You are watching: How to fold a paper into thirds
Fold double to attain quarter markings at the record bottom.Fold along the line with the top corner and the third of these marks.The vertical lines v the first two marks intersect this inclined heat at thirds, which enables the final foldings.
(Photo by Ross Millikan listed below - if the image aided you, you can up-vote his too...)
Here is a snapshot to go through Hagen von Eitzen"s answer. The horizontal lines room the result of the very first two folds. The diagonal line is the 3rd fold. The heavy lines room the points at thirds because that folding right into the envelope.
This is both practical (no extra creases) and an exact (no guessing or estimating).
Roll the file into a 3-ply tube, through the ends aligned:
Pinch the document (crease the edge) whereby I"ve attracted the red line
Use the pinch note to present where the folds should be
This solution works just with a paper of paper having facet ratio of sqrt(2) (as A4 has).Only two extra crease required.
Roll in to a cylinder until both edges room opposite to each other.Fold the points whereby the edges touches the paper. (Squish the cylinder native left and also right)
Approximation method:Assume a 120 level angle and fold as presented below.For accuracy, complement edge-side to any of the various other two sides.
Beyond the more geometric methods defined so far, over there is an iterative algorithm (in practice, as specific as any kind of exact method) due to Shuzo Fujimoto that ns think no one mentioned. In fact, the following technique can be generalized to any kind of shape, dimension and number of foldings.
Let me signify $d_l$ the street from the left next of the document to the first mark ~ above the left and also $d_r$, the street from the ideal side come the right mark. To simplify, assume that the lenght of the next you want to divide is 1.
Make a an initial approximation because that $d_l$. You desire $1/3$, however imagine you take $1/3+\varepsilon$ ($\varepsilon$ being some error to the right or come the left). Thus, on the appropriate you now have actually $2/3-\varepsilon$.
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Next, divide the right component into two for the very first approximation of $d_r$ (by taking the right side that the paper to your first pinch; again, just a pinch). This gives you $d_r=1/3-\varepsilon/2$, thus, a far better approximation!
Now, repeat this procedure on the left. ~ above the left component you now have $2/3+\varepsilon/2$. Take it the left side of the paper to the 2nd pinch to attain a 2nd approximation the $d_l = 1/3+\varepsilon/4$. Note that, after 2 pinches, you have diminished your initial error come a quarter!
If your initial guess was precise enough, you will certainly not need to continue more. But, if you need much more precission, you just have to repeat the process a couple of times more. For instance, an initial error the $\varepsilon=1$cm reduce to much less than 1mm (0.0625mm) after ~ repeating the iteration twice.