Tutorial: Recreating a Soccer Ball

(using Alias|wavefront PowerAnimator, though it works just as well in Maya or Rhino)

There are 20 hexagons and 12 pentagons in a (traditional) soccer ball, with five hexagons around each pentagon and three pentagons alternating with three hexagons surrounding each hexagon. Curves are the only possible geometric shape for such a surface.

A pentagon or a hexagon’s radius (measured from any vertex to the center) is the same length as one of its sides. The soccer ball’s radius of curvature, R, can be calculated from its side length, L.

A sphere’s radius R can be calculated from its equatorial perimeter P using the formula two pi R = P, where pi is a value close to 3.141592. The ball’s perimeter is 15 times its length, L since it is based on a grid of hexagons. Since there are 360 degrees in a circle, each side L of the polygon has a circumference of 360 degrees divided by 15 degrees or 24 degrees.

If the flat length L’ of the sides of the polygons (hexagon and pentagon) is taken as 1 unit (any), then the length of its side on the curved surface of the ball (L) will be greater.

The ball’s radius, R, can be found using the trigonometric sine of an angle expression: sin(24o) = L’ / R, or 1 / 0.4067366430758 = 2.458593335574 units. Since the perimeter P is proportional to 2 pis (360 degrees), we can calculate the length L of an arc of circumference using the same expression: L = 2 pis (24 o / 360 o) R = 1.029852953906 units.

Modeling: Let’s construct the soccer ball by intersecting a sphere with either hexagons or pentagons, each of which has its own section and curvature radius. The arcs of revolution are the source of these peaks.

To begin the hexagon, switch to the Front view and use the grid magnet (Alt key) to position a circle with a radius of 1 and 6 parts from a primitive (Objects folder) at the origin (coordinates). Now, in the Right view, place six CVs (Control Vertices) as follows: the first with a magnet on the upper Edit Point of the circle (Ctrl key); the second 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. With the spline, we can set its pivot to the origin (XForm folder; Pivot icon) by typing the a0 0 0 commands. Revolving the curve along the Y axis produces a 12-segmented surface (Surface folder; Revolve icon). Then, you may create a new angle from the circle using a template (ObjectDisplay > Toggle Template > Alt+T). Now, relocate the designed surface (after clearing its Construction History) to a new relative position r0, that is, R, the radius of the ball minus 0.3, the height of the sphere cap. Since we will be rotating the cap around the ball’s center (the origin of coordinates), we must now shift the pivot from its current surface location to the source (a0 0 0).

We calculate the offset angles concerning the original position to obtain the locations of the hexagons that make up the soccer ball.

The centers of the hexagons are rotated about the X-axis by an amount equal to half the hexagon’s apothem a for positions close to the ball’s equator. Based on Pythagoras’ theorem, the value of the apothem of a hexagon is a =.8660254037844 if a2 = L’2 – (L’/2)2. For a hexagon with side length L’, the angle proportional to the apothem is a 24° / L’ = 20.78460969083o. Half of this angle, 10.39230484541 degrees, is thus represented by the apothem a.

Apply a relative rotation of r10.39230484541 to the first hexagon’s surface (XForm folder; Rotate icon) over the X axis. Use the Edit > Duplicate Object menu item to make a copy of the first hexagon’s capping sphere, and then spin the document around the X-axis by an angle equal to twice (2) the apothem and of the hexagon (in this case, 41.56921938165).

An angle of 3/4 the distance L between the vertices of the hexagon, or 1.5 L = 1.5 24° = 36°, is used to rotate the location of the third hexagon relative to the first hexagon and the Z axis. And above the X-axis, an angle equal to a. Then, let’s make a copy of the first surface and rotate it by a factor of -20.78460969083 0 36 to make a third copy.

For the fourth hexagon, replicate the third capping shape and rotate the resulting surface by -41.56921938165 degrees about the X-axis.

To create the other surfaces, we make copies of the first four (Edit > Duplicate Object) and rotate them by 72 degrees about the Z axis. The remaining hexagonal caps for the soccer ball’s surface will be generated by doing so.

Pentagon surface construction is very similar to hexagonal surface construction, except a lower beginning reference circle radius must be used. Both polygons’ side L’ must be identical. Sin(36o) = (L’/2) / r is used to determine the radius r since 36o is equal to half the angle of the arc that represents a single side of the pentagon (360o / 5 = 72o). The radius, then, is deduced as follows: r = (L’/2) / sin(36o) = 0.5 / 0.5877852522925 = 0.850650808352 meters. To make a pentagon in the front view, first, create a 5-sided circle out of a primitive and position it at the grid’s origin using the magnet. Now, in the Right perspective, place an interest at the first point and set the second point at r0.1 0 relatives to the first point, the third point at 0.05 -.1, the fourth at 0.05 -.3, the fifth at 0.1 -.3, and the final point at a0.3 0 absolute coordinates to create a spline with CVs. Once we have the spline, we can set its pivot to the origin by typing a0 0 0. Now, we may rotate the curve along the Y-axis to create a 10-segmented surface. Next, make a pattern using the circle and generate a turn. Now, relocate the designed cover to the relative point r0 = 2.158593335574 (after clearing its Construction History). Since we will be rotating the cap around the ball’s center (the origin of coordinates), we must now shift the pivot from its current surface location to the source (a0 0 0). We calculate the offset angles concerning the original position to determine the areas of the pentagons that make up the soccer ball. The X coordinate of the pentagon’s center is rotated by 90 degrees to place it at the poles of the sphere.

The initial pentagon’s surface will be copied and rotated 90 degrees around the X-axis. The remaining pentagons are turned over the Z axis by an angle equal to 3/4 the distance between the hexagon’s vertices, or 1.5 L = 1.5 24° = 36°, and concerning the X coordinate by an angle proportional to the apothem a’ of the pentagon plus one half the apothem an of the hexagon. According to Pythagoras’s theorem, the apothem of a pentagon is equal to.6881909602356 units if and only if a’2 = r2 – (L’/2)2. For a pentagon with side length L’, the angle proportional to the apothem a’ is a’ 24° / L’ = 16.51658305o. Turn the second pentagon’s face through a relative angle of r26.9088879 over the X-axis. The other pentagon surfaces on the upper hemisphere of the soccer ball can be generated by duplicating the second pentagon four times and rotating it 72 degrees around the Z axis. When we repeat the pentagon surfaces with a Scale of -1 along the Z axis (Mirror), we get a reflection of the lower hemisphere. This is accomplished by selecting the Group option under the Edit menu.

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