George Hart wrote: Funsters, As Chief of Content at the Museum of Mathematics, http://momath.org/, which will open in New York City in 2012, I am in the midst of designing dozens of hands-on exhibits which convey the richness of math to the public (of all ages). The best exhibits are usually interactive, open ended, and visually engaging. We need to touch on many branches of mathematics beyond the school curriculum. I have been gathering ideas from many sources, but have not sent out a call to this math-fun list. So I'd be interested to hear suggestions for novel exhibits that you feel would help make MoMath the coolest hands-on museum anywhere while conveying mathematical ideas and ways of thinking. Georgehttp://momath.org/http://georgehart.com/ P.S. Gyroid sculpture barn raising this weekend in D.C.: http://www.georgehart.com/DC/ --------- Birollers! (Steinmetz solids) Maybe wooden, softball sized. Trough wants to be quite rigid. [math-fun] tetraroller Thursday, October 9, 2008 5:28 AM From: "rwg@sdf.lonestar.org" <rwg@sdf.lonestar.org> To: "math-fun" <math-fun@mailman.xmission.com> Recall (from a couple of years) that a "biroller" is the intersection of two perpendicular circular cylinders and takes an intriguing choatic path if rolled diagonally down a shallow, inclined (bathtub-like) trough. A "triroller" is the intersection of three perpendicular cylinders, with perhaps slightly less interesting rolling behavior. In either case, the roller can only change direction if it momentarily rests on the intersection of two ridges. Now consider the intersection of the four cones circumscribing a regular tetrahedron. (http://gosper.org/tetraroller1.gif, http://gosper.org/tetraroller2.gif). Instead of tipping on one of the triple points of ridges, it chooses left or right as it rolls up an initially nonexistent ridge. I'm not sure what would be the most entertaining surface, but these might make an even more hellish alternative to Gilbert & Sullivan's "elliptical billiard balls". The intersection ridges are in fact elliptical, and the projection along an altitude of the tetrahedron reveals quite probably a Reuleaux triangle (Wankel). The vector from the center to one of the triple points is exactly -3/5 times the vector to the opposite vertex. --rwg [Older mail] The volume formula for the perpendicular intersection of two cylinders of radius r can be found by replacing pi by 4 in the volume of the inscribed sphere. It is said that Archimedes found this thirteen centuries before Newton's calculus. Did he also know that the same 4/pi trick gives the surface area? If not, perhaps he noticed the universal prismatoid volume formula A + 4 A + A t m b V = h --------------, 6 (= one panel of Simpson's rule, with A:=area, t:=top, m:=middle, b:=bottom, h=height), which works on all the elementary solids, and deduced that it also works for the "biroller". (Is there a classical name for this solid?) I've said this before, but a fairly dense, fairly hard biroller (say, lathed from 2" aluminum bar stock) makes an intriguing chaotic toy, when rolled diagonally down a shallow, inclined trough (like a bathtub, but longer and shallower). The roller oscillates with gradually decaying amplitude until a downswing pauses on one of the two poles, whereupon it rocks with renewed amplitude, downslope on a perpendicular arc, taking a completely unpredictable time to reach the bottom. This was shown to me by an adolescent David Silver ca. 1970. --rwg PS, a Macsyma command to draw a biroller is block([plotnum1:2,f],[sin(t),sin(t),cos(t)],f:(1-u)*%%+u*[-1,1,1]*%%, makelist(f:part(f*[-1,1,1],[2,1,3]),k,1,4),plotsurf(%%,t,0,%pi,u,0,1)) [Answering private query:] Maybe it's time we make a triroller, I think it would be interesting, but harder and less interesting than the biroller. It would come down way faster, having twice the turning opportunities, which would only be desirable if the biroller had too much friction and missed turns by decaying too fast. OK, so when did you learn enough calculus to figure out the volume? --Bill It may be Mitchell Riley who needs thanking for telling me the Steinmetz name for these.