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Gilson: Polyhedra - A New Approach
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Peter Kane



Joined: 25 Oct 2009
Posts: 79
Location: S.E England

PostPosted: Thu Sep 27, 2012 3:41 pm    Post subject: Gilson: Polyhedra - A New Approach Reply with quote

Noticed a new book on Amazon the other day, so dipped into my piggy bank. Its by Bruce R. Gilson and the "new approach" is to consider polyhedra from the perspective of axial symmetry. It only arrived a couple of hours ago, so I can't provide a review yet, but I thought I'd let you all know there was a new book on the street. In fact, the Preface is dated 16th Sept 2012, so it is pretty new. I don't think it really matters whether it IS or ISNT a new approach; at first glance, it isn't just a rehash of other publications, so that's enough for me.

Anyway, at 12, its only the price of a few beers, so you could always take a risk. If you don't like it, you can always shred it in a bucket with a couple of pounds of sugar, a packet of yeast and some hops and brew your own beer: "A pint of your fine quastitruncatedsmallstellated mild, please landlord".

I hope that I can provide a proper review in due course, but I'm happy to support his effort anyway.

Pete K
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guy



Joined: 11 Feb 2008
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PostPosted: Thu Sep 27, 2012 7:59 pm    Post subject: Reply with quote

Thanks for the heads-up. I just ordered my copy before replying Smile
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Peter Kane



Joined: 25 Oct 2009
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PostPosted: Fri Sep 28, 2012 6:31 am    Post subject: Reply with quote

guy wrote:
Thanks for the heads-up. I just ordered my copy before replying Smile


You're welcome.

Pete K
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oxenholme



Joined: 16 Jan 2008
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PostPosted: Tue Oct 02, 2012 8:43 am    Post subject: Reply with quote

I've ordered one too...
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guy



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PostPosted: Fri Oct 05, 2012 6:23 pm    Post subject: Reply with quote

OK, I have now read Gilson's book.

Sadly I can recommend it only to the most hardened polyhedron enthusiast (like me) Wink . Others will find it badly wanting.

Its approach purports to be based on symmetry, and to focus on polyhedra having a single major axis of symmetry, such as a regular prism. It is aimed at two kinds of reader, those with a small recollection of school geometry and those with a more mathematical eye.

Broadly speaking the first half covers some general theory while the second half classifies some axially-symmetric polyhedra. Perhaps understandably, but disappointingly, the discussion is confined to convex polyhedra. The illustrations were prepared using Adrian Rossiter's excellent Antiprism software.

A brave attempt is made to systematise and describe various classes of axially-symmetric polyhedron, using a naming system based on those of Conway and Johnson. There is some interest here, and at least for me some new polyhedra. I rather enjoyed the "globoids" on Pages 91-2.

But from here on it's bad news all the way. Like so many self-published works, the production desperately misses the guiding hand of a professional editor, in both the text and the mathematics. In what follows, I illustrate this judgement with just a few of its many flaws.

The title is misleading. There is nothing new about an approach based on symmetry, rather the only novelty lies in the chosen rabbit-hole. The content satisfies neither target readership.

Chapters are badly organised, the text verbose and overly pedantic in some places but skimpy and short on detail in others, illustrations are too few and poorly prepared for the print medium. References are hinted at in the text but seldom given - there is no list at the end.

The mathematical discussions are lacking in rigour and full of results, reasonable in themselves, which do not follow from badly-chosen definitions. Symmetry is introduced in Chapter 1 but relegated to the sidelines thereafter. The Platonics are introduced not through principles of symmetry but the old-fashioned assembling of faces round a vertex. The notion of symmetry groups and transitivity is skimped on, there is no mention of flags. The archaic definition of an Archimedean polyhedron allows Miller's Mistake (the pseudo-rhombicuboctahedron) which then requires fumbling handwaving to disappear from the list, while the two pseudo-rhombicosidodecahedra also allowed by the definition are not even mentioned. Table 7 ends up using the letter n to denote both a symmetry group variable (as in a symbol including n-fold symmetry) and various other unrelated counts - confusing and unforgivable in any mathematics text. The discussion of duality is informed but badly presented. Appendices on trig and vector algebra fail to supply all that is missing from the text.

With much guidance and greater focus on the main subject matter, this book might have been turned into something worth reading. Sadly, it lacks both.

(FYI I intend to post a review on Amazon, based on this one. Copyright fanatics please note).
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oxenholme



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PostPosted: Mon Oct 29, 2012 10:58 am    Post subject: Reply with quote

The mathematics in the book is lost on me.

I do very much like the look of the Heptagonal Pentagonised Globoid depicted in Fig 48, but sadly there do not appear to be any clues on how to construct the symmetrical but otherwise irregular pentagons used therein.
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Peter Kane



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PostPosted: Tue Oct 30, 2012 5:27 pm    Post subject: Reply with quote

I may be wrong, but I think that you can start by drawing a pentagon that has similar angles to the 14 pentagons around the equator of the figure. I don't think that you need to be precise, so long as you make it symmetrical. Make 14 of these and arrange them into a ring. The remaining pentagons that make up the upper and lower pyramids are them defined by the sides of the equatorial pentagons, plus a fairly arbitrary length for the longer sides.

Don't forget to publish a pic !

Pete K
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oxenholme



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PostPosted: Wed Oct 31, 2012 1:24 pm    Post subject: Reply with quote

I think Chapter 24 contains some equations or whatever they are - it keeps on saying it's simple - for the dimensions of the two types of pentagon. I wouldn't have a clue how to use them.

I wonder if one way would be to make a couple of seven sided "pyramids" , and to mount them on a pencil or the like, orienting them and sliding them apart until the equatorial pentagons are coplanar. I could take the measurements, then make it neatly using 7 different colours of card. I assume it is possible with 7, or would 14 be necessary?

Listen out at the weekend for swearing coming from up north...

Hoping Guy might look in and take pity on my ignorance and offer some guidance!
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Peter Kane



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PostPosted: Wed Oct 31, 2012 3:27 pm    Post subject: Reply with quote

oxenholme wrote:
I think Chapter 24 contains some equations or whatever they are - it keeps on saying it's simple - for the dimensions of the two types of pentagon. I wouldn't have a clue how to use them.

I wonder if one way would be to make a couple of seven sided "pyramids" , and to mount them on a pencil or the like, orienting them and sliding them apart until the equatorial pentagons are coplanar. I could take the measurements, then make it neatly using 7 different colours of card. I assume it is possible with 7, or would 14 be necessary?

Listen out at the weekend for swearing coming from up north...

Hoping Guy might look in and take pity on my ignorance and offer some guidance!


What do you want to make the equatorial pentagons coplar to - I don't see how it can be coplanar or in a parallel plane to any of the larger petagons. For example, starting with the white equatorial pentagon, how would you navigate to the coplanar pentagon ?

I'll listen out at the weekend, but you can always hear swearing coming from up north ! Wink

I tried reading through the next chapter last night, but fell asleep !

Pete K
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oxenholme



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PostPosted: Thu Nov 01, 2012 6:37 am    Post subject: Reply with quote

Peter Kane wrote:

What do you want to make the equatorial pentagons coplar to - I don't see how it can be coplanar or in a parallel plane to any of the larger petagons. For example, starting with the white equatorial pentagon, how would you navigate to the coplanar pentagon ?

I'll listen out at the weekend, but you can always hear swearing coming from up north ! Wink

I tried reading through the next chapter last night, but fell asleep !

Pete K


I was less than precise in what I wrote!

What I meant was to slide the two pyramids until the apex and base edges of the equatorial pentagons are coplanar. For the white equatorial pentagon the apex edges are on the upper pyramid, the base on the lower. It becomes a flat pentagon only when the upper and lower are the right distance apart.

However, it occurs to me that maybe all the vertices of the polyhedron in the picture are on the circumsphere, and the whole thing needs to be very specific?

Is that in Chapter 24 co-ordinate geometry? Whatever it is, we never did it at school.
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guy



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PostPosted: Thu Nov 01, 2012 9:00 pm    Post subject: Reply with quote

Hi,

This is not an easy figure to construct. Some dimensions are arbitrary, but there are some determined relationships too.

I find the maths to be awkwardly developed. Personally, if I were working out coordinates I would use polar in the XY plane and Cartesian in the Z. And if you don't understand what I mean by that, don't go near the maths!

To make a card model, I think I would draw some constructions and calculate others.

First let's label the vertices so we know what we are talking about.
Top, A. Next there are four rings of 7 vertices per ring. Call these, working down from B to E, B1-7, C1-7, D1-7 and E1-7. I'll worry later about how these all match up. And bottom, F.

So, some observations:
Symmetry requires that all of C1-7 and D1-7 are equidistant from the centroid. This means that all edges CnDn lie in vertical planes.

You can make the various rings as far apart as you like (as long as you keep the symmetry). But it helps if the separation between B and C must be fairly small, as this determines the distance to A which is a lot greater.

I'm not sure of the easiest way from here, and must dash now. Anyway, hope this helps.

[edited to delete some mistakes]
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Peter Kane



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PostPosted: Fri Nov 02, 2012 8:31 am    Post subject: Reply with quote

At first reading, I thought that this was pretty straightforward to construct. I assumed that there was plenty of leeway in the choice of the equatorial pentagons, but a first quick model has proven me wrong. I'll have to read the details more carefully next time.
The maths in the subsequent section is a mixture of school trigonometry, vector maths, coordinate geometry and some other techniques. It also uses material from earlier chapters and from one of the appendices. I'll try to slog through it, but having built a prototype, I don't think the model alone will be worth the effort. Mine looks like a golf umbrella after Hurricane Sandy hit. Back to the drawing board - literally.

Pete K
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Peter Kane



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PostPosted: Sat Nov 03, 2012 3:55 pm    Post subject: Reply with quote

Hmm, don't know what happened there. I was mid-way through a post. I'll start again. I have created a mock-up model that I think illustrates the process that is used to create this polyhedron:

Each trapezium on the central yellow band is merged with a blue triangle (which is part of a pyramid which is not shown). The problem is that the trapezium faces are almost vertical unless the bands are very close together , hence the band must be very thin, otherwise the pyramid becomes very high (because each face of the pyramid shares an edge with each blue triangle). I guess this is what Guy pointed out in his last post. The slope of the pyramid can be reduced by shortening the height of the blue triangles too. Probably, the ones I used were way too tall, but I just built it to better visualize what is going on. Next step is to calculate the edges of the pyramid.

Pete K
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oxenholme



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PostPosted: Sun Nov 04, 2012 5:22 pm    Post subject: Reply with quote



And here is my dreadful attempt at some equatorial pentagons.

I used two pieces of card, sort of heptagonal with the pentagons round the periphery.

Gloriously inaccurate, but it's more exploratory than anything else.

Next step is to stick a skewer through the centre holes and to try to determine the length of the remaining pentagons. It is fairly easy to determine the base and the lower left and right sides.
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oxenholme



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PostPosted: Sun Nov 04, 2012 9:40 pm    Post subject: Reply with quote



Last edited by oxenholme on Mon Nov 05, 2012 4:39 pm; edited 2 times in total
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