Beadwork

Flat Peyote Octagon

Earlier this year I was challenged to make a flat peyote octagon. Most beaders are probably familiar with peyote triangles, which use the increase pattern 2-2-2, and  hexagons, which uses the increase pattern 2-1-0:

BeadMechanics_PeyoteHexagon

Diane Fitzgerald shows how to make all the flat polygons up to a hexagon in her book Shaped Beadwork. But what is the increase pattern for an octagon?

octagon

To answer the question I decided to make a mathematical model of the beadwork! (It’s actually just some simple geometry – so please don’t run away at this point – there is beadwork here too!)

Lets start by thinking about what a peyote polygon looks like. The easiest one to think of is a triangle:

BeadMechanics_PeyoteTriangle2

We can split the triangle up into 3 segments or “cells” – the parts in-between the three herringbone (or 2-2-2) increases. These are themselves isosceles triangles:

  BeadMechanics_PeyoteTriangle      triangle

You can see that each of the “cells” grows by 2 beads width each row – counting the total width of the row, rather than just the number of beads you would use to actually bead the row.

If you think about any regular polygon, you can see that it can be split up into these triangular “cells” too:

polygons

The angle at the corner of each of the triangular cells (the one at the centre of each polygon) will be equal to 360 degrees divided by the number of sides – this will be useful later on!

Since any polygon can be built out of these cells this is a good point to start with the mathematical model. Here is a diagram of a cell – to make the drawing easier I’ve used the increase pattern for a triangle, but I’ll extend it to other shapes in a minute:

BeadMechanics_Triangle0

The width (= length) and height (= diameter) of the beads will be important. I am going to call them w and h. For delicas we can look them up on Miyuki’s website, which states that the width of size 11 delicas is 1.3 mm and the height is 1.6 mm.

BeadMechanics_Triangle1

I want to know how many extra beads are needed each row to make the cell the right shape to make an octagon. Lets start in an ideal world where we can add the exact same number of beads every row, even if it’s a fraction of a bead (in reality to actually bead the shape you’d add beads one row then none the next to end up at the same number on average rather than per row). I’m going to call the extra number of beads needed “x“.

We know the width w of the beads, so if any given row is width R then the row above it will be width R plus the number of extra beads x times w:

BeadMechanics_Triangle2

As this is peyote the height difference between these two rows will be half the height of the beads, h/2:

BeadMechanics_Triangle3

What we want to know is what is x to keep the cell the right shape. To answer this we just need some trigonometry (see it is useful!). The row height and width differences worked out above are the height and width of the two orange triangles shown in the diagram below:

BeadMechanics_Triangle4

Since the cell is symmetric these two triangles are identical: both are height h/2 and together they must be width xw, so each one is half that, or xw/2. We also know that the angle at the inner point of the triangle will be 360 degrees divided by the number of sides of the polygon. For now lets say that there are n sides (for the octagon n will be 8). This gives us the angle at the bottom of the cell, which can be used to work out the angle of the orange triangle too (using symmetry and similar triangles – something else you probably thought would never be useful!):

BeadMechanics_Triangle5

That gives a right-angle triangle with an angle and two sides, and all that’s needed is some trigonometry to determine that the number of extra beads needed per row for a polygon with n sides is:

x = (h/w) tan(180°/n)

Simple (honest!) – but does this simple calculation actually work in practice?

If we use the width and height of size 11 delicas then for a triangle (n = 3) the formula above says that you need 2.1 extra beads per row – almost spot on for the triangle increase of 2 beads per row! And for a hexagon (n = 6) it says you need 0.7 beads extra per row. If you look at a hexagon you can see that overall the 2-1-0 increase adds 2 beads width every three rows – so on average the increases is 2/3 = 0.67 beads per row – which is again almost identical to the formula. So it seems to work!

Now the interesting part: what is the number of beads needed for an octagon with n = 8? If you put the numbers in to the formula it gives about 0.5 beads extra needed per row… which unfortunately is going to be a bit difficult.

It’s only really possible to increase at in increments of 2 (or more) beads per row, as 1 or 0 is just a normal peyote stitch (or possibly a decrease). So to get on average 0.5 beads per row you’d need to increase 2 beads every 4 rows, which wouldn’t really work – not only would it be a bit to far between increases to approximate the gradual increase, but if you add 2 in a row, then you must add 1 the next, then 0, then 1, which leaves you trying to do an increase on top of a bead not a space, which isn’t going to work well. So, instead of sitting around waiting for the invention of half-delicas I decided to try a different approach: an asymmetric alternating increase pattern:

  • Start with 4 beads in the first row.
  • Then alternate 2-1-0 and 0-2-1 increases around each side – so the eight 2 increases aren’t all in the same row.
  • This gives slightly too many beads per row so to keep the piece flat one set of increases needs to be skipped at about row 10.

Here’s a photo of my first attempt at a flat octagon using this method:

BeadMechanics_Octagon1

I used blue and red for the alternating increases and green for the normal peyote stitches in-between. I started with 4 beads, then added another 4 total in next round (all in red), then increased by 2 in each space in the third round (this is where the first blue beads appear). The next round the blue “increases” are just 1 bead, and in-between them the red increases are 2 beads. After that I continue on alternating the 2-1-0 increase pattern until row 10, where I skip a 2 increase in the red beads, and just use 1 bead instead (you can see where there are two single red beads stacked on top of each other. Then I continued on with in-step 2-1-0 increases for a few rows until I ran out of thread.

The centre has gone a bit wonky – mostly this was a result of poor tension as for some reason I could only find size 00 thread, which was just too thin to maintain the tension in the centre! It’s fairly symmetrical though, despite the alternating increases, and also – more importantly – flat!

Having located a more appropriate thickness of thread I made a second attempt with a few more rows:

BeadMechanics_Octagon2

This uses the same increase pattern, but now the increases are all in silver and the normal peyote between them in alternating blue and green until row 10, which is in orange (and is where one set of 2 increases is skipped). All the rows beyond this (where the increases are in-step) are in red. The benefit of the skipped increase at row 10 (apart from keeping the shape flat!) is that you end up with identical edges, which is helpful both for the overall symmetry of the shape and if the octagon is going to be stitched to other shapes.

I’m much happier with the second attempt – it’s more symmetrical and also still flat, so all-in-all a successful experiment applying mathematical modelling to beadwork!

 

 

Beadwork, Polyhedra, Tutorials

Rhombicosidodecahedron Hyparhedron Variation

Here’s an interesting variation on a hyparhedra – a rhombicosidodecahedron which uses both warped squares and hexagons.

rhombicosdodec_hypar_var_verrier2

A rhombicosidodecahedron is an expanded dodecahedron with rings of squares and triangles surrounding the pentagon faces. This means that this beaded version can be thought of as an expanded version of Hypernova! Here they are side-by-side:

rhombicosdodec_hypar_var_verrier3.JPG

As you can see it’s pretty big! It took a lot longer to bead than I thought it would, but I’m very happy with how it turned out.

The idea for this started with a geomag 1.5 scale rhombic hexacontahedron by Rafael Millán, which I came across earlier this year. At about the same time I was working on warped square hyparhedra, and I realised that this polyhedron would be possible as a hyparhedron using a combination of both warped squares and warped hexagons. Essentially it’s a variation on the warped square rhombicosidodecahedron hyparhedron where the three warped squares making up the triangular faces are replaced by a single warped hexagon.

I’ve wandered into this idea before with the truncated tetrahedron hyparhedron – on the left in the photo below is the warped square hyparhedron version and on the right is the variation with the triangular faces replaced with warped hexagons:

truncated_tetra_hypar_verrier

It’s interesting to see that it also works with a larger shape. I’m still working on the plain hyparhedra version of the rhombicosidodecahedron, but it will be great to see them side-by-side when finished too!

This one took too long to create a tutorial, but here’s a layout diagram if you want to attempt it! In total it needs 20 warped hexagons and 60 warped squares. The warped squares sit over the edges of the pentagons, with the peaks and the corners and the valleys meeting at the centre. The diagrams below show the top half of the rhombicosidodecahedron. On the left is the position of one warped square outlined in red – the increases are shown as dashed lines and the peaks marked with circles. On the right a whole set of warped squares is shown around the top of the shape.

The warped hexagons join it all together. They sit at the centres of the triangular faces and are “zipped” to the warped squares. The diagram on the left shows how one warped hexagon joins to one warped square. The diagram on the right shows a set of warped hexagons around the top of the shape.

I made the warped hexagons completely, and then joined the warped sqaures to them. The angles in the square faces are quite tight, so I tried to always start the join towards them and end it at the pentagon side, so there was more space to work. As ever with these shapes, a curved beading needle is essential!

It took me a while to get my head around the shape, but it eventually clicked! Just ask if you have any questions about it!

rhombicosdodec_hypar_var_verrier1

Beadwork, Polyhedra

Five Intersecting Tetrahedra

Earlier this year Diane Fitzgerald posted a geometric challenge: make a beadwork version of the origami model of five intersecting tetrahedra. After many months of work here’s my version!

BeadMechanics_IntersectingTetra2

I’m not the first person to make this – Kris Empting-Obenland gets the credit for that! Her beautiful version has been accepted to this year’s Bridges conference – you can see it in the Bridges 2019 gallery here!

Five intersecting tetrahedra is an interesting compound shape that has some similarities to a dodecahedron. I’m always interested in geometric challenges so I decided to see what I could do. My first attempt was with 30mm bugles, which worked surprisingly well!

BeadMechanics_IntersectingTetra1

If you want to try making one of these I recommend working from a video showing the construction of the origami model, like the one here. However, I definitely recommend using similar colours to those shown in the video – and not what I did, which was mostly similar colours but in a different order which got very confusing!

My next step was to take a few ideas I had been working on following on from my Sunburst dodecahedron, which uses Sue Harle’s tubular diagonal peyote technique.  I realised I could use this technique here to make individual beams for the sides of the tetrahedra, meaning the construction would be very similar to the origami version – and hopefully that it would be easy(ish) to join together. Here’s my first attempt at a tetrahedron using this method:

BeadMechanics_IntersectingTetra_InProgress1

It worked! My next step was to work out the dimensions of the beams so that the compound would work – if the tetrahedra are too small then you can’t interlink them, if they’re too big then it won’t hold its shape. Because the geometry of the beams here is very slightly different to the origami version they need to be a very slightly different size so the model will fit together correctly. So there was a brief interlude from the beadwork for some maths to work out the exact size, which turns out to be a ratio of width to mid-beam length of 1:13.5441. I checked this calculation three times and then asked someone else to check it, as I had nightmares of making 5 tetrahedra that were all slightly too small to fit together!

The next step was to start beading. I spent a lot of time measuring and re-measuring the first few beams to check the size (I’m very glad I invested in some digital calipers a while ago!) but eventually managed to make a full size tetrahedron – 1 down, 4 to go!

BeadMechanics_IntersectingTetra_InProgress2

I decided to use just one colour, which also gave me an excuse to use some Crystal Marea delicas that I’d been wanting to make something with for ages!

The next step was to make the second tetrahedron. This one was easy to assemble around the first, and it seemed like I was making reasonable progress.

BeadMechanics_IntersectingTetra_InProgress3

However, experience from the bugle bead version (or rather, experience unpicking many mistakes in the bugle bead version) left me feeling a bit apprehensive when it came to the next tetrahedron – and I ended up leaving the project to languish for a few months.

To try and make it easier to pick up I tried making another bugle bead version – this time all in the same colour. However, this ended up being so difficult it didn’t really help my confidence at all! I decided to just make all the beams for the remaining three tetrahedra then set aside a long weekend to try putting them together. Here are all the pieces ready for the final assembly:

BeadMechanics_IntersectingTetra_InProgress4

I was dreading this bit, so decided to just tack the beams together at the ends with separate thread to make it easy to take apart if (when) I made a mistake. To my surprise though it went together fine – I’d like to think it was all the practice with the bugle bead versions, but I think it might just have been luck! Here’s the initial assembly, before I stitched the beams together properly – it’s a bit of a tangle!

BeadMechanics_IntersectingTetra_InProgress5

But it gradually sorted itself out into something more symmetric and geometric!

BeadMechanics_IntersectingTetra_InProgress6

At this point it became clear that the beams were the right size – which was a huge relief! Then it was just a matter of stitching in the last few threads to finish the piece and complete the challenge!

BeadMechanics_IntersectingTetra4

And this definitely was a challenge! But I’m glad I attempted it as it pushed me to develop a few new construction ideas, and even though at times failure felt inevitable I did enjoy the process! Also, at about 65g of delicas this is easily the largest piece I’ve ever made!

And as for that second bugle bead version? I did eventually manage to get it to work – and it’s now one of my favourite pieces!

BeadMechanics_IntersectingTetra3

Beadwork, Polyhedra, Tutorials

Truncated Octahedron Hyparhedron

Here are some brief instructions for the truncated octahedron hyparhedron. This is actually a pretty simple shape to make. It’s just 4-hats joined together with a few extra warped squares. If you know how to zip together warped squares to make a star you can use the same method here! My warped squares are 7 rows in total, and I make them out to row 6 then use row 7 to zip to any other squares as needed.

truncated_octahedron_beadmechanics

First join four warped squares with 2 brown sides and 2 green sides into an upside-down 4-hat – that is, with all the points in the centre pointing downwards. This will be one of the square faces you can see in the photo above.

trunc_octa1_beadmechanics

Here’s a diagram for the individual warped squares that make up the 4-hat:

warped_square

Make the first square completely all the way out to row 7 and stitch in the threads (the photo is in red and white instead of brown and green – sorry!):

TruncatedOctahedronStep1_BeadMechanics

Now make a second square out to row 6 and join it to the first square on one brown (or red!) side as part of row 7 as shown (note I’m working anticlockwise around the square):

TruncatedOctahedronStep2_BeadMechanics

Finish the round and weave in the end – you should now have two squares joined together like this:

TruncatedOctahedronStep2b_BeadMechanics

Make a third square out to row 6 and again join to one of the others on one brown (/red) side as part of row 7 as shown:

TruncatedOctahedronStep3_BeadMechanics

When this square is completed it will look like this:

TruncatedOctahedronStep3b_BeadMechanics

Make a fourth square out to row 6 and this time join it to the two remaining brown (/red) edges from the previous squares using row 7:

TruncatedOctahedronStep4_BeadMechanics

When this square is complete you will have a finished 4-hat like the one below!

TruncatedOctahedronStep4b_BeadMechanics

Here it is from the side – the centre points downwards (so technically it’s an upside-down 4-hat!):

truncated_octahedron1_beadmechanics

Make five more of these so that you have six identical 4-hats in total. The warped squares here are all edges of a square face on the finished shape.

Now make a completely green warped square out to row 6 (I use the same silver diamond pattern as before, but all the sides just have the same background colour). Step up for row 7 and zip it on all sides to two of the 4-hats, as shown on the left of the diagram below. The centre pyramid of both 4-hats should be pointing downwards. (Note that I’ve shown this new warped square in blue rather than green!) The new warped square is an edge of a hexagonal face. To show the shape flat I’m going to draw the warped squares slightly distorted (as on the right of the diagram) from this point onwards.

trunc_octa2_beadmechanics

Here are two 4-hats and a warped square ready to be joined together:

TruncatedOctahedronStep5a_BeadMechanics

Here are the first two sides being joined together:

TruncatedOctahedronStep5b_BeadMechanics

And here’s the piece from the other side showing last two sides being joined together:

TruncatedOctahedronStep5c_BeadMechanics

When the join is complete the beadwork will look like this:

TruncatedOctahedronStep5d_BeadMechanics

Here’s another in-progress photo from slightly later in the construction outlining how this square fits between two of the upside-down 4-hats:

truncated_octahedron3_beadmechanics

(Note though that this particular photo used a slightly different order for joining the squares than the instructions here!)

Repeat this step three more times to join three more 4-hats around the first, as shown in the diagram below:

trunc_octa3_beadmechanics

Now join in four more warped squares around the edge of the shape connecting some of the remaining edges of the 4-hats as shown below:

trunc_octa4_beadmechanics

The diagram above looks pretty distorted but in reality the warped squares will fit easily into place.

Turn the beadwork over. There will be a space for the remaining 4-hat, which should be joined in using 4 more green warped squares, as shown below:

trunc_octa5_beadmehanics

Once all these joins are finished the hyparhedron is complete. Sorry the instructions are a bit brief but if you have any questions just ask and I will try and help!

 

 

© Copyright 2019 Patricia Verrier. All rights reserved.

These instructions are for personal use only. Please contact me if you require more information.

Beadwork, Polyhedra

Hyparhedra

Hyparhedra are polyhedra made from hypars – hyparbolic paraboloids – a shape more commonly known in the bead world as a warped square. I was introduced to the idea of hypars and the academic work on them a while ago by the Contemporary Geometric Beadwork project.

Polyhedra made from hypars were named hyparhedra by Erik Demaine, who wrote a paper (“Polyhedral Sculptures with Hyperbolic Paraboloids”, Demaine, Demaine & Lubiw, 1999) on a method of making any polyhedra from paper folded hypars. There are details of the method and photos of the shapes here. It’s based on making “k-hats” – pyramids made from hypars – and using them as the faces of polyhedra. They demonstrated how to make a variety of shapes including all the Platonic Solids – that’s the five possible polyhedra made entirely out of the same regular shape: the tetrahedron, the cube, the octahedron, the dodecahedron and the icosahedron.

You may also notice on that page that one of the first shapes is recognisable to the beader as a warped square star! (In their paper they mention that the earliest example of a warped square star in origami is from 1958!) This method also works with beadwork hypars – Joke van Biesen and Kim Heinlein have both made beautiful beadwork versions of the hyparhedra cube from the paper.

The photo below shows a beadwork 4hat – four warped squares joined together on two sides each to make a 4-sided pyramid. Each of the warped squares is half cream and half red, so that the pyramid in the centre ends up one colour and the edges another. I’ve outlined the warped squares in green on the right to highlight how they’re joined together.

k_hat_BeadMechanics k_hat2_BeadMechanics

I’ve been playing around with a slight variation of the idea for some time now. Instead of the method using k-hats as the faces, I fold one warped square over each edge of a polyhedron and join them so that their points are at the vertices and centres of each face. This method seems to be fairly well known in the origami world, but I haven’t found anything about it in the mathematical literature so far. I have however seen numerous independent discoveries of a dodecahedron made this way in the beadwork world – the infamous rhombic hexacontahedron! I think the dodecahedron is the only Platonic solid that will work this way, so I decided to try using this method to make various Archimedean solids.

So far I’ve completed a truncated tetrahedron, the dual of a cuboctahedron (a rhombic dodecahedron) and a truncated octahedron:

Hyparhedra1_BeadMechanics

I initially didn’t think that these would work as their dihedral angles are quite small (that is, they have “sharper” edges compared to larger shapes), but the beadwork turned out to be more flexible than I thought! Here’s a photo of the truncated tetrahedron from a different angle showing how much some of the squares will distort:

hyparhedra7_BeadMechanics

Here’s an more detailed illustration of the method using the truncated octahedron – a shape with 6 square faces and 6 hexagonal faces. The photo below shows the outline of the truncated octahedron over the hyparhedra:

hyparhedra2_BeadMechanics

And this photo shows how a warped square has been placed over an edge with the two “upwards” points at the corners and the two “downwards” points at the centre of the faces:

hyparhedra3_BeadMechanics

Interestingly the shape formed by this hyparhedra is quite similar to the third stellation of the rhombic dodecahedron. Which brings me back to the rhombic dodecahedron hyparhedra:

hyparhedra4_BeadMechanics

This shape was interesting – trying to put warped squares over the edges of a cuboctahedron didn’t work as the surface ends up curving in the wrong direction. The squares want to be the opposite way around – so that their upwards points are in the centre of the faces not at the vertices. This just creates the “dual” polyhedron – the shape you get if you join the centres of the faces of a polyhedron together. In this case it’s a very pointy rhombic dodecahedron!

I’ve started a snub cube which seems to be working out the same way and looks like it will end up as it’s dual shape too. I have a rhombicuboctahedron in progress too but I’m not sure if it will work as itself or as it’s dual yet.

I originally started all this with a truncated icosahedron. However I couldn’t decide on colours so ended up with a lot of individual pieces and not much completed. This is the main reason I’ve switched to using the picasso delicas for all these shapes – it gives me a few colours options without being overwhelming. So I’m restarting the truncated icosahedron in red and brown to match the two other truncated shapes I’ve made so far!

The biggest piece I have in progress though is a rhombicosidodecahedron, which is working well but is also going to be very big! Here is the piece I’ve completed so far (about a sixth of the total) compared to the other shapes:

hyparhedra5_BeadMechanics

There’s only one other Archimedean solid larger than this – a snub dodecahedron which needs 150 warped squares! I’m wondering if this one will work as its dual since this has happened with some of the other shapes with a lot of triangular faces.

There are also some Archimedean solids with 8 or 10 sided faces, which will not be flat when made this way with warped squares (since warped squares make flat hexagons when joined on two sides). However, I’ve tried joining 8 warped squares together as the start of a truncated cube and I think it will still work!

The different shapes and surface curvatures that can be generated by this hyparhedra method are interesting – part of my reason for working through all of these systematically is to gain a better understanding and feeling for how to make more general surfaces so I can create more complex shapes. I’ve already used the ideas in several other shapes – for example the rick rack dodecahedron and variation are joined together with warped squares making up the edges of the underlying polyhedra. It was the hyparhedra idea here that helped me work out how make these two pieces!

If you’d like to try making a hyparhedra here are some brief instructions on how to make the truncated octahedron!

Beadwork, Tutorials

Warning about Phishing Websites

Please be aware that a number of phishing websites have come to light over recent days that have copied a large number of beadwork and craft listings from Etsy, apparently in order to scam people out of payment information. Unfortunately some of my tutorial listings appear on some of these sites. These are not genuine listings!

My tutorials are only available from my Etsy shop, Interweave and here on my website.

You can find a full list of those available on the tutorials page.

Finally, please be careful when following links from sites such as Pinterest – always check that the link is genuine before clicking! Be(ad) safe out there!

Beadwork, Polyhedra

Rick-Rack Polyhedra Variations

While I was making the first rick-rack dodecahedron I had an idea for a slight variation made by joining the rick racks together point-to-point instead of edge to edge. Since this would require a join between three edges, I first thought that I could use a warped hexagon instead of a warped square. However, I was completely wrong about that! The angle of the warped hexagon was far too small. After a bit of trial and error I found that three warped squares joined into a pyramid on two of their sides resulted in a triangular shape with the right surface angle. I then spent the better part of a day trying out different colour combinations and patterns to end up at the conclusion that they looked best just all in the dark blue colour. It’s a bit of extra work making the three squares for each join but they look great – the small pyramids create this extra spikey structure around the rick racks that I really like. Here’s the finished piece:

BluebellTwo

If you compare it to the original you can see the differences – in the new shape two rick-rack points meet at each join where in the original three points meet at each join:

rickrack_dodecahedron_2_beadmechanics

You can see the difference in how the two are constructed in the diagram below – on the left is the variation and on the right is the original:

The two polyhedra – the original and the variation – correspond to a dodecadodecahedron (no that’s not a typo!) and a ditrigonal dodecadodecahedron. (Although the triangular faces on the new shape are convex not concave, but close enough!). The original shape is also similar to a hyperbolic dodecahedron (which I believe is a dodecahedron that tessellates in hyperbolic space, rather than is entirely hyperbolic itself).

I was also asked if it was possible to make smaller shapes such as tetrahedra or cubes out of rick-racks and warped squares. The crucial factor here is the angle of the warped squares – their maximum angle (the angle between the ‘V’ shape each one forms) is about the same as the angle needed to join the rick-racks together on a dodecahedron. They’ll squash down enough to join together rick-racks at a smaller angle, which should make shapes with higher dihedral angles (i.e. less pointy edges) possible, such as truncated icosahedra, but not those with small dihedral angles such as tetrahedra. Fortunately there’s an easy solution! You can make something something very similar to a warped square by joining two triangles together one on side, as shown on the below on the left, compared to the original warped square join on the right:

Unlike the square however, the triangles will bend to any angle along this join, which should make any regular or semi-regular polyhedra possible! To test the idea I made a tetrahedron using this method:

BluebellTetra

The rick-racks here have three peaks instead of five, but otherwise it’s made in exactly the same way as the original dodecahedron, just joining each one together with the pairs of triangles instead of with a warped square. The resultant shape is very pointy (and hard to photograph!) but much quicker to make!

I’ve now made a set of podcasts with various different numbers of sides so I can make a whole set of these polyhedra!

BluebellRickRacks2