Fibonacci bricks

Turtle · 03-13-2008

Fibonacci on da brain lately.

I looked around and didn't find this subject/perspective; not to say it's not extant.

Let's get to it.

Fibonacci set: {1 1 2 3 5 8 13 21 34 55 89 144...}

The ratio between any two adjacent elements (greater to lesser) approximates phi more closely as the values increase, and each such pair contructs a Golden rectangle.

Make a list of the areas of these pairs:
1*1=1
1*2=2
2*3=6
3*5=15
5*8=40
...
The ratios of these areas is ~phi+1 e.g. 40/15= 2.66=~phi+1

That's the basic commonly known part, but what I tried was using Fibonacci numbers to make bricks, i.e. Golden bricks.

Resultant values are volumes.

So:
1*1*2=2
1*2*3=6
2*3*5=30
3*5*8=120
5*8*13=520
8*13*21=2184
13*21*34=9282
21*34*55=39270
34*55*89=166430
55*89*144=704880
...

The ratio quickly settles to ~4.2 and then the fractional part bobs up and down as it heads to...what? What happened to phi? Where is this ratio headed? (Besides blockheaded.

) What can we build with a set of these bricks? What can we not?

Turtle · 03-14-2008

Here I have models of the first two Fibonacci bricks. Clearly, it is possible to construct a cube from 4 of the smallest/first brick, but is it possible to contruct a cube using only the 2nd Fibonacci brick?

CraigD · 03-14-2008

Quote:

Originally Posted by Turtle

The ratio quickly settles to ~4.2 and then the fractional part bobs up and down as it heads to...what? What happened to phi? Where is this ratio headed?

It heads for $2 \phi +1 = \sqrt{5}+2$

Though you couldn’t build anything physical with them, you could define “Fibonacci hyperbricks” of more than 3-dimensions by just extending the rule Turtle used to define Fibonacci tiles and bricks.

There’s appears to be a pattern to the ratio of the values, areas, volumes, and hypervolumes that consecutive terms approach for Fibonacci sequences, tile, bricks, and hyperbricks:
For 1 dimension, it’s $1 \phi +0 = \frac{\sqrt{5}+1}2 \dot= 1.6180340$

For 2 dimensions, it’s $1 \phi +1 \dot= 2.6180340$
For 3 dimensions, it’s $2 \phi +1 \dot= 4.2360680$
For 4 dimensions, it’s $3 \phi +2 \dot= 6.8541100$
…
For 12 dimensions, $144 \phi +89 \dot= 321.99689438$

In general for n dimensions, the ratio approaches $F_n \phi + F_{n-1}$ , where $F_n$ is the n-th term of the standard Fibonacci sequence.

Wow!

Proving this, and thus possibly knowing why it’s so, seems a good bit harder what I did to notice it, which was just write a little program to crank out Fibonacci hyperbrick hypervolumes, and compare them to the value of different $\phi$ (phi) –containing expressions.

Before getting all proof-y, it would be interesting to see if it holds for the $\phi$ s for Fibonacci sequences other than the standard 2-fib (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987), such as the 3-fib (1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927) and greater –fib sequences.

CraigD · 03-14-2008

Quote:

Originally Posted by Turtle

Clearly, it is possible to construct a cube from 4 of the smallest/first brick, but is it possible to contruct a cube using only the 2nd Fibonacci brick?

Sure: just lay out 36 of them, 6 wide, 3 deep, and 2 high, to get a 6x6x6 cube.

The trick will work for any rectangular prism with rational-length sides.

Turtle · 03-14-2008

Quote:

Originally Posted by CraigD

Sure: just lay out 36 of them, 6 wide, 3 deep, and 2 high, to get a 6x6x6 cube.

The trick will work for any rectangular prism with rational-length sides.

Very nice. I took the approach that since I had integral unit volumes, then any solution for a cube would be a cube. So, the next cube volume after 8 (the four first order bricks) that divides by 6 (~~the largest dimension~~ the volume of the second order brick) is 216, or 6^3, which is 6*36, so 36 second order bricks to make a cube.

Similarly, one needs 900 third order bricks to make a cube, and 225 fourth order bricks to make a cube. My work attached below, and a photo of the models of the first four Fibonacci bricks.

Also notable, one can make a second order brick out of 3 first order bricks, make a third order brick out of 5 second order bricks, make a fourth order brick out of 8 third order bricks, and so on.

Now start thinking about how many different ways to make these constructions, that is, without reference to specific different orientations, it is possible to make a second order brick with first order bricks 2 different ways.

Including specific reference to individual orientations, we can color bricks in different ways and the possibilities start getting rather daunting.

~~Off to have a look at some Fibonacci triangles & tetrahedrons~~.

[

That didn't take long. Those Fibonacci triangles are thin.

]

Turtle · 03-14-2008

Quote:

Originally Posted by CraigD

...Though you couldn’t build anything physical with them, you could define “Fibonacci hyperbricks” of more than 3-dimensions by just extending the rule Turtle used to define Fibonacci tiles and bricks.

There’s appears to be a pattern to the ratio of the values, areas, volumes, and hypervolumes that consecutive terms approach for Fibonacci sequences, tile, bricks, and hyperbricks:
For 1 dimension, it’s $1 phi +0 = frac{sqrt{5}+1}2 dot= 1.6180340$

For 2 dimensions, it’s $1 phi +1 dot= 2.6180340$
For 3 dimensions, it’s $2 phi +1 dot= 4.2360680$
For 4 dimensions, it’s $3 phi +2 dot= 6.8541100$
…
For 12 dimensions, $144 phi +89 dot= 321.99689438$

I'm not so sure of the 'can't build anything physical' aspect, but I agree on the pattern. Before I went past the 3rd dimension, that fraction .2360680 had me thinking it was headed to the answer of that cylinder problem.

Quote:

Originally Posted by Craig

In general for n dimensions, the ratio approaches $F_n phi + F_{n-1}$ , where $F_n$ is the n-th term of the standard Fibonacci sequence.

Wow!

Proving this, and thus possibly knowing why it’s so, seems a good bit harder what I did to notice it, which was just write a little program to crank out Fibonacci hyperbrick hypervolumes, and compare them to the value of different $phi$ (phi) –containing expressions.

Before getting all proof-y, it would be interesting to see if it holds for the $phi$ s for Fibonacci sequences other than the standard 2-fib (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987), such as the 3-fib (1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927) and greater –fib sequences.

I love it when you get all poofy

, erhm, all proofy with my concepts.

Here's another interesting observation: 4 orthogonal Fibonacci tree segments bound every Fibonacci brick; 2 up, 2down, 2 left, 2 right. First video is my straw construction of the tree, the second video is Modest's computer model.