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