Worst Origami Fold?

E_Bingham · Post by **E_Bingham** » June 19th, 2005, 3:44 pm

Sinking multiple layers, round corners; ala Lang's 'Cicada'. Nightmare!

Euam

dani luddington · Post by **dani luddington** » June 19th, 2005, 6:05 pm

sooo where do we find the woodland elf diagram? book? internet? sincerely, dani

malachi · Post by **malachi** » June 19th, 2005, 7:33 pm

I believe it was most recently published in Jay Ansill's Origami Sourcebook. Part of that book was published in the past as Mythical Beings

http://spinflipper.com/origami/sff/show ... an&pic=229

dani luddington · Post by **dani luddington** » June 19th, 2005, 9:07 pm

thanks for the info, i will have to give that model a try. sincerely, dani

wolf · Post by **wolf** » June 21st, 2005, 3:26 am

...or any other odd division of the square, for that matter.

I used to hate it too, until I came across the method utilising Haga's theorems. The best part of it is that you don't even need to think about the mathematics behind it; it's a nice turnkey formula.

Here is Koshiro's writeup on this:
http://www.origami.gr.jp/People/CAGE_/divide/05-e.html

Since the majority of models require thirds or fifths (or any even multiples of these two numbers), all you need to learn are the first two: bringing the corner to the 1/2 mark, and bringing it to the 1/4 mark, as shown on the first figure on the page above. After doing this a few times for square grids, doing the division would become an automatic thing, just like any other folding technique.

DavidW · Post by **DavidW** » July 17th, 2005, 12:32 am

Precreasing can be a pain! But a pleasure when it pays off later down the road.

What I really hate is not the unfold completely, but the second time you have to unfold completely!

platypusguy · Post by **platypusguy** » July 27th, 2005, 4:59 am

you wver tried folding into sevenths, oh god, it took me like an hour just to do that.

wolf · Post by **wolf** » July 27th, 2005, 5:11 am

platypusguy wrote:you wver tried folding into sevenths, oh god, it took me like an hour just to do that.

One trick I learnt from another folder is to first fold the square into eights, then cut off the eighth strip on two adjacent edges. You do end up with a slightly smaller square, but then you can always start off with a larger square to compensate for this.

Post by **TheRealChris** » July 27th, 2005, 7:10 am

7th???
as easy as breathing:

DavidW · Post by **DavidW** » July 27th, 2005, 5:09 pm

platypusguy wrote:you wver tried folding into sevenths, oh god, it took me like an hour just to do that.

Why didn't you just use a ruler?

DavidW · Post by **DavidW** » July 27th, 2005, 5:13 pm

Chris, where are the landmarks to do step 3? Without them the only natural thing that pops into mind is angle bisection which gives you thirds not sevenths.

Post by **origami_8** » July 27th, 2005, 7:25 pm

DavidW wrote:where are the landmarks to do step 3? Without them the only natural thing that pops into mind is angle bisection which gives you thirds not sevenths.

It is an angle bisector. I would very like to know how you got thirds with it, if I try I get 5/7 on the left side and 2/7 on the right side or 5/7 on the top and 2/7 on the bottom.

DavidW · Post by **DavidW** » July 27th, 2005, 8:21 pm

Line 1 is described by y=1-x

Line 2 is described by y=x

Line 3 is described by y=x/2

Any point on line 1 takes the form (x,1-x)
Any point on line 3 takes the form (x,x/2)

So naturally the intersections of lines 1 and 3 occur when 1-x=x/2 or x=2/3. That is the coordinate of the intersection of lines 1 and 3 is (2/3, 1/3).

Now how do you get 2/7?

JMcK · Post by **JMcK** » July 27th, 2005, 8:48 pm

DavidW wrote: Line 3 is described by y=x/2

No, it isn't. If line 3 went from the bottom left corner of the square to a point exactly half way up the right hand side it would be described by y = x/2, but since it's an angle bisector it hits the right hand side below the mid point.

origami_8 wrote: I would very like to know how you got thirds with it, if I try I get 5/7 on the left side and 2/7 on the right side or 5/7 on the top and 2/7 on the bottom.

I've just tried this method and I got a significant bit less than 5/7 on the left side and a bit more than 2/7 on the right. (And obviously the same for the top and bottom.)

DavidW · Post by **DavidW** » July 27th, 2005, 9:18 pm

Well okay that was a mistake. I have it done right now

Clearly it hits the other side of the paper by the tan(22.5 deg).

Alright we know that the tan(x)=2*tan(x/2)/(1-tan^2 (x/2))

So inverting that equation we get tan(x/2) = {sqrt{1+tan^2(x)}-1over tan(x)}

Now we know that if x=45 deg, then tan(x)=1 so then using that formula above tan(22.5 deg)=sqrt(2)-1. That's the slope, instead of m=1/2 it's m=sqrt(2)-1.

So going back the equation should be 1-x=(sqrt(2)-1)x or x=1/sqrt(2) and y=1-1/sqrt(2).

Those numbers look like (5/7, 2/7), but they're not, they are actually (1/sqrt(2),1-1/sqrt(2)).

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