Talk:Dodecahedron

I don't think Dodecaeder should be merged with dodecahedron. I think there should be two separate pages: one dealing with the Platonic solid and one dealing with the weird antique. Merge request deleted and "see also" added. Robinh 21:00, 26 May 2005 (UTC)

It's been a long time since I saw the episode, but I'm pretty sure that the polyhedra in the Star Trek episode were not dodecahedra but something else, either cuboctahedra or rhombicuboctahedra. --Matt McIrvin 04:04, 31 July 2005 (UTC)

There are some pictures here; it's a cuboctahedron. Should probably delete the trivia item. --Matt McIrvin 04:11, 31 July 2005 (UTC)

Dodecahedron Canonical coordinates

Can someone explain or update the canonical coordiantes:

"Canonical coordinates for the vertices of a dodecahedron centered at the origin are {(0,±1/φ,±φ), (±1/φ,±φ,0), (±φ,0,±1/φ), (±1,±1,±1)}, where φ = (1+√5)/2 is the golden mean."

if there are 20 vertices, then why are there not 20 coordinate sets (x, y, z)? Zalamandor 23:23, 8 December 2005 (UTC)

• The 20 vertices are hidden in the sign permutations. (±?,?,?) has two values. (±?,±?,?) has 4 (2x2) value permutations and (±?,±?,±?) has 8 (2x2x2) permutations. So the above paragraph lists 4+4+4+8=20 vertices. Tom Ruen 23:55, 8 December 2005 (UTC)

The dodecahedron is one of the most complex, completely symetrical, geometric three-dimensional figures.

• I understand the permutations but this coordinate set seems wrong: (±1,±1,±1). These describe a cube with edge length = 2 units, centered at the origin. Let's reduce that to the upper square: (±1,±1,1). No matter how you rotate a dodecahedron, it's obviously impossible to get four vertices to end up at these coordinates since with this body, you always have five points in a plane and the angle between any two vertices is != 90 degrees. What's wrong? --Digulla (talk) 12:46, 16 February 2009 (UTC)
  • The four vertices do not belong to the same face. Consider the compound of five cubes. Its convex hull is a regular dodecahedron. Notice the two vertices "above" the prominent green face; if the green cube is in canonical position, these two are among the twelve whose coordinates contain φ. —Tamfang (talk) 07:01, 17 February 2009 (UTC)
    • If someone else wonders the same thing, the vertices which lie on the cube are the "outer" vertices of the two pentagons forming the "roof" of the sample image: Find the three topmost points; two vertices of the cube are the leftmost and the rightmost point of those three. If you follow the line down from the topmost point, then the next two vertices of the cube are to the left and right. The four bottom vertices are the ones connected next to the lowest edge. See this image: --Digulla (talk) 08:56, 19 February 2009 (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 00:55, 4 March 2006 (UTC)

Suggest correcting the dihedral angle in the table from arccos(-1/5) to arccos(-1/sqrt(5)) or equivalent. cadull 21:16, 10 March 2006 (UTC)

Electrical resistance

I removed this long unsupported statement under "uses":

• If each edge of a dodecahedron is a one-ohm resistor, the resistance between adjacent vertices is 19/30 ohm, and that between opposite vertices is 7/6 ohm.

I stumbled upon this paper [1], but couldn't clearly confirm or contradict the uncited statement above. So here it is if anyone cares! Tom Ruen 02:37, 29 September 2006 (UTC)

It's right there in Table I (page 643). It also gives the resistances between vertices which are neither adjacent nor opposite. —Keenan Pepper 06:01, 30 September 2006 (UTC)

Dihedral angle in the frame

The dihedral angle formula is wrong in the frame on the side of the article and I could not find how to modify this frame. (Unsigned comment)

The correct formula is π − 2ψ where ψ is defined by:

cos(ψ) = csc(π / 5)cos(π / 3) = cos(π / 3) / sin(π / 5)

(This is from Coxeter's Regular Polytopes.)—Tetracube 17:54, 13 November 2006 (UTC)

It's fixed now. Someone was trying to be a little too fancy with templates when they made that frame. As a result, it is very hard to edit. The correct page to edit is Template:Reg_polyhedra_db. -- Fropuff 18:05, 13 November 2006 (UTC)

Uses Vs Trivia

I renamed the "Uses" section because only one of the three facts listed can be considered an "use" of a Dodecahedron (the die). I'm not sure that "Trivia" is a good name, another possibility is "curious facts". Ossido 16:29, 30 December 2006 (UTC)

This following section was removed. I moved it here for reference.

Regular dodecahedra in the arts, sciences, and popular culture

Roman dodecahedron

• Small, hollow bronze Roman dodecahedra dating from the 3rd century A.D. have been found in various places in Europe. Their purpose is not certain.
• A dodecahedron sits on the table in M. C. Escher's lithograph print "Reptiles" (1943), and a stellated dodecahedron is used in his "Gravitation". In Salvador Dalí's painting of The Sacrament of the Last Supper (1955), the room is a hollow dodecahedron.
• One of the characters in The Phantom Tollbooth, a children's novel from 1961, is named Dodecahedron and is a man with 12 faces.
• The Dodecahedron was the mysterious power source for an underground city in the Doctor Who episode "Meglos" (1980).
• In "Blood Feud", an episode of The Simpsons, Lisa attempts to teach Maggie the word dodecahedron.
• In Carl Sagan's novel Contact, the transport device constructed to the plans transmitted by the alien intelligence is dodecahedral.
• "Dodecaheedron" (misspelled, possibly intentionally, with an extra "e") is the title of a song by Aphex Twin.
• Plato in the dialogue Timaeus c.360 B.C associated platonic solids with the four classical elements. Aristotle added a fifth element, aithêr (aether in Latin, "ether" in English) and postulated that the heavens were made of this element, but he had no interest in matching it with Plato's fifth solid.
• In 2003, an apparent periodicity in the cosmic microwave background led to the suggestion, by Jean-Pierre Luminet of the Observatoire de Paris and colleagues, that the shape of the Universe is a finite dodecahedron, attached to itself by each pair of opposite faces to form a Poincaré homology sphere. ("Is the universe a dodecahedron?", article at PhysicsWeb.) During the following year, astronomers searched for more evidence to support this hypothesis but found none.
• The 20 vertices and 30 edges of a dodecahedron form the map for an early computer game, Hunt the Wumpus. In the seminal 1980s computer game Elite, the more advanced "Dodec" class space stations took the form of dodecahedra. The save points in the Castlevania games, Castlevania: Symphony of the Night (for the PSX) and Castlevania: Harmony of Dissonance (GBA) are shaped like dodecahedrons. In the Nintendo 64 game Paper Mario, the mountains in the background of Toad Town are dodecahedra.
• The regular dodecahedron is often used in role-playing games as a twelve-sided die ("d12" for short), one of the more common polyhedral dice.
• Desk calendars are occasionally made in the shape of a dodecahedron, usually from a die-cut folded card, with one month on each face.

There are a few face-transitive (with congruent but irregular faces) dodecahedrons missing from the list

• The ones missing are the octahedral pentagonal dodecahedron, the tetrahedral pentagonal dodecahedron, and the trapezoidal dodecahedron.
• If you look up "Isohedron" on Mathworld, you can see a rotating models of each of them.
• As a side note, I would love to know if the pyritohedron is the same thing as either the octahedral or tetrahedral pentagonal dodecahedron. —Preceding unsigned comment added by 63.72.235.4 (talk) 15:15, 10 February 2009 (UTC)

(I took the liberty of adding the relevant link.) It's hard to tell from the small picture, but yes, I think the octahedral pentagonal dodecahedron is the "pyritohedron" (a name I dislike: it's the whole body, not the faces, that resembles a pyrite crystal). It has T_h symmetry. —Tamfang (talk) 17:27, 12 February 2009 (UTC)

Skeletal Platonic solids

I removed the following text from the article, added by an IP, because it wasn't added in the right place and was uncited; I wasn't sure what to do with it:

Skeletal Platonic Solids

If the regular poyhedra are considered as skeletal 3-dimensional figures with flexible vertices and unidimensional edges, two of them. the hexadron(cube) and the dodecahedron exhibit properties different from the other three in that they can be folded or collapsed; The hexahedron first into a double square then a line. and the dodecahedron into a triangle. This can be demonstrated by constructing them out of drinking straws and twistable inserts

Somebody please verify this, find a citation, and integrate it into the article. Thanks.—Tetracube (talk) 03:18, 30 March 2009 (UTC)

It's correct that the hexahedron, or cube, and the dodecahedron, if you make them out of drinking straws and twistable inserts, can be collapsed, but what they collapse into is wrong. --Professor M. Fiendish 12:07, 22 August 2009 (UTC)

"2 acos(36)"

The formula "2 acos(36)" seems to make no sense. Should it be 2 acos(36°), or is it completely wrong? -- 05:42, 19 October 2009 (UTC) —Preceding unsigned comment added by 80.168.224.185 (talk)

The argument to acos must be a value between -1 and 1, otherwise it is undefined. I don't know where this formula came from, but it looks suspiciously wrong.—Tetracube (talk) 17:03, 19 October 2009 (UTC)

If it makes any sense, it should be 2cos(36°). Tom Ruen (talk) 19:29, 19 October 2009 (UTC)

It is a pretty trivial claim, more about a pentagon than as a polyhedron, unclear value. Ugh, the more I look the more the whole article seems a largely unorganized collection of unrelated facts. Tom Ruen (talk) 00:02, 20 October 2009 (UTC)

insphere: factors

In the new section "Dimensions" (thanks), the radius of the insphere is given as $\frac{a}{20} \sqrt{250 +110\sqrt{5}}$. It could be put in lower terms as $\frac{a}{2} \sqrt{ \frac{5}{2} + \frac{11}{2\sqrt{5}} }$. Any strong preferences either way? —Tamfang (talk) 17:50, 23 February 2010 (UTC)

One could also describe the diameter of the insphere as $\frac{a \phi^2}{\sqrt{3-\phi}}$, which is even simpler. It turns out that nearly all of the dimensions of the regular pentagonal dodecahedron can be written with similar brevity, all using the golden ratio. I can't cite a source, because I've done the derivations myself. In any case, I think someone needs to start a general discussion about the relationship between the platonic dodecahedron and the golden mean. I propose there is enough material showing the relationship between the two to start a separate related page. Brycehathaway (talk) 01:57, 12 January 2011 (UTC)

I'm trying to reconcile any of the above expressions with Coxeter's $\frac{a}{2} \sqrt{\frac{\phi^5}{\sqrt{5}}}$ and I keep getting stray factors of 2. I am prone to copying-errors in algebra ... —Tamfang (talk) 05:55, 22 January 2011 (UTC)

Note that I wrote "diameter" above, not radius. Tamfang's recent edit to the dimensions is incorrect because of this misunderstanding; please remember to always verify changes before committing them. Also, I am not sure why this user chose to remove the reference to the regular pentagon's apothem. Let me summarize all of the radii in a geometrically meaningful form: Given a regular pentagonal dodecahedron with edge length one, the distance from the center to vertex is the product of the golden mean with half the length of the long diagonal of a unit cube, the distance from the center to the midpoint of an edge is the product of the golden mean with half the golden mean, and the distance from the center to the midpoint of the face is the product of the golden mean with the length of a unit regular pentagon's apothem. I care less about how that is represented than that it is represented, because doing so connects the dimensions of this shape to related shapes along with the golden mean. Considering this shape has a stronger affinity with geometry than algebra, the focus of these formulas should shift accordingly. I will leave it to Tamfang to make corrections, since this user edited my additions to the dimensions section and I would rather avoid edit wars. Brycehathaway (talk) 00:27, 24 January 2011 (UTC)

Thanks for explaining my error.

The formula using the apothem felt out of place among the simplifications using φ, in that it makes more work for anyone who happens not to know already what that value is. It would make more sense imho to give a literal value, using nothing more arcane than φ, and then say how this value relates to the apothem. —Tamfang (talk) 05:52, 24 January 2011 (UTC)

I agree with you on this, Tamfang. I will update the section accordingly. Brycehathaway (talk) 20:57, 24 January 2011 (UTC)

filling a sphere

Is it really bigger than an icosahedron when made to fit in a sphere? References/Math? Simanos (talk) 15:32, 1 March 2010 (UTC)

Using the numeric values given for volume and circumradius in the respective articles, I get the ratios 0.6649 for the dodecahedron, 0.6055 for the icosahedron. I tried it with the algebraic formulae but made a booboo somewhere (getting a dodec >3 times as big as its circumsphere). —Tamfang (talk) 20:20, 2 March 2010 (UTC)

Regarding the supression of the Dodecahedron?

According to Carl Sagan during his TV series Cosmos - Knowledge of the Dodecahedron was kept hidden and suppressed by the Pythagoreans who discovered it. Should this be added into the history section on the shape? Omega2064 (talk) 19:03, 4 July 2010 (UTC)

If you can cite a reliable source for it, yes.—Tetracube (talk) 14:43, 7 July 2010 (UTC)

Field of view of

If I have done my math right, the maximum field of view occupied by a face as seen from the center of the dodecahedron is 2*acos[1.1135/1.40126], or just under 75.76 degrees. This might be of interest in selecting the field of view needed for the lens of an omnidirectional camera setup, for instance. With a typical aspect-ratio rectangular sensor such a 35mm film only the angle between a vertex and the center of the opposite side needs to fit on the shorter side of the sensor (24mm for 35mm film), so the calculation of the longest 35mm-equivalent lens that can fit a whole pentagon in the frame from the center of a dodecahedron is 1/(tan[(acos[1.1135/1.40126] + acos[1.1135/1.30902] )/2 ] /12) = 17.43mm max lens length. (the 12 in the calculation is = 1/2 the film width)

Some stats useful for doing calculations for other sensors: if the height is 24mm as stated before, then the side of the pentagon is just under 15.6mm, the width (penagon's diagonal) is under 25.24mm, and the diameter of the circumscribing circle is less than 26.54mm. A square sensor should be able to use a longer lens.Enon (talk) 23:48, 24 February 2011 (UTC)

Are these left-handed Cartesian coordinates?

To draw a dodecahedron myself, I had to reverse the Golden Ratio with its inverse in the Cartesian coordinates, or change the rotation of the indices. As I don't believe I'm having this problem drawing other things, I'd appreciate it if someone would check what's here. I'm using a right-handed coordinate system. Short of that, I wasn't clear on what tag to use, as I couldn't find a section-limited expert needed tag. So if someone improves on how I tagged this, that's great too. Thanks! Marc W. Abel (talk) 01:40, 3 May 2011 (UTC)

The coordinates are on the MathWorld page[2], and I have no reason to doubt them. Left-handed coordinates would make no difference. You can see the regular dodecahedron with the 8 vertices of the cube, plus 2 vertices above each of the square faces, forming wedges over each square. The side triangles of the wedges merge with side trapezoids as regular pentagons, as seen in the related irregular Pyritohedron#Geometric_freedom. Tom Ruen (talk) 05:26, 3 May 2011 (UTC)

What are you using to "draw"? —Tamfang (talk) 05:59, 3 May 2011 (UTC)

At a guess you might be drawing only one side of you polygons, whether polygon is displayed depends on which face is pointing towards the viewer, if its the notional "front" face then the polygon is displayed, if not the polygon is not displayed, see backface culling. Changing the order of the vertices changes the notion of which is the front face and hence the visibility.--Salix (talk): 14:40, 3 May 2011 (UTC)

Surface area - how is the formula obtained?

The article gives a formula, but does not explain how it comes to be in the first place. I have never studied such a formula, and my attempt to obtain it from the formula I did study (how to obtain the area of a regular polygon) gives me a formula that looks completely different. Here's the reasoning I followed:

• We know that the area of any regolar polygon is equal to P*a/2, where P is the perimeter and a is the apothem.
• In the case of a pentagon, P equals to 5*L, where L is the length of a side, so the surface area of a pentagon is 5*L*a/2
• We also know that the dodecahedron has 12 sides, so a first attempt to a surface area formula for the dodecahedron is 12*5*L*a/2 = 6*5*L*a = 30*L*a
• The apothem of the pentagon equals L*tan(54°)/2, so, if we replace this expression to a in the previous formula, we get 30*L*L*tan(54°)/2, which can be simplified to 15*L²*tan(54°).

The formula so obtained does not even resemble the one in the article. So, where does that one come from? Devil Master (talk) 22:36, 23 November 2011 (UTC)

I'm guessing the reduction comes from Exact trigonometric constants#36°: pentagon. Tom Ruen (talk) 22:59, 23 November 2011 (UTC)