Hexagons and pentagons on a standard soccer ball

Question

How would one go about calculating the diameter of the circumscribing sphere formed when the soccer ball is fully inflated? Suppose, for example, the curved side lengths where all the hexagons and pentagons join on the sphere's surface are 2.5 inches.

Answer 1

From the Wikipedia page on Truncated Icosahedron

we have the formula

$ r_u = \dfrac{a}{2} \sqrt{1 + 9 \varphi^2 } = \dfrac{a}{4} \sqrt{58 + 18 \sqrt{5}} = 2.47801866 a $

where $r_u$ is the radius of the circumscribing sphere and $a$ is the (straight) edge length of the truncated icosahedron.

Now the angle subtended by the edge at the center is

$ \theta = 2 \sin^{-1} \bigg( \dfrac{ a }{2 r_u } \bigg) = 0.40633789 \text{ (radians)}$

Therefore, the curved edge length is

$ c = r_u (0.40633789) $

i.e.

$ r_u = 2.461006 c $

Now, we're given by $ c = 2.5 \text{ inches} $, therefore,

$ r_u = 2.461006 (2.5) = 6.152515 \text{ inches} $