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seen Dec 16 '13 at 10:25

Nov
10
comment Balance and proportions
Great! I was almost there but what I did was 1/2x + 1/2y = 1 and got stuck. . .
Oct
29
comment Derive a formula of a specific curve
Thanks Daniel!!
Oct
29
comment Derive a formula of a specific curve
Great! Do you have any resources on how to come up whit a solution like this? I want to learn instead of spamming question on math.stackexchange :D
Oct
28
comment logarithmic function between two points
Edit. I think I got it. . .
Oct
28
comment logarithmic function between two points
edited but still in the dark :(
Oct
28
comment logarithmic function between two points
you mean f(x)=a*log(x/c+b)?
Oct
12
comment Exponential equation and point on curve
So X will be X = LOG Base{[logx0-x1(y0-y1)^x0/y0] * logx0-x1(y0-y1)} of Y
Oct
12
comment Exponential equation and point on curve
Uhm I don't think I fail but I was seeking too much abstaction. I wanted something that cold be use for a exponential curve between two point A and B. If I'm not mistaken for a generic point (X,Y) on the exponential curve between (X0,Y0) and (X1,Y1) the function must be: Y = [logx0-x1(y0-y1)^x0/y0] * logx0-x1(y0-y1)^X
Oct
12
comment Exponential equation and point on curve
You solution is brilliant. Can you show the process that you used to find this solution?
Sep
29
comment Jumping from lower ground
Thanks a lot, will try to implement this and let you know if works. Finger crossed :D
Sep
29
comment Jumping from lower ground
Same heght stands for same plane of the character. Let assume that our character is in S(Sx,Sy,Sz) so the right jump spot must be J(Jx,Sy,Jz) so Sy e Jy must be equal on the same plane. The char is still because I must process the force in two steps: first move to jump spot and stop, Then when I say "Yes you can" jump with our fixed V :D