Parametric curve for a tennis ball seam

Question

I hadn't noticed until now tennis balls were symmetric about more than one axis. Which lead me to think there could be an elegant way of expressing the seam as a parametric curve, but I haven't found any yet.

I've tried using products of sines and cosines, but none of them even approximate the curve and it's getting rather ugly.

This is not homework, by the way.

Answer 1

You might be interested in an article entitled "Generalized Baseball Curves: Three Symmetries and You're In!" published in Loci/JOMA. The article is available online at http://www.maa.org/publications/periodicals/loci/generalized-baseball-curves-three-symmetries-and-youre-in.

We look at the class of spherical curves that have the same symmetries as the curve that forms the seam of a baseball. Also some of the references in this article try to address your question. While we don't directly answer your question by finding the explicit parametric equations of a baseball curve, we do propose a curve (see Figure 7 in the article) that does fit the seam of a real-life baseball very well.

Answer 2

Let's aim for a simple approximation. Define the curve by $r = r_0$, $\theta = (\pi/3)\sin 2\phi$.

OK, what is that and how did we get it?

We want a curve on the surface of a sphere $r = r_0$. We want the curve to go up and down twice as it revolves around the $z$-axis. We want the curve to be symmetrical in that we could turn it over, shift it halfway around, or mirror-reflect it and it would fit onto itself.

We know that $f(\phi)=\sin 2\phi$ goes up and down twice as $\phi$ goes from $-\pi$ to $+\pi$. We know that $f(\phi)=\sin 2\phi$ has several symmetries such as $f(\phi)=-f(-\phi)$, $f(\phi)=-f(\phi-\pi/2)$, $f(\phi)=f(\pi/2-\phi)$. So we're getting there. Now we want to wrap this $f$ around a sphere somehow. One way is if we imagine the sphere has an equator like Earth's equator and $f$ is proportional to the elevation angle $\theta$ above or below the equator, like latitude on Earth. Let's say that we want $\phi$ to be at maximum, two thirds of the way to the "North Pole", and at minimum, two thirds of the way to the "South Pole". That's one sixth of the circumference of the sphere, which is an angle in radians of $\pi/3$. That will be the proportion: $\theta = (\pi/3)f$.

We have a plan. To be quite clear, it's appropriate to explain how we relate the radius $r$, the elevation angle $\theta$, and the parameter $\phi$ of $f$ (thus also of $\theta$) to Cartesian coordinates.

In $xyz$-space, we can define spherical coordinates in the following way, which is one of the usual ways:

Let $r = \sqrt{x^2 + y^2 + z^2}$.

• Then $r(x_0,y_0,z_0)$ measures the distance from the point $(0,0,0)$ to the point $(x_0,y_0,z_0)$.

Let $\phi = \arctan(y/x)$.

• Then $\phi(x_0,y_0,z_0)$ measures the azimuth angle $(1,0,z_0)$, $(0,0,z_0)$, $(x_0,y_0,z_0)$. It's like measuring the longitude at which you're standing on Earth's surface. From (literally) another point of view, it's like measuring the compass direction in which you're facing, if you're standing on Earth's surface (not at a pole).
  The number $1$ is arbitrary. If we would write any other positive number instead of $1$, the angle measure would be the same.
• This is an angle in the plane where $z=z_0$.
• The azimuth angle measures zero when $x_0$ is positive and $y_0=0$.
• Assuming that $x_0$ is positive, the azimuth angle begins to increase as $y_0$ begins to increase from $0$.

Let $\theta = \arctan(z/r)$.

• Then $\theta(x_0,y_0,z_0)$ measures the elevation angle $(x_0,y_0,0)$, $(0,0,0)$, $(x_0,y_0,z_0)$.
• This is an angle in the plane where $(x,y)$ is in any proportion $k$ to $(x_0,y_0)$, that is, $(x,y)=k(x_0,y_0)$.
• The elevation angle measures zero when $z=0$.
• The elevation angle begins to increase as $z$ begins to increase from $0$.