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A helicoid has the following parametric equation:

$$ r(u,v)=\left \langle v\cos u,v\sin u,cu \right \rangle,\qquad u,v,c\in\mathbb{R}. $$

In ruled form,


it has the directrix and the rulings

$$\begin{align} \alpha(u)&=\left \langle 0,0,cu \right \rangle,\\ \Lambda(u)&=\left \langle \cos u,\sin u,0 \right \rangle, \end{align}$$


I want to create a circular helicoid whose directrix is not a vertical line but a circle,

$$ \alpha(u)=\left \langle R\cos u,R\sin u,0 \right \rangle, $$

and whose rulings rotate on the plane spanned by $\left \langle -\cos t,-\sin t,0 \right \rangle$ and $\left \langle 0,0,1 \right \rangle$ and not on the plane spanned by $\left \langle 1,0,0 \right \rangle$ and $\left \langle 0,1,0 \right \rangle$.

I need a hint on how to begin approaching this problem.

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2 Answers 2

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Let $a>b>0$. Then $$\gamma_m:\quad \theta\mapsto \left\{\eqalign{\rho&:= a+ b\cos\theta \cr z&:=b\sin\theta\cr}\right.\qquad(0\leq\theta\leq 2\pi)$$ is a circle of radius $b$ in an (abstract) $(\rho, z)$ meridian half-plane. The center of $\gamma_m$ is at distance $a$ from the $z$-axis. We rotate this circle around the $z$-axis by introducing a second rotation variable $\phi$ and letting $$x=\rho\cos\phi, \quad y=\rho\sin\phi\qquad(0\leq\phi\leq 2\pi)\ .$$ In this way we get a parametric representation of a torus $T\subset{\mathbb R}^3$ whose "soul radius" is $a$ and whose "thickness radius" is $b$: $$T: \quad [0,2\pi]^2\to{\mathbb R}^3\ ,\qquad (\theta,\phi)\mapsto\left\{\eqalign{x&=(a+b\cos\theta)\cos\phi \cr y&=(a+b\cos\theta)\sin\phi\cr z&=b\sin\theta\cr}\right.\quad .$$ You obtain a "helicoidal curve" $\gamma_h$ on this torus if you keep $\phi$ as parameter running from $-\infty$ to $\infty$ and let $\theta$ be a fixed linear function of $\phi$. Thereby you have to introduce an extra parameter $k\ne0$ which dictates how fast $\gamma_h$ winds around $T$ as $\phi$ varies. Up to a fixed rotation around the $z$-axis the resulting parametric representation of $\gamma_h$ is given by $$\gamma_h:\quad {\mathbb R}\to{\mathbb R}^3\ ,\qquad \phi\mapsto\left\{\eqalign{x&=\bigl(a+b\cos(k\phi)\bigr)\cos\phi \cr y&=\bigl(a+b\cos(k\phi)\bigr)\sin\phi\cr z&=b\sin(k\phi)\cr}\right.\quad .$$

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Note that for each $v$ the intersection of the helicoid with a coaxial cylinder (spiral) in cylindrical coordinates $(v,u,z)$ has equation $z=u$, or more generally $z=\alpha u$ i.e. a "straight line through the origin".

Therefore, it makes sense to define a "circular helicoid" in toroidal coordinates $(r,u,v)$ by the equation $v=\alpha u$.

$$\begin{aligned}x= & \left(R+r\cos\alpha u\right)\cos u\\ y= & \left(R+r\cos\alpha u\right)\sin u\\ z= & r\sin\alpha u \end{aligned}$$

enter image description here

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funnily enough if you replace the expression for $z$ with $r\sin\alpha u +\beta u$ you get something that looks like a "helicoidal helicoid" –  Valentin Oct 24 '12 at 21:39

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