Mathematics in Everyday Life

How many times have I opened a box of tissues by removing the elliptical cover on the top of the box? Every time I do it, I'm aware of its elliptical shape but I never paused to consider the resulting ellipse of cardboard that I held in my hand. It would be discarded as rubbish. Today however, I paused and really looked at what I had in my hand (see Figure 1).

Figure 1

There's even a little semi-circular tab on the right that can be depressed to facilitate the removal of the cover. I'd never noticed that before. Turning the cardboard ellipse over reveals blank cardboard on which I marked in the major and minor axes and measured their lengths, to the nearest millimetre (see Figure 2).

Figure 2

These measurements enable calculation of the eccentricity \(e\) of the ellipse and so in this case, with \(a=28\) and \(b=62.5\) where \(a\) and \(b\) are the lengths of the semi-minor and semi-major axes respectively, we have:$$e=\sqrt {1-\frac{a^2}{b^2}}=\sqrt {1-\frac{28^2}{62.5^2}} \approx 0.894$$This of course is highly elliptical, especially if it's compared with the eccentricities of the planets of the solar system (see Figure 3).

Figure 3

As can be seen in Figure 3, Mercury and Pluto have the most eccentric orbits but much less eccentric than my cardboard ellipse. The other planets have elliptical orbits that would be hard to distinguish from circles if their proportions were displayed on a cardboard cut-out similar to that shown in Figure 2. Coincidentally, there is a centaur with an eccentricity of 0.894 as the table shown in Figure 4 reveals. The academic paper that the table was taken from is quite an interesting but I won't go into here but this is the link, the same as the one shown in Figure 4.

Figure 4

As explained in Figure 4, centaurs are planetesimals with perihelia (closest distance to the Sun) exterior to the orbit of Jupiter and aphelia (farthest distance from the Sun) interior to the orbit of Neptune. The most famous of the centaurs in Chiron, the first to be discovered in 1977 but the somewhat less famous C/2012 H2 (McNaught) does have an orbit that exactly matches that of the cardboard ellipse shown in Figures 1 and 2. Figure 5 provides a little more information about this object.

Figure 5

To calculate the length \(F\) from the centre of the ellipse to the two foci, the following formula can be used involving once again the lengths of the semi-minor and semi-major axes:$$F=\sqrt{b^2-a^2}=\sqrt{62.5^2-28^2} \approx 55.9$$These foci for the cardboard ellipse are shown in Figure 6.

Figure 6

The mathematics in this post is very basic but that was my intention. Though basic, the shape of the cardboard ellipse is nonetheless reflected in the shape of a particular centaur's orbit and it's pretty cool to find a connection between an everyday household item and the solar system in which we are immersed.