removed

http://i112.photobucket.com/albums/n174/oxenholme/equatorialpentagons.jpg

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 ...

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 ...

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 ...

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.

I've ordered one too...

Six Dodecahedra

My second attempt. My first attempt was cut short when it got demolished by a football flying across the room!

To my surprise I have made the compound of Five Icosahedra - it must have been about twenty years ago.

It's on the bottom row in this picture, second from the right...

I've been fascinated by polyhedra ever since 1961 or thereabouts when I got introduced to them at school.

From time to time I post photos of my own polyhedra on here (via Photobucket) - Peter Kane shows some of his too.

Please tell us about some of the polyhedra that you've been making, and post ...

This polyhedron is one of 5 that I intend to make.

Its tetrahedral equivalent is a stellation of the regular dodecahedron. I made it last year.

Its octahedral and cubic equivalents are both stellations of the pentagonal icositetrahedron. I made both a year or two ago.

Its icosahedral equivalent ...

Is the mitre angle the angle you would need on the edges if you were to make the polyhedron from a material thick enough to need mitring?

The Cube's dihedral angle is 90 degrees, so the mitre angle is half - i.e. 45 degrees.

My last computer died a death, so I haven't got Stella at the moment, so ...

Five Icosahedra looks a pain to make in such a manner that the end result is rigid.

I fancy having a go. The problem being that the "trough" double tabs (thirty of them) are not continuous colourwise. It will take a lot of care adding reinforcement without getting balsa cement where it shouldn't be ...

I've made the compounds of 5 dodecahedra and 5 rhombic triacontahedra, but not 5 icosahedra.

Sadly yet another computer has bitten the dust, so I no longer have access to Stella, but I've made several polyhedral compounds with coplanar pairs of faces. I've always split the common area down so that ...

I made that snub many moons ago using plain white card.

Seeing the colour representations above has whet my appetite to having a go at making it in colour.

I would need six colours for the pentagrams and pentagons, another ten colours for the icosahedral triangles. Would I get away with the ...