Is there a neat way to write the parameterization of this tennis-ball-seam-like curve on the sphere?

Question

I have been experimenting with drawing curves on the surface of a sphere (of radius 1). In order for it to lie on the sphere, every point $(x,y,z)$ must satisfy: $$ \begin{equation} \tag{1} \label{Eq1} x^2 + y^2 + z^2 = 1 \end{equation} $$ I became interested in drawing something vaguely resembling the seam on a tennis ball:

My first guess was to try (for some parameter $\theta$ between $0$ and $2\pi$), $x(\theta)=cos(\theta)$ and $y(\theta)=sin(\theta)$, before I realised that's obviously just a circle and from $(\ref{Eq1})$ we get $z(\theta)=0$ (oops). So instead, I replaced $sin(\theta)$ with a triangular wave:

Then, using $(\ref{Eq1})$, I know $z(\theta)$ must satisfy:

$$ \begin{equation} \tag{2} \label{Eq2} z(\theta) = \pm\sqrt{1 - x^2(\theta) - y^2(\theta)} \end{equation} $$

If I just take the positive solution for all $\theta$, then my curve is discontinuous. However, if I alternate + and - (or vice versa) for each of the four quadrants, then I get a nice smooth curve like I want:

However, I can't figure out if there's a neat equation for describing my $z(\theta)$. It doesn't look so complicated in the graph below. Can someone tell me if there's a neat way of describing it (as some function of $\theta$)?

Sorry for the long question and many thanks for your help!

Answer 1

Here’s an idea. Parameterize the curve you want in spherical coordinates, then use the standard conversion to write the curve parametrically in cartesian coordinates. I don’t see a simple way to get the curve in the form $z=f(x,y)$, but this should be a useful technique.

Here’s my try, using a parameter range of $t=0$ to $t=2\pi$. The azimuth angle ($\theta$) of your curve goes one full turn around the $z$-axis, so let the azimuth angle be $t$. The declination angle ($\phi$) goes above and below the $\phi=\pi/2$ equator by less than $\pi/2$ and completes one sinusoid as the azimuth angle goes around. So let $\phi=\pi/2+\pi\sin(t)/3$. Because the curve must stay on the unit sphere, the radius $\rho$ is constant and equal to $1$. (This last detail is the one that makes using spherical coordinates so useful!) So a parameterization of the curve in spherical coordinates is

$$\left(\rho,\theta,\phi)=(1,t,\pi/2+\pi\sin(t)/3\right).$$

The coordinate conversion from spherical to $(x,y,z)$ coordinates is (always) $$(x,y,z)=\left(\rho\sin(\theta)\cos(\phi),\rho\sin(\theta)\sin(\phi),\rho\cos(\theta)\right),$$

so your curve in parametric cartesian coordinates is

$$\left(\sin(t)\cos\left(\frac{\pi}2+\frac{\pi\sin(t)}3\right),\sin(t)\sin\left(\frac{\pi}2+\frac{\pi\sin(t)}3\right),\cos(t)\right).$$

Here’s a picture of that curve superimposed on the unit sphere. (I might have labeled the axes out of order, but hopefully you get the idea.)

After working this out, I googled “tennis ball seam curve parametric equations” and there are a lot of places to look for more ideas.