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I am trying to parameterize the surface of a 3D helix tube like below.

enter image description here

As known, a parametric curve only requires one parameter $t$

$x=R\sin(t)$, $y=R\cos(t)$, $z=p t$,

where $R$ is the radius, and $p$ represents the pitch.

However for a surface, I think I need at least two parameters $t_1$ and $t_2$ to represent the 3D closed surface of a 3D helix tube.

Here, I need $x$, $y$, and $z$ in the parametric equations to be "decoupled". i.e., I am expecting something like

$x=f_1(t_1, t_2)$, $y=f_2(t_1, t_2)$, and $z=f_3(t_1, t_2)$.

What are the expressions?

Sibbs Gambling
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  • Could you try to clarify what surface it is exactly that you're trying to describe? – Danu May 23 '14 at 15:27
  • How do you go from a straight line to a cylinder? – Bernhard May 23 '14 at 15:33
  • @Danu Updated. Thanks! – Sibbs Gambling May 23 '14 at 15:38
  • @Bernhard Thanks for the heads-up. I understand that they are essentially the same thing, but once I decouple $x$, $y$, and $z$, I only get $1/4$ of my surface. It is not closed. – Sibbs Gambling May 23 '14 at 15:41
  • I have answered this question already before. Start with a toroid, one special case is : $ (3 + \cos u ) \cos v ,(3 + \cos u) \sin v, \sin u + 0.5 v $. The last term $ 0.5 v $ is the twistor added to the torus. – Narasimham Aug 13 '15 at 20:57

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