
nxm
I Just Want Real Love
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- Joined
- Jul 5, 2025
- Posts
- 14
Proving facepill>heightpill
T=αF+(1-α)[η,λ,ε].[H,L,W]
α=f(F)≠k*F
To keep it simple I'll aprox with α=-kF+c
If the face is a 10, you don't mind about face (as it's non linear I guess there's a non zero weight at 10, I'll say a 30%), so a 3 in body would be around a 5 overall.
If face is a 1 you really mind about it so let go 90% so even a 10 in body wouldn't put a 1 face over a 2 overall
In that case k = 0.066, c= 0.966 so α=0.966-0.066*F
I prefer shorties, so η should be non linear, with a maximum on H optimal. I purpose the following term.
η(10-(H-1.Ho)²)
ε would be negative for my preference, but YMMV
There's another discussion with λ as some people would value length at different heights differently, here I purpose a 9/4 cycle negative sin function to start, but that would keep the peaks all the same weight.
λ(h)=-Λsin(9π/2Hh)
Nose, belly and feet shouldn't be huge, the other two add up but you might need a different freq sin to adjust the weight on those two, I'll leave it out for simplicity.
T=cF-kF²+(1-c+kF).((10-(H-1.Ho)²)+(int(Λ*sin(9π/2H h)*L(h) dh, 0,H))-εW).
Choose c, k, η, Ho, Λ, ε for your preference, maybe a different λ(h), and fill with F, H, L(h) and W of the subject to find the total points.
I don't agree with the strictly negative derivative of α(F), I think a 9 or 10 in F does make the α higher than at 6, although maybe not as high as a 1 or 2. Maybe something along the lines of
α(F) = e-p*F + q*F²
So α(0)=1 but α(10)>α(6).
I wouldn't go for a 0.75 constant α.
T=αF+(1-α)[η,λ,ε].[H,L,W]
α=f(F)≠k*F
To keep it simple I'll aprox with α=-kF+c
If the face is a 10, you don't mind about face (as it's non linear I guess there's a non zero weight at 10, I'll say a 30%), so a 3 in body would be around a 5 overall.
If face is a 1 you really mind about it so let go 90% so even a 10 in body wouldn't put a 1 face over a 2 overall
In that case k = 0.066, c= 0.966 so α=0.966-0.066*F
I prefer shorties, so η should be non linear, with a maximum on H optimal. I purpose the following term.
η(10-(H-1.Ho)²)
ε would be negative for my preference, but YMMV
There's another discussion with λ as some people would value length at different heights differently, here I purpose a 9/4 cycle negative sin function to start, but that would keep the peaks all the same weight.
λ(h)=-Λsin(9π/2Hh)
Nose, belly and feet shouldn't be huge, the other two add up but you might need a different freq sin to adjust the weight on those two, I'll leave it out for simplicity.
T=cF-kF²+(1-c+kF).((10-(H-1.Ho)²)+(int(Λ*sin(9π/2H h)*L(h) dh, 0,H))-εW).
Choose c, k, η, Ho, Λ, ε for your preference, maybe a different λ(h), and fill with F, H, L(h) and W of the subject to find the total points.
I don't agree with the strictly negative derivative of α(F), I think a 9 or 10 in F does make the α higher than at 6, although maybe not as high as a 1 or 2. Maybe something along the lines of
α(F) = e-p*F + q*F²
So α(0)=1 but α(10)>α(6).
I wouldn't go for a 0.75 constant α.