rwg>(http://gosper.org/tetraroller1.gif, http://gosper.org/tetraroller2.gif)
PS, to compute the volume of a tetraroller, I need to run the prismatoid formula on a wedge cut off the base of a cone. This requires finding two saggitae. Since osage orange is most highly prized for making bows, it is therefore
Can't work. The saggita and radius are linear functions of the vertical coordinate. If the segment area were a polynomial function of height, we could solve the transcendental Kepler's Pizza equation (www.tweedledum.com/rwg/pizza.htm). Did anyone else try for the volume formula? I got a couple of "challenge integrals" that defeated Mma and Macsyma. But this is fairly nice: dissect the tetraroller into twelve congruent conical wedges. These are "tetrahedroids", with one edge joining a tetrahedron vertex to the centroid, and the opposite "edge" being an elliptical arc, the intersection of two cones. For a tetrahedron centered at the origin with edge length sqrt3, apex on the z axis, one of the three tetrahedroids sharing the apex has parametric edge formula E(t) = [cos(t)*(sqrt(cos(t)^2+3)-cos(t))/2,sqrt(3)*(sqrt(cos(t)^2+3)-cos(t))*sin(t)/4, cos(t)*(sqrt(cos(t)^2+3)-cos(t))/(2*sqrt(2))], |t|<acos 1/5. Now a volume element(t) is vol_tetrahedron(E(t),E(t+dt),apex,origin), a big mess that to first order simplifes down to sqrt(2)*sqrt(3)*(2*cos(t)*sqrt(cos(t)^2+3)-2*cos(t)^2-3)*dt/64 . Integrating -acos(1/5)<t<acos(1/5) (times 12) gave 3*(sqrt(6)*asin(sqrt(6)*(2*sqrt(19)-1)/25)/2-3*sqrt(19)/25+3/50) ~1.76132660170417, nearly thrice the tetrahedron's sqrt(6)/4 ~0.6123724356958. It would be nice to get an independent check of this, and then use it as the rhs of the challenge integrals. --rwg PS, did anybody look at intersecting four cylinders instead of cones? I'm not even sure how many faces this makes (http://gosper.org/face.jpg). Likewise the intersections of 6 and 10 cylinders corresponding to the higherhedra. Nice workouts for somebody's solid-intersector software. True or False Quickie: The volume of a tetrahedron is determined by the areas of its faces.