual cube.


 * That's right!!!.WFPM (talk) 02:30, 8 May 2010 (UTC)
 * Do your beliefs about symmetry, or something, prevent you from following the indentation convention which allows readers to distinguish one 'voice' from another? —Tamfang (talk) 19:52, 8 May 2010 (UTC)

Unfortunately, that definition does not work in even numbers of dimensions. An alternative approach would be to define duality to be volume-preserving, in which case it neatly generalises to all dimensions. In that case, the edge length of the cube would actually be larger, by a factor of cbrt(3/sqrt(2)). I prefer this version, as it seems more natural -- it works in any number of dimensions and doesn't alter the 'size' of the polytope. Calcyman (talk) 16:06, 6 May 2010 (UTC)


 * But still, the point is that there are several differing operations you can perform on a polyhedron/polytope which will produce what can be rightfully called its "dual", and so regardless of which definition you pick, you still have to state which one it is in order for blanket statements on edge lengths to make sense. The thing is, the operation of taking a dual is a topological operation, not an Euclidean geometric operation in the sense of preserving lengths or angles or some other Euclidean property. As such, it doesn't map uniquely to a single realization within Euclidean geometry. The early conception of duals disregarded lengths, because one was dealing with the Platonic solids, and any instance of the object at any non-zero scale was considered to be the same as the archetype itself. So whether you inverted the cube by its in-sphere, mid-sphere, or circumscribing sphere, the result was still regarded as identical to the octahedron, even though you're actually dealing with different instances of the octahedron at different sizes. But the point in studying such relationships was the symmetry and the topology, and so parameters such as sizes, edge lengths, etc., are disregarded.


 * Nowadays, with the advent of abstract polytopes, many of those things cannot even be realized in Euclidean space, let alone have any meaningful definition of "edge length". Yet it is a theorem that every (abstract) polytope has a well-defined, unique dual, and the operation of taking the dual is self-invertible. In other words, it's really the topology that matters, not so much the actual sizes of the objects in question. So statements about edge lengths or other specific measurements, while they may not be wrong per se (if one states exactly which instance of the dual operation is being used), are really not that relevant to the subject at hand, which is concerned with the topological relationships between the polyhedra rather than their specific Euclidean measurements.&mdash;Tetracube (talk) 18:07, 6 May 2010 (UTC)


 * I agree. The topological nature of this operation has been explored in the form of dual graphs. Tilings of hyperbolic space undoubtedly have dual polytopes, but they cannot be defined in Euclidian space. However, defining the definition of 'dual' topologically (rather than geometrically) would allow such statements as: "The parallelopiped is the dual of the regular octahedron", since parallelopipeds are topologically equivalent to cubes. Would you consider this to be acceptable, or should a dual necessarily have the same automorphism group of the object in question? Calcyman (talk) 20:46, 6 May 2010 (UTC)


 * As I said, the concept of dual polyhedra first arose in the context of regular and semiregular/uniform polyhedra. The concept has been generalized several times since then, but it's not entirely clear how later definitions map back to the original concepts---see esp. Grunbaum's insightful article Are My Polyhedra the Same as Your Polyhedra?. So it depends on the context in which you're making such statements. An abstract polytopist would definitely agree that the regular octahedron is dual to the parallelopiped -- in fact, he makes no distinction between the parallelopiped and the cube, as both are identical according to the definition of an abstract polytope, and both are, in fact, regular (according to the definition of regularity wrt. to an abstract polytope). However, in the context of Platonic and Archimedean solids (in the traditional, or classical, sense), one would tend to exclude irregular variations of the basic, regular shapes from consideration, so one wouldn't really compare a non-regular parallelopiped to a regular octahedron, although one might do so to a general octahedron (not necessarily regular). Due to the complicated history of the development of the theory of polyhedra and polytopes, a lot of these terms and definitions are quite context-sensitive; something to keep in mind when addressing this issue.


 * Now, as far as this article is concerned, it mainly describes the regular octahedron, and so its context is mainly classical in scope. So while it's not wrong to say that the octahedron is dual to the parallelopiped (if you use the topological definition of dual), the convention is to regard the regular representative of the dual, which is the cube. Again, this underlines the point I was trying to make, which is that you will end up with different results depending on the context you're working in (i.e., which method of the dual operation you use, or indeed, which definition you're working with), and so it's important to state your assumptions before making blanket statements like "the edge length of X is Y times the edge length of Z", or "A is the dual of B". I'm not against making statements about the edge length of duals, but one should state what assumptions he is working under, when making such statements. That's really the bottom line.&mdash;Tetracube (talk) 21:17, 6 May 2010 (UTC)

WFPM's models

 * Moved from previous section for clarity. —Tamfang (talk) 22:40, 9 May 2010 (UTC)

Note that if you only add progressive wrap series lengths plus a one sided layer addition to an octahedral you will still have an octahedral, so they expand from the center out..WFPM (talk) 21:05, 2 May 2010 (UTC) And they have a "least angular momentum" orientation direction of the axis of rotation, which might be a significant feature. And I appreciate being given that name "center of symmetry" for that axis.


 * I have no idea what you just said, and why this is relevant here. Why are you trying to push your original research on nuclear models?&mdash;Tetracube (talk) 15:51, 5 May 2010 (UTC)

Okay. Now with regard to the octahedron, It does have a central axis. And if you want to rotate it, I was wondering about which orientation of rotation would result in the least amount of angular momentum, which I think would be the axis that you called the axis of symmetry. So please excuse me.WFPM (talk) 18:52, 5 May 2010 (UTC) But maybe I'm wrong and it has 3 equal directions of symmetry, and if so, I would like to know that.


 * That is a valid and relevant question. Partial answer: the regular octahedron has three axes of fourfold rotational symmetry, four threefold axes and six twofold axes; the moment of inertia is presumably different for these three kinds of axis.  It shouldn't be hard to compute, but I haven't needed to know how in thirty years. ;) —Tamfang (talk) 20:38, 5 May 2010 (UTC)


 * I'll bet it's least for the fourfold axes. Maproom (talk) 20:41, 5 May 2010 (UTC)

Well I am concerned about it as a real physical entity, rather than as a mathematical abstraction or a computer program, and note that it evidently has a much better organization of it's spin possibilities than the cube and of the sphere, which is used a lot in spin situations without showing any orientation. And I can only see one axis of symmmetry, so that must be it.WFPM (talk) 22:55, 5 May 2010 (UTC) And if I knew the axis pf rotation, I could then start talking about it's corner impact stability, as if I almost knew what I was talking about, and which others avoid by not having any irregular structured rotational entity. So I can understand that it's easier to consider it as abstract entity, if you're not going to use it for anything.WFPM (talk) 00:09, 6 May 2010 (UTC) But I'm an applications Engineer, and very seldom worry about geometric details, unless there's an application.


 * Do I want to know what "better organization of [its] spin possibilities" means? —Tamfang (talk) 17:39, 6 May 2010 (UTC)
 * No, you don't.&mdash;Tetracube (talk) 18:07, 6 May 2010 (UTC)

Well consider your spinning image. It's spinning! But is that the way it would spin around some oblique axis? I don't think so. So how would it spin? Probably around some axis of symmetry. So what is the logic of the spin of a real physical octahedral structure. But rather than worry about that, most people just assume it to be some kind of spinning spheroid and say it has to be spinning up or down. And for an individual particle that's not too bad. But when you have multiple particles it might be a significant factor in it's behavior. And since spinning particles have stored angular momentum and kinetic energy, I'm forced to to try to figure the relationship of the spin properties to their angles of spin, and try to figure out the lowest stored energy mode and maybe some incremental values. That's what an inquiring mind will do to you.WFPM (talk) 21:09, 6 May 2010 (UTC) And incidentally I store my models in a cubic volume of space within a cut down 2 liter milk carton and so I don't think your image of an octahedron within a spacial cubic volume with all 6 corner of the octahedron in contact with the center of the cube faces can represent the true relationship of a cubic container to an enclosed octahedron.WFPM (talk) 16:25, 7 May 2010 (UTC)


 * Certainly that is how it would spin around the particular axis arbitrarily chosen for that animation. How else would it spin about that axis? — A rotation axis other than a symmetry axis has the advantage of providing a greater variety of views. — This article is not about physical particles.
 * What gave you the idea that the picture of an octahedron and its dual cube was meant to show efficient packing? (Heh, if there is a worse orientation for packing an octahedron in a tangent cuboid, I'd like to know it.) —Tamfang (talk) 00:12, 8 May 2010 (UTC)

When I put the "octahedral" model into the cubic container, the bottom of the octahedral rests on the bottom of the cube and the cover closes on the top. But the corners of the model fit into the side corners of the cube at the midpoint vertacle distance. If I used a cube of a size that would allow the octahedron to contact the 4 sides of of the cube at the center, then the two other ends of the octahedron wont reach to the top and bottom center faces, and the octahedron drops down and leaves a distance of space above the top of the top of the octahedron. And if I cut the box down to fit then I no longer have a cubic volume of storage space. Is that what you call a cuboid?WFPM (talk) 00:38, 8 May 2010 (UTC) It certainly isn't an efficient way to store an octahedron, because it uses twice the base area and volume of space as is needed. And the spin of an object in space has to be about an axis through the center of mass. And if the axis of rotation doesn't extend through the more remote part parts of the object, you can note an inefficiency with relation to it's stored Mvr value, which is more apparent when considering the rotation of say a toothpick.WFPM (talk) 02:24, 8 May 2010 (UTC)


 * By 'cuboid' I meant any rectangular box, though I see that some use cuboid to mean any hexahedron whose faces are all quadrilateral. — This article is not concerned with efficiency of rotation (whatever that means). —Tamfang (talk) 19:52, 8 May 2010 (UTC)

Okay. But you're showing a cube? with the ends in contact with the center of all 6 faces of the cube. And if it's an octahedron the two Corners of the 4 sided end of the octahedron would not be in contact with the faces of the cube.WFPM (talk) 20:15, 8 May 2010 (UTC)


 * "the two Corners of the 4 sided end"?? I give up trying to understand your confusion. —Tamfang (talk) 02:39, 9 May 2010 (UTC)

When I put my octahedron models into a close fitting cube, the 4 sided central part of the structure slips down to the central distance between the top and the bottom of the cubic container volume. And the top and bottom points of the octahedron touch the centerpoint of the cubes' top and bottom faces. But the corners of the 4 central section are in the corners of the side sections of the cube, and not in contact with the centers, as shown in the image. So we have the pyramid bottom section of the octahedron upside down in the bottom of the cubic volume, and the top half pyramid of the octahedron located upright in the top half section of the cubic volume. And only 2 of the octahedron's corners are touching the faces of the cube. In the larger cubic volume, where the 4 corners of the central section coincide with the centers of the side faces of the cube, the top and bottom corners don't extend in distance far enough to touch the faces of the larger cube.WFPM (talk) 14:24, 9 May 2010 (UTC)


 * Sure, you can arrange an octahedron within a cube (or cuboid) in that way. But, as I think I said before, that picture has nothing to do with "close fitting".  It's not about efficient packing, it's about the definition of a dual polyhedron. If that's beyond you, consider that the arrangement with corners touching all faces of the cube is the most symmetrical arrangement of an octahedron within a cube, because any corner can be substituted for any other. —Tamfang (talk) 22:40, 9 May 2010 (UTC)

There is, of course a polyhedron that will fit inside a cube and whose corners contact the centers of all 6 faces of the external cube. But that polyhedron has the shape of a smaller cube, and not an octahedron. And isn't that significant?WFPM (talk) 00:23, 11 May 2010 (UTC)


 * Huh, what? A cube has 6 faces, and therefore 6 centers of faces. Therefore, your polyhedron must have exactly 6 corners. A cube does not have 6 corners; it has 8. Therefore, your polyhedron cannot possibly be a smaller cube. In fact, it has to be an octahedron.&mdash;Tetracube (talk) 00:35, 11 May 2010 (UTC)

Yes!, Your right and I was wrong. I have to extend the length of the triangles such that the third axis through the face centers has the same length as the other two. And I was previously thinking about a general case of a dual pyramid structure and the octahedron is a special case of that. So please excuse my meanderings and I appologize for interfering in your general discussion.WFPM (talk) 01:19, 11 May 2010 (UTC)

Boat
Should some mention of the boat octahedron (http://mathworld.wolfram.com/Boat.html) be made? 130.102.158.19 (talk) 08:28, 19 October 2014 (UTC)
 * I see, it's an augmentation of 3 regular tetrahedra. Here's some images, boat is first one obviously, others would be concave decahedron and dodecahedron. I see Mathworld has no sources listed. Tom Ruen (talk) 13:30, 19 October 2014 (UTC)
 * Biaugmented_tetrahedron.png Triaugmented_tetrahedron.pngQuadaugmented_tetrahedron.png

A Commons file used on this page has been nominated for deletion
The following Wikimedia Commons file used on this page has been nominated for deletion: Participate in the deletion discussion at the. —Community Tech bot (talk) 17:54, 18 May 2019 (UTC)
 * Rhombicdodecahedron.jpg

Link to German language
I intended to add a menu link to de:Oktaeder, but get an error "Site link dewiki:Oktaeder is already used by item Q12557050", which is []. Would anybody experienced fix this, pls? -- 185.69.244.210 (talk) 20:35, 10 January 2021 (UTC)

Hey, cool, I just tried reporting that to https://www.wikidata.org/wiki/Wikidata:Interwiki_conflicts/Unresolved/2021#regular_octahedron_(Q12557050)/_? and while researching. I find this -- 87.79.110.19 (talk) 03:56, 14 September 2021 (UTC)