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For an university assignment I have to visualize some curves in 3 dimensional space.

Until now I've implemented Bézier, helix and conical spiral.

Could you give me some advice about some famous curves in geometry history?

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The baseball seam maybe? Does it actually have a name? – Willie Wong Dec 23 '10 at 2:14
@Willie: I don't remember it being "officially" named either. Anyway, I found two more interesting links... – J. M. Dec 23 '10 at 2:27
...and I mentioned two more historically interesting space curves in the answer to this question. – J. M. Dec 23 '10 at 2:36
The curve that solves the brachistochrone problem is historically important, though not much to look at. – Simon Jul 31 '11 at 2:45

Space-filling curves, in particular the Hilbert curve (in 3D): alt text

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How about Viviani's curve? It is the intersection of a sphere with a cylinder. Viviani was a student (or disciple) of Gallileo. Wikipedia has a beautiful image:

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I'd mention the curve of Archytas, the space curve formed by the intersection of a horn torus and a cylinder:

curve of Archytas

The parametric equations for the curve of Archytas are

$$\begin{align*}x&=\frac{\cos^4 t}{(1+\sin^2 t)^2}\\y&=\frac{\cos\,t\sin\,2t}{(1+\sin^2 t)^2}\\z&=\frac{\sin\,2t}{\sqrt 2(1+\sin^2 t)}\end{align*}$$

Archytas of Tarentum used this curve to help him solve the Delian problem.

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Though is it is not 3D, the Clothoid or Cornu Spiral is an amazing curve. It surely can be made 3D by adding a simple extra parameter $z(t)=t$. It has infinite length but converges to two points in the plane. It has several applications in optics and road engineering, for example. An it looks quite nice:

enter image description here

I found a 3D plot too: enter image description here

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Let $t \in [0,2\pi]$. And let $r(t)=\frac{1}{2-\sin (2t)}[\cos(3t),\sin(3t),\cos(2t)]$. Then $r(t)$ parametrizes a 3-dimensional curve with no three-tangent plane (I must admit I'm not sure of the English terminology here) - that is, every tangent plane of the curve meets the curve in no more than two points.

(I have no clue how to prove such things, however)

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Which plane, the plane spanned by the tangent and normal vectors, or the plane spanned by the tangent and binormal vectors? – J. M. Dec 23 '10 at 1:27
I must admit I'm not sure. The (popularized) text in which I saw this, used the explanation that "if you build a three-dimensional model of this curve and place it on a table, then no matter how you place it, only one point will touch the table at any time.". – Fredrik Meyer Dec 23 '10 at 13:12

If I may be allowed to give another example, there is the spherical epitrochoid, the locus of a circle rolling on another circle. It was studied by Johann Bernoulli and Franz Reuleaux in the context of the kinematics of gear systems.

Letting $p$ be the radius of the fixed circle ("pitch circle"), $r$ the radius of the rolling circle, $hr$ the distance of the tracing point from the center of the rolling circle, and $\omega$ the inclination of the rolling circle from the x-y coordinate plane, the parametric equations for the spherical epitrochoid are:

$$\begin{align*}x&=(p+r\cos\;\omega)\cos\;t-h r\left(\cos\;\omega\cos\left(\frac{p}{r}t\right)\cos\;t-\sin\left(\frac{p}{r}t\right)\sin\;t\right)\\y&=(p+r \cos\;\omega)\sin\;t-h r\left(\cos\;\omega\cos\left(\frac{p}{r}t\right)\sin\;t+\sin\left(\frac{p}{r}t\right)\cos\;t\right)\\z&=r\sin\;\omega\left(1-h\cos\left(\frac{p}{r}t\right)\right)\end{align*}$$

Here is the case $p=1$, $r=\frac15$, $h=1$, $\omega=\frac{2\pi}{3}$:

three-dimensional epitrochoid

Here is the case $p=1$, $r=\frac15$, $h=1$, $\omega=\frac{\pi}{6}$:

three-dimensional epitrochoid

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wow! thanks all, everyone got very interesting answer i've found also slinky that is quite funny cause it is also a toy

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i've just found this link that seems pretty good – nkint Jun 27 '12 at 10:05

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