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
-------------------------------------
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