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hbghlyj
Posted at 2022-2-11 23:38:12
Last edited by hbghlyj at 2023-3-4 00:59:00回复 5# isee
如何理解Möbius环的参数方程
\begin{align*}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}}\\\end{align*}for $0\leq u<2\pi$ and $-1\leq v\leq 1$. This produces a Möbius strip of width 1, whose center circle has radius 1, lies in the $xy$-plane and is centered at (0,0,0). The parameter $u$ runs around the strip while $v$ moves from one edge to the "other".
当$u=0$时得到的参数曲线$\left(1+\frac v2,0,0\right)$是从$\left(\frac12,0,0\right)$到$\left(\frac32,0,0\right)$的$x$轴上的一条线段.
当$u=2π$时得到的参数曲线$\left(1-\frac v2,0,0\right)$是从$\left(\frac32,0,0\right)$到$\left(\frac12,0,0\right)$的$x$轴上的一条线段.
所以说这是两条线段反向粘在一起了!
当$u=π$时得到的参数曲线$\left(-1,0,\frac v2\right)$是从$\left(-1,0,\frac12\right)$到$\left(-1,0,-\frac12\right)$的平行于$z$轴的一条线段.
当$v=0$时得到的参数曲线$\left(\cos u,\sin u,0\right)$是$xy$-平面上的圆. |
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色k
Posted at 2018-11-30 20:34:58
