This does include a universal pattern at the bottom as long as you don’t mind a bit of Maths!
Having had a few people interested in the possibility of a placemat in similar style to my Celtic Coaster, being the person I am, I wasn’t able to stop considering it.
My first thought was that it would be a lot of work and very fiddly. However……….
Imagining a placemat made in a similar way to the coaster as being made of similar width strips. I wondered how many that would need to be and in the end decided that about sixteen would equal a foot. (Not sure I am right here but it was a good place to start.)
Now the number of separate strips needed for a piece of Celtic plaitwork depends on the whether, on counting the number of bumps on each side (including the corners), the two numbers have a common factor.
No common factor: one piece (as in my odd numbered Celtic bookmarks.)
If there is a common factor that determines how many separate strips there are.
For a square, both sides are the same so you need that number of strips. Hence the style of my coasters.
I tend to think of placemats as being rectangular so having decided on sixteen colours for one foot (30cm). I chose to draw out one that was 16 x 32 bumps.
Now to be similar to my coasters each colour must be different. Sixteen was about the number of colours I used for my spectrum blanket so I coloured each strip in similar colours and produced this. Now one of the things I wanted with my coasters was for no adjacent overlaps to be the same colour.
I think, if you look closely, you will see that this breaks this rule in a vertical strip down the centre.
Of course a square placemat would not. (Each strip of the 16 x 32 placemat would need about 360 chains and there would be sixteen of them to weave together. That is a similar number of chains to those needed for each of my bookmarks.)
Now one person had asked about a matching coaster and obviously this could not match with all those colours so I looked to see what would happen if you repeated the colours of my first (non rainbow) coaster (As represented in my drawing programme) and got this. Even more matching adjacent overlaps.
Even if it was square. Some people might like the patterns it gives rise to but it wasn’t really what I wanted.
So I decided to work out how to make a bigger version of my coaster with the same shape strips but just wider.
As it was just a trial effort, I used some of my acrylic yarn that I had no specific plans for, as the cotton yarn is more expensive and I wasn’t sure I had enough anyway. Acrylic is much stretchier though and so needs more TLC to get it into shape. However I hope it will give you the general idea of what is possible.
For symmetry I decided to just make the strips three times wider and see how large it ended up. This would mean nine trebles (US-dcs) for each cross-over and over one hundred chains for each strip.
The thing that surprised me was to find that when adding further rows it takes two added rows to equal the width of one row on its own. So I ended up with five rows and not three! (And 114 starting chains see formula below.) Since each strip is approximately the same size you should need something less than 25g of each colour. As the whole thing weighed just under 85g. More of course if it was cotton. I used a 4.5mm hook for the starting chain and then a 4mm hook for the stitches. I tend to crochet quite tightly.
Here it is with a plate.
Although I was doing all this primarily for other people it has proved quite useful, as one of my first thoughts was to use plaitwork to make a cushion cover and now I have the tools to plan such a cover – watch this space!
For a square coaster, placemat etc.
If N is the number of bumps down the side (including the corners), N is also the number of strips and so also the number of colours needed if each strip is a different colour.
As an aside: I think that N is best if it is even, as if it is odd you get a square shape in the middle which I think stands out too much.
For an even number the first half of the shapes are the same as the second half and they blend together more. However the formula works for all values of N.
In all the following (US readers read ‘double crochet’ where I say ‘treble’)
Then for each strip if only one row wide:
The number of chains to start = 12(N-1) + 6. This includes the two extra chain needed for the first treble equivalent.
For thicker strips:
If m is the number of rows. I think m works best if it is odd from the point of view of symmetry. (If you chose an even numbered m you will have to adjust for any halves you get. I suggest rounding down as crochet is stretchy.)
The number of chains to start = 12[1 +(m-1)/2](N-1) + 6. This includes the two extra chain needed for the first treble equivalent.
Hope you remember your BODMAS!
I have even come up with a formula pattern for any size you might want to make.
For strips only one row wide:
Work the following, using values of ‘t’ from 1 to N.
(Following on from my remark about an even N, when N is even you can just make two each of the first N/2 shapes, which is what I did for the coaster.)
Treble into 4th chain from hook, 5tr into one chain,
Then one treble into each chain for 6(t-1) chains. (Yes this gives zero for the first ‘t’), 5trs into one chain.
Then one treble into each chain for 6(n-t) chains. 5trs into one chain.
Then one treble into each chain for 6(t-1) chains. 5trs into one chain.
Then one treble into each chain for 6(n-t)-2 chains. (This corrects for the first two trebles made at the beginning which will give a join that is underneath.)
A more general formula that will also work for thicker strips:
If N is the number of strips (colours) and m is the number of rows.
The first rows come from:-
Work the following, using values of ‘t’ from 1 to N-1. Then repeat t=1.
Treble into 4th chain from hook, Then one treble into each chain for [(m-1)/2] chains, 5tr into one chain,
Then one treble into each chain for 6(m+1)(t-1)/2 chains. (Yes this gives zero for the first ‘t’), 5trs into one chain.
Then one treble into each chain for 6(m+1)(n-t)/2 chains. 5trs into one chain.
Then one treble into each chain for 6(m+1)(t-1)/2 chains. 5trs into one chain.
Then one stitch into each chain for (6(m+1)(n-t)-(m+3))/2 chains. (This corrects for the extra trebles made at the beginning which will give a join that is underneath. If when you put the plait together the join is not underneath then you have the strip the wrong way up. I always presume that the right side is the front of the first row.)
For the rows after that work one tr into each tr except for the turns. (Remember to start with 3ch, miss the first stitch, and work the last tr into the top of the 3ch on the previous row.)
For 180deg turns, on the second row I worked into the 10 stitches of the turn as follows – (tr, tr, 2tr, 2tr, 2tr, 2tr, 2tr, 2tr, tr, tr) (16)
On the third row I worked into the central sixteen stitches of the above as follows – (tr, tr, tr, 2tr, tr, 2tr, tr, 2tr, 2tr, tr, 2tr, tr, 2tr, tr, tr, tr). (22)
Hopefully you can see a pattern here. I felt it was similar to working a circle, (or see below.)
For 90deg turns I simply worked 5trs into the central treble of the five of the previous row and one treble into all the others.
Pattern for 180deg turns continued
How the stitches increased for the fourth and fifth rows.
Over 22 stitches: (tr, tr, tr, tr, 2tr, tr, tr, 2tr, tr, tr, 2tr, 2tr, tr, tr, 2tr, tr, tr, 2tr, tr, tr, tr, tr.) (28)
Over 28 stitches: (tr, tr, tr, tr, tr, 2tr, tr, tr, tr, 2tr, tr, tr, tr, 2tr, 2tr, tr, tr, tr, 2tr, tr, tr, tr, 2tr, tr, tr, tr, tr, tr.) (34)
Caveat: Although I have checked and double checked my figures and formulae, I do make mistakes, and if anyone thinks they have found one, I am always grateful to be told so I can investigate.