Finding the volume of a rugby ball (prolate spheroid)
With the rugby union World Cup currently underway I thought I’d try and work out the volume of a rugby ball using some calculus. This method works similarly for American football and Australian rules football. The approach is to consider the rugby ball as an ellipse rotated 360 degrees around the x axis to create a volume of revolution. We can find the equation of an ellipse centered at (0,0) by simply looking at the x and y intercepts. An ellipse with y-intercept (0,b) and x intercept (a,0) will have equation:
Therefore for our rugby ball with a horizontal “radius” (vertex) of 14.2cm and a vertical “radius” (co-vertex) of 8.67cm will have equation:
We can see that when we plot this ellipse we get an equation which very closely resembles our rugby ball shape:
Therefore we can now find the volume of revolution by using the following formula:
But we can simplify matters by starting the rotation at x = 0 to find half the volume, before doubling our answer. Therefore:
Rearranging our equation of the ellipse formula we get:
Therefore we have the following integration:
Therefore our rugby ball has a volume of around 4.5 litres. We can compare this with the volume of a football (soccer ball) – which has a radius of around 10.5cm, therefore a volume of around 4800 cubic centimeters.
We can find the general volume of any rugby ball (mathematically defined as a prolate spheroid) by the following generalization:
We can see that this is very closely related to the formula for the volume of a sphere, which makes sense as the prolate spheroid behaves like a sphere deformed across its axes. Our prolate spheroid has “radii” b, b and a – therefore r cubed in the sphere formula becomes b squared a.
Prolate spheroids in nature
The image above [wiki image NASA] is of the Crab Nebula – a distant Supernova remnant around 6500 light years away. The shape of Crab Nebula is described as a prolate spheroid.
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Some years ago I was developing calculus formulae [can’t seem to do it anymore] for boat volumes. You can’t use pi in formulas for boat immersed volumes because boats are pointed. Pointed in either the vertical plane, or the horizontal plane or both. I found that instead of pi/4 [.785], to simply use 0.67 . This works for pretty much any sailboat shape since bottoms and waterplanes are often arc-shaped, [unless it has a very blunt bow angle, like 90 degrees], and the factor moves from .67 to .785 .
Now an American football is not an ellipse, it is definitely pointed at both ends, I suggest to use 0.67 instead of pi.