Montroll's Five-sided square

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Re: Montroll's Five-sided square

Postby Falcifer » October 27th, 2017, 10:25 am

I can't say I'm surprised about not having an exact solution.

The sequence from ReferenceFinder, as you say, gives an error of 0.00001104889..., which is relative to a unit square. So it's off by 1.1mm with a square 100m x 100m. With typical paper (15cm - 45cm) you're looking at about 1/200 mm error, which I dare say is magnitudes smaller than the human error involved.

Montroll's sequence, which is apparently good enough for the model, gives you (2√2 - 1)/4, and an error of 0.00047 (0.2mm with a 45cm square), which is 42.5 times greater than the other solution.

I think it'd be an interesting project to see just how many folds it would require to reach a precise reference point, not just for the above point, but others. I know the source code for ReferenceFinder is available, so I'm sure with enough computing power and some programming know-how, someone could allow the program to run into much higher ranks. Potentially printing an output every time it finds a solution with a lower error than the current lowest. That way it's not printing every solution, and it's not running indefinitely trying to find an extremely high ranked solution.

Though, paper isn't perfect and neither are people. Even if you fold accurately, lining up a corner with an edge isn't always precise. Constant folding and unfolding will slightly change the dimensions as the fibres are stretched and flattened.

Ultimately, it's a case of determining how important the precision of the reference point is vs. how many folds you want to do to get there, and just how much more precision you're gaining with the extra folds. As well as if the extra precision will even be noticeable compared to the imprecision of folding.
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