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