Do these equations create a helix wrapped into a torus?

Question

I don't have any graphing software capable of plotting this, but it makes sense to me.

First a circle:

$$ x = r\cos\theta \\ y = r\sin\theta $$

Then a sine wave wrapped into a circle:

$$ x = (r+\cos\theta)\cos\theta\\ y = (r+\sin\theta)\sin\theta $$

And finally, add in the z-coordinate:

$$ z = \sin\theta $$

Is this correct? If not, what's wrong with it?

$$ x = (r+\cos\theta)\cos\theta = r\cos\theta + \cos^2\theta\\ y = (r+\sin\theta)\sin\theta = r\sin\theta + \sin^2\theta\\ z = \sin\theta $$

See http://math.stackexchange.com/questions/192067/parameterization-of-helical-torus — Thomas, Mar 08 '13 at 09:22

score 8 · Answer 1 · answered Mar 08 '13 at 09:20

8

Here's the parametrization for a helix of $n$ winds wrapped around a torus of major radius $R$ and minor radius $r$: $$\begin{align*} x&=(R+r\cos(nt))\cos(t)\\ y&=(R+r\cos(nt))\sin(t)\\ z&=r\sin(nt) \end{align*}$$

Making this into a function in Mathematica,

HelixPlot[R_, r_, n_] := 
 Show[ParametricPlot3D[{(R + r Cos[t]) Cos[u], (R + r Cos[t]) Sin[u], 
    r Sin[t]}, {t, 0, 2 Pi}, {u, 0, 2 Pi}, PlotPoints -> 30, 
   Mesh -> None, PlotStyle -> Opacity[0.3]], 
  ParametricPlot3D[{(R + r Cos[n*t]) Cos[t], (R + r Cos[n*t]) Sin[t], 
    r Sin[n*t]}, {t, 0, 2 Pi}, PlotPoints -> 30, 
   PlotStyle -> {Thick, Black}]]

we can toy with different settings.

HelixPlot[6, 2, 5] produces

enter image description here

HelixPlot[6, 1, 10] produces

enter image description here

HelixPlot[6, 5, 20] produces

enter image description here

answered Mar 08 '13 at 09:20

Zev Chonoles

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So basically, my equations are correct? – user912 Mar 08 '13 at 09:22
Basically yes, but your equations don't allow one to vary the minor radius of the torus or the number of winds. – Zev Chonoles Mar 08 '13 at 09:23
I was just wanting to make sure that the basic shape was right, before modelling simulations on it. :P – user912 Mar 08 '13 at 09:25

score 4 · Accepted Answer · answered Mar 08 '13 at 09:15

4

You need two radii to decribe a torus. Let's call them $a$ and $b$. Then the parametric equations of the torus are: $$x = (a + b\cos u)\cos v$$ $$y = (a + b\cos u)\sin v$$ $$z = b\sin u$$

Then, to get a helical curve, set $v = ku$, where $k << 1$.

Here's the result with $a=3$, $b=1$, $k=0.05$: enter image description here

answered Mar 08 '13 at 09:15

bubba

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Okay, but for a helix of radius 1, doesn't that just simplify to my equations? – user912 Mar 08 '13 at 09:16
A torus has two radii. If you set them both equal to the same value, you get a sphere. – bubba Mar 08 '13 at 09:25
I think you mean a circle, but yes, I'm aware of that. Thanks for the help. :) – user912 Mar 08 '13 at 09:26
I see. With my notation, you are setting $a=r$ and $b=1$. So, we agree. But, also, you are setting $k=1$, so your curve will not be very helix-like -- it will only wind around the torus once. – bubba Mar 08 '13 at 09:29
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No, I meant sphere. – bubba Mar 08 '13 at 09:30
I just needed a basic idea first, so I can mess around with the properties. Thanks for your help. – user912 Mar 08 '13 at 09:30
The surface is a sphere; the curve is a circle. – bubba Mar 08 '13 at 09:31
1

You don't get a sphere when $a=b$, you get a horn torus. You get a sphere of radius $b$ when you require that $a=0$. – Zev Chonoles Mar 08 '13 at 09:33
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I don't quite understand, could you perhaps make another image to illustrate what you're saying? – user912 Mar 08 '13 at 09:33
1

Here's the picture from Wikipedia. – Zev Chonoles Mar 08 '13 at 09:35
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@zev -- you're right. Sorry. – bubba Mar 08 '13 at 09:35
@ZevChonoles That's when the inner radius is zero and the outer radius is positive, but if the radii are equal it should be a circle. – user912 Mar 08 '13 at 09:39
There is a central circle to the torus of radius $a$. The torus is the surface of distance $b$ from the central circle. When you set $a=0$ and let $b>0$, you get the surface of distance $b$ from a point (the central circle has radius 0), which is a sphere of radius $b$. When you set $b=0$ and let $a>0$, you get a circle of radius $a$. – Zev Chonoles Mar 08 '13 at 09:41
@user912: By the way, using HelixPlot[4,1,1], here is a plot of what your equations produce. Only using one wind makes it hardly recognizable as a helix :) – Zev Chonoles Mar 08 '13 at 09:43
@ZevChonoles I know it's only one, but it can easily be expanded to include variables for number of turns, as was shown in one of the answers. – user912 Mar 08 '13 at 09:44

Do these equations create a helix wrapped into a torus?

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