So kicking in My crazy math:

C= 2π*r

if r is 2(or something close to it) then:

Lower Right

r = 2

Uper Left

3 = (2* 3,14159265359 * 2)/4

Lower Left

3/3 = 1

And the Uper Right can be read as 2*(C/4)

In this case 6.

this would give us

1 |--

2 --|

3 -|-

4 ???

5 ???

6 -||

7 ???

8 ???

Now , like in base 10 we only have 9 digits (the last one is 10 right?) so lets kill this 8 and make it zero

0 ---

1 |--

2 --|

3 -|-

4 ???

5 ???

6 -||

7 ???

Now maybe i'm hunting for patterns doing this but does it sound possible that this is the mistery that we need to solve?

Also i forgot to say but most numbers we're rounded up....

- - - - - Additional Content Posted / Auto Merge - - - - -

C= 2π*r

if r is 2(or something close to it) then:

Lower Right

r = 2

Uper Left

3 = (2* 3,14159265359 * 2)/4

Lower Left

3/3 = 1

And the Uper Right can be read as 2*(C/4)

In this case 6.

this would give us

1 |--

2 --|

3 -|-

4 ???

5 ???

6 -||

7 ???

8 ???

Now , like in base 10 we only have 9 digits (the last one is 10 right?) so lets kill this 8 and make it zero

0 ---

1 |--

2 --|

3 -|-

4 ???

5 ???

6 -||

7 ???

Now maybe i'm hunting for patterns doing this but does it sound possible that this is the mistery that we need to solve?

Also i forgot to say but most numbers we're rounded up....

- - - - - Additional Content Posted / Auto Merge - - - - -

Have a look in my calcsActually it is

DecimalBinary0 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111

Left>R or L<Right

- - - 0 - - -

- - | 1 | - -

- | - 2 - | -

- | | 3 | | -

| - - 4 - - |

| - | 5 | - |

| | - 6 - | |

| | | 7 | | |

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