# Finding parametric curves on a sphere

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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For any curve $\gamma$, the curve $\frac{\gamma}{||\gamma||}$ lies on the unit sphere. – Antonio Vargas May 3 '12 at 19:44
Also, this is the first result when googling "spiral on sphere". – Antonio Vargas May 3 '12 at 19:46
The inverse of the stereorgaphic projection will take any planar curve to a curve on the sphere. – Yuri Vyatkin May 3 '12 at 20:10
For 2: see this question. – J. M. May 4 '12 at 3:42

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.

(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

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

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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 $a=5$:

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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I like mine better. Look how the cross hairs get the red and orange dots.

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It's difficult to see how this addresses the question "How does one map a plane curve to a sphere?" Could you please add some explanation of how these graphics are relevant? – Andrew D. Hwang Nov 17 at 0:05
I was responding to his second question about spirals and the sphere. It also works on the torus. – user1698948 Nov 17 at 3:09
Also, the other example is not natural. You're mixing two coordinates that use pi, and the last coordinate cancels pi so you're using arctan of a rational. You can use arctan and make spheres with rationals, just don't mix it with pi like that. When using arctan you need to multiply by one of the variables and your range should be integers. My way also explains why lissajous figures have rose curves on the x/y plane when wrapped around a sphere. – user1698948 Nov 17 at 15:57