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A Helix is parameterized as $\langle R \cos(t), R \sin(t), \alpha t\rangle$ and one can visualize it as "wrapping" around a cylinder of radius R. I would like to accomplish the same thing but wrapping around a torus(or one can think of bending the cylinder into a torus).

$\langle R_1 \cos(t) + R_2 \cos(\beta t), R_1 \sin(t), R_2 \sin(\beta t)\rangle$ is sort of a solution but is not quite right as "sides" of the enclosed hypothetical torus are flat.

A seemingly better result is $\langle(R_2 + \cos(t)) \cos(\beta t), \sin(t), (R_2 + \cos(t))\sin(\beta t)\rangle$ and it looks almost right but it seems there might need to be a special relationship between $\beta$ and $R_2$ because some turns seem to get out of whack slightly.

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Since $\langle \cos(x)(R_1 +R_2 \cos(y)), \sin(x)(R_1 +R_2 \cos(y)), R_2 \sin(y) \rangle$ is a nice parametrization of a torus, I suggest $\langle \cos(t)(R_1 +R_2 \cos(\beta t)), \sin(t)(R_1 +R_2 \cos(\beta t)), R_2 \sin(\beta t) \rangle$.

After reading your post again, I see that isn't substantially different from your second try.

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Yeah, its identical. But to get a uniform "wrapping" of torus by the helix I think one has to potentially know the arclength of the helix and make sure it is a multiple of $2\pi R_2$. – Jubao Sep 6 '12 at 20:35
With uniform do you mean you want to achieve equal line density everywhere? In the "inner" part of the torus the curve should than go more straight and in the outer art more skew. Or does your reference to arclength mean that you want the curve to be geodesic? Or when you say "multiple of" does that mean you want the curve to close (that happens iff $\beta$ is rational)? – Hagen von Eitzen Sep 6 '12 at 20:42
If the arclength is not a multiple of $2\pi R_2$ I believe the helix will not meet up with it's starting point. But by uniform I mean a sort of symmetry similar to the torus(which has two axis of symmetry). Essentially any rotation of the torus about it's axis should be equivalent. (although it might not be exact due to the helix having an offset) – Jubao Sep 6 '12 at 21:48
@Jubao, you're looking for something like this curve, then? – J. M. Sep 6 '12 at 22:35
@J.M. Yes, that looks like it. – Jubao Sep 7 '12 at 1:44

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