# The Mobius edge.

1. Jul 4, 2008

### Alfi

The Mobius edge.
Not the two dimensional object, the one dimensional curve.

btw - I am just a re-awakening old guy trying to relearn math from the ground up. :) just so you know your questioner.

I think I have worked out the parametric for a mobius strip. I get the classic x,y,z, coordinates for a strip with two variables. R constant at 1.0 for me.
x = (R + L*Cos(A/2)) * Cos(A)
y = (R + L*Cos(A/2)) * Sin(A)
z = L*Sin(A/2)
A, ranging from 0 to 360 degrees, L ranging from -Lmax to +Lmax,
I use an R of 1 and a L approaching +/- 0.50

My problem, is trying to restrict the L quantity till it is just a single curved line. Eventually I hit a division by zero error in the graphing programs.

I can make it draw up to some point (?) , like L= 0.5 +- .00001 and then I end up with just a very narrow two dimensional strip, not a line.

Is there a simple equation to define the line or am I chasing a wrong path?

Thanks.

2. Jul 4, 2008

### Hurkyl

Staff Emeritus
You mean -- you are trying to draw the curve that represents the edge of a Möbius strip?

Wouldn't that just be the curve you get from plotting for each A the two L-values Lmin and Lmax? If you want to get a single parametrization for the entire curve (rather than drawing the two curves I suggested), do you see why it's not enough to just the polar angle (A) range from 0 to 360?

3. Jul 4, 2008

I do

4. Jul 5, 2008

### Alfi

I'll try to figure out another way to ask the question.
Thanks.

5. Aug 15, 2008

### Alfi

http://www.cosmosmagazine.com/news/1457/moebius-strip-riddle-solved-last [Broken]
( not exactly a good source but it's all I got *shrug* Sorry.

Since 1930, the Möbius strip has been a classic poser for experts in mechanics. The teaser is to resolve the strip algebraically – to explain its unusual shape in the form of an equation.

I am interested in the equation that describes the edge only.
I don't have the training or the knowledge to figure it out myself ( I've tried .. I'm out of my league and I know it )

Not a general equation! , Just in the specific case of the center circle radius being unity ( x^2 + Y^2 = 1 ) and the width or the separation between the curve and the center circle at 0.5 of the radius.
Think of a torus with a curved line (mobius edge) drawn on the outer surface.

I just don't understand why the stress or curve or whatever would not be constant through out the line. The picture they show in the link is not the idealized torus or mobius strip that i envision.

I have built dozens of models in various forms to try and visualize how a torus and the mobius edge are related. My work room is getting littered with the damn things. :)

To me, in my very limited understanding, the edge ( the drawn curved line on the surface of the torus) must be a subset of the torus.
If I cut the torus along the edge ( top side to bottom side ... and once again I prove that I don't know the correct terminology ) for 360 degrees, the torus remains intact. It does not fall into two pieces. The part that gets cut out ( width of the blade ) is a mobius strip. If I had a very thin blade the area cut out is two dimensional, length and width only, but needs to be three dimensional in it's description.

I have probably made a dozen errors in my questions. Please forgive me for my ignorance .

*crosses fingers that I explained my desire in some kind of meaningful way *

Last edited by a moderator: May 3, 2017
6. Aug 22, 2008

### disregardthat

wait a min

Last edited: Aug 22, 2008
7. Aug 22, 2008

### disregardthat

Try this parametric equation with Matlab or something similar with 0<=t<2pi

[ (L./(2.*pi)+W.*cos(t./2)./2).*cos( t ) , (L./(2.*pi)+W.*cos(t./2)./2).*sin( t ) , W.*sin( t./2 )./2 ]

L is the length of the paper and W is the width. Remember that 2*pi*W<L

This möbious band is twisted with a constant "bend" and the centre line of the paper strip is located on a circle of radius L/(2pi) on the xy plane.

The equation is this:

$$\vec{r}(\theta)=[ (\frac{L}{2\pi}+\frac{W}{2}\cos(\frac{\theta}{2}))\cos( \theta ) , (\frac{L}{2\pi}+\frac{W}{2}\cos(\frac{\theta}{2}))\sin( \theta ) , \frac{W}{2}\sin( \frac{\theta}{2} )]$$

Last edited: Aug 22, 2008