Parametric formula for figure 8 mobius strip

Question

I'm making 3D prints with Mathematica, and am interested in a parametric formula for a mobius strip that is in the form of a figure 8, rather than simply a circle with a twist in it. Can someone help me with that? Thanks Rick Russell

Answer 1

It is on Wikipedia:

$x(u,v)= \left(1+\frac{v}{2} \cos \frac{u}{2}\right)\cos u$

$y(u,v)= \left(1+\frac{v}{2} \cos\frac{u}{2}\right)\sin u$

$z(u,v)= \frac{v}{2}\sin \frac{u}{2}$

http://en.wikipedia.org/wiki/M%C3%B6bius_strip#Geometry_and_topology

Answer 2

F[a_, t_] := a {1, Sin[t]} Cos[t]/(1 + Sin[t]^2)
G[a_, t_] := Reverse[{-1, 1} #/Sqrt[#.#] & [D[F[a,u],u]]] /. u -> t
ParametricPlot3D[Append[F[3,t] + v Cos[t/2 - Pi/4] G[3,t],
     v Sin[t/2 - Pi/4] - Sin[t]/2], {t, 0, 2Pi}, {v, -1/2, 1/2}]

gives the following surface:

Explanation: F[a, t] is a vector-valued function for the parametric equation of a lemniscate in the $xy$ plane, with scaling parameter $a$: $${\boldsymbol f}(t) = (x(t), y(t)) = \left( \frac{a \cos t}{1+\sin^2 t}, \frac{a \cos t \sin t}{1 + \sin^2 t} \right), \quad t \in [0,2\pi).$$ G[a, t] gives a unit normal vector to $\boldsymbol f$. Then we observe that by rotating the unit normal vector in the plane perpendicular to the tangent vector to $\boldsymbol f$, such that the unit normal completes a rotation of $\pi$ radians for $0 \le t < 2\pi$, we achieve a half twist. This is accomplished by taking $\cos (t/2 - \pi/4) \boldsymbol g(a,t)$ in the $x$ and $y$ components, and $\sin (t/2 - \pi/4)$ in the $z$ component; where we used a phase shift of $\pi/4$ to get the surface to exhibit rotational symmetry. Then $v$ just scales the magnitude of this vector. Finally, we included a $-\tfrac{1}{2}\sin t$ term in the $z$ component so that the surface avoids self-intersection for $v \in [-1/2, 1/2]$.

Something fun happens when we animate the phase: