Hi, I am playing around with Möbius strips in Matlab and had a strange problem I cannot resolve. I am only interested in vectorized solutions, not for- or while-loops, please. The graph almost looks like a Möbius strip, but the edges are not joined. Can anyone see what the problem is?
Code below:
%Begin
clf
clear all
u = linspace(0,2*pi,100);
v = linspace(-0.5,0.5,100);
%Parametrization, vectorized
x=cos(u)'+diag((diag(v)*cos(u/2)'))*cos(u)';
y=sin(u)'+diag((diag(v)*cos(u/2)'))*sin(u)';
z=v'*sin(u/2);
%Plotting: figure 1 and 2 are quite a lot off. figure 3 almost looks like a
%Moebius strip except the edges are not joined.
figure(1)
surf(x,y,z)
figure(2)
mesh(x,y,z)
figure(3)
plot3(x,y,z)
%Code that works and actually produce a Moebius strip
syms e r;
s = cos(e)+r*cos(e/2)*cos(e);
d = sin(e)+r*cos(e/2)*sin(e);
f = r*sin(e/2);
figure(4)
ezsurf(s,d,f, [0, 2*pi, -0.5, 0.5])

 Akzeptierte Antwort

Rahul K
Rahul K am 7 Mär. 2017

2 Stimmen

Use meshgrid to generate matrices for both your parameters, rather than using vectors.
u = linspace(0,2*pi,100);
v = linspace(-0.5,0.5,100);
[u,v] = meshgrid(u,v);
Now that u and v are matrices, you don't have to worry about transposing or using diag etc. Leave the parametric equations in their explicit form (i.e. don't expand the brackets), remember to use element-wise multiplication, and everything works out.
x = (1+v.*cos(u/2)).*cos(u);
y = (1+v.*cos(u/2)).*sin(u);
z = v.*sin(u/2);

2 Kommentare

John Klint
John Klint am 7 Mär. 2017
Thanks, great answer :)
Neil
Neil am 8 Apr. 2023
Bearbeitet: Neil am 8 Apr. 2023
Thank you for this example. I was trying to work out how to plot my version of a Klein Strip (which is a 4D equivalent of a Mobius Strip and similar to a Klein Bottle though more symmetric). Your example simplified the whole process for me. Very much appreciated.

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