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 - - - - -
Have a look in my calcs
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 - - - - -
Actually it is
Decimal Binary 0 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 | | |
Have a look in my calcs
Last edited: