| (with Alias|wavefront PowerAnimator, but fits Maya or | | | | Rotate (XForm folder; Rotate icon) the surface of the |
| Rhino easily) | | | | first hexagon, in relative, and over the X coordinate, a |
| Theoretical basics: A (classical) soccer ball is made up | | | | r10.39230484541 angle. Duplicate the sphere cap of |
| of 20 hexagons and 12 pentagons, distributed so that 5 | | | | the first hexagon (Edit menu; Duplicate Object option), |
| hexagons surround each pentagon, and each hexagon | | | | and rotate in relative coordinates the copy around the |
| is surrounded by 3 pentagons alternated with 3 | | | | X axis by an equivalent angle to twice (2) the |
| hexagons. A surface like this cannot be geometrically | | | | apothem a of the hexagon 41.56921938165. |
| flat, it will always be curve. | | | | The position of the third hexagon is shifted (rotated) |
| Both pentagons and hexagons have sides of equal | | | | over the first, and the Z axis, an angle equal to 3/4 the |
| length, and this distance is the same of its radius (from | | | | distance between the hexagon's vertexes; that is 1.5 |
| any vertex to its center). Given the side length L, we | | | | · L = 1.5 · 24° = 36°. And over the X axis an |
| can obtain the radius of curvature R of the soccer ball. | | | | angle corresponding to the apothem a. Let's duplicate |
| To deduce the radius R of the sphere we use the | | | | then the first surface, and rotate the third copy in |
| relation between it and the perimeter P of its equator | | | | relative -20.78460969083 0 36. |
| 2 · pi · R = P, where the pi number is approximately | | | | The fourth hexagon, duplicate this time the third cap, |
| 3.141592. Due to the layout of hexagons over the ball | | | | and rotate this fourth surface -41.56921938165 over |
| surface, you can see that the perimeter is equal to 15 | | | | the X axis. |
| times the length L. Therefore, if the circle has 360 | | | | The remaining surfaces, we obtain them duplicating |
| degrees, then each side L of the polygon corresponds | | | | (Edit menu; Duplicate Object option) the four we |
| to 360º / 15 = 24º of circumference. | | | | already built, with a rotation over the Z axis by 72º, |
| If we take as 1 unit (any) the flat length L' of the | | | | and a number of 4 duplications. This will generate the |
| polygons' sides (hexagon and pentagon), then the | | | | rest of sphere caps corresponding to hexagons, and |
| length of its side on the curved surface of the ball (L) | | | | thus closing the soccer ball surface. |
| will be higher. | | | | 2. The process to build the surfaces of pentagons is |
| Using the expression for trigonometric sine of an angle, | | | | similar, but must take into account that the initial |
| we can calculate the radius R of the ball sin(24º) = L' | | | | reference circle radius will be smaller than that of the |
| / R, so R = L' / sin(24º) = 1 / 0.4067366430758 = | | | | hexagon. The side L' must be the same for the two |
| 2.458593335574 units. We can also infer the length L | | | | polygons. Therefore, the radius r is calculated using |
| of an arc of circumference, using the same expression | | | | sin(36º) = (L'/2) / r, where 36° matches half the |
| as for the perimeter P, since P is proportional to 2 · pi | | | | angle of the arc that corresponds to a one side of the |
| · (360 degrees), L = 2 · pi · (24 º / 360 º) · R | | | | pentagon (360º / 5 = 72º). Thus, the radius r = (L' |
| = 1.029852953906 units. | | | | 2) / sin(36º) = 0.5 / 0.5877852522925 = |
| Modeling: Let's build the soccer ball upon intersections | | | | 0.850650808352 units. The pentagon, in the Front view, |
| with a sphere, whose sections and radii of curvature | | | | first create a circle of 5 sections from a primitive, and |
| are different, depending on hexagons or pentagons. | | | | place it in the origin with the grid magnet. From the |
| These caps are generated from revolution curves. | | | | Right view, now draw a spline with CVs: the first point |
| 1. First the hexagon, in the Front view, create a circle of | | | | with a magnet on the upper Edit Point of the circle; |
| radius 1 and 6 sections from a primitive (Objects | | | | place the second point shifted, in relative coordinates, |
| folder), and place it in the origin (coordinates) with the | | | | to the position r0.1 0; the third point of the curve at 0.05 |
| grid magnet (Alt key). From the Right view, now draw | | | | -.1; the fourth at 0.05 -.3; the fifth at 0.1 -.3; and the sixth |
| a spline with CVs (Control Vertex): the first point with | | | | and last in absolute coordinates at position a0.3 0. |
| a magnet on the upper Edit Point of the circle (Ctrl | | | | Once we have the spline, we place its pivot in the |
| key); the second point is placed with a shift, in relative | | | | origin with command a0 0 0. Revolution now the curve |
| coordinates, to the position r0.05 0; the third point of the | | | | over the Y axis and generate a surface of 10 |
| curve at 0.05 -.1; the fourth at 0.1 -.3; the fifth at 0.1 -.4; | | | | sections. Then template the generating curve and the |
| and the sixth and last in absolute coordinates at | | | | circle. Now move the generated surface (removing its |
| position a0.3 0. Once we have the spline, we place its | | | | Construction History) to the relative position r0 |
| pivot in the origin (XForm folder; Pivot icon) with | | | | 2.158593335574. In this position, move the pivot again, |
| command a0 0 0. Revolution now the curve over the | | | | now from the surface to the origin: a0 0 0; because |
| Y axis and generate a surface of 12 sections | | | | when we rotate the cap, we will do it on the center of |
| (Surface folder; Revolve icon). Then template the | | | | the ball (the origin of coordinates). To get the positions |
| generating curve and the circle (ObjectDisplay menu; | | | | of the pentagons that form the soccer ball, we |
| Toggle Template option or Alt+T keys). Now move | | | | calculate the offset angles with respect to the original |
| the generated surface (removing its Construction | | | | position. Positions of the pentagons at the poles of the |
| History) to the relative position r0 2.158593335574, | | | | sphere have their center shifted by a 90º angle, in |
| which is the ball radius R minus the height of the | | | | the X coordinate. Duplicate the surface of the first |
| sphere cap 0.3. In this position, move the pivot again, | | | | pentagon, rotating it 90º over the X axis. The position |
| now from the surface to the origin: a0 0 0; because | | | | of the rest of pentagons is rotated, with respect to the |
| when we rotate the cap, we will do it on the center of | | | | X coordinate, an angle proportional to the apothem a' |
| the ball (the origin of coordinates). | | | | of the pentagon, plus one half the apothem a of the |
| To get the positions of the hexagons that form the | | | | hexagon; and over the Z axis an angle equivalent to 3 |
| soccer ball, we calculate the offset angles with | | | | 4 the distance between the hexagon's vertexes; that |
| respect to the original position. | | | | is 1.5 · L = 1.5 · 24° = 36°. If the apothem a' of |
| Positions of the hexagons near the equator of the ball | | | | the pentagon is, by Pythagoras theorem, a'^2 = r^2 - |
| have their center shifted by an angle, in the X | | | | (L'/2)^2, then a' = .6881909602356 units. The angle |
| coordinate, proportional to half the apothem a of the | | | | proportional to the apothem a', with respect to the |
| hexagon. If the apothem a of the hexagon is, by | | | | 24º angle that corresponds to the pentagon's side |
| Pythagoras theorem, a^2 = L'^2 - (L'/2)^2, then a = | | | | length L', is a' · 24° / L' = 16.51658305º. Rotate the |
| .8660254037844 units. The angle proportional to the | | | | surface of the second pentagon, in relative, and over |
| apothem a, with respect to the 24º angle that | | | | the X coordinate, a r26.9088879 angle. To generate |
| corresponds to the hexagon's side length L', is a · | | | | the remaining surfaces of the pentagons, on the upper |
| 24° / L' = 20.78460969083º. Therefore, half the | | | | hemisphere of the soccer ball, duplicate the second |
| apothem a represents half of this angle: | | | | pentagon, with a rotation over the Z axis by 72º, and |
| 10.39230484541º. | | | | a number of 4 duplications. |