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Re: Cube inside a torus
Posted:
May 30, 2012 10:58 AM
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> 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 cross-section 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=(Ro-Ri)/2
Arrange the cube, of side 'd', in the torus with sides parallel to the coordinate axes, with faces at x-d 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 (x-d, 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.
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