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Deleted member 17791
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SMV=A
Ratios=B
Colouring=C
A=B*C
If 10/10 is the maximum possible SMV, lets assume that with plane colouring, your maximum SMV (face only) is 8/10. So the best colouring (north atlantid) would have a multiplier of 1.25.
A=8*1.25
A=10
So each colouring would have a maximum possible looks level. I'm not sure what the multiplier for all would be yet, you'll have to look at an example of a phenotype with perfect/near perfect colouring to find the multiplier. I already know that north atlantid already has the maximum possible multiplier.
For a north atlantid, their looks level for each level of ratios is as follows.
1=1.25
2=2.5
3=3.75
4=5/10
5=6.25
6=7.5
7=8.75
8=10
Lets assume that gingers have a looks multiplier of 0.75. Basically as strong as north atlantid but in the opposite direction.
1=0.75
2=1.5
3=2.25
4=3
5=3.75
6=4.5
7=5.25
8=6
As you can see, it takes having just 6.66 ratios to even be considered a 5/10 as a ginger while a north atlantid only has to have 4 ratios to be a 5/10.
A ginger can only ever be a 6/10 while a north atlantid can be a 10. That's a difference of 4 points which is night and day.
Why is this useful? It shows that colouring becomes more important at higher looks levels. someone with 1 ratios would only deviate about 0.5 points, which is tiny and not very noticeable but is massive once you have high level ratios. This is good because it shows phenotype and race doesn't matter when you're ugly, blue eyes wont ascend an ugly eye area but will massively improve a top tier eye area.
This is better than simply adding a point or two depending on colouring because a north atlantid with 1 ratios will be turned into a 2 or even a 3 which is obviously not reflective of reality, a severely deformed atlantid isn't going to be rated a 3/10 but they might be rated a 1.25/10.
Ratios=B
Colouring=C
A=B*C
If 10/10 is the maximum possible SMV, lets assume that with plane colouring, your maximum SMV (face only) is 8/10. So the best colouring (north atlantid) would have a multiplier of 1.25.
A=8*1.25
A=10
So each colouring would have a maximum possible looks level. I'm not sure what the multiplier for all would be yet, you'll have to look at an example of a phenotype with perfect/near perfect colouring to find the multiplier. I already know that north atlantid already has the maximum possible multiplier.
For a north atlantid, their looks level for each level of ratios is as follows.
1=1.25
2=2.5
3=3.75
4=5/10
5=6.25
6=7.5
7=8.75
8=10
Lets assume that gingers have a looks multiplier of 0.75. Basically as strong as north atlantid but in the opposite direction.
1=0.75
2=1.5
3=2.25
4=3
5=3.75
6=4.5
7=5.25
8=6
As you can see, it takes having just 6.66 ratios to even be considered a 5/10 as a ginger while a north atlantid only has to have 4 ratios to be a 5/10.
A ginger can only ever be a 6/10 while a north atlantid can be a 10. That's a difference of 4 points which is night and day.
Why is this useful? It shows that colouring becomes more important at higher looks levels. someone with 1 ratios would only deviate about 0.5 points, which is tiny and not very noticeable but is massive once you have high level ratios. This is good because it shows phenotype and race doesn't matter when you're ugly, blue eyes wont ascend an ugly eye area but will massively improve a top tier eye area.
This is better than simply adding a point or two depending on colouring because a north atlantid with 1 ratios will be turned into a 2 or even a 3 which is obviously not reflective of reality, a severely deformed atlantid isn't going to be rated a 3/10 but they might be rated a 1.25/10.
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