Re: [math-fun] Tricylinder puzzle
Dan Asimov>Perhaps I should define my terms. The solid angle at a point on a convex surface S is the limit of the solid angle subtended by the surface in an epsilon-neighborhood N(eps) of the point, as epsilon -> 0. And that is the area on the unit sphere of the set of all perpendicular unit vectors to the supporting hyperplanes* to S on N(eps) (when the vectors are translated to the origin). --Dan ________________________________________________________ * < http://en.wikipedia.org/wiki/Supporting_hyperplane > On Feb 2, 2015, at 10:27 AM, Daniel Asimov <asimov@msri.org> wrote: Consider the tricylinder T -- the intersection of three cylinders each of unit radius in 3-space with perpendicular axes. Let S denote its surface bd(T). PUZZLE: Find the solid angle in closed form at a point of S common to the 3 cylinders. --Dan The tricylinder, a.k.a. triroller, a.k.a. Steinmetz solid, is the union of four copies of the convex hull of the three ellpses {{Cos[t], Sin[t], Sin[t]}, {Sin[u], Sin[u], Cos[u]}, {Sin[v], Cos[v], Sin[v]}} which, when t=u=v=-π/4±π/2 coincide at the two points ±(1,1,1)/√2 , where their tangent vectors are proportional to (1,-1,-1), (-1,-1,1), (-1,1,-1) and (-1,1,1),(1,1,-1),(1,-1,1). The angle between any pair of these (at the same vertex) is acos(-1/3), the tetrahedral bond angle, gving a solid angle of π. So four Steinmetzes should fit snugly around a single point. --rwg
On Thu, Feb 5, 2015 at 6:12 PM, Bill Gosper <billgosper@gmail.com> wrote:
Dan Asimov>Perhaps I should define my terms.
The solid angle at a point on a convex surface S is the limit of the solid angle subtended by the surface in an epsilon-neighborhood N(eps) of the point, as epsilon -> 0.
And that is the area on the unit sphere of the set of all perpendicular unit vectors to the supporting hyperplanes* to S on N(eps) (when the vectors are translated to the origin).
--Dan ________________________________________________________ * < http://en.wikipedia.org/wiki/Supporting_hyperplane >
On Feb 2, 2015, at 10:27 AM, Daniel Asimov <asimov@msri.org> wrote:
Consider the tricylinder T -- the intersection of three cylinders each of unit radius in 3-space with perpendicular axes. Let S denote its surface bd(T).
PUZZLE: Find the solid angle in closed form at a point of S common to the 3 cylinders.
--Dan
The tricylinder, a.k.a. triroller, a.k.a. Steinmetz solid, is the union of four copies of
[ the convex hull] No, the figure has these edges, but is plumper than that.
of the three ellpses {{Cos[t], Sin[t], Sin[t]}, {Sin[u], Sin[u], Cos[u]}, {Sin[v], Cos[v], Sin[v]}} which, when t=u=v=-π/4±π/2 coincide at the two points ±(1,1,1)/√2 , where their tangent vectors are proportional to (1,-1,-1), (-1,-1,1), (-1,1,-1) and (-1,1,1),(1,1,-1),(1,-1,1). The angle between any pair of these (at the same vertex) is acos(-1/3), the tetrahedral bond angle, giving a solid angle of π. So four Steinmetzes should fit snugly around a single point. --rwg
I think this is still right. If so, Nice one, Dan. If not, Nice one, Dan. --rwg
Yes, you just 3D-checkerboard the steinmetzes like you do regular dodecahedra, leaving "endosteinmetzes" gosper.org/endosteinmetz.png (front and back removed) instead of endododecahedra in alternate cells. --rwg On Sat, Feb 7, 2015 at 9:04 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Feb 5, 2015 at 6:12 PM, Bill Gosper <billgosper@gmail.com> wrote:
Dan Asimov>Perhaps I should define my terms.
The solid angle at a point on a convex surface S is the limit of the solid angle subtended by the surface in an epsilon-neighborhood N(eps) of the point, as epsilon -> 0.
And that is the area on the unit sphere of the set of all perpendicular unit vectors to the supporting hyperplanes* to S on N(eps) (when the vectors are translated to the origin).
--Dan ________________________________________________________ * < http://en.wikipedia.org/wiki/Supporting_hyperplane >
On Feb 2, 2015, at 10:27 AM, Daniel Asimov <asimov@msri.org> wrote:
Consider the tricylinder T -- the intersection of three cylinders each of unit radius in 3-space with perpendicular axes. Let S denote its surface bd(T).
PUZZLE: Find the solid angle in closed form at a point of S common to the 3 cylinders.
--Dan
The tricylinder, a.k.a. triroller, a.k.a. Steinmetz solid, is the union of four copies of
[ the convex hull] No, the figure has these edges, but is plumper than that.
of the three ellpses {{Cos[t], Sin[t], Sin[t]}, {Sin[u], Sin[u], Cos[u]}, {Sin[v], Cos[v], Sin[v]}} which, when t=u=v=-π/4±π/2 coincide at the two points ±(1,1,1)/√2 , where their tangent vectors are proportional to (1,-1,-1), (-1,-1,1), (-1,1,-1) and (-1,1,1),(1,1,-1),(1,-1,1). The angle between any pair of these (at the same vertex) is acos(-1/3), the tetrahedral bond angle, giving a solid angle of π. So four Steinmetzes should fit snugly around a single point. --rwg
I think this is still right. If so, Nice one, Dan. If not, Nice one, Dan. --rwg
Here are four steinmetzes (the foremost cut away) surrounding a gridpoint: <http://gosper.org/4stein.png> gosper.org/4stein.png their centers forming a regular tetrahedron, as in a 2x2x2 checkerboard. NeilB may have a better pic. Dan, did you know all along that steinmetzes did this? --rwg On Sat, Feb 7, 2015 at 11:46 PM, Bill Gosper <billgosper@gmail.com> wrote:
Yes, you just 3D-checkerboard the steinmetzes like you do regular dodecahedra, leaving "endosteinmetzes" gosper.org/endosteinmetz.png (front and back removed) instead of endododecahedra in alternate cells. --rwg
On Sat, Feb 7, 2015 at 9:04 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, Feb 5, 2015 at 6:12 PM, Bill Gosper <billgosper@gmail.com> wrote:
Dan Asimov>Perhaps I should define my terms.
The solid angle at a point on a convex surface S is the limit of the solid angle subtended by the surface in an epsilon-neighborhood N(eps) of the point, as epsilon -> 0.
And that is the area on the unit sphere of the set of all perpendicular unit vectors to the supporting hyperplanes* to S on N(eps) (when the vectors are translated to the origin).
--Dan ________________________________________________________ * < http://en.wikipedia.org/wiki/Supporting_hyperplane >
On Feb 2, 2015, at 10:27 AM, Daniel Asimov <asimov@msri.org> wrote:
Consider the tricylinder T -- the intersection of three cylinders each of unit radius in 3-space with perpendicular axes. Let S denote its surface bd(T).
PUZZLE: Find the solid angle in closed form at a point of S common to the 3 cylinders.
--Dan
The tricylinder, a.k.a. triroller, a.k.a. Steinmetz solid, is the union of four copies of
[ the convex hull] No, the figure has these edges, but is plumper than that.
of the three ellpses {{Cos[t], Sin[t], Sin[t]}, {Sin[u], Sin[u], Cos[u]}, {Sin[v], Cos[v], Sin[v]}} which, when t=u=v=-π/4±π/2 coincide at the two points ±(1,1,1)/√2 , where their tangent vectors are proportional to (1,-1,-1), (-1,-1,1), (-1,1,-1) and (-1,1,1),(1,1,-1),(1,-1,1). The angle between any pair of these (at the same vertex) is acos(-1/3), the tetrahedral bond angle, giving a solid angle of π. So four Steinmetzes should fit snugly around a single point. --rwg
I think this is still right. If so, Nice one, Dan. If not, Nice one, Dan. --rwg
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Bill Gosper