1. Is there some general method for finding such curves? Let's say I have a planar curve, how can I project it onto a sphere?

  2. I am interested in a curve that starts at the south pole of a sphere, then wraps it in spiral motion and ends at the north pole. Is it possible to construct?

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    $\begingroup$ For any curve $\gamma$, the curve $\frac{\gamma}{||\gamma||}$ lies on the unit sphere. $\endgroup$ May 3, 2012 at 19:44
  • 2
    $\begingroup$ Also, this is the first result when googling "spiral on sphere". $\endgroup$ May 3, 2012 at 19:46
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    $\begingroup$ The inverse of the stereorgaphic projection will take any planar curve to a curve on the sphere. $\endgroup$ May 3, 2012 at 20:10
  • $\begingroup$ For 2: see this question. $\endgroup$ May 4, 2012 at 3:42

2 Answers 2


For your first question, suppose $\gamma$ is any curve. Then the norm of $\frac{\gamma}{||\gamma||}$ is $\frac{||\gamma||}{||\gamma||} = 1$, so the image of $\frac{\gamma}{||\gamma||}$ lives on the unit sphere in your space.

Your second question can be answered with the help of stereographic projection. We'll take a curve in the plane and project it onto the unit sphere.

enter image description here

(Image from Wikipedia.)

If we describe the plane with the polar coordinates $(R,\Theta)$, and the sphere with the coordinates $(\varphi,\theta)$, where $\varphi$ is the zenith angle and $\theta$ the azimuth, then the map from the plane to the sphere is given by

$$ \begin{align} \varphi &= 2 \arctan\left(\frac{1}{R}\right), \\ \theta &= \Theta. \end{align} $$

If you'd like, this can be converted to Cartesian coordinates by

$$ \begin{align} x &= \cos \theta \sin \varphi, \\ y &= \sin \theta \sin \varphi, \\ z &= \cos \varphi. \end{align} $$

As an example, if we take the logarithmic spiral defined in polar coordinates by $R = e^{\Theta/8}$, the Cartesian parametric curve describing its image on the sphere is

$$ \gamma(t) = \left(\begin{array}{c} \cos t \,\sin \left(2 \arctan \left(e^{-t/8}\right)\right) \\ \sin t \,\sin \left(2 \arctan \left(e^{-t/8}\right)\right) \\ \cos \left(2 \arctan \left(e^{-t/8}\right)\right) \end{array}\right). $$

enter image description here

  1. Do you really mean circle, rather than sphere?

  2. A parametrization with constant rate of increase along the $z$ axis is $(x,y,z)=\left( \sqrt{1-t^2}\cos (a \pi t), \: \sqrt{1-t^2} \sin (a \pi t), \: t \right), \: 0<t<1$, where the $a$ parameter controls the number of revolutions. This is with $a=5$:

Spiral with a=5

You can derive this by noting that $x \propto \cos(\cdot) $ and $y \propto \sin (\cdot)$. If $z$ is to increase in a linear fashion from -1 to +1 we can let $z=t$ for $t=-1 \dots 1$. Finally, since the spiral is to lie on the surface of a sphere $x^2+y^2+z^2=1$, so it follows that $x=\sqrt{1-t^2}\cos (a \pi t)$ and $y=\sqrt{1-t^2} \sin (a \pi t)$

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    $\begingroup$ This figure is for the case $-1<t<1$ not for $0<t<1$. $\endgroup$
    – Tarek
    Dec 7, 2016 at 20:14

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