On 08-Feb-99 Hubert Mantel wrote:
But that's too easy. The question was meant to be: Find an equation f(x,y,z)=0 so that all solutions of the equation form the surface of a torus. To be honest: I don't know the solution. I even don't know if this equation exists ;) -o) Hubert Mantel Goodbye, dots... /\\
Well, how about: Suppose the torus is swept out by a circle of radius r1 whse centre is carried round a circle of radius r0 (r1 < r0 for a proper torus). Let u = x/r0, v = y/r0, w = z/r1, C = r1/r0. Then u^2 + v^2 = (1 + C sqrt(1 - w^2) )^2 (Consider r = r0 + r1 cos q, z = r1 sin q, x = r cos p = r0 cos p + r1 cos p cos q y = r sin p = r0 sin p + r1 sin p cos q and eliminate the angles p, q. p is the angle in the x-y plane from a fixed direction to the centre of the sweeping circle; q is the "angle of elevation" from the centre of this circle to a point on the torus as seen along the line from the origin to the centre of this circle; r is the distance from the origin to the point in the x-y plane vertically beneath the point on the torus.) I think it's right ... Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <Ted.Harding@nessie.mcc.ac.uk> Date: 08-Feb-99 Time: 20:34:37 ------------------------------ XFMail ------------------------------ - To get out of this list, please send email to majordomo@suse.com with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at <A HREF="http://www.suse.com/Support/Doku/FAQ/"><A HREF="http://www.suse.com/Support/Doku/FAQ/</A">http://www.suse.com/Support/Doku/FAQ/</A</A>> and the archiv at <A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html"><A HREF="http://www.suse.com/Mailinglists/suse-linux-e/index.html</A">http://www.suse.com/Mailinglists/suse-linux-e/index.html</A</A>>