找回密码
 快速注册
搜索
查看: 27|回复: 0

[几何] Möbius strip作为可展曲面

[复制链接]

3149

主题

8386

回帖

6万

积分

$\style{scale:11;fill:#eff}꩜$

积分
65391
QQ

显示全部楼层

hbghlyj 发表于 2023-3-4 09:51 |阅读模式
长边为$2\pi R$, 短边为$2r$的矩形条, 一个短边旋转半圈, 粘合到另一端, 形成的曲面的只有1个外边缘, 为原矩形的两条长边粘在一起, 长度等于$4\pi R$, 所以等于中心圆长度的两倍.
常见的Rotoidal representation$\left\{\begin{array}{l}x=\left(a+u \cos \frac{v}{2}\right) \cos v \\ y=\left(a+u \cos \frac{v}{2}\right) \sin v \\ z=u \sin \frac{v}{2}\end{array}\right.$不能展开为矩形条: 可以计算它的边缘的长度, 大于中心圆长度的两倍.
The Möbius surface or half-twist surface is the non-developable ruled surface generated by the rotation of a line on a plane turning on itself around one of...
mathcurve

作为可展曲面 首先由Wunderlich得到:
Translation of W. Wunderlich’s “On a Developable Möbius Band”
Attempting to reproduce the equilibrium depiction of a band of finite width, using a rational-algebraic developable, Wunderlich then extends Sadowsky’s results by presenting perhaps the first successful model of a closed, analytic, developable Möbius band with associated thinness bounds.

surface envelope of the rectifying planes (i.e. the rectifying developable) of the rational curve $\cases{x=\frac{a t+b t^{3}+c t^{5}}{1+d^{2} t^{2}+2 d e t^{4}+e^{2} t^{6}} \\ y=\frac{d t+e t^{3}}{1+d^{2} t^{2}+2 d e t^{4}+e^{2} t^{6}} \\ z=\frac{-f}{1+d^{2} t^{2}+2 d e t^{4}+e^{2} t^{6}}}$ with $a=\frac{1}{2}, b=\frac{1}{3}, c=\frac{1}{6}, d=\frac{2}{3}, e=\frac{1}{3}, f=\frac{4}{5}$
import graph3;
size(200,IgnoreAspect);
size3(200,IgnoreAspect);
currentprojection=orthographic(camera=(1.5,0.3,2),up=Z,target=(0.5,0,0),zoom=0.8);
real r=2, w=1;
real x(real u, real v){return (r+v/2*cos(3pi*u))*cos(2pi*u);};
real y(real u, real v){return (r+v/2*cos(3pi*u))*sin(2pi*u);};
real z(real u, real v){return (v/2*sin(3pi*u));};
triple f(pair p){return (x(p.x,p.y),y(p.x,p.y),z(p.x,p.y));};
draw(surface(f,(0,-w),(1,w),nu=9,Spline),cyan);

上图为Rotoidal representation (代码来自Klaus-Anton的aops博客)
The shape of a Möbius strip - University College London
The simplest geometrical model for a Möbius strip is the ruled surface swept out by a normal vector that makes half a turn as it traverses a closed path. A common paper Möbius strip (Fig. 1) is not well described by this model because the surface generated in the model need not be developable, meaning that it cannot be mapped isometrically (i.e., with preservation of all intrinsic distances) to a plane strip. A paper strip is to a good approximation developable because bending a piece of paper is energetically much cheaper than stretching it. The strip therefore deforms in such a way that its metrical properties are barely changed.
It is reasonable to suggest that some nanostructures have the same elastic properties. A necessary and sufficient condition for a surface to be developable is that its Gaussian curvature should everywhere vanish.
Given a curve with non-vanishing curvature there exists a unique flat ruled surface (the so-called rectifying developable) on which this curve is a geodesic curve [10]. This property has been used to construct examples of analytic (and even algebraic) developable Möbius strips [29, 24, 25, 21].

Rotoidal representation是不可展的, 在中心圆Gaussian curvature为$-1/4$ MSE - Construction of the Möbius band
As for why the "usual" parameterization is not developable, I am not very good at calculating Gaussian curvature for a parameterization, so I can't say directly, but an easy way to see that it cannot lay flat is to consider a cut at $u=0$. The center line has length $2π$, but the edge lines each have length $2π(1+v^2_0/32+O(v^4_0))$, and there is no way for a flat surface to have both sides longer than the center. If anything, it is probably developable to a helicoid, but not a plane.
– Mario Carneiro, Aug 30, 2014 at 19:38
MSE
Möbius bands, unstretchable material sheets and developable surfaces
A remarkable exception to this trend is the comparatively early work of Halpern & Weaver [24], who used homotopy methods to prove that a flat rectangular strip of half-width $b$ and length $l$ admits an isometric immersion as a Möbius band in three-dimensional Euclidean point space if and only if $πb<l$ and, moreover, conjectured that such a strip can be isometrically embedded as a Möbius band in three-dimensional Euclidean point space only if the more restrictive inequality $2\sqrt3b<l$ holds. It is noteworthy that Sadowsky [1,2] imposed the latter inequality to ensure the viability of his construction.

手机版|悠闲数学娱乐论坛(第3版)

GMT+8, 2025-3-4 15:59

Powered by Discuz!

× 快速回复 返回顶部 返回列表