A common 6-in.-radius soccer ball contains 12 pentagons arranged so that every pentagon is separated from the next by the same arc length as one of the spherical (great circle segment) sides of the regular hexagons. As the hexagons are regular, this is the same arc length as one of the sides of the pentagons, as the pentagons also border the hexagons.

Calculate the arc length of a pentagon's side of a new soccer ball using the same radius and instead of one line of separation between pentagons, use two lines of separation between pentagons and consider every new line with a distance equal to a side of a pentagon. (See picture)

Note: The endpoints of the mentioned lines intersect with the surface of the soccer ball or sphere.

First figure the distance from the center of one pentagon to the center of the next. To do this, consider the net of arcs connecting the centers as the spherical projection of the edges of a regular icosahedron, a solid having 20 triangular faces. Consider one spherical triangle face. The arc length of one side is the arc length between pentagons. Each corner angle of this triangle is 72 degrees, as five fit together to completely surround the vertex.

Then we use the law of cosines for spherical triangle, for angles:

cos 72 = -cos²72 + sin²72 cos S,
where S is the side length sought.

This solves to S=63.424948... degrees or 1.10714871779... radians, or in this instance 6.6428923... inches.

If x is the arc length of one side of each pentagon, then 2x is the arc length between tips of adjacent pentagons, as the latter is subtended by two separate chords each the same length as that subtending the side of the pentagon.

If we take a pentagon and divide it into 5 isosceles triangles, and then divide each isosceles triangle into two right triangles, we get the side of one of those right triangles opposite its 36-degree angle as x/2 degrees. The hypotenuse, which runs from the center of the pentagon to a vertex, is (S - 2x) / 2, whether measured in degrees or radians. Algebraically this equals S/2 - x.

Using the spherical law of sines, sin(S/2-x)/sin(90) = sin(x/2)/sin(36), using degree measure.

Using Excel's solver, using the radian equivalent, x comes out to be about 0.298135 radians, which equates to 1.7888... inches given the 6-inch radius of the ball.

Posted by Charlie on 2003-12-04 16:31:53