folding 3rds, 5ths, 7ths, etc on diag.

wquiroz225 · Post by **wquiroz225** » January 13th, 2010, 5:32 am

How is this accomplished? I did an internet search but could only find a solution for folding perpendicular to the edge. I've tried dividing each edge's total length (L / 7 = x) by 7, then the combined length of the two by 7 (2L / 7 = x), and lastly tried the measuring the total diagonal length and divide by 7, with different answers. I used all answers to try and determine the valley and mountain folds, but none of them worked. I have yet to try dividing 90 degrees by 7 yet because I dont have a protractor available to accurately plot 11.8 degrees anyway. Any clue how to do this?

If it helps I am folding Lang's scorpion found in complete origami. I checked the forum and there was one topic I saw that covered this, but the link was broken. I am not sure if it showed how to solve the diagonal issue or if it was perpendicular anyway.

Hopefully I am making this harder then it is!
Thanks guys.
Quiroz

Joe the white · Post by **Joe the white** » January 13th, 2010, 6:27 am

I'm not a math guy, so I've often used the visual landmarks other folders have found (such as Robert Lang). After a bit of digging (go, go magic librarian skills), the answer you seek is on page 36 of this PDF file on Robert's website: http://www.langorigami.com/science/hha/ ... ctions.pdf

I'd suggest browsing around the site if you haven't before

.

wquiroz225 · Post by **wquiroz225** » January 13th, 2010, 7:01 am

Thanks again! At this rate I'm going to owe you a birthday present.

Glad I stayed awake in math class

Even so, a good bit of what he wrote went right over my head. The diagrams he provided are within reach of my IQ however

The problem with using land marks with this model is there are none. It's step 3 of the model and you have half a water bomb base with the right side extended as a plain square with no creases. There's nothing to guide off of. I think I got this one now though.

I have looked at that same file you provided before, but it looked so complicated that I just glossed over it at the time, and forgot all about it. Felt like I needed a graphics calculator to understand it all.

thanks again!

kehom · Post by **kehom** » January 15th, 2010, 7:19 pm

I don't know if I understood you correctly.
What you want is to properly divide an angle in equal parts?
Well, if that is the case, then this is what I do:
1 - try to compute the coordinates of a point that should work as a reference (pattern wise).
2 - once with the coordinates, I throw that into Lang's Reference Finder
3 - experiment which of the sequences works best for the model I'm trying to fold.

I did that for some things when I was trying to fold this Biplane, from The Complete Book of Origami. There are some places where you have to divide an angles in 3rds. Well, I have some sequences to have the references and I'm about to diagram those.
If you are interested in knowing how I found the coordinates to those points, I can write in here (though I doubt I will be able to express myself only through words).

ahudson · Post by **ahudson** » January 16th, 2010, 2:56 am

Use the Fujimoto Approximation. Don't even think of using something else, the fujimoto approximation works just as well and can be adapted to fit any situation: angle division, length division, ...other kinds of division too I suppose?

Every time I bring it up, nobody seems to know what it is or how to use it, so I drew a diagram to explain:

http://dl.dropbox.com/u/232756/sevenths.pdf

kehom · Post by **kehom** » January 17th, 2010, 2:03 pm

ahudson wrote:Use the Fujimoto Approximation. Don't even think of using something else, the fujimoto approximation works just as well and can be adapted to fit any situation: angle division, length division, ...other kinds of division too I suppose?

The choice is yours:
Go bisect freak or do some "math + reference finder"!

Anyway, thanks for the diagrams explaining the method!