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I want to know if a 3D spiral, that looks like this:

Spiral

can be approximated to any sort of geometric primitive that can be described with a known equation, like some sort of twisted cylinder I suppose.

Thanks

RNs_Ghost
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3 Answers3

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Equation for a helix: $$x(t) = R \cos t, \quad y(t) = R \sin(t), \quad z(t) = at.$$

If you actually want a surface, then use the above to write $$(x-x(z/a))^2+(y-y(z/a))^2 = r^2$$ or $$(x-R \cos(z/a))^2+(y-R \sin(z/a))^2 = r^2$$ where $r$ is the radius of the "tube" and $R$ is the winding radius. In your particular picture you will have $a >> 1$ and $r, R = o(1)$.

abnry
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  • Does he just want a line or a shape with volume? – ClassicStyle May 21 '14 at 23:02
  • I have him both. The first is just the line. The second is surface thought of as a circle whose position moves depending on z. – abnry May 21 '14 at 23:03
  • Thanks, yep - wanted to be able to dictate the width of the "tube". – RNs_Ghost May 21 '14 at 23:38
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    The implicit equation is bad because a helix is twisted so the circular section is not parallel to the $xy$-plane. But for helicoid, we can have an implicit equation: $x\sin \frac{z}{b}=y\cos \frac{z}{b}$ and for torus $(x^{2}+y^{2}+z^{2}+a^{2}-b^{2})^{2}=4a(x^{2}+y^{2})$. – Ng Chung Tak Jan 31 '16 at 19:23
  • I agree with @NgChungTak. The circular cross-sections should IMO be orthogonal to the curve - not parallel to the $xy$-plane. If $a$ is not large, then the distortion becomes clear (props for observing the need to have $a>>1$ though). See my old answer. – Jyrki Lahtonen Apr 03 '18 at 05:46
  • Your switching between parametric and explicit equations is confusing. Is there a way to use only a parametric equation? – posfan12 Oct 02 '19 at 03:50
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The tube thickness can also be varied as a function of any desired parameter ( radius, axial length, theta, arc length along tube etc.), the following image is that of of a helix on a cone. Code line is in Mathematica.

HelixOnCone

Narasimham
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These are the parametric equations of a corkscrew \begin{equation*} \begin{aligned} x&=a \cos(\theta)\\ y&=a \sin(\theta)\\ z&=a \theta\tan(\alpha) \\ \end{aligned} \end{equation*}

Where $a$ is constant, the radius of the cylinder, and $\alpha$ is the constant angle made by the tangent with the $x, y$ plane.

Rene Schipperus
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