Theoretical basics : A (classical) soccer ball is made up of 20 hexagons and 12 pentagons, distributed so that 5 hexagons surround each pentagon, and each hexagon is surrounded by 3 pentagons alternated with 3 hexagons. A surface like this cannot be geometrically flat, it will always be curve.
Both pentagons and hexagons have sides of equal length, and this distance is the same of its radius (from any vertex to its center). Given the side length L, we can obtain the radius of curvature R of the soccer ball.
To deduce the radius R of the sphere we use the relation between it and the perimeter P of its equator 2 · pi · R = P, where the pi number is approximately 3.141592. Due to the layout of hexagons over the ball surface, you can see that the perimeter is equal to 15 times the length L. Therefore, if the circle has 360 degrees, then each side L of the polygon corresponds to 360º / 15 = 24º of circumference.
If we take as 1 unit (any) the flat length L' of the polygons' sides (hexagon and pentagon), then the length of its side on the curved surface of the ball (L) will be higher.
Using the expression for trigonometric sine of an angle, we can calculate the radius R of the ball sin(24º) = L' / R, so R = L' / sin(24º) = 1 / 0.4067366430758 = 2.458593335574 units. We can also infer the length L of an arc of circumference, using the same expression as for the perimeter P, since P is proportional to 2 · pi · (360 degrees), L = 2 · pi · (24 º / 360 º) · R = 1.029852953906 units.
Modeling : Let's build the soccer ball upon intersections with a sphere, whose sections and radii of curvature are different, depending on hexagons or pentagons. These caps are generated from revolution curves.
1. First the hexagon, in the Front view, create a circle of radius 1 and 6 sections from a primitive (Objects folder), and place it in the origin (coordinates) with the grid magnet (Alt key). From the Right view, now draw a spline with CVs (Control Vertex): the first point with a magnet on the upper Edit Point of the circle (Ctrl key); the second point is placed with a shift, in relative coordinates, to the position r0 .05 0; the third point of the curve at 0 .05 -.1; the fourth at 0 .1 -.3; the fifth at 0 .1 -.4; and the sixth and last in absolute coordinates at position a0 .3 0. Once we have the spline, we place its pivot in the origin (XForm folder; Pivot icon) with command a0 0 0. Revolution now the curve over the Y axis and generate a surface of 12 sections (Surface folder; Revolve icon). Then template the generating curve and the circle (ObjectDisplay menu; Toggle Template option or Alt+T keys). Now move the generated surface (removing its Construction History) to the relative position r0 2.158593335574, which is the ball radius R minus the height of the sphere cap 0.3. In this position, move the pivot again, now from the surface to the origin: a0 0 0; because when we rotate the cap, we will do it on the center of the ball (the origin of coordinates).
To get the positions of the hexagons that form the soccer ball, we calculate the offset angles with respect to the original position.
Positions of the hexagons near the equator of the ball have their center shifted by an angle, in the X coordinate, proportional to half the apothem a of the hexagon. If the apothem a of the hexagon is, by Pythagoras theorem, a^2 = L'^2 - (L'/2)^2, then a = .8660254037844 units. The angle proportional to the apothem a, with respect to the 24º angle that corresponds to the hexagon's side length L', is a · 24° / L' = 20.78460969083º. Therefore, half the apothem a represents half of this angle: 10.39230484541º.
Rotate (XForm folder; Rotate icon) the surface of the first hexagon, in relative, and over the X coordinate, a r10.39230484541 angle. Duplicate the sphere cap of the first hexagon (Edit menu; Duplicate Object option), and rotate in relative coordinates the copy around the X axis by an equivalent angle to twice (2) the apothem a of the hexagon 41.56921938165.
The position of the third hexagon is shifted (rotated) over the first, and the Z axis, an angle equal to 3/4 the distance between the hexagon's vertexes; that is 1.5 · L = 1.5 · 24° = 36°. And over the X axis an angle corresponding to the apothem a. Let's duplicate then the first surface, and rotate the third copy in relative -20.78460969083 0 36.
The fourth hexagon, duplicate this time the third cap, and rotate this fourth surface -41.56921938165 over the X axis.
The remaining surfaces, we obtain them duplicating (Edit menu; Duplicate Object option) the four we already built, with a rotation over the Z axis by 72º, and a number of 4 duplications. This will generate the rest of sphere caps corresponding to hexagons, and thus closing the soccer ball surface.
2. The process to build the surfaces of pentagons is similar, but must take into account that the initial reference circle radius will be smaller than that of the hexagon. The side L' must be the same for the two polygons. Therefore, the radius r is calculated using sin(36º) = (L'/2) / r, where 36° matches half the angle of the arc that corresponds to a one side of the pentagon (360º / 5 = 72º). Thus, the radius r = (L'/2) / sin(36º) = 0.5 / 0.5877852522925 = 0.850650808352 units.
The pentagon, in the Front view, first create a circle of 5 sections from a primitive, and place it in the origin with the grid magnet. From the Right view, now draw a spline with CVs: the first point with a magnet on the upper Edit Point of the circle; place the second point shifted, in relative coordinates, to the position r0 .1 0; the third point of the curve at 0 .05 -.1; the fourth at 0 .05 -.3; the fifth at 0 .1 -.3; and the sixth and last in absolute coordinates at position a0 .3 0. Once we have the spline, we place its pivot in the origin with command a0 0 0. Revolution now the curve over the Y axis and generate a surface of 10 sections. Then template the generating curve and the circle. Now move the generated surface (removing its Construction History) to the relative position r0 2.158593335574. In this position, move the pivot again, now from the surface to the origin: a0 0 0; because when we rotate the cap, we will do it on the center of the ball (the origin of coordinates).
To get the positions of the pentagons that form the soccer ball, we calculate the offset angles with respect to the original position.
Positions of the pentagons at the poles of the sphere have their center shifted by a 90º angle, in the X coordinate. Duplicate the surface of the first pentagon, rotating it 90º over the X axis.
The position of the rest of pentagons is rotated, with respect to the X coordinate, an angle proportional to the apothem a' of the pentagon, plus one half the apothem a of the hexagon; and over the Z axis an angle equivalent to 3/4 the distance between the hexagon's vertexes; that is 1.5 · L = 1.5 · 24° = 36°. If the apothem a' of the pentagon is, by Pythagoras theorem, a'^2 = r^2 - (L'/2)^2, then a' = .6881909602356 units. The angle proportional to the apothem a', with respect to the 24º angle that corresponds to the pentagon's side length L', is a' · 24° / L' = 16.51658305º.
Rotate the surface of the second pentagon, in relative, and over the X coordinate, a r26.9088879 angle. To generate the remaining surfaces of the pentagons, on the upper hemisphere of the soccer ball, duplicate the second pentagon, with a rotation over the Z axis by 72º, and a number of 4 duplications.
The surfaces of the pentagons regarding to the lower hemisphere, we get them grouping (Edit menu, Group option) and duplicating them, with a Scale of -1 to the Z axis (Mirror).
