SpheroidTruncatedIcosahedronFilter!
June 9, 2006 6:57 PM Subscribe
My daughter wants to decorate a sphere in a football pattern. How do we determine the correct side length for the pentagons and hexagons, given a sphere of diameter n?
As joe's_spleen already knows, the ball is a regular truncated icosahedron. As such, it has 20 hexagons and and 12 pentagons.
I assume you've seen these, and I know they don't answer your question, but to head off future commenters, here are the mathworld page on truncated icosahedra and the wikipedia article.
posted by gleuschk at 7:24 PM on June 9, 2006
I assume you've seen these, and I know they don't answer your question, but to head off future commenters, here are the mathworld page on truncated icosahedra and the wikipedia article.
posted by gleuschk at 7:24 PM on June 9, 2006
Here is a suggestion for a rough approach: using a real soccer ball, work out how many of the little shapes it takes to do a full circumference of the ball. Each shape will therefore represent "x" degrees out of the full 360. This number of degrees will be stable regardless of the size of the ball, so you can scale it up or down at will with a simple calculation.
The approach is a bit rough because of the way the shapes are tessellated. It would work best if the pentagons & hexagons are close to identical in size.
On preview, head explodes. Returns to study.
posted by UbuRoivas at 7:30 PM on June 9, 2006
The approach is a bit rough because of the way the shapes are tessellated. It would work best if the pentagons & hexagons are close to identical in size.
On preview, head explodes. Returns to study.
posted by UbuRoivas at 7:30 PM on June 9, 2006
Best answer: So, the mere fact you know it's called a truncated icosahedron seems like it would lead you to the answer. But from what I can tell:
Ballpark estimate
From the wikipedia on the icosahedron, it tells us that given an icosahedron with edges of length a, the radius of an inscribed sphere (meaning, a sphere big enough to fit inside the icosahedron and touch all sides) is equal to ~75.5% of a. Which would mean that for a regular icosahedron, given a diameter of n, the length of every icosahedron edge is roughly equal to 1.325*n/2. So now you can make a nice icosahedron.
Understanding that a truncated icosahedron is just lopping off a 5-sided pyramid from the icosahedron vertexes, the surface exposed by this is a 5-sided pentagon. the pentagon and hexagons could be most any size depending on where you lop it off, but it would seem intuitive and aesthetic that equilateral triangle face is split by s/3, so that the equilateral triangle becomes 6/9 its original size in the form of a hexagon. In other words the "lopped off" missing triangle end pieces of the original equilateral triangles are equal the 6 triangles that make up the hexagon, and they originally each continued on to meet at a point somewhere above the new pentagon face.
So this tells us that the new edges of our six-sided hexagons, and thus of our 5 sided pentagons, is equal to 1/3 the length of the icosahedron's original edge. Thus, I'd say the rough math is that each edge of the 12 pentagons and 20 hexagons is equal to (roughly) 1.325(n/2*3), or about 21-22% of the diameter of your sphere.
More precise math
That napkin math aside, according to the Wiki on the truncated icosahedron itself, the edge is length 2 where the radius squared is equal to 9*golden ratio + 10. This would make this theoretical truncated icosahedron equal to ~4.95 radius, or diameter of ~9.9, with an edge length of 2. The ratio of the edge to diameter is about 20%, or close to the number derived above.
Conclusion
So- if you're not being super crazy precise, you could say that of a sphere of diameter n, make 20 hexagons and 12 pentagons each with a length of slightly more than one-fifth of n. That should, given the inaccuracy of children and scissors, fit pretty snugly.
posted by hincandenza at 7:37 PM on June 9, 2006 [1 favorite]
Ballpark estimate
From the wikipedia on the icosahedron, it tells us that given an icosahedron with edges of length a, the radius of an inscribed sphere (meaning, a sphere big enough to fit inside the icosahedron and touch all sides) is equal to ~75.5% of a. Which would mean that for a regular icosahedron, given a diameter of n, the length of every icosahedron edge is roughly equal to 1.325*n/2. So now you can make a nice icosahedron.
Understanding that a truncated icosahedron is just lopping off a 5-sided pyramid from the icosahedron vertexes, the surface exposed by this is a 5-sided pentagon. the pentagon and hexagons could be most any size depending on where you lop it off, but it would seem intuitive and aesthetic that equilateral triangle face is split by s/3, so that the equilateral triangle becomes 6/9 its original size in the form of a hexagon. In other words the "lopped off" missing triangle end pieces of the original equilateral triangles are equal the 6 triangles that make up the hexagon, and they originally each continued on to meet at a point somewhere above the new pentagon face.
So this tells us that the new edges of our six-sided hexagons, and thus of our 5 sided pentagons, is equal to 1/3 the length of the icosahedron's original edge. Thus, I'd say the rough math is that each edge of the 12 pentagons and 20 hexagons is equal to (roughly) 1.325(n/2*3), or about 21-22% of the diameter of your sphere.
More precise math
That napkin math aside, according to the Wiki on the truncated icosahedron itself, the edge is length 2 where the radius squared is equal to 9*golden ratio + 10. This would make this theoretical truncated icosahedron equal to ~4.95 radius, or diameter of ~9.9, with an edge length of 2. The ratio of the edge to diameter is about 20%, or close to the number derived above.
Conclusion
So- if you're not being super crazy precise, you could say that of a sphere of diameter n, make 20 hexagons and 12 pentagons each with a length of slightly more than one-fifth of n. That should, given the inaccuracy of children and scissors, fit pretty snugly.
posted by hincandenza at 7:37 PM on June 9, 2006 [1 favorite]
Fuck- not one but two people responded while I was typing up that thesis?! Damn me and my needless verbosity!!! I thought I'd be the first responder.... :(
posted by hincandenza at 7:38 PM on June 9, 2006
posted by hincandenza at 7:38 PM on June 9, 2006
Er, make that 4 people! dammit... :(
posted by hincandenza at 7:40 PM on June 9, 2006
posted by hincandenza at 7:40 PM on June 9, 2006
On the positive side, you demonstrated some nice mathematical ability. On the negative side, it seems you can't count ;P
posted by UbuRoivas at 7:49 PM on June 9, 2006
posted by UbuRoivas at 7:49 PM on June 9, 2006
Response by poster: Brilliant. That seems to work well.
For the curious, she is doing a project for school on the World Cup, and hit on the idea of putting a booklet inside a model soccer ball - you open up the halved ball like the covers of a book. She bought a polystyrene ball and then asked me how to draw a soccer ball pattern on it. Thanks to hincandenza we've created pentagon and hexagon stencils of the right size and they're covering the surface beautifully. Hincandenza, you've made a ten year old very happy.
posted by i_am_joe's_spleen at 8:53 PM on June 9, 2006
For the curious, she is doing a project for school on the World Cup, and hit on the idea of putting a booklet inside a model soccer ball - you open up the halved ball like the covers of a book. She bought a polystyrene ball and then asked me how to draw a soccer ball pattern on it. Thanks to hincandenza we've created pentagon and hexagon stencils of the right size and they're covering the surface beautifully. Hincandenza, you've made a ten year old very happy.
posted by i_am_joe's_spleen at 8:53 PM on June 9, 2006
Thanks i_a_j_s! Your kudos brightened an otherwise rather crappy Friday. :) Glad I could help!
posted by hincandenza at 1:07 AM on June 10, 2006
posted by hincandenza at 1:07 AM on June 10, 2006
Just last night I opened my copy of American Scientist to find... "The Topology and Combinatorics of Soccer Balls." Highlights:
- An official soccer ball must be a sphere with a circumference obetween 68 and 70 cm, with at most a 1.5% deviation from sphericity when inflated to a pressure of 0.8 atm.
- The earliest soccer championship in the US, in 1863, used a ball with eight panels of vulcanized rubber glued together.
- This year's World Cup ball is made from 14 synthetic-leather panels and is technically a truncated octahedron. Scandalous!
posted by Mapes at 7:41 AM on June 10, 2006 [1 favorite]
- An official soccer ball must be a sphere with a circumference obetween 68 and 70 cm, with at most a 1.5% deviation from sphericity when inflated to a pressure of 0.8 atm.
- The earliest soccer championship in the US, in 1863, used a ball with eight panels of vulcanized rubber glued together.
- This year's World Cup ball is made from 14 synthetic-leather panels and is technically a truncated octahedron. Scandalous!
posted by Mapes at 7:41 AM on June 10, 2006 [1 favorite]
This thread is closed to new comments.
(Um, the problem is that I do not have the maths to do this in reverse. I am just procrastinating when I should be studying for an exam in a couple of hours...)
posted by UbuRoivas at 7:15 PM on June 9, 2006