Need help with locating crease & points on paper

[Ori-JD]

Hello everyone,

I have designed a crease pattern for a model using adobe illustrator, but I have a problem; I am unable to locate and fold the necessary creases on a physical sheet of paper.

Essentially what I have an isosceles triangle located within a square with angles 45, 67.5 and 67.5. The apex point of the triangle touches the upper edge of the paper, the left angle touches the left side of the paper and the right angle touches the axial crease of the square sheet. This square is actually part of a larger crease pattern, but I need to be able to locate any of the points where the triangle touches the square edges or centre axial crease in order to actually make the model.



If someone can help me achieve the creases to make this triangle (or locate the triangle points on the square paper) it would be really appreciated. Thanks.

[Splunge]

Hi,

It would probably help, if you could explain to us, how you found the reference points of the triangle within the square, when drawing it in Illustrator.

Or did you construct the square around the triangle?

[Ori-JD]

I haven't found the reference point for the triangle; I simply drew a triangle with the corresponding angles that I needed and aligned the left and apex angles with the edges of the square, and then I dragged the triangle until the right angle aligned with the axial crease of the square.

Here's some more of the crease pattern with the triangle highlighted.

Note that this is not the complete pattern on a square sheet of paper - this is just a segment of the design, but I can't finish it because I can't locate the triangle without printing the crease pattern out and folding along the lines.

[Splunge]

Assuming my calculations are right, the base of the triangle divides the square into an upper rectangular with a height of approximately 54.7% of the original square. That's somewhere between 5/9 and 6/11.
Folding a diagonal from the bottom right corner of the square to the bottom left corner of the triangle would result in an angle of approximately 24.4°.

At the moment I doubt that there will be an exact solution by folding.

But maybe someone else comes up with a better solution...

[Bodo]

Here is a exact solution
https://twitter.com/bodo_ori/status/891566365911461888
I hope its understandable.
1:Fold both diagonals
2:Fold one angle bisector
3:Fold the paper in half
4: connect the points of the 22.5 line and the horizontal line
5: fold a line through the point where the diagonal and the crease from step 4 meet.

[Splunge]

Here is a exact solution

Wow, respect! How did you get there?

[alando]

Use Robert Lang Reference Finder

[Ori-JD]

Thank you so much Bodo, that solution works perfectly!

[anonymous person]

If you ever need to find a reference like this again, it is best done using the method of similar shapes.

You see at the bottom edge where the vertical crease line emerges, on the left hand side you have two "square" diagonals and on the right hand side you have one diagonal emerging at an angle of 22.5 degrees to the horizontal. Hence, the location of the vertical crease is found at the intersection of a line with gradient 1/2 and a line making an angle of 22.5 degrees to the horizontal.