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Glen Whitney 2024-02-07 05:06:02 +00:00
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User-Stories.md

@ -4,6 +4,8 @@ Brief summaries of activities one might try/problems one might solve with dyna3.
- P12 Rigid Hexagon from the Playground in Math Horizons Vol 2 no.1. Specifically, let $ABCDEF$ be six points in space such that $AB=CD=EF=a$, $BC=DE=FA=b$, and $AD=BE=CF=a+b$. Then $ABCDEF$ is a planar equiangular hexagon. So in particular it is rigid, which hopefully dyna3 should detect even though fewer distances are specified than generically produce rigidity. A more extreme version of this is specifying $AB=BC=CD=DE=EF=1$ and $AF=5$, constraining all six points to lie on a line. A question one might explore with dyna3, for which I do not know the answer, is whether there are also (interesting) specifications of exactly 7 or 8 distances that also entail rigidity. - P12 Rigid Hexagon from the Playground in Math Horizons Vol 2 no.1. Specifically, let $ABCDEF$ be six points in space such that $AB=CD=EF=a$, $BC=DE=FA=b$, and $AD=BE=CF=a+b$. Then $ABCDEF$ is a planar equiangular hexagon. So in particular it is rigid, which hopefully dyna3 should detect even though fewer distances are specified than generically produce rigidity. A more extreme version of this is specifying $AB=BC=CD=DE=EF=1$ and $AF=5$, constraining all six points to lie on a line. A question one might explore with dyna3, for which I do not know the answer, is whether there are also (interesting) specifications of exactly 7 or 8 distances that also entail rigidity.
- In the other direction, consider equiangular, equilateral, not-necessarily-planar hexagons, see https://arxiv.org/pdf/1105.5046.pdf. These generally have a one-dimensional configuration space, possibly with additional isolated point(s?). This is true even though the system might seem overdetermined. And according to Dan Piker, an equilateral right-angled heptagon _also_ has a one-dimensional configuration space (or at least the space has at least one one-dimensional component),
- A slightly farfetched one: put three or more "pins" in a plane (or in space). Constrain an additional point in the plane as the "pencil" such that the length of a string looped around the pins and the pencil in some way (there are multiple configurations) has constant length. Find the locus of positions of the pencil. (Generalizations of ellipse-drawing; note James Clerk Maxwell considered these loci in his youth and apparently wrote an article or report of some kind on his findings.) - A slightly farfetched one: put three or more "pins" in a plane (or in space). Constrain an additional point in the plane as the "pencil" such that the length of a string looped around the pins and the pencil in some way (there are multiple configurations) has constant length. Find the locus of positions of the pencil. (Generalizations of ellipse-drawing; note James Clerk Maxwell considered these loci in his youth and apparently wrote an article or report of some kind on his findings.)
- Somewhat less farfetched, another natural generalization of ellipse-drawing: find the locus of a point P such that the surface area of tetrahedron $ABCP$ is a constant. (In the plane, an ellipse is the locus of a point $P$ such that the perimeter of $ABP$ is constant.) - Somewhat less farfetched, another natural generalization of ellipse-drawing: find the locus of a point P such that the surface area of tetrahedron $ABCP$ is a constant. (In the plane, an ellipse is the locus of a point $P$ such that the perimeter of $ABP$ is constant.)