Alexander Volovics
Mon, 8 Feb 1999 18:59:36 -0500
On Mon, 08 Feb 1999 Ted Harding wrote: >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 ;) >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 you can express this equation in a slightly more simple form, namely: (sqrt(x^2 + y^2) - a)^2 + z^2 = b^2 where a>b Namely revolve a circle of radius b about an exterior line in its plane, that is distance a from the center. Gives a ring shaped torus. At last the mailing list is becoming interesting :-) Alexander ------------------------------------- Alexander Volovics Dept of Methodology & Statistics Maastricht University, Maastricht, NL ------------------------------------- - To get out of this list, please send email to [EMAIL PROTECTED] with this text in its body: unsubscribe suse-linux-e Check out the SuSE-FAQ at http://www.suse.com/Support/Doku/FAQ/ and the archiv at http://www.suse.com/Mailinglists/suse-linux-e/index.html