Möbius band-like strip

67 Views Asked by At

A möbius band can be parametrized by the following

$ x = (R+r\cos(1/2\theta))\cos(\theta)\\ y = (R+r\cos(1/2\theta))\sin(\theta)\\ z = r\sin(1/2\theta) $

with $R = 1, r \in [0,1], \theta \in [0,2\pi]$

However what is this manifold called?

$ x = (R+r\cos(\theta))\cos(\theta+a)\\ y = (R+r\cos(\theta))\sin(\theta+a)\\ z = r\sin(\theta),\\ a \in [0,2\pi] $

When I plot it it looks like a möbius strip, here $a = 0$

enter image description here

1

There are 1 best solutions below

0
On

Here consider a curve $c_1$ $$((2 +\cos\ t/2)\cos\ t,(2+\cos\ t/2)\sin\ t,\sin\ \frac{t}{2}) $$ where $t\in [0,2\pi] $.

It is joining from $(3,0,0)$ to $(1,0,0)$.

Define $c_2(t)=2(\cos\ t,\sin\ t,0)$.

Hence Mobius band is $$ [c_2(t)-c_1(t) ]s + c_1(t) $$ where $t\in [0,2\pi),\ s\in [0,2]$