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I am trying to work out a "helix in a helix" mathematically. Intuitively I think of this as a steel cable, which is made up of a number of smaller steel cables all bound together in spiral. If I wanted to find the length of one of the individual cables, it would be bound in a spiral in the smaller cable and then those cables bound in a larger spiral cable. I know that if I wanted to do a helix whose ends meet, I would use the parametrization $$((a+b\cos(\omega{t}))\cos{t},(a+b\cos(\omega{t})\sin{t},b\sin(\omega{t})),t=0..2\pi$$ I've been trying to map out in my head how to, instead of curling the helix, making the helix travel in the path of a helix. I've achieved it partially with $$((a+b\cos(\omega{t}))\cos{t},(a+b\cos(\omega{t})\sin{t},t),t=0..\infty$$ But this doesn't keep the smaller helix in tact, and turns it into a sine wave helix. I've also tried $$((a+b\cos(\omega{t}))\cos{t},(a+b\cos(\omega{t})\sin{t},tb\sin(\omega{t})),t=0..2\pi$$ But this gives me sort of a nautilus shape where the helix curls into itself and increases in size and curls around into itself. What am I missing?

EDIT: Also, what if we wanted to do this again, like a 'helix in a helix in a ... in a helix'?

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  • $\begingroup$ I'm not sure what you want. Are you looking for curve that wraps around the helix surface in this answer? Or would you like the "big helix" itself to wrap around a torus instead of a cylinder as it usually does? Or do you want both? I think I saved the Mathematica code I used to produce those images, so... $\endgroup$ – Jyrki Lahtonen Aug 18 '13 at 15:44
  • $\begingroup$ close. I'm looking for a path along the surface of that helix. It's analogous to the helix along the path of a torus except instead of a torus it's another helix $\endgroup$ – Eleven-Eleven Aug 18 '13 at 15:53
  • $\begingroup$ I think I got it. Gimme a minute. I'll try with images first and then add formulas :-) $\endgroup$ – Jyrki Lahtonen Aug 18 '13 at 15:56
  • $\begingroup$ Thanks. I was using Maple and i got close with the sine wave helix, but I'm just not sure how to procede... $\endgroup$ – Eleven-Eleven Aug 18 '13 at 15:57
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As in an earlier answer I use the local frame along the helix to help parametrize the desired curve. From that answer I reuse: a parametrization for a helix along the $x$-axis $$ \vec{r}(t)=(ht,R\cos t, R\sin t). $$ Its tangent vector $$ \vec{t}=\frac{d\vec{r}(t)}{dt}=(h,-R\sin t,R\cos t). $$ Its normal vector $$ \vec{n}(t)= \frac{\frac{d\vec{t}}{dt}}{\left\Vert\frac{d\vec{t}}{dt}\right\Vert}=(0,-\cos t,-\sin t). $$ And it binormal vector $$ \vec{b}(t)=\frac1{\Vert\vec{t}\Vert}\vec{t}\times\vec{n}=\frac{1}{\sqrt{R^2+h^2}}(R,h\sin t,-h\cos t). $$ This is, of course, orthogonal to both $\vec{t}$ and $\vec{n}$.

The tube around the helix (with radius $a$) then has a parametrization $$ S(t,u)=\vec{r}(t)+a\vec{n}(t)\cos u+ a\vec{b}(t)\sin u $$ with $t$ ranging over as many loops as you wish, and $u$ ranging over the interval $[0,2\pi]$.

To get a curve looping around the helix along that surface we simply set $u=kt$, where $k$ indicates the number of rotations around the tube per single rotation of the tube around the $x$-axis.

Here's the image of the resulting curve with $R=3$, $h=1$, $a=1$ and $k=12$. For clarity I included both the tube as well as the curve.

Helix on a helix

The parametrization of that thin "cable" on the tube surface is (with the above values for the constants) $$ \left\{\begin{array}{ccl} x&=&t+\frac{3\sin 12 t}{\sqrt{10}},\\ y&=&3\cos t-\cos t\cos 12t+\frac{\sin t\sin 12t}{\sqrt{10}},\\ z&=&3\sin t-\sin t\cos 12t-\frac{\cos t\sin 12t}{\sqrt{10}}. \end{array}\right. $$

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  • $\begingroup$ Nice, now what does that look like algebraically? $\endgroup$ – Eleven-Eleven Aug 18 '13 at 16:08
  • $\begingroup$ @jyrki, that is perfect. You used the tangent, normal, and binormal vectors too which is what I'm teaching myself in Colley's vector calculus! Awesome. $\endgroup$ – Eleven-Eleven Aug 18 '13 at 16:25
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    $\begingroup$ @Christopher: Then it behooves me to warn you that in general you need to exercise a bit more care. You see that I didn't use the natural parameter here. That's why I had to normalize the vectors carefully to get unit vectors. Here that didn't result in any distortions, because on a helix the arc length is a constant multiple of the parameter $t$. This is not always the case, and writing everything in terms of $s$ may be somewhat taxing. $\endgroup$ – Jyrki Lahtonen Aug 18 '13 at 16:31
  • $\begingroup$ Is there a generalization for more iterations of this process? I do want to play with this '2-helix' a bit before diving into 'n-helices' but is te process similar? $\endgroup$ – Eleven-Eleven Aug 18 '13 at 16:57
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    $\begingroup$ In principle, yes, the same process should work. I just don't want to find the natural parameter of our latest curve :-). You may have seen that with some curves it is possible that the normal vector has a discontinuity (i.e. jumps from one side to the other) at the points where the radius of curvature goes to zero. It's been over 25 years since I last read about it, so I don't remember the terminology very precisely. Anyway, if that happens, you may need to make some more modifications. $\endgroup$ – Jyrki Lahtonen Aug 18 '13 at 17:00

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