Date: Mon, 16 Oct 95 14:06:06 CDT From: rusin (Dave Rusin) To: t.mueller@mail.rhein-ruhr.de Subject: Re: Torus - equation In article <45mmgt\$eo9@alpha.fact.rhein-ruhr.de> you write: >Does someone know the equation of a torus? > >The best one would be a vector-equation. All right, then move from the origin to a point p lying on a circle by adding a vector v = cos(a) e1 + sin(a) e2 (here e1, e2, e3 are standard orthogonal basis vectors). On the plane F containing v and e3 draw a circle of smaller radius centered at the origin; add any vector from this circle to v, and you'll get to a circle around p. For example, you could take the new radius to be 1/2. Then you'll get to the point cos(a) e1 + sin(a) e2 + (1/2)( cos(a)cos(b) e1 + sin(a)cos(b) e2 + sin(b) e3 ) Letting a and b vary from 0 to 2pi will trace out the whole torus. You can eliminate a, b, and c here to get an implicit equation for the torus: this one is (x^2+y^2+z^2-5/4)^2 + 4 z^2 - 1 = 0. The general torus is obtained in a similar way by varying the lengths of the two vectors (as well, of course, as by choosing different orthonormal bases {e1, e2, e3} and by lateral translation). dave