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Calculations of Japanese 4 in 1 Cubes and its Derivatives
Article © MAIL User: haKet

Calculations of Japanese 4 in 1 Cubes and its Derivatives

Introduction

After having spent several time knitting cubes of the Japanese 4in1 family, I started thinking about how I could get the exact number of rings used to create the cube, except by counting them one by one. So I began to work on an equation, which could easily reveal the desired fact. Right in the beginning a little difficulty occurred in form of two different layers. Each layer has its own orientation resulting in a different amount of vertical and horizontal rings. Based on this inequality, the alternating combination of the layers lead to two slightly different types of cubes.

Layers

The following sketch shows the different layers L_1 and L_2.

Image: layers.jpg

Considering the horizontal rings ( O ) as a determining fact the values a and b refer to the number of those on one axis. If you square those values you will receive the total amount of horizontal rings in the particular layer. The arrows answer the question of how to combine the layers to get a cube. As a consequence of the diagonal combination of the layers you can substitute b with (a+1).

With the following equation the total of the layers can be calculated.

L=a+b=a+(a+1)=2a+1 (1)

The number of horizontal rings is a^2 in L_1 and b^2 = (a+1)^2 in L_2

The number of vertical rings v is surprisingly equal for L_1 and L_2. Another interesting fact, I found mere empirically, is the correlation between the number of horizontal and vertical rings (2).

v=a^2+b^2-1 (2)

v=a^2+(a+1)^2-1
v=a^2+a^2+2a+1-1

v=2a^2+2a (3)

Furthermore you can easily see that the total amount of rings for each layer is the sum of horizontal and vertical rings.

L_1=a^2+v (4)

L_2=b^2+v (5)

Now you have to solve those equations to get the total number of rings in a particular layer.

L_1=a^2+v
L_1=a^2+2a^2+2a

L_1=3a^2+2a (6)


L_2=b^2+v
L_2=(a+1)^2+2a^2+2a
L_2=a^2+2a+1+2a^2+2a

L_2=3a^2+4a+1 (7)



Types of Cubes

With the previous results you can start to think of the types of cubes possible to build. Regarding the alternating set up of the pattern you can figure out two types.

Type I has the configuration [a|..|a] and six equal outer surfaces of a.

Type II has the configuration [b|..|b] and four equal outer surfaces of a plus two equal and parallel outer surfaces of b.

If you think of the smallest cube possible you now know that we need a total of three layers. One variant to combine them is in the configuration of [a|b|a] to get a Type I cube and a Type II cube would have the configuration [b|a|b].


Now we will calculate #, the total amount of rings per cube. All variables we need are given above.

Type I

#_"Type1"=a*L_1+b*L_2 (8)

#_"Type1"=(a+1)*L_1+a*L_2
#_"Type1"=(a+1)*(3a^2+2a)+a*(3a^2+4a+1)
#_"Type1"=3a^3+2a^2+3a^2+2a+3a^3+4a^2+a

#_"Type1"=6a^3+9a^2+3a (9)

If you have doubled or tripled all of your rings you will have to multiply the whole equation with the factor z

#_"Type1" =(6a^3+9a^2+3a)*z (10)

The calculation for the Type II cubes works in the same way, except that the factors a and b are switched in the first row.

Type II

#_"Type2"=a*L_1+b*L_2 (11)

#_"Type2"=a*L_1+(a+1)*L_2
#_"Type2"=a(a^2+2a)+(a+1)*(3a^2+4a+1)
#_"Type2"=3a^3+2a^2+3a^3+4a^2+a+3a^2+4a+1

#_"Type2"=6a^3+9a^2+5a+1 (12)

#_"Type2"=(6a^3+9a^2+5a+1)*z (13)



Another interesting fact is the total ring difference between cubes of Type I and Type II with the same value of a. So we need to identify (9) with (12)

#_"Type1" =#_"Type2"

6a^3+9a^2+3a=6^3+9^2+5a+1

#_difference => 0=2a+1=L

At last we have to recognize the factor z to get the exact difference for cubes with multiple rings per connection.

#_difference => 0=z*(2a+1)=z*L (14)

Now we know that the total ring difference between Type I and Type II cubes is consistent with the total number of layers, used to build the cube, multiplied with z (14).


Example

At last we will try to figure out what properties a cube of Japanese-4-in-1 has. The only given facts are that the number of horizontal rings in layer 1 is 3 and all rings were tripled.

a = 3
z = 3

Type I

#_"Type1"=(6a^3+a^2+3a)*z
#_"Type1"=(6*3^3+9*3^2+3*3)*3

#_"Type1"=756

Type II

#_"Type2"=(6a^3+9a^2+5a+1)*z
#_"Type2"=(6*3^3+9*3^2+5*3+1)*3

#_"Type2"=777

Difference

#_difference=z*(2a+1)=z*L
#_difference=z*(2*3+1)=3*7

#_difference=21


As a result we now know that under given conditions Type 1 cubes consist of 756 rings and Type 2 cubes of 21 rings more, in other words a total of 777.




If you have any further questions or need a PDF of this article please send me a pm

Philipp J. Thoss 2010
Original URL: http://www.mailleartisans.org/articles/articledisplay.php?key=575