How to locate reference points?

[hobbestheprince]

Brimstone: In a thread about Eric Joisel's Self-Made Man, a member named FlareglooM wrote the following:

"The picture of the crease pattern is 432x720 pixels. That is a ratio of 1:1+2/3 So an example for the paper size is: 15 by 25 cm or 30 by 50 cm."

What I would like to know is the math that was used to get the size of paper you would need. Then, I can use the equation for any CP I run into ... especially for models that call for a rectangle to be used.

[FlareglooM]

Scale one of the sides by the other and the other one by itself.. So one side becomes 1 and the other one becomes a number.

So for the CP of 432x720px, divide both by 432, so you get 1x1.66667, in other words for every 1 to the right you go 1.6667 up -> 1:1+2/3.

If you did it the other way you would have 1:3/5 which is basically the same, but I just find it easier one having equal to 1 and the other 1 bigger than 1 (in other words dividing by the lowest number)

[hobbestheprince]

And from that ratio how did you determine the paper size?

[FlareglooM]

You now have 1:1.666667. Since we are talking about surfaces I will use the following analogies:
On a grid: 1 square to the right means 1.66667 squares up.
Or if you want it in cm, for every 1 cm to the right you go 1.66667 cm's up.

So let's say we want the small side to be 30 cm. That means you have to go 30 times to the right in steps of 1 cm. But for every step you go 1.6667 cm up. You also do this 30 times. In total you will go up 30 times *1.6667 cm up each time = 50 cm.

So a 30 cm by 50 cm will work.

That's just to get the understanding. Now we want to use one side to be 12 cm. Using the 1:1.6667, we say that's the left side multiplied by 12. But you must also multiply the right side by 12 since for each 1 to the right you go 1.6667 up. So you will get 12:20. In other words a 12 cm by 20 cm rectangle.

[hobbestheprince]

Ah, I understand perfectly. Thank you for the explanation. So, you just picked the number 30 cm then ... this is easier to understand than I thought.

[FlareglooM]

Yeah I just picked it out as an example , you can choose every number you want. The numbers I used were just easy to work with. Glad I could explain

[hobbestheprince]

You definitely helped me, I can't thank you enough for the guidance.

So I guess this is typically used for rectangles then, eh? I only say that because a cp made from a square is still going to be a square, it just depends on how big or small you want your model.

[hobbestheprince]

Yeah I just picked it out as an example , you can choose every number you want. The numbers I used were just easy to work with. Glad I could explain

I was thinking about this during the afternoon. Let's take Joisel's Hobbit, which is 1:1.25. Now, going over to the right 1 cm would have me go up 1.25 cm ... if I were to mark those, would that be the very first valley fold in the boxpleated grid?

[FlareglooM]

I don't think so, it could be if it was designed that way, but let me put it like this:
We take a 32 cm by 40 cm rectangle. (so 1:1.25)
The square (bottom part of the CP if I got the right CP in front of me) in this rectangle is 32 cm x 32 cm. You can basically make any division in that square so you can make every width of that pleat.

By inspection for Joisel's Hobbit CP: You have 5 quarters, the lowest and the top two are both divided 8th's. Conclusive: the total square in that CP (bottom part) has a division of 32. With the example 32cmx40cm rectangle we started, this means that height of a single pleat in the square is basically going up 1 cm, or for the width 1 cm to the right.

But if those pleats would have been 24th's , the pleats would appear after every 1.3333 cm.

Also if you went 1 to the right and 1.25 up your pleat width's wouldn't be equal in both directions, but they are in Joisel's CP.

[hobbestheprince]

I don't think so, it could be if it was designed that way, but let me put it like this:
We take a 32 cm by 40 cm rectangle. (so 1:1.25)
The square (bottom part of the CP if I got the right CP in front of me) in this rectangle is 32 cm x 32 cm. You can basically make any division in that square so you can make every width of that pleat.

By inspection for Joisel's Hobbit CP: You have 5 quarters, the lowest and the top two are both divided 8th's. Conclusive: the total square in that CP (bottom part) has a division of 32. With the example 32cmx40cm rectangle we started, this means that height of a single pleat in the square is basically going up 1 cm, or for the width 1 cm to the right.

But if those pleats would have been 24th's , the pleats would appear after every 1.3333 cm.

Also if you went 1 to the right and 1.25 up your pleat width's wouldn't be equal in both directions, but they are in Joisel's CP.

Darn, I was hoping to find that first pleat with a little clever math so I can measure most of it out. Is something like that even possible? I know how to divide up the dwarf, but that is because it's a square. I would have loved to find a neat little shortcut to get either the first valley or mountain fold, and just go from there.

[hobbestheprince]

Hm, I just thought of this. Using the same example, the model 32x40 grid ... couldn't I just use the length of my paper (we'll just say 25) divided by 20 (half of 40) and that number would give me how far I go down until I do a valley fold. Then, turn the model over and divide those in half?

[FlareglooM]

If you take the long side of your paper and divide it in 20th's and then turn it over and divide those you will have 20 pleats (so alternating 20 mountains and 20 valleys and thus 40 lines).

You also say that you want to find the first valley/mountain and then continue from there. If I understand you right you use a previous pleat to find the next one.
I think it would be easier to fold a grid first, or parts of the grid, without putting much attention to the crease direction.. My personal reason for doing that is because long creases tend to deviate easily while pleating. And if you finished the grid you can "correct" the directions.

I am hoping I understood you right

[hobbestheprince]

If you take the long side of your paper and divide it in 20th's and then turn it over and divide those you will have 20 pleats (so alternating 20 mountains and 20 valleys and thus 40 lines).

You also say that you want to find the first valley/mountain and then continue from there. If I understand you right you use a previous pleat to find the next one.
I think it would be easier to fold a grid first, or parts of the grid, without putting much attention to the crease direction.. My personal reason for doing that is because long creases tend to deviate easily while pleating. And if you finished the grid you can "correct" the directions.

I am hoping I understood you right

Here's an example:

Standard copy paper is 21.6 cm x 27.9 cm. Let's say I want to do 28 pleats down the long side. Well, couldn't I just take 27.9 divided by 14, which is 1.99 or roughly 2 cm. I would then make a mark on the side of the paper, scoring it as reference. From there, I do a regular valley fold. Then, I go back and forth in regular pleating fashion until I end up with the number of pleats I need.

[FlareglooM]

Oh like that , could do that. Though the more pleats you make this way, the easier the pleats width will going to deviate.

[Brimstone]

Here's an example:

Standard copy paper is 21.6 cm x 27.9 cm. Let's say I want to do 28 pleats down the long side. Well, couldn't I just take 27.9 divided by 14, which is 1.99 or roughly 2 cm. I would then make a mark on the side of the paper, scoring it as reference. From there, I do a regular valley fold. Then, I go back and forth in regular pleating fashion until I end up with the number of pleats I need.

In my experience this is a bad method. It "amplifies" the error rather than containing it. With every new iteration of the pleat, the error increases.

Of course you could be a hell of a good folder and this wouldn't apply to you.