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Origami boxes
I was folding a few origami boxes at the weekend and decided to try my pentagonal box again, and realised that I could no longer remember how to do it!
Fortunately, once I got home, I was able to work it out again with the help of a previous LJ posting I'd made.
So, this is mostly a reminder to myself if I forget again, but also for anyone else who really wants to know.
Take the base angle of the isosceles triangle required to make the sections for the lid. eg. for a pentagon, 360 divided by 5 is 72, so the top angle of each triangle is 72 degrees, which by requiring 180 degrees in each triangle, means that the base angles are 54 degrees each.
To be able to make that fold at 54 degrees requires a fold to line up with the first-fold triangle. This fold will be 90-54 =36 dregrees from the 'vertical'. Using some moderately messy geometry I don't want to have to remember in a hurry, I can show that halving this angle (36/2) will give the angle needed for the fold of the first-fold triangle.
The first fold (and all others for that matter) in origami should really be done without a protractor. Fortunately, 'cos 18' is very close to '1/3'. To get a right-angled triangle where two of the sides (not the hypotenuse) are 1 and three in relative length, fold the top of one side into three (roll loosely and flatten out and shuffle it along until it's a neat three part fold with two short creases (you can crease all the way if you want, but you don't need to). Then fold a line between one of the creases and the corner that would be closest to the other end of the crease if you'd folded it all the way. That's your starting triangle. After that point, continue pretty much as you would for a hexagonal or triangular box.
BTW, I don't know if this particular box design has been published. I know there's a pentagonal box in a book I haven't bought yet, but I don't know if it uses the same or a different technique. If you can actually figure out what I'm talking about here, feel free to make the box, but don't copy elsewhere. You can link, just don't copy. Some day, I'll try and do a diagram, which will be much, much clearer. The advantage of writing down the math is that I should be able to do boxes with other irregular numbers of sides, as long as I can find a trig function that allows an easy relationship between the sides of the first-fold triangle.
Fortunately, once I got home, I was able to work it out again with the help of a previous LJ posting I'd made.
So, this is mostly a reminder to myself if I forget again, but also for anyone else who really wants to know.
Take the base angle of the isosceles triangle required to make the sections for the lid. eg. for a pentagon, 360 divided by 5 is 72, so the top angle of each triangle is 72 degrees, which by requiring 180 degrees in each triangle, means that the base angles are 54 degrees each.
To be able to make that fold at 54 degrees requires a fold to line up with the first-fold triangle. This fold will be 90-54 =36 dregrees from the 'vertical'. Using some moderately messy geometry I don't want to have to remember in a hurry, I can show that halving this angle (36/2) will give the angle needed for the fold of the first-fold triangle.
The first fold (and all others for that matter) in origami should really be done without a protractor. Fortunately, 'cos 18' is very close to '1/3'. To get a right-angled triangle where two of the sides (not the hypotenuse) are 1 and three in relative length, fold the top of one side into three (roll loosely and flatten out and shuffle it along until it's a neat three part fold with two short creases (you can crease all the way if you want, but you don't need to). Then fold a line between one of the creases and the corner that would be closest to the other end of the crease if you'd folded it all the way. That's your starting triangle. After that point, continue pretty much as you would for a hexagonal or triangular box.
BTW, I don't know if this particular box design has been published. I know there's a pentagonal box in a book I haven't bought yet, but I don't know if it uses the same or a different technique. If you can actually figure out what I'm talking about here, feel free to make the box, but don't copy elsewhere. You can link, just don't copy. Some day, I'll try and do a diagram, which will be much, much clearer. The advantage of writing down the math is that I should be able to do boxes with other irregular numbers of sides, as long as I can find a trig function that allows an easy relationship between the sides of the first-fold triangle.
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I used to (when bored) be able to take an A4 sheet of paper and make 4 bassett hounds from the squares that could reasonably be created. 1 large, 1 medium, 2 small. (I used to think of them as a little family.) I actually tried to remember how to fold a Bassett Hound today, and could I even remember the first base shape I need? Could I buggry. Will have to go look it up again. *bah*
p.s. Do you know if Henry got his badge ok? (The Alchemist/Seamstress film memorial one I gave to you at the DWCon.)
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Glad he liked it.