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"semi-regular" should be defined; right now it is not clear what the difference between Johnson and Archimedian solids is. Furthermore, I take it that the platonic solids also count as Archimedian? --Axelboldt The first graph is wrong it is nt the largest — Preceding unsigned comment added by 212.166.157.122 (talk) 18:37, 24 November 2018 (UTC)[reply]

I don't think so. semi-regular, i think has to do with the fact that multiple kinds polygons can meet at a vertex And platonic solids are definitely not Archimedean. (since there are five platonic solids, and thirteen archimedean ones....)

I think not being face-uniform just means there are more than one kind of regular polygon used. And I guess that being vertex-uniform means that there is a rotation that can move any one chosen vertex to the place of any other chosen vertex, while also mapping all other verticies to the place of either itself or another vertex. I hope so, because if I guessed wrongly, then my pictures and coordinates probably aren't of Archimedean solids. Ксйп Cyp 20:24, 31 Jul 2003 (UTC)
Hey, that was a reply in less than 2 years... How's that for wiki-fast replies? Ксйп Cyp 23:16, 31 Jul 2003 (UTC)
Suppose to the above comment on rotation, mirroring also counts, since otherwise I can't see how a truncated cuboctahedron or truncated icosidodecahedron would count. Ксйп Cyp 10:13, 1 Aug 2003 (UTC)

A platonic solid is made up of only one type of regular polygon, and is vertex transitive. An Archimedian solid is any vertex transitive solid and is made up of two or more regular polygons. Obviously, these two can't be the same. And a Johnson solid is any solid that is made up of two or more regular polygons, that aren't vertex transitive. Vertex transitive means that any vertex, all the same polygons meet, in the same order and number. Hope that clears it up. Timeroot (talk) 00:33, 24 June 2008 (UTC)[reply]

Not quite. Vertex-transitivity means that the solid has a symmetry group under which any vertex maps to any other. The congruency of vertex figures follows from this, but not the other way around. The elongated square gyrobicupola (aka "false rhombicuboctahedron") has a triangle and three squares at each vertex, but the sixteen vertices of its octagonal prism are not in the same symmetry class as the eight vertices of the 'tops' of the cupolae. —Tamfang (talk) 04:02, 9 July 2008 (UTC)[reply]

The last sentence uses the term "regular vertex" without defining it. AxelBoldt 09:50, 2 Oct 2003 (UTC)

I worked a bit on the definition of Archimedean solids, but it's still not satisfactory. Right now, it's not clear why the Elongated Square Gyrobicupola is excluded. AxelBoldt 10:13, 2 Oct 2003 (UTC)

Found http://www.math.washington.edu/~king/coursedir/m444a03/as/polyhedra-links.html

Contains a lot of sections, seems to be in the format of mentioning an Archimedean solid site, someone reviewing it, and selecting their favourite solid. There's at least 2 about this page. I hope copying the comments about this site to here would be considered fair use or something... Κσυπ Cyp   19:15, 18 Jan 2004 (UTC)

Jennifer Brosten reviews http://en2.wikipedia.org/wiki/Archimedean_solid

This website was nice because it has the general idea of the Archimedean solids in a rather concise manor though it also offers more in depth information on polyhedras and all of the different Archimedean Polyhedra. The table showing the different Archimedean Polyhedra was well done, because it illustrated the figures while also giving useful information about the vertices, faces and how they meet. The illustrations make it so that you can see the 3D aspect, whereas many sites only show the front without being able to see what is happening at the back of the object as well. The web site does not stop at giving the general information of the different solids. If you click on the names of the solids, you are taken to a new web page which is devoted strictly to that solid.


My favorite Archimedean polyhedra would have to be the Icosidodecahedron. What made me pick this shape was first it's name, because it's kinda fun to try to say. The Icosidodecahedron is made up of 20 triangular faces and 12 pentagon faces. There are a total of 60 edges and 30 vertices. At each vertex, there are 2 triangles and 2 pentagons meeting. They go triangle-pentagon-triangle-pentagon.

Mine would have to be one with both icosahedral-group and octahedral-group symmetry. Does any such polyhedron, Archimedean or not, exist? —The Doctahedron, 68.173.113.106 (talk) 21:50, 22 November 2011 (UTC)[reply]

Mary Moser reviews http://www.ezresult.com/article/Archimedean_solid (Note: That is a not so good mirror of Wikipedia, without working pictures... Ironically, the only complaint about the article is that the pictures don't work.)

I was able to find the above website and saw that it had all the basic information presented clearly as well as some interesting history. I really appreciated that throughout the sight vocab words are linked to further definitions and explenations. There is a lot of potential in this sight unfortunately it seems the pages that are supposed to provide images of the polyhedra are not working, (at least I was unable to view them). Also I would like to see descriptions relating them to the platonic solids (essentially how we get the Archimedean polyhedron by truncating the platonic polyhedron).

I think a good example of the pictures and information my initial sight is missing can be found at http://www.ul.ie/~cahird/polyhedronmode/favorite.htm. I especially liked the animation showing the truncations for some of the polyhedra.

While each of the polyhedra we are discussing is really interesting and fun to explore, the assignment is to choose one favorite so I choose truncated cuboctahedron. It has 26 faces (12 squares, 8 hexagons and 6 octagons), 72 edges, and 48 vertices.

Educational toy

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Jovo [[1]] is a toy that is ideal for constructing Archimedean solids. Can such a link be in the Wikipedia, or is it too commercial? --80.162.63.207 17:01, 5 Feb 2005 (UTC)

Inscribed Archimedean polyhedra

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NOTE: This text for a new and uncompleted section was removed from article!

Edit here as you like and move back when it's done!
Tom Ruen 21:43, 14 January 2006 (UTC)[reply]

When the Archimedean polyhedra are inscribed in a sphere, they occupy the following percentages of that sphere's volume:

  • truncated tetrahedron 40.134%
  • cuboctahedron 56.270%
  • truncated cube 57.682%
  • truncated octahedron XX%
  • rhombicuboctahedron XX%
  • truncated cuboctahedron XX%
  • snub cube XX%
  • icosidodecahedron XX%
  • truncated dodecahedron XX%
  • truncated icosahedron XX%
  • rhombicosidodecahedron XX%
  • truncated icosidodecahedron XX%
  • snub dodecahderon XX%
I hope someone who has the percentages will add them here. I was attempting to make the Archimedean article include some of the same things that are in the Platonic article. Unfortunately, removing the list to the discussion page instead of leaving it on the article page means that it will probably languish here, and not get finished, but we'll see. Jimaginator 13:06, 19 January 2006 (UTC)[reply]

Here is how you could calculate those percentages: Go onto the page for that solid, and scroll to the part about the cartesian co-ordinates. Take the three numbers, square each of them, and take their sum (for the truncated tetrahedron, these would yield 11, for example). Now take the square root of that. Now, in that same section, it will say "For a side length of --." The truncated tetrahedron, for example, will have "the square root of 8". Divide this side length number by the previous number you got (the square root of 11), and put that into the formula for volume, which will be else where on the page. Now, divide that result of that formula by 4/3pi, and multiply by 100. This is your percentage, and it will work for any Archimedean solid. I already did a few. Timeroot (talk) 17:57, 24 June 2008 (UTC)[reply]

Wythoff construction

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All the Archimedean solids can be constructed by Wythoff construction of a spherical tiling.

Blue and yellow

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Hi,
I've just created the following table, and I think the bottom line would look much more logical with blue and yellow exchanged. The trigons and hexagons, which are blue at the moment, correspond to the yellow trigons in the two big files above. The squares (or trigon pairs in the snubs), yellow at the moment, have no corresponding faces in all the files above. They just correspond to the vertices in the two big files. Thus I think, these "new" faces should have the "new" color. Any thoughts? Watchduck (talk) 23:33, 7 August 2010 (UTC)[reply]



Yes. —Tamfang (talk) 03:01, 9 August 2010 (UTC)[reply]

That would make sense. Who made these pictures, anyway? —The Doctahedron, 68.173.113.106 (talk) 21:45, 22 November 2011 (UTC)[reply]

With almost 8 years delay I have done that now. Watchduck (quack) 01:42, 16 April 2018 (UTC)[reply]

Why are there only 13 Archimedean solids?

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Could someone add a section on why there are only 13 Archimedean solids please, and explain it well? Thanks, please do it asap Akhi666 (talk) 18:55, 11 October 2014 (UTC)[reply]

Figurate numbers?

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Just as platonic solids have formulas for generating figurate numbers Figurate_number, are there any formulas for generating numbers of Archimedean solids? — Preceding unsigned comment added by R3hall (talk • contribs) 05:47, 30 April 2016 (UTC)[reply]

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Images

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I propose replacing the solids and vertex figures (and nets) by more consistent images. This is partly the same topic I already brought up in the Blue and yellow section above. I made the proposed changes in the article (edit) and in the template (edit) where the images for the individual articles come from. The old vertex figure images show the complete faces around the vertex. I don't think that is necessary when they are shown in context with the full solids — which is always the case in these articles. And the vertex figure is a single polygon after all.

@Tomruen: I request that you do not immediately revert my changes here and in the template, and instead leave them visible for a while, so others can see them and offer their opinion. I am quite sure that you will be against this change, but it would be good to get more than two opinions on this. (Apparently that was already too much to ask for.)

I found it absurd to waste vertical space in the transparent column for these animation links, so I did that thing with the gear icons. It's just a proposal, and even someone who finds it horrible should not find it horrible enough to revert my edit because of that. Just say so below, and a better solution will be found. Watchduck (quack) 01:44, 16 April 2018 (UTC)[reply]

The partial polygon faces are confusing. There's nothing to argue here. That's a fact. Tom Ruen (talk) 01:48, 16 April 2018 (UTC)[reply]
Snub cube
A wider discussion on illustrating vertex figures in general has now been started at Talk:Vertex figure#Images of solids. I have commented there and posted some example diagrams. — Cheers, Steelpillow (Talk) 11:32, 16 April 2018 (UTC)[reply]
Vertex figure images similar to the one shown on the right are among the possible alternatives. See the general discussion mentioned above. Watchduck (quack) 17:45, 16 April 2018 (UTC)[reply]
Compare the illustration of cuboctahedron and rhombicosidodecahedron vertex figures in Figure 21.1 (p. 289) in The Symmetries of Things by Conway et al. Watchduck (quack) 23:30, 26 July 2018 (UTC)[reply]
This is also dealt with in the linked discussion. — Cheers, Steelpillow (Talk) 17:58, 27 July 2018 (UTC)[reply]
(Comment by Tomruen copied here by Watchduck from the linked discussion, because my answer is specific to this article.)
TSoT also shows vertex figures with number labels for each p-gon. Tom Ruen (talk) 00:40, 27 July 2018 (UTC)[reply]
with labels
new (2020)
I am not opposed to including labels. How about something like the image on the right? (Shown as 140px like in the infobox.) Watchduck (quack) 16:35, 27 July 2018 (UTC)[reply]
The following reply of mine also needed to be copied across from the other discussion because it is also relevant here.
@Watchduck: You are cherry-picking snippets to try and back your case. The text in TSOT accompanying the illustrations on p.289 states categorically that "the vertex figure of the cuboctahedron (Figure 21.1 left) and icosidodecahedron are rectangles whose sides are numbered ...". The drawing shows a slice through a polyhedron, the "pyramid" is merely that part to one side of the slice, with the overall polyhedral form remaining intact. It does not, as you claim, merely "show a pyramid cut from the solid". Note the close correspondence between Fig 21.1 and the cuboctahedron drawing I mentioned in the other discussion.
I am not sure if polychora have been mentioned here, but if a pyramid does occur as a vertex figure then it is of some four-dimensional polychoron. It is, as I have said, mathematical madness to use the same pyramid in an attempt to illustrate the lower-dimensional vertex figure of a polyhedron. The central point adds nothing to the description here, save confusion.— Cheers, Steelpillow (Talk) 20:19, 27 July 2018 (UTC)[reply]
No need to outdent. I doubt this belongs here, because it's not specific to Archimedean solids. With your quote from TSoT ("states categorically" etc.) you imply that our disagreement is factual, and not just stylistic. I consider this a straw man done in bad faith, because in the other discussion I have already made clear that we do not disagree about what a vertex figure is ("not proposing to illustrate a pyramid" etc.). Watchduck (quack) 21:13, 27 July 2018 (UTC)[reply]
The problem is that when the central vertex is included, what the image shows is indeed a pyramid, regardless of whether or not it is proposing to illustrate that. Double sharp (talk) 09:32, 18 September 2018 (UTC)[reply]

This should be solved now. I created a set of vector files that show the full faces and the outline of the vertex figure. 01:58, 10 December 2020 (UTC)

Capitalization

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@Wcherowi: About colons, Wikipedia:Manual of Style/Capital letters#Initial letters in sentences and list items says, "The same {not capitalizing the word immediately following a punctuation mark} usually applies after colons, although sometimes the word following a colon is capitalized, if that word effectively begins a new grammatical sentence, and especially if the colon serves to introduce more than one sentence." I think that the length of the sentence that follows the colon warrants capitalization. Also, while the section does not mention parentheses, the sentence inside the parentheses is a complete sentence, so I assume that it should be capitalized. Care to differ or discuss with me? The Nth User 02:36, 4 November 2018 (UTC)[reply]

You have an arguable point concerning the colon and perhaps a better solution would be to eliminate the colon altogether as the sentence containing it is borderline run-on. As for the parenthetical remark I would point to the Sentences and brackets section of MOS:PAREN which states, "A sentence that occurs within brackets in the course of another sentence does not generally have its first word capitalized nor end with a period (full stop) just because it starts a sentence". --Bill Cherowitzo (talk) 04:50, 4 November 2018 (UTC)[reply]
@Wcherowi: I didn't see the information at MOS:PAREN in Wikipedia:Manual of Style/Capital letters#Initial letters in sentences and list items, so maybe it should be added, but now that I know that the MoS has a guideline about it, I am going to go by that. As for the colon, I agree that the sentence should be broken up, but I am not sure that the colon is the best place to do it. Even if it was broken at the colon, the sentence would still be somewhat long.

Some authors give a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (each vertex looks the same from close up), so only a local isometry is required, but then they omit a 14th polyhedron that meets this weaker definition, elongated square gyrobicupola (or pseudo-rhombicuboctahedron), but not the stronger definition and only list the 13 Archimedean solids.

The obvious place to split this sentence is at the switch from discussing which isometry definition authors use to discussing whether or not authors include the pseudo-rhombicuboctahedron. This results in two sentences of about the same length.

Some authors give a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (each vertex looks the same from close up), so only a local isometry is required. However, these authors omit a 14th polyhedron that meets this weaker definition, elongated square gyrobicupola (or pseudo-rhombicuboctahedron), but not the stronger definition and only list the 13 Archimedean solids.

Because the part before the colon isn't that long, the colon probably doesn't have to be replaced with a period. Keeping the colon, moving the naming of the polyhedron fowards a few words, and adding an i.e. to the beginning of the statement in the parentheses results in this.

Branko Grünbaum (2009) pointed out a widespread error in the literature on Archimedean solids: some authors give a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (i.e. each vertex looks the same from close up), so only a local isometry is required. However, these authors omit a 14th polyhedron, the elongated square gyrobicupola (or pseudo-rhombicuboctahedron), that meets this weaker definition but not the stronger definition and only list the 13 Archimedean solids.

Incidentally, because the colon would now be introducing two sentences (because neither the fact that most authors use the weaker definition nor the fact that most authors do not include the pseudo-rhombicuboctahedron amounts to an inconsistency or contradiction individually), there would be an even stronger case for capitalizing the word some. Care to differ or discuss with me? The Nth User 04:16, 6 November 2018 (UTC)[reply]
These are definite improvements, but I still think that this can be done without a colon. For example,

Branko Grünbaum (2009) pointed out that some authors give a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (i.e. each vertex looks the same from close up), so only a local isometry is required. Grünbaum decries a common error in the literature, namely, these authors omit a 14th polyhedron, the elongated square gyrobicupola (or pseudo-rhombicuboctahedron), that meets this weaker definition but not the stronger definition and thus only list the 13 Archimedean solids.

--Bill Cherowitzo (talk) 16:45, 6 November 2018 (UTC)[reply]

@Wcherowi: As a whole, I like your version, but there are some things that I would change, like making the point of the first sentence more obvious sooner and removing the word decries, which I think is too strong. How about this?

Branko Grünbaum (2009) pointed out an error where some authors give a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (i.e. each vertex looks the same from close up), so only a local isometry is required. The contradiction arises when these authors omit a 14th polyhedron, the elongated square gyrobicupola (or pseudo-rhombicuboctahedron), that only meets this weaker definition. In order to be consistent, the authors who omit the pseudo-rhombicuboctahedron (and most do) must use the stronger, global isometry-based definition instead of the weaker, local isometry based definition.

I think that summarizing the inconsistency in a way that offers a resolution to the problem. Care to differ or discuss with me? The Nth User 04:07, 7 November 2018 (UTC)[reply]
We are getting closer, but I do have some problems with this version. First of all, it is not an error to use a weaker definition, the error only arises if you use the weaker definition and don't include the 14th example. Also, this is not a contradiction, it is a bona fide error. I used "decries" only because I didn't want to repeat the use of "points out" in two adjacent sentences, any other wording would suffice. --Bill Cherowitzo (talk) 19:40, 7 November 2018 (UTC)[reply]
How about something like
Branko Grünbaum (2009) observed that a 14th polyhedron, the elongated square gyrobicupola (or pseudo-rhombicuboctahedron), meets a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (i.e. each vertex looks the same from close up), so only a local isometry is required. Grünbaum pointed out a frequent error in which authors define Archimedean solids using this local definition but omit the 14th polyhedron. If only 13 polyhedra are to be listed, the definition must use global symmetries of the polyhedron rather than local neighborhoods.
? —David Eppstein (talk) 19:54, 7 November 2018 (UTC)[reply]
Thanks David, that looks good to me. --Bill Cherowitzo (talk) 20:11, 7 November 2018 (UTC)[reply]
My only suggestion is that the last sentence be rephrased to If the weaker, "local" definition is to be used, the 14th polyhedron, the pseduo-rhombicuboctahedron, must be included. because that is what Grünbaum has suggested. Care to differ or discuss with me? The Nth User 01:23, 9 November 2018 (UTC)[reply]
I don't really feel strongly about this, but I chose the contrapositive formulation because our article generally takes the position that there are only 13, despite what Grünbaum said. His suggestion to use 14 and the weaker definition have not been widely accepted. So your formulation looks like "if counterfactual thing X, then counterfactual thing Y" (a vacuous truth) while the way I wrote it looks like "if (thing we already say is true in the article), then (other thing we already say is true in the article)". They're logically equivalent but I think it's less confusing that way. —David Eppstein (talk) 01:40, 9 November 2018 (UTC)[reply]
@David Eppstein: You have a valid point. What about this?

Branko Grünbaum (2009) observed that a 14th polyhedron, the elongated square gyrobicupola (or pseudo-rhombicuboctahedron), meets a weaker definition of an Archimedean solid, in which "identical vertices" means merely that the faces surrounding each vertex are of the same types (i.e. each vertex looks the same from close up), so only a local isometry is required. Grünbaum pointed out a frequent error in which authors define Archimedean solids using this local definition but omit the 14th polyhedron. Grünbaum suggests that since the weaker, "local" definition is used more often, the pseduo-rhombicuboctahedron should be counted as an Archimedean solid. However, most authors only list 13 polyhedra, which means that they must use the definition that utilizes global symmetries of the polyhedron rather than local neighborhoods.

Care to differ or discuss with me? The Nth User 04:04, 10 November 2018 (UTC)[reply]
Your version is internally inconsistent. It says that the local version is used more often, and then it says that most authors must use the global definition. Which is it, and what is the evidence for the claim about the frequency of use of different definitions? —David Eppstein (talk) 04:22, 10 November 2018 (UTC)[reply]
I think that that is the error that Grünbaum points out: Most authors use the weaker definition but omit the pseudorhombicuboctahedron. Please tell me if that is not what you mean. Care to differ or discuss with me? The Nth User 04:30, 10 November 2018 (UTC)[reply]
If your last sentence says that most authors should and do use the global definition, it contradicts the earlier sentence that says that most authors use the local definition. If it says that most authors should use the global definition (but don't), it is an inappropriate insertion of opinion, because it would be equally valid for those authors to switch to the 14-polyhedron count and stick with the local version. And in any case, "most" means "more than half" and would need to be backed by a reliable source that actually surveys authors and counts how many of them follow each variant, which we don't have. —David Eppstein (talk) 04:55, 10 November 2018 (UTC)[reply]
My last sentence means that must authors do not include the pseudorhombicuboctahedron, so they must use the stronger definition for consistency, while the earlier sentence says that Grünbaum thinks that all authors should include the pseudorhombicuboctahedron. Neither of these is my opinion. Also, the article already says that most authors omit the pseudorhombicuboctahedron. Care to differ or discuss with me? The Nth User 05:02, 10 November 2018 (UTC)[reply]
You're still falling into the trap of dictating to authors which solution they should take rather than merely pointing out that some of them got it wrong. —David Eppstein (talk) 05:55, 10 November 2018 (UTC)[reply]
In that case, just remove the last sentence and put the rest in. Care to differ or discuss with me? The Nth User 23:49, 10 November 2018 (UTC)[reply]

User:The Nth User/Extended Archimedean solid table

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What do you think? I think that it helps to show underlying pattern and definitely belongs somewhere, but not necessarily here. Care to differ or discuss with me? The Nth User 03:44, 18 December 2018 (UTC)[reply]

Snub Cube Errata

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In the "Classification" data table for the 13 solids, the Volume shown for Snub Cube, 7.889295, is incorrect. The correct value, 7.8894774, is available at various websites. My own calculation with standard precision Excel comes to 7.88947739998, so 7.8894774 is a deceptively nice approximation. The Sphericity value, =(pi*(6V)^2)^(1/3)/A, appears to have been calculated with correct surface area, A, and the incorrect volume listed. Consequently, the Sphericity listed, .9651814, is also incorrect. I have snub cube area as 19.85640646 and sphericity as .96519625.

I am confident of this errata but would prefer one of the primary authors confirm and make the change. Change Snub Cube volume, was 7.889295, to 7.8894774. Change Snub Cube sphericity, was .9651814, to .9651963. JFSather (talk) 19:15, 19 October 2021 (UTC)Jim Sather [1] [2] [3][reply]

References

  1. ^ https://mathworld.wolfram.com/SnubCube.html
  2. ^ https://www.geometrycode.com/free/polyhedra-math-tables-for-platonic-and-archimedean-solids/
  3. ^ https://rechneronline.de/pi/snub-cube.php

Johnson

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The Archimedean solids are a subset of the Johnson solids,

Oh? Usually the latter are defined as strictly convex polyhedra with regular faces that are not uniform. —Tamfang (talk) 04:45, 13 September 2023 (UTC)[reply]

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