Reference Help

[Brimstone]

I tested the result provided by RF and it wasn't working for me. Then I realised the diagonal over which the rectangle and square lie on, is no the one shown by RF but the other one. Now it works perfectly.

[Baltorigamist]

That's a prime example of my tendency to overlook simple solutions to reference problems. I was familiar with the reference all along; I just didn't think to try it for some reason.

[Baltorigamist]

Sorry for the double post, but I thought it only right to share the finished CP for the model.

[Brimstone]

Sorry for the double post, but I thought it only right to share the finished CP for the model.

Cool, now looking forward to the folded model.

[PleatBox]

Excuse me but is there a post that talks about references like these? I cannot understand all the weird equations and stuff

[Brimstone]

Excuse me but is there a post that talks about references like these? I cannot understand all the weird equations and stuff

Just go ahead and ask here or start a new thread for the specific reference you want, it won't matter.

[PleatBox]

Thanks! I am really new to this, so it might take a bit of time to understand it!

First of all, what are the numbers about? how do they relate to finding ref. points? Can you kindly show me the process of using, and coming up with these? A help from the experienced folders will be incredibly helpful!

I believe understanding how these work will help me in origami so much!

[Brimstone]

Thanks! I am really new to this, so it might take a bit of time to understand it!

First of all, what are the numbers about? how do they relate to finding ref. points? Can you kindly show me the process of using, and coming up with these? A help from the experienced folders will be incredibly helpful!

I believe understanding how these work will help me in origami so much!

I'll do my best and maybe someone else will complement it.

You assume that the lenght of the side of the square is 1 and you consider your start point as the bottom left corner which in turn will have coordinates 0,0 and like wise, top right corner will have coordinates 1,1.

As an example if you need to fold the square in half vertically, the crease will go from 0.5, 0 (middle of the bottom side) to 0.5, 1 (middle of top side).

Since the Pythagoras' theorem states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides, the hypotenuse will be of sqr (1+1) since the square of 1 is 1 again and each side value is 1 as stated before.

With these three values you can reckon many things and if you need a crease at a specific location but you don't know how to obtain the references to fold it exactly there, you can input the coordinates of the crease into a program developed by Robert Lang called ReferenceFinder.

Hope this helps and if it is not what you were looking for, let us know.

[PleatBox]

Thanks! This gave me a general understanding of the subject!

So now how do the ratios come into play? For example, the ratios that Baltorigamist posted here, 2:2+sqrt2, how do i turn it into a number that I can put into the paper?

I'm so sorry if I am asking noobish questions, I am not really good at understanding math in text form only, and I am not a really good english speaker

[ShuaiJanaiDesu]

We use ratios like "2:2+sqrt2" for simplicity/exactness so we can avoid using long decimals.

Some simple examples:


If we didn't know how to fold this Preliminary Base, we would want to know the circled point as a ref. point.
With geometry(or simply intuition), the circled point is the center, so the ref. point can be written as above. We can say "1:1" or "0.5:0.5". We would need to use the latter if we were to use RefFinder.


Now for a Kite Base, the circled point's reference point is shown above. It can be written as "sqrt2:1" or "0.58...:0.42...".

Now the question is, 'why don't we use '1' as the length of the side and compute the 'exact' length?' To do this, it is shown with Exact*. Notice that the fractions get really big, even for a simple ratio like "sqrt2:1"

Another question might be, 'what is up with the 'sqrt2'? Why do those appear often?' That has to do with using 22.5 degrees. As shown below, Crease Patterns with 22.5 degree folds have many of these triangles which contain 'sqrt2' lengths.


(As you can see, you can easily calculate the length of the long side be imagining the red line. The red line is the hypotenuse of the right triangle with sides (1-1-?) so the hypotenuse must be sqrt2. And by symmetry, the length between the '22.5deg' and the red line is also sqrt2. So the long side is 1+sqrt2)

So now how do the ratios come into play? For example, the ratios that Baltorigamist posted here, 2:2+sqrt2, how do i turn it into a number that I can put into the paper?

So to answer your 2nd question, you need to find the 'total length' of the ratio(which can be done by replacing the colon(:) with plus signs(+) ) then divide each part of the ratio with the 'total length'

For my first example: "1:1"
total length = "1+1" (replace : with +) = 2
therefore: decimal ratio is 1/2 : 1/2 => 0.5 : 0.5

For "2:2+sqrt2":
total length = (2+2+sqrt2) = (4 + sqrt2)
therefore: decimal ratio is 2/(4+sqrt2) : (2+sqrt2)/(4+sqrt2) => 0.369 : 0.631

P.S. Sorry for the long post. I hope I didn't explain it in a confusing manner...Or if I explained something you already understood, sorry! If you have more question feel free to ask

[PleatBox]

Wow! thanks for the explanation! thanks for giving me illustrations and all that, really helpful!

So now, I know how to read those ratios, now I have to know how to make those ratios in the first place.
On the kite base, how were you able to calculate the circled point? And how about the length between the circled point and the upper right point?

I tried to use pythagoras' theorem, since the vertical line on the right is equal to 1(lets call it a), and the 22.5 degree line is equal to 1+sqrt2(called c), I can then calculate b, the horizontal line right? I did the math but it doesn't work, probably because I'm doing something wrong

What I did:
(1+sqrt2)^2 - (1)^2 = b^2
√(b^2) = √(4.828...)
b = 2.19...

Obviously the problem is that b is bigger than 1, when it shouldn't be. What error am I making here?
Again, sorry for the lack of knowledge in math

[Brimstone]

I tried to use pythagoras' theorem, since the vertical line on the right is equal to 1(lets call it a), and the 22.5 degree line is equal to 1+sqrt2(called c), I can then calculate b, the horizontal line right? I did the math but it doesn't work, probably because I'm doing something wrong

I highlighted your mistake, 1+sqrt2 is not the 22.5° line (hypotenuse), it is the adjacent side (what you called b).

[ShuaiJanaiDesu]

Brimstone answered what was wrong and I don't really use pythagorean theorem (at least for this problem) since 22.5 degree right triangle has side lengths of 1 , 1+sqrt2 and sqrt(4+2sqrt(2)). The edges of the paper are perpendicular so you will never get a value like sqrt(4+2sqrt(2)) (Unless you aren't working with a pure 22.5 degree CP). Of course, you can use it to check your work (like you did) but to figure out the ratios initially, I'd suggest not using it.


For the Kite Base in particular, these are two ways to figure out the ratio.
1. Look at the folded shape. The top edge of the original square paper, creates two sides of a right isosceles triangle (solid blue triangle). Therefore, we can let the hypotenuse = sqrt2 and leg = 1. So the edge of the Paper can be divided into sqrt2:1

2. We can also try to break up the crease pattern into easier shapes(isosceles triangles and squares). We can label the red triangle as shown and since the two yellow triangles are identical, we can see that the hypotenuse of the red triangles equals one side of the blue square. Therefore, the side of the paper is sqrt2:1

Personally, I use Paint/Adobe Illust. and draw on the CP to find the ratios/ref. points. I basically draw lines and shapes like I did in the 2nd way above, till I figure out the 'ratio' of a side of the paper.
(And then actually folding the ref. point is a whole other step... Although, this step can be avoided by using Reference Finder, so it's not too bad.)

[PleatBox]

Why is the red triangle measured 1-1-sqrt2? Shouldn't be the numbers be a bit lower, since 1 is the length of the square's sides, and sqrt2 is the main diagonal of the square?

Also, what if I want to find the coordinates for a point inside the paper?, for example, if I fold a paper into thirds, the creases will intersect the main diagonal right? how will I find the references for those?

[ShuaiJanaiDesu]

Why is the red triangle measured 1-1-sqrt2? Shouldn't be the numbers be a bit lower, since 1 is the length of the square's sides, and sqrt2 is the main diagonal of the square?

Read my post (2 posts before) that starts with:

We use ratios like "2:2+sqrt2" for simplicity/exactness so we can avoid using long decimals.

Look at the very first picture. I used the ratio "1:1" (instead of "0.5:0.5") which would mean the square's side is 2. I used "1:1" instead of "0.5:0.5" because it's looks much simpler. Similarly, the "Exact" for the second picture is "sqrt2:1" = "1.414:1" which are not the same numbers as "0.58:42". And the reason is said right below the picture.

Basically, the way I go about figuring out ratios is to pick the shortest line segment in the CP and call it '1'. Then I calculate the rest of the segments. (At this point the edge of paper can be any number greater than 1) At the very end, you can convert the ratio into a decimals so that side length is 1. (This last conversion step was discussed at the end of the same post)

================================

Also, what if I want to find the coordinates for a point inside the paper?, for example, if I fold a paper into thirds, the creases will intersect the main diagonal right? how will I find the references for those?

If the ref. point is on the main diagonal, then its X coordinate is the same as its Y coordinate. (kinda like the Preliminary Base) If it's on the edge of the paper, X or Y is going to be either 0 or 1. If not, you'd have to do 2 separate calculations (one for X and one for Y).