Montroll's Five-sided square

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qtrollip
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Montroll's Five-sided square

Post by qtrollip »

Does anybody know whether there is an existing way of getting the reference points without using the guestimating procedure (as used for his Starfish in "Animal Origami for the Enthusiast")?

Thank you
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Falcifer
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Re: Montroll's Five-sided square

Post by Falcifer »

If it helps, the point where the crease meets the edges is at 0.457576875, assuming the length of the paper is 1. To be more precise, it's √2 + 1 - √(1 + 2√2)

I'm not sure whether it's okay to post pictures of the output of ReferenceFinder, but you can put the number in [ or even the whole expression - (2^0.5) + 1 - ((1 + 2*(2^0.5))^0.5) ] and it'll give you what you want.

Personally, with obscure references like that, I usually measure them.

Hope this helps.
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Re: Montroll's Five-sided square

Post by qtrollip »

Thank you Falcifer.
I eventually worked out a way of getting it with reference points that are very accurate.

I'll go use those coordinations in Refereence Finder and see how it compares.

Thank you again
Quentin
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Re: Montroll's Five-sided square

Post by Brimstone »

I didn't know Montroll had a "5 sided square", I was aware of his "6 sided square" but not of this. He seems to have stretched this conept more than I thought possible.
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Falcifer
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Re: Montroll's Five-sided square

Post by Falcifer »

qtrollip wrote:Thank you Falcifer.
I eventually worked out a way of getting it with reference points that are very accurate.

I'll go use those coordinations in Refereence Finder and see how it compares.

Thank you again
Quentin
No problem, Quentin. I actually enjoy the mathematical side of origami, so it's nice to get the chance to play around with that.

I notice that the first few results in ReferenceFinder are less accurate (but probably not detrimentally so), while a later solution has 0 error, but requires an extra crease.

Also, for anyone who doesn't know, there's some information and downloads for Robert Lang's ReferenceFinder here. Available for Windows, Mac and Linux.
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Re: Montroll's Five-sided square

Post by dinogami »

qtrollip wrote:Does anybody know whether there is an existing way of getting the reference points without using the guestimating procedure (as used for his Starfish in "Animal Origami for the Enthusiast")?
Isn't it the same reference point that Montroll found more systematically later, as in, for example (pulling a random Montroll book off the shelf) steps 2-6 in his "Gorilla" model from _African Origami_?
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Falcifer
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Re: Montroll's Five-sided square

Post by Falcifer »

dinogami wrote:Isn't it the same reference point that Montroll found more systematically later, as in, for example (pulling a random Montroll book off the shelf) steps 2-6 in his "Gorilla" model from _African Origami_?
It's very similar. ReferenceFinder actually gives a (different) way of finding the exact same point using only three folds. And while it's close enough that it won't make a difference - it equates to about 0.5mm difference when using a 1m square - it's not the exact same reference point.

But it's definitely good enough, I would think.
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Re: Montroll's Five-sided square

Post by qtrollip »

Ah, thanks Dino
Here is the solution I came up with using only geometry (I suck at algebra). Obviously a lot more creases so bigger margin for error. Anyway, it was just an experiment.

Falcifer, I havent checked referencefinder yet, but thank you. I'm glad all that stuff makes sense to you.

Image

Quentin
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Re: Montroll's Five-sided square

Post by origamiguy »

Quentin,
John has been working a new book tentatively titled :"origami Math" where he describes his "On The Edge" reference point finding technique. I saw some of the pages about the 5-sided square, but do not know the folding sequence for the Gorilla with which to compare it. I can say that your method is different than what Montroll has diagrammed for the "Origami Math" book.

To get the first reference point like yours with paper starting as square:
1. Fold and unfold to get the half then quarter mark on the top right side.
2. Using the quarter mark as pivot point, fold and unfold the bottom right corner to the left edge.
3. Fold and unfold the diagonal from bottom left corner to top right corner.
4. Fold the bottom left corner up along the center diagonal crease line and crease at the mark created in step 2.

Hope that's clear, if not we could do a Skype session sometime.

BTW you attending any conventions this year Quentin?
Cheers,
Brian K. Webb
http://www.eorigamipublishing.com
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Re: Montroll's Five-sided square

Post by qtrollip »

Hi Brian
Thanks, yes I got it.
So simple that way, John is a genius for sure.

Conventions... Japan in August and I'm going to try PCOC as well (September). How about you? Will you be at PCOC, NYC or Centerfold?

Thanks again
Greetings
Quentin
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Re: Montroll's Five-sided square

Post by origamiguy »

Quentin - I am going to NYC and Centerfold. As of right now, PCOC is not possible.
Cheers,
Brian K. Webb
http://www.eorigamipublishing.com
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Re: Montroll's Five-sided square

Post by dinogami »

origamiguy wrote:...but do not know the folding sequence for the Gorilla with which to compare it. I can say that your method is different than what Montroll has diagrammed for the "Origami Math" book.
It's a technique he's used in several models across several books. Basically, if you start with the square oriented as a diamond and both diagonals pre-creased:

(1) fold the lower point (corner) to the center point, crease to one edge (let's say the right edge), and unfold;
(2) bisect the left point (corner), creasing only where it intersects the lower right edge;
(3) fold the lower point up again such that the crease from (1) contacts the crease from (2).
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Re: Montroll's Five-sided square

Post by bilborigami »

origamiguy wrote: John has been working a new book tentatively titled :"origami Math" where he describes his "On The Edge" reference point finding technique. I saw some of the pages about the 5-sided square
Are you referring to "Origami and Math: simple to complex", by J.Montroll? I don't have it.

In this thread two folding sequences have been described. Both of them give (I did the math)
1/sqrt(2) -1/4=0.4571..., which is a fairly good approximation to the exact point (given in another post by Falcifer). And a third sequence is pointed out by Falcifer as being obtained in Reference Finder with 0 error.

Are any of those three sequences discussed or described in Montroll's "Origami and Math: simple to complex"?
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Re: Montroll's Five-sided square

Post by bilborigami »

Falcifer wrote: I notice that the first few results in ReferenceFinder are less accurate (but probably not detrimentally so), while a later solution has 0 error, but requires an extra crease.
The 0 error of the rank 4 solution that Reference Finder provides is, indeed, 0.0000110489... So, it is not the exact one (though it is really close!).

I wonder wether J.Montroll provides an exact solution in his math book (just for fun).
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Re: Montroll's Five-sided square

Post by bilborigami »

bilborigami wrote: I wonder wether J.Montroll provides an exact solution in his math book (just for fun).
I answer to myself (I purchased the book): he doesn't. He just provides the sequence described by origamiguy, which is 1/sqrt(2) -1/4, a fairly good approximation.

Three messages in a row is quite a monologue, so I give up until some else shows interest ;-)
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