Any soccer fan that watches World Cup or Euro Cup on TV, is offered to theanimations of thesoccer ballshown before many games, which ended up being so famous in 1990s. As a soccer fan,and also a perboy discovering computer system graphics, I was curious around making something favor what I check out on TV. First I had actually to figure out exactly how to draw the round. I had actually an old, worn-outTango from World Cup"1978 in Argentina to use as a model in front of me.Today it"s feasible to find graphical tools that can be supplied to make suchimages, and draw a nice photo without discovering much around the sphere itself. This article is not around them. I will certainly emphasis on the mathematical part of the job. Once this component is addressed, it"s possible to attract a decent photo making use of incredibly basic computer graphics methods.

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The timeless soccer ball is made of the collection of leather pieces of 12 continual pentagons (that are generally painted black) and 20 regular hexagons (painted white). The pentagons areplaced at the vertices of imaginary icosahedron (one of thePlatonic perfect solids), and also separated one from an additional through the hexagons. The pentagons and hexagons border each various other, and also therefore have the very same side sizes.
I am not claiming that this is the just way to attract the round, yet knowingwhere precisely the points that specify the pentagons (and the hexagons) are in 3D, have the right to obtain us started.Knowing these points, I can attract a spbelow, reform the seam areas (this is wright here pentagons andhexagons border each other) and also paint matching areas as I wish. First, let"s begin via figuring out collaborates of the icosahedron vertices. For simplicity, we collection round radius to be 1, and facility of the round to be at (X=0,Y=0,Z=0). Also, let"s place the optimal vertex, T at (0,1,0), and also bottom at (0,-1,0). For simplicity, we place one of the vertices neighboring T (I will certainly call it A), at Z=0. What are the other 2 works with of A, X and Y? To calculate them, it would certainly be sufficient to know ∠TOA, the angle between imaginary lines connecting the center of the round (O) with T and via A. I will certainly usage the reality that all ranges in between neighboring vertices are the same, therefore∠TOA is the very same as ∠TOB, ∠AOB, ∠AOC and ∠BOC. Let"s speak to the ∠TOA beta (β). Then collaborates of A, B and also C are the following:
Asin βcos β0
Bsin β cos72°cos βsin β sin72°
Csin β cos36°-cos βsin β sin36°
We have actually adhering to equations:
Subtracting one from another, we get:sin²β×(cos36° - cos72°) = 2×cos²β
ortg²β×(cos36° - cos72°) = 2Using the truth that
and also
we gettgβ = 2
and also finally:
sin β = 2/√5 and also cos β = 1/√5
As we have actually the value for beta, it"s incredibly simple to represent all the remaining vertices. Five vertices neighboring the height vertex T, will all have actually the very same Y coordinate as A, cos β. All of them areon the circle on the aircraft ABD, and all are the corners of the continuous pentagon through bounding radius of sin β. Enumerating them as N from 1 to 5, we haveX (N) = sin β × cos (N×72°) Z (N) = sin β × sin (N×72°)Finding the other 5 vertices (of the reduced hemisphere) is trivial: icosahedron is symmetrical.As such, just take negative of each coordinate of the 6 vertices we"ve uncovered, and you obtain the continuing to be six ones.
The second action is to find the specific coordinates of all corners of the babsence pentagons on the ball. The full variety of such pentagons is 12, and also each is centered at among 12 the icosahedron vertexes we simply uncovered. Each pentagon has actually five corners, so the full number of these points is 60. Each is connected to the three closest other such points with lines that coincide via seams on the round. To find how much the corners of these pentagons are from the pentagon center, let"s research the sphere a little bit.Let"s call two corners on 2 babsence pentagons, focused at T and also A, which lie in plane TOA, respectively U and V (refer to the image on the left). Let"s also contact angle TOU alpha. Alpha is the angle that we are going to uncover. Keep in mind that ∠TOU is equal to ∠VOA.
Wsin(α) cos72°cos(α)sin(α) sin72°
For calculations purposes, I will use γ = (90° - β).Then: Vx = sin (β-α) = cos (γ+α) Vy = cos (β-α) = sin (γ+α) Distance UV is the exact same as UW (they alives of a consistent hexagon). Because of this, we deserve to compose the followingequations:UV = UWUV² = (sin (α) - cos (γ+α))² + (cos (α) - sin (γ+α))²UW² = (sin (α) - sin (α) × cos 72°)² + (sin (α) × sin 72°)²2 - 2sin (α) × cos (γ+α) - 2cos (α) × sin (γ+α) = 2sin²(α) - 2sin²(α) × cos72°2 - 2sin (γ+2α) = 2sin²(α) × (1 - cos 72°) = (1 - cos (2α)) × (1 - cos 72°) 2 - 2sin (γ) × cos (2α) - 2cos (γ) × sin (2α) = 1 - cos 72° - cos (2α) + cos 72° × cos (2α)(1 + cos 72°) + cos (2α) × (1 - cos 72° - 2sin (γ) = 2 cos (γ) × sin (2α)Now, square both components of the equation. We gain a quadratic equation via cos(2α) as unknown and a = 4 × cos²(γ) + (1 - cos72° - 2×sin(γ))² = (55 - 13√5)/8 b = 2 × (1 + cos 72°) × (1 - cos72° - 2sin (γ)) = (5 - 7√5)/20 c = (1 + cos72°)² - 4×cos²(γ) = (15√5 - 93)/40Solving it, we get:cos (2α) = (345 + 32√5)/545α ≅ 20.07675128°
Now we recognize the works with of all 12 vertices and also the angle betweenthe facility of the black pentagon (coinciding through a vertex) and this pentagonedge. For U and V, the coordinates currently have the right to be calculated usingthe table displayed above. For the remaining 58 points that are corners of ball"s blackpentagons, we can find works with in several different methods.Each of these 58 points, simply like U and also V, additionally belongs totwo hexagons that sepaprice 2 bordering pentagons. I will certainly callcenters of such pentagons T" and also A". Each (T", A") pair has its own pair of (U", V").
Method one would certainly be to uncover the straight transformationof (T, A) into (T", A"). Then, by applying the exact same transdevelopment to well-known U and V, we find works with of U" and V".Method two is to use device of 3 equations
. Plane OT"A" is defined by the following equation:(T"y×A"z - A"y×T"z)×x + (A"x×T"z-T"x×A"z)×y + (T"×A"y - A"x×T"y)×z = 0Due to the fact that U" belongs to this plane too:
(T"y×A"z - A"y×T"z)×U"x + (A"x×T"z-T"x×A"z)×U"y + (T"x×A"y - A"x×T"y)×U"z = 0(1)
Multiplying vectors OT" and OU" we get:
T"x×U"x + T"y×U"y + T"z×U"z = cos(α)(2)
Multiplying vectors OA" and also OU" we get:
A"x×U"x + A"y×U"y + A"z×U"z = cos(β - α)(3)
Method 3 is maybe the easiest one. It requires a straightforward trick. I will specify the allude wbelow lines TA and OU intersect as X. Since both TA and also OU belengthy to the plane Z=0, Xz=0. What I am trying to find is the ratio TX/TA. Knowing this proportion, it is possible to uncover X" (intersection of T"A" and also OU") only making use of collaborates of T" and also A". PointX" (simply as X) is not on the ball"s spbelow, yet normalizing vector OX" gives us coordinates of U". Equation for TA:x × (Ay - Ty) + y × (Tx - Ax) + (Ty×Ax - Tx×Ay) = 0orx × (cos (β) - 1) - y × sin (β) + sin (β) = 0
Equation for OU is simply:x × cos (α) = y × sin (α)From these 2 equations, we quickly find:Xx = sin (α)×sin (β) / (sin (α) + sin (β-α))Xy = cos (α)×sin (β) / (sin (α) + sin (β-α))
TX/TA = sin (α) / (sin (α) + sin (β-α)) = 1/3 (Surprise!)
At this suggest, we are ready to draw a simple, however accurate picture ofthe soccer sphere. No shading or ray tracing. I will certainly use only the the majority of basicgraphics functions, choose line(), circle() and polygon(). Tright here area couple of hurdles to overcome. Segments, such asUV and also UW, being projected on the spright here, come to be components of ellipses, through centers in O. But the focuses of these ellipses are not positioned on horizontal or vertical lines, to be quickly drawn through traditional Ellipse or Arc calls, present in the majority of graphical libraries.I decided to break-up each such segment into a variety of smaller sized components (utilizing technique equivalent to the Method Three in previous section), and also attract a collection of short right lines that reexisting them.

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This additionally helps to hide lines that are on the ago of the round and also are invisible.If both ends of the segment have negative Z coordinate, this segment is not attracted.Anvarious other difficulty is paint the babsence pentagons on the sphere. Using callssuch as floodfill would certainly just work once pentagon is cshed to the centerof the visible part of the round. But if pentagon is on the edge, and just partof it is visible, it often would certainly not produce the desired result.I was able to fix this puzzle in the following way. Calculate the sequence A of all points that reexisting (accurately enough) the black pentagon.For instance, if each seam (such as UW) were break-up right into 10 components, there would be50 points. It"s important to have all 50 points in a appropriate sequence,going clockwise or counter-clockwise roughly the pentagon.Check if all points in A are visible (have positive Z coordinate). If yes, simply contact Polygon feature that draws the polygon and also fills it via a wanted shade, passing A as a parameter. Go to action one and also repeat for next black pentagon.If just some, yet not every one of the 50 points in A are visible, save the visible points (they always are sequential in the original array of 50), in a one-of-a-kind array B. Then calculate the 2 "edge" points that sepaprice visible part of the pentagon from invisible, and also include these two to B.Calculate the number of points that recurrent ball"s edge between these two points, and also additionally include them to B. The variety of these extra points have to depfinish on the size of the "edge" segment, to recurrent it smoothly as component of the circle.Call Polygon on B. Go to action one and also repeat for the next black pentagon.The complying with Java applet encapsulates this logic.
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Source on GitHubClick on the round and drag the mouse to rotate. links:Polyhedron of Soccer BallsThe Icosahedronwww.mathleague.com/help/geometry/polygons.htm#hexagonMorphing PolyhedronThe History of the Soccer BallYuri Yakimenko, 2021