
Re: Cube inside a torus
Posted:
May 30, 2012 10:58 AM


> Let R = 2 and r = 1 be respectively the major radius > and minor radius of a torus. Determine the edge > length a of the largest cube that fits in the given > torus.
This interesting problem seems not to have received a response yet.
Let Rc be the centroidal radius and rc the crosssection radius, then equation of torus is
(Rc  sqrt(x^2+y^2))^2 + z^2 = rc^2
If Ro and Ri are the outer and inner radii of the torus, then Rc= (Ro+Ri)/2 and rc=(RoRi)/2
Arrange the cube, of side 'd', in the torus with sides parallel to the coordinate axes, with faces at xd and x; y=+d/2; z=+d/2.
The max cube will sit with outer corners touching inside of torus, at C1,C2; and mid inner side tangent to torus, at M
Corner points C1,C2 are (x, +d/2, d/2) Mid tangent point M is (xd, 0, d/2)
For Rc=2, rc=1 2<x<3 and 0<d<2, x=2.641, d=1.370
For Ro=2, Ri=1 Rc=3/2, rc=1/2 3/2<x<2 and 0<d<1, x=1.829, d=0.691
Regards, Peter Scales.

