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) <[EMAIL PROTECTED]>
Date: 08-Feb-99 Time: 20:34:37
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