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

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: $$$$\tag{1} \label{Eq1} x^2 + y^2 + z^2 = 1$$$$ 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:

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

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!

• Even the triangular wave which is the start of your construction doesn't have any particularly nice expression, so I don't see any reason why the 3d curve would. Commented May 29, 2018 at 20:01
• Congratulations for your refined questions. Commented May 29, 2018 at 20:01
• Commented May 29, 2018 at 20:57
• To add a slight bit of specificity to @HagenvonEitzen's pair of related questions: this answer links to a journal article on "Generalized Baseball Curves".
– Blue
Commented May 29, 2018 at 21:02

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.

• Thanks for your detailed answer. I tried the idea in the link (i.e. to set $\theta=(\pi/3)sin2\phi$), but it doesn't work for me. It just draws like a flower on the top of the ball. I'm not a mathematician, so I probably just can't understand the notation. It's frustrating because it seems to have nice properties - like splitting the ball into two identical pieces. Commented May 30, 2018 at 11:16
• Or, if I'm interpreting their $\theta$ and $\phi$ the wrong way around, then I feel like their assumptions are wrong - in particular that the elevation should do 2 oscillations. I think $\phi=(\pi/3)cos\theta$ looks better than $\phi=(\pi/3)sin2\theta$. Commented May 30, 2018 at 11:34
• $\thete$ and $\phi$ are sometimes reversed, so maybe that’s it. Commented May 30, 2018 at 13:29