User talk:Double sharp

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Whoo yeaah![edit]

That's awesome Neptunium passed its GA! Thanks for all the work you put into it and for dealing with the nomination process. I'm sorry I haven't been on lately, I've been crazy busy in real life and it looks like it's probably going to be that way for a while unfortunately (like probably through at least the end of August, maybe longer; not really sure right now) But yeah, when things do calm down again, I'll definitely be up for doing some more element articles if you're not to busy then either. Thanks again for all your help and hard work with the article! Thingg 00:06, 12 July 2014 (UTC)

Thank you! It wouldn't have passed without your help.
I think the next ones I'll do are flerovium, thorium, and iron, but they'll have to wait for some time. (Fl will definitely be first: I'm almost done, with one section not yet rewritten. Th and Fe will take a considerable amount of time.) Double sharp (talk) 04:50, 12 July 2014 (UTC)
Easy actinides might be Md, No, and Lr, as there is not much to say. Double sharp (talk) 13:28, 31 July 2014 (UTC)
I did Md. No and Lr should be about as easy, although there would be more to say about history due to the Transfermium Wars (which were honestly really the Transmendelevium Wars). Double sharp (talk) 19:34, 8 August 2014 (UTC)
I started No in my userspace. Double sharp (talk) 19:43, 11 August 2014 (UTC)
I finished No and Lr, both awaiting GA review. Double sharp (talk) 12:42, 22 August 2014 (UTC)

Period 1 element and the others[edit]

I noticed that you redirected articles on the Period elements. This includes Period 1, which is the lead article in Wikipedia:Featured topics/Period 1 elements. So before I take this topic for review, I want to know if there was a consensus to redirect these to Period (periodic table). I don't want it to be that the redirects were unjust or anything. GamerPro64 18:34, 20 July 2014 (UTC)

See WT:ELEM#Notability of as yet unsynthesized superactinides, where I explained the rationale. Kwamikagami agreed (and in fact proposed this earlier, but didn't carry it out for period 1 because it was a GA), and after some discussion DePiep appears to support this decision. Double sharp (talk) 05:18, 21 July 2014 (UTC)
Yes, we can understand that as a consensus. About featured topic: as Double sharp remarked, the article was low in quality. It did not actually describe the period, it was listing two individuals. Therefor, the article was of ow quality, and likely too low to fit the Wikipedia:Featured topics standard. -DePiep (talk) 11:47, 21 July 2014 (UTC)
And also, it seemed very unlikely to me that it could ever describe the period instead of two individuals, because H is certainly the most oddball element on the periodic table, and there's not much to talk about periodic trends if there are only 2 elements. (For the periods further down, the trends are either largely the same or are distinct enough in various sections that they should go to their own articles, e.g. transition metal, lanthanide, actinide.) Double sharp (talk) 11:51, 21 July 2014 (UTC)
Key for the Featured topics nomination status seems to be the GA status of the article (now a redirect). That could have been reduced some steps with reason as described now, and so pull the FT ground (it just happened to join two FA's). Wikipedia:Featured topics/Noble gases, the other FT in this area, looks better suited. -DePiep (talk) 12:14, 21 July 2014 (UTC)
I'm afraid I don't quite understand. Could you rephrase that, please? Double sharp (talk) 12:26, 21 July 2014 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── Period 1 element was the naming/lead article for Wikipedia:Featured topics/Period 1 elements. The whole set was three articles together (with H and He). A requirement for FT is that all articles are GA/FA (Wikipedia:Featured topics criteria). OK so far.

But we saw (you pointed out) that the content of Period 1 element page was not that substantial. More of a bad listing. So for that reason alone, we could have pulled the GA status from that page. And without GA, the page looses its right to form a FT, and this FT would have ceased to be. Had we done that route, the FT status would not have helped the page either (keeping it in existence). Simply, I doubt that its FA status was deserved. -DePiep (talk) 17:48, 21 July 2014 (UTC)

Well with all this in mind, I started an FTRC for the topic. It can be found here if anyone is willing to participate. GamerPro64 19:01, 21 July 2014 (UTC)

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Fixed. Double sharp (talk) 12:28, 22 July 2014 (UTC)

Neptunium; unun- elimination[edit]

Re neptunium. As I said, that would happen after the FAC is over. It's slowly coming to its end, I think, and I'll try to start the review after that. But I must say: if I haven't finished by the end of the first decade of august, it'll have to wait until late September or even October :( Really sorry, but I have real difficulties to go through. (I don't think such a situation will happen, though, the end of the FAC is near (I think) but have that in mind)

Re unun- names elimination. I've written a RfC proposal draft (here; unfortunately, it seems like we'll have to go through a RfC), I'd love you to go check it, see what's wrong, correct grammar if necessary. I've written about it on the project's talkpage, no one reacted :( (I think because I forgot to ping everyone, and changes to our WT:ELEM page happen very often nowadays, so nobody saw the moment when it was the latest change). I hope you can help me bring some attention, could you?--R8R (talk) 08:09, 28 July 2014 (UTC)

I'll take a look. But I think you should write more about how a periodic table cell should look like for an unnamed element, like E113 (we discussed this before: it is unclear otherwise). Double sharp (talk) 08:31, 28 July 2014 (UTC)

Hodgepodge[edit]

About the period 1's: I disagree that the rants by Nergaal can count as a disagree argument (I read not one sentence of substance). Also ththat page was not the place to discuss. And this is rewarding the distractions by that same editor, instead of channeling a talk. And when "there is no consensus any more" that should be an outcome in a more proper way. Anyway, next steps: rm the GA. -DePiep (talk) 09:08, 30 July 2014 (UTC)

Even if you don't count Nergaal, there is stil R8R's statements: given that only 3 supported, there is no consensus, I think. (Even at 3-1, I would revert to the previous status and discuss, even if the change had been done while it was 3-0. See Talk:Piano Sonata No. 14 (Beethoven) for an example: first RM was 2-1, moved. Then others complained a bit later, once it was moved. So it shows that the original 2-1 not-quite-a-consensus was not a representative sample of the community, and so it was later reverted after a long discussion.) So I think that my earlier action should be reverted, since there is (as you say) no longer a consensus for it.
I see two options here: either start a GAR, or try to fix it ourselves first. R8R Gtrs has stated how he thinks it can be done: we would do well to listen to him. What he says has substance, and he explains how he would make it a good article. But the article title should still be period 1 (periodic table), or something similar, to make it clear that the current content has no place in the new article.
I think what we should do is divide them not into individual elements (except maybe period 1), but instead into blocks, and not discuss individual elements one-by-one. See extended periodic table for an example of what I mean. Double sharp (talk) 11:53, 31 July 2014 (UTC)
Changing your !vote AFTERWARDS because the outcome has changed? Why would I spend time on this? Why reasoning at all? -DePiep (talk) 21:34, 31 July 2014 (UTC)
@DePiep: It's purely procedural: I can't really vote for the topic to be delisted, since the lead article is back and its GA status has not (yet) been revoked. If and when that happens, I will vote for delisting. On that page we are voting about removing the topic's FT status, not whether the period 1 element article is worthy of its GA status. Double sharp (talk) 17:29, 1 August 2014 (UTC)

truncating even denominator polygons[edit]

Dear Double sharp

It is strange not knowing who or where you are, but it seems we have a fundamental disagreement over how to truncate even denominator polygons, as is required to construct a number of polyhedra

I see a polygon's number defined as n/d, where n is the number of vertices and d the number of the next vertex one visits to form the polygon's edges: for instance 5/2 has five vertices and one visits every second vertex to form the pentagram; similarly 5/3, its inverse is formed by visiting every third vertex and effectively going around the same pentagram in reverse direction

Taking one of our disputed polyhedral examples, my personal view is that a 10/2 has ten vertices and one visits every second one to form the polygon, which thus consists of two pentagons interlocked, but at 180 degrees to each other, maintaining the ten distinct vertices

The result of truncating a 5/2, however, whilst also comprising two pentagons, has them coincident, so that the ten theoretical vertices are arranged in five pairs, the edges are doubled up and indeed so are the faces, hence the concept of a double pentagon for this phenomenon

Physically taking a 5/2 pentagram and cutting its corners off, one arrives at a point where the truncated edge is the same length as the truncating edge, as one would with a normal polygon being truncated, and this is simply a double coincident pentagon, not a 10/2; A double pentagon is what is actually shown on all the illustrations of the polyhedra, where although you list a 10/2, I cannot find any such polygon consisting of two interlocked but out of phase pentagons

You have referred me back to 'refs', which I have not looked at as yet, but they are likely to be wrong as this error goes right back to one that the great Professor Coxeter made in his famous 1953 Uniform polyhedra paper, where he dismissed as non-uniform any polyhedron requiring the truncation of an even denominator polygon; This does not mean that the truncation cannot be done, he simply did not know how to handle it using the concept of the double polygon

I do not want to start what I believe is known as a Wikipedia War, so I will not attempt to edit out your work, but simply hope you can perhaps appreciate that I might have a point: the version I spent most of yesterday editing to make what I consider is a correct presentation of this material is not lost, as it sits two down the list below your 'corrections' and can easily be reinstated if you are in agreement

My intention is to discuss this with you further and reach common ground, so I trust you will respond in similar vein

Polystar, its an anagram (Patrick Taylor, Ipswich UK) not sure about these tildes? have I done it right Polystar (talk) 09:06, 2 August 2014 (UTC)

Yes, you've done the tildes right.
Like I said, it's because the meaning has changed. The modern interpretation of {10/2} is actually what you mean by a truncated 5/2, and looks exactly like a double pentagon. I think this is because others (e.g. B. Grünbaum) have also realized that Coxeter made a mistake here, but (perhaps unfortunately) decided to correct him by reassigning his terminology to the correct figures. I agree with you that the truncation can always be done, although it sometimes produces degenerate figures like {10/2} with coincident vertices and edges.
Coxeter's usage of {10/2}, which is incorrect, is that you take every vertex of a decagon and connect it to the second one after it. This, as you say, traces the two-pentagon compound, with the pentagons out of phase by 180 degrees.
However the modern usage of {10/2} is what you call 25: it means that you start with one vertex, join it to the second one after, and then repeat the process with the vertex you are now on until you have drawn ten edges. This means that half the original vertices are not visited, and the other half are visited twice: and the resulting polygon is two coincident pentagons. Klitzing makes this clear on his website, when he says that {6/2} is a "doubly wound hexagon that looks like a single triangle": this can only be 23. He explicitly states that it "does not mark the Star of David" (two triangles out of phase with each other, the hexagram).
The Grünbaumian terminology has the advantage that it means that the rule that truncating a polygon doubles its Schläfli symbol always holds, even if the denominator is an even number. So a truncated pentagram {5/2} forms a doubled pentagon {10/2}.
I recommend this paper by Branko Grünbaum: p.466 shows diagrams of the polygons that in the modern usage are symbolized {6/0}, {6/1}, {6/2}, and {6/3}. {6/2} is exactly what you call 23 in this terminology.
If this was too long and unclear, I hope this makes the above clearer: yes, truncating a pentagram leads to a double pentagon, which is the current meaning of {10/2}, not the non-coincident 2-pentagon compound. Double sharp (talk) 13:06, 2 August 2014 (UTC)

Hello again

I have had a look at your reasoning for calling the double polygons with which we are concerned 6/2, 10/2 and 10/4 and can see that it all depends on how you define the polygon in the first place:

If you start at one vertex and then follow on around the requisite number of vertices repeating serially the operation of say going to the second next vertex, you do indeed trace out a double polygon, leaving every other vertex out of the picture; However if you start at every vertex at the same time and proceed in parallel you get the compound versions, star of david etc.

Not sure how to resolve this, but the difficulty is that your way takes the {n/d} name away from every case of compound polygon where both n and d are even, so that we have a set of polygons that do not have a nomenclature, and note that this is not every compound as 6/3, 9/3, 12,3 etc. or any others with odd denominators are not so affected

My main difficulty however is that when a mobius triangle is doubled up to form the basis for mainly double versions, it works in the polyhedra for (3 3 3/2), (5 5 3/2), (3 3 5/2), (4 4 3/2) and in my view for (3 3 4/2), but when we come to the tilings and try (6 6 3/2) it works, but (3 3 6/2) does not if the polygons are named your way

This last schwarz triangle requires the 'star of david' as a face in these double tilings, along with its truncation the double hexagon, which you would call 12/2, and its inverse which I would call {6/4} and the inverse truncation, which to me is a double 'star of david', that you would call 12/4: all a bit of a mess

The simplest of these to imagine is 3 6/2 |3 which has a vertex with two hexagons, a triangle and a star of david producing a double covering of the plane: what do I now call that star of david if you have taken the 6/2 name for a double triangle?

It seems to me that the Schwarz triangles favour my way of seeing these polygons Polystar (talk) 08:11, 3 August 2014 (UTC)

{n/d} works the same way for any d: e.g. {9/3} traces three coincident triangles in Grünbaum's nomenclature. The compounds have nomenclature as well: the Star of David is 2{3} or {{3}}, the Star of Goliath (Coxeter's {9/3}) is 3{3}, and two out-of-phase pentagrams are 2{5/2} or {{5/2}}.
Now I'm confused. Is a double Star of David supposed to mean two coincident Stars of David? I would call that 2{6/2}, I think.
How can 3 6/2 | 3 contain Stars of David? Its vertex configuration is 6.6/2.6.3, so it just looks like a trihexagonal tiling with "triangles" alternating between being {3} and {6/2}. I don't think compounds can be truly generated with Schwarz triangles – how does one make a stella octangula this way, for example? (Schläfli symbol 2{3,3}.) In your example the angle defined by 6/2 is 2π/6: wouldn't that trace the same way as π/3, but twice? Then how does 2{3} get into it? I'm confused: could you explain again what you meant? Double sharp (talk) 12:46, 3 August 2014 (UTC)


I can see how your nomenclature works now you have explained that what I call {9/3} is your 3{3}, but was this necessary to change to have the term {9/3} available for a triple triangle, three coincident, that does not arise in the normal course of events, whereas double ones do result from truncating even denominator polygons: a question then arises: what do you call a truncated 3{3}?; to me it is {18/3}, truncation of {9/3}, comprising three out of phase hexagons, so is it 3{6}, leaving your {18/3} symbol to describe three coincident hexagons?

A double 'Star of David' is just that, two coincident Stars of David, four triangles in total, two in phase, two out, and thus difficult to put into your terminology, but it is the natural result of truncating a {6/4} ( 2{3/2} in your system?), which is the inverse of {6/2} in my terminology, 2{3} in yours

You ask about the stella octangula which to me is simply derived from a Schwarz triangle as 3| 4/2 3, the {4/2} faces ( 2{2} in your terminology) being the glue that binds the two tetrahedra together

I would like now to insert an illustration showing 3 6/2 |3, but cannot work out how as this system seems to take text only; If you would care to email me at polystar@ntlworld.com I can email my scan back to you. There are other more complex tilings which also include this {6/2}Polystar (talk) 16:22, 4 August 2014 (UTC) star of david face (2{3}) and some quasitruncations including the double star face: 6/4 2 6| is one of the simpler ones

Yes, you got the specifics of "my" (actually Grünbaum's, I think) system correct. I think the use of {9/3} had to be changed so that all fractions would work the same way. It would look odd and inconsistent to have {12/2} and {12/3} work differently, for example.
You can actually insert illustrations into Wikipedia. Instructions can be found at Wikipedia:Uploading images and Wikipedia:Picture tutorial. I'd love to see your picture of 3 6/2 | 3 this way, where others can also see it and comment.
Ah, OK, so the 2{2} faces act as "glue" in the same way the digon faces generated by the Wythoff construction for non-compound polyhedra act as "glue", right? Thank you: I learned something new and very cool! I'd like to know some more: can the other regular compounds (5 or 10 tetrahedra, 5 cubes, 5 octahedra) also be generated via Schwarz triangles? Double sharp (talk) 15:48, 5 August 2014 (UTC)

periodic table by quality[edit]

I updated it, and I will try to add your updates on the actinides if you like? --Stone (talk) 20:03, 4 August 2014 (UTC)

@Stone: Actually I just finished doing them. :-) Thanks for E119 and E120!
I've been meaning to do this rerating for a while. I'm starting from the end and working toward the beginning. I'll try to be very harsh when an article is lacking in an area, like citations. A-class would be granted by comparing with already A-class articles, or by just seeing if it's comprehensive. The way I see it A is like FA in comprehensiveness, and GA is like FA in structure and style. Double sharp (talk) 20:07, 4 August 2014 (UTC)
Gold B to C as worst rating from WP Physics. Double sharp (talk) 20:10, 4 August 2014 (UTC)
I did the same years ago. It is fun to compare and to evaluate. My problem was I had to do it two times because in the first run I was to critical to the end and to friendly in the beginning. Good luck for the 120 elements marathon.--Stone (talk) 20:12, 4 August 2014 (UTC)
@Stone: Thank you for your good wishes. I'll try to be consistently nasty, because in my experience a C→FA is felt as more glamorous than B→FA or GA→FA (I've never seen anyone do the last one).
I'm not looking at the GAs/FAs in this first round, even though there are some I am not too happy with quality-wise. Those will need a GAR/FARC and I don't immediately have the time to set them up. And honestly for some of them I could try to fix them myself (e.g. Cf). Double sharp (talk) 20:15, 4 August 2014 (UTC)
P.S. Due to lack of time I may pause it halfway and then continue tomorrow, perhaps somewhere around element 40 or 50. I could get here so quickly because so much of periods 6 and 7 is GA/FA. Double sharp (talk) 20:33, 4 August 2014 (UTC)

I finished it! :-D

Anyone up for an actinide GT? We need Th, No, and Lr. I can do the last two (each will probably take a week). Double sharp (talk) 19:30, 12 August 2014 (UTC)

I did the last two. Double sharp (talk) 22:25, 21 August 2014 (UTC)

Tilings discussion part 2[edit]

Incomplete tilings page 38.tif

— Preceding unsigned comment added by Polystar (talkcontribs) 20:43, 5 August 2014 (UTC)

I do not think that worked, but the image is somewhere in commons if you can retrieve itPolystar (talk) 20:46, 5 August 2014 (UTC)

OK, I found it and fixed the link above. I will take a look first and think about it before responding. Double sharp (talk) 20:52, 5 August 2014 (UTC)

DYK for Neptunium[edit]

 — Crisco 1492 (talk) 01:22, 6 August 2014 (UTC)

Your GA nomination of Mendelevium[edit]

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Your GA nomination of Mendelevium[edit]

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even denominator polygons[edit]

You seem a bit busy with radioactive elements at present so I have taken the opportunity to collect my own thoughts and summarise these matters from my own particular point of view:

My derivation of symbols for the various polygons is based on {n/d} representing a series of n equidistant points arranged in a circle, a polygon being produced by placing an edge from each of these to the dth next one around that circle. Thus in this system a {6/2} is a ‘Star of David’, comprising two, albeit out of phase, overlying triangles.

The derivation adopted by Grünbaum is different, but also involves n points in a circle, but each edge towards the dth next one is placed in turn, starting from where the last one finished. In this system {6/2} is a pair of coincident triangles that share vertices and edges and only half of the n points get visited at all. It is this double triangle, the natural result of truncating the inverse triangle {3/2}, which I call {2^3}.

In terms of simple symmetry considerations, I think Grünbaum’s choice of symbols is counter-intuitive as the Star of David that he calls 2{3} possesses sixfold rotational symmetry, whilst the double triangle that he calls {6/2} has only threefold rotational symmetry.

My derivation of {6/2} as the symbol for the Star of David also relates to the process of inscription, as within every polygon there are layers of internal edges to be found that run between non-adjacent vertices of the external polygon, say {x}, so that {x/2} is the first inscription, {x/3} the next, and so on going deeper towards the centre and beyond where the inverse polygons are to be found.

This is essentially the same process as that of stellation, which occurs outside a polygon, where edges are extended to produce a similar series of polygons: {x/2} formed at the first crossing of extended edges, {x/3} at the next and so on again.

Indeed stellation is the process that occurs when one takes an icosahedral Möbius triangle (3 2 5) and doubles it up to produce the Schwarz triangle (3 3 5/2), or even trebles it to produce (5 2 5/2), producing star polyhedron families with densities of 2 and 3 respectively, but both involving {5/2} star polygon faces, which no-one disputes.

This process has analogues in the world of octahedral symmetry where the (3 3 4/2) Schwarz triangle produces a 2-dense family that includes polyhedra such as the Stella Octangula, 3| 4/2 3, and the 3-dense (4 2 4/2) family is that of the cross polyhedra {4,4/2} and {4/2,4}, along with their intermediate truncations. Although completely analogous to their icosahedral cousins, these are all generally dismissed as mere compounds but all include the crossed digon face that I call {4/2}, which can be regarded as either the first inscription or first stellation of a square. Treating this as a face in its own right allows it to be regarded as the glue that binds the compounded solids together and I can see no reason to call this face by Grünbaum’s name, 2{2}, since it possesses fourfold symmetry.

There are also analogous families in the field of tilings, where the (3 2 6) Möbius triangle can be doubled up to produce (3 3 6/2) and indeed trebled to produce (6 2 6/2), all with consistent results parallel to the icosahedral examples, but now set in the plane. The {6/2} these imply is definitely a stellated hexagon, the Star of David polygon with its six distinct vertices and sixfold symmetry, rather than a double coincident triangle.

Going beyond these and into the realms of quasitruncation, there are tilings that require a truncated {6/4} to be incorporated. In my system this is a {2^6/2} and comprises four triangles overlaid, two coincident pairs out of phase with each other, or alternatively an overlaid pair of Stars of David. This again has sixfold symmetry and I am not sure whether Grünbaum’s system can cope with it in a sensible way? 81.151.237.61 (talk) 12:18, 7 August 2014 (UTC)

That system copes with your last polygon as 2{6/2}.
What you say makes a lot of sense. The one point I am not quite comfortable with, however, is your choosing the call t{3/2} "23". This breaks the rule that truncating a polygon doubles its Schläfli symbol, viz. t{3} = {6}, t{4/3} = {8/3}, t{5/3} = {10/3}, t{7/3} = {14/3}. Nevertheless you raise a good point that t{3/2} doesn't have sixfold symmetry, which 2{3} does. I wonder why there is this conflict between attempting to extend both rules. There must be an interesting story here. Double sharp (talk) 12:50, 7 August 2014 (UTC)

even denominator polygons[edit]

Sorry forgot to log in first, the above is from me Polystar (talk) 12:21, 7 August 2014 (UTC)


OK with 2{6/2}, which is not very different to "26/2" (the doubling implied here is different: Grunbaum's sets a pair of coincident triangles out of phase with another coincident pair, my system makes coincident two sets of already out of phase triangles) The point here is probably what counts as the Schläfli symbol?; In the examples you give of odd denominator polygons, truncation doubles the numerator, so that effectively the value of the fraction within the brackets is doubled and the symmetry of the truncated polygon is apparent For even denominator polygons Grunbaum's system does the same with the numerator, implying extra vertices, which are not there as in fact they have actually been doubled up; My system also doubles the value inside the brackets, but by halving the denominator, which can always be done to an even number and leaves you a result that reflects the symmetry of the polygon produced by the truncation; (double is indicated by the superscript 2 before the number, which has no algebraic function, which is why I put it there, rather than after the number!)

Polystar (talk) 18:16, 7 August 2014 (UTC)

I dunno about whether the doubled vertices are still there. I always thought of {6/2} (Grünbaum's notation is used in this post) as having six vertices, conicident in pairs: so that if you called it ABCDEF (vertex naming), the coincident pairs would be AD, BE, and CF. I suppose you could picture it this way as a series of ditrigons, i.e. hexagons whose sides alternate between length 1 and length x. At x = 0 you have a triangle, which is progressively truncated to become a hexagon at x = 1, and you get the dual triangle as x → ∞. In both limiting cases, the cases just before them have noncoincident vertices: so maybe there really are two vertices, that are just in the same place. Of course if you go in the negative direction, you first get triangles dangling off the original triangle, which shrinks as x decreases. Somewhere along this line I think you get {6/2}. Or have I just rediscovered quasi-, hyper-, and antitruncation? Double sharp (talk) 18:34, 7 August 2014 (UTC)

even denominator polygons[edit]

I think you are talking about quasitruncation here, i.e. the truncation of an inverse polygon

Normal truncation can be seen as the insertion of a new edge halving the external angle of the original polygon, the truncating edge being grown until all the edges, both truncated and truncating alternately, are equal in length; For a triangle {3}, that external angle is 120 degrees, which is halved to 60 degrees in its truncation the hexagon {6}

For an inverse triangle {3/2}, the edges follow on from each other in reverse direction, but the external angle is 240 degrees if measured in the same direction as originally; Thus its truncation has six edges which have an external angle of 120 degrees, the same as for a normal triangle; inserting such extra truncating edges has to be done by forming the triangles you describe outside the original one, rather than inside as for the normal truncation; simply growing them this way they can never be equalised in length, however equalising them by shrinking the original truncated edges allows equality to be reached in the form of a double triangle with alternating truncated and truncating edges as you go round (each double edge has one of each present)

A similar process with external triangles will work for any inverse polygon and is quite instructive in getting from an inverse square {4/3} to its truncation the octagram {8/3}; Enjoy Polystar (talk) 20:06, 7 August 2014 (UTC)

But if the vertex count has doubled when truncating {4/3}, then I don't see why it shouldn't do the same when truncating {3/2}. Maybe this is a false analogy and I am mistaken, but at first glance I would expect both cases to behave the same way as abstract polytopes, although an accident of geometry happens to make t{3/2} look degenerate. That would also kind of explain t{3/2}'s lack of apparent sixfold symmetry, as t{4/3} is a fourfold-symmetric subsymmetric construction of the octagram (which of course naturally has eightfold symmetry). The inability to realize {6/2} with sixfold symmetry is odd, though, and makes me think that there may be something deeper going on here. Double sharp (talk) 19:30, 8 August 2014 (UTC)

Alkali metals[edit]

Duue to a recent (yesterday) bereavement I will have only limited time to give to wikipedia so I am posting a wikibreak. Axiosaurus (talk) 12:06, 8 August 2014 (UTC)

Oh, I'm very sorry to hear that. Take as much time as you need off from WP. Double sharp (talk) 14:27, 8 August 2014 (UTC)

August 2014[edit]

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Fixed. Double sharp (talk) 19:42, 11 August 2014 (UTC)

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Radium[edit]

Hi, I had a longer search for the melting point of radium. The 700°C are 1910s work from one of the Curies, I think. There might be a better source from Kirby. If this is really 700°C or 960°C we should state this also?--Stone (talk) 22:04, 21 August 2014 (UTC)

@Stone: Yes, I know about this discrepancy. Kirby basically throws up his hands and writes that it is either 700°C or 960°C. I do not know which is right. The CRC quotes 700, but I do not really trust them for rare radioactive elements (e.g. their Ac value is also wrong, or at least outdated, according to Kirby). A clue comes from this book, which says that Ra melts at 700°C and volatilizes at 960°C. Still doing some reading on the matter. But in general for elements beyond Bi you seem to get different values for fundamental properties depending on who you read. I stumbled into this situation for Np as well. Ah well. Probably best to just state in the article that it melts at either 700 or 960°C and then have a note explaining the situation. Double sharp (talk) 22:16, 21 August 2014 (UTC)
OK, I wrote that note in the article. Double sharp (talk) 22:25, 21 August 2014 (UTC)

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Fixed Double sharp (talk) 12:45, 22 August 2014 (UTC)