> On Jul 21, 2025, at 12:12 PM, wanderer via Origami > <[email protected]> wrote: > > I was prepping to fold the $ wreath by Dennis Walker and not knowing how > simple or complex or intricate the folding was, i thought I’d make > $-proportional pieces from 3” squares. I folded a 6” sq in half, then lightly > folded bottom corner at 45deg, then bisected that angle to create 22.5deg > fold, and where that last crease touches the edge away from the pivot point, > is where one folds to, to then get a $-proportional piece. I needed 8 pieces > for the model and so i valley folded then mountain folded —- zigzag zigzag > etc —- only to realize that the 6” paper height was perfectly split into > 1/5ths. Is this true in terms of geometry or is it something that’s possible > when dealing with the physicality of the paper? See the attached photo > showing the pre-creases i made and then the valley-mountain folds that show > the exact 1/5ths. Thank you.
It’s close to 1/5 of the square height (= 0.200), but not exact; the exact height of the bottom rectangle relative to the square height is (sqrt(2)–1)/2, which is about 0.207. The fractional height compared it its width is thus 0.414 (rather than 0.400 if it had been a perfect fifth). But you’re right about it being a decent approximation of a US dollar, which is 66.3 x 156 mm, which gives it a fractional height of 0.425. This is close enough to 0.414 that there are lots of dollar folds using 22.5° geometry that can make use of this near-coincidence of geometry. One examples is the $2 Crane that, coincidentally, I just taught at OrigamiUSA’s Convention a few hours ago. https://origamiusa.org/classes/convention/2025/2-crane-and-lily Best, Robert
