2011-06-30, 08:24 AM
I did this some time back on Asiasoft forums... I'll put it in the spoiler here. It's kinda long(winded).
Hadriel
Spoiler
I did this a long time ago. I was even about to use Lagrangian on this when I realised that it's such a simple linear relationship that I don't even need to bother about this.
I'll see if I can find the post or not.
Ok found it.
D = A(4I + L) where weapon multiplier = 1 and A is MA/100
dD = (4I+L)dA + 4AdI + AdL
or since we prefer to keep L as a constant,
dD|L = (4I + L)dA + 4AdI
Then from here you can throw in whatever you want, noting that dA must be expressed as the ratio of the potential gain to the current amount. Now if you look closely, if you assume L to be negligible e.g. pure INT guys... then it follows that if dA = dI, then you obviously need the same fractional change! i.e.:
delD/delA |I,L = 4I + L, and
delD/delI |A,L = 4A
and therefore,
delI/delA |L = (4I + L) / 4A
which approximates I/A.
The above result can also be easily obtained by noting that:
dD = (4I+L)dA + 4AdI [keeping L constant]
is an exact differential, which implies that
del(4I+L)/delI |A = del(4A)/delA |L
=> 4 = 4
i.e. their fractional rates of change are equal. The above result is obtained because we have defined the relationship between I and A (and L) by the Legendre transformation
D = A(4I + L)
which basically is the original equation!
So to cut the long story short, every MA is worth totalINT/totalMA amount of INT. E.g. I have 800INT and 125MA, so every MA is equivalent to 800/125 = 6.4INT.
The effect of potential EQ does not change the outcome because %INT, %ATT and %DMG all equally affect the equation ASSUMING you're just adding MA or INT. If you're deciding between a %INT and a %MA eq, then it depends on what you are currently wearing because +% equipment adds additively, i.e.
(1+%D)D = (1+%A)A[4(1+%I)I + (1+%L)L]
So if you already have MW or some %INT eq and you are considering between a %INT or a %MA eq, then %MA will be better. If you don't have anything of those and are buying them new, then for the same %, they make no difference.
It doesn't take anymore complicated math to show why stacking %INT is not as good as distributing %MA and %INT. Let's try this: if you could add .1 to %A or %I and you have a current %I only of .15, then which would be larger?
If I add .1 to the already existing .15 from MW, then
D'/D = A[4(1+.15+.1)I] / A[4(1+1.15)I] = 1.25/1.15
Whereas if I add .1 to MA, then
D''/D = (1+.1)A[4(1+.15)I] / A[4(1+1.15)I] = 1.1
And if you punch your calculator, 1.25/1.15 = 1.087 < 1.1
This will ALWAYS be true because (1+%I) and (1+%A) are all greater than 1, and if you didn't know, squares have a larger area per unit surface area than rectangles. Don't believe? Try differentiating it yourself. The perimeter is how much % you can give, and the area is the amount of damage you can deal. Same goes for %D.
CONCLUSION: Each MA is worth TotalINT/TotalMA amount of INT, and this effect is independent of %potentialgear for the case of direct addition of INT and/or MA.
AND: Spread out your %MA, %INT and %DMG. Currently not sure if %stat adds directly to %INT or is a secondary external multiplier.
*edit* I found the old thread where I properly did everything nicely short and sweet. Do try it on WolframAlpha. Don't know what it is? Go Google it.
http://forums.asiasoftsea.net/showthread.php?t=903623
I did this a long time ago. I was even about to use Lagrangian on this when I realised that it's such a simple linear relationship that I don't even need to bother about this.
I'll see if I can find the post or not.
Ok found it.
Quote:Magic Damage = MA*(4*INT + LUK)/100Now for the proper math. I just came back from math exam, and am feeling pissed.
Let's take my old ArchMage to compare. I'll hypothetically remove his Maple Earrings for now.
INT: 647
LUK: 132
MA: 241
An increase in 5 MA gives me 5/241 = 2.075% damage increase.
If I increase my INT by 14, then
56/(647*4 + 132) = 2.059% damage increase.
Which means, yes, scroll those Elemental Piercing earrings!
By the way, always compare % changes.
Actually I was thinking of trying to model the damage formula using PDE and Lagrangian multipliers to solve... but halfway I realised that it's kind of heavy machinery... you don't need to use a sledgehammer to kill an ant when you can just squish it.
D = A(4I + L) where weapon multiplier = 1 and A is MA/100
dD = (4I+L)dA + 4AdI + AdL
or since we prefer to keep L as a constant,
dD|L = (4I + L)dA + 4AdI
Then from here you can throw in whatever you want, noting that dA must be expressed as the ratio of the potential gain to the current amount. Now if you look closely, if you assume L to be negligible e.g. pure INT guys... then it follows that if dA = dI, then you obviously need the same fractional change! i.e.:
delD/delA |I,L = 4I + L, and
delD/delI |A,L = 4A
and therefore,
delI/delA |L = (4I + L) / 4A
which approximates I/A.
The above result can also be easily obtained by noting that:
dD = (4I+L)dA + 4AdI [keeping L constant]
is an exact differential, which implies that
del(4I+L)/delI |A = del(4A)/delA |L
=> 4 = 4
i.e. their fractional rates of change are equal. The above result is obtained because we have defined the relationship between I and A (and L) by the Legendre transformation
D = A(4I + L)
which basically is the original equation!
So to cut the long story short, every MA is worth totalINT/totalMA amount of INT. E.g. I have 800INT and 125MA, so every MA is equivalent to 800/125 = 6.4INT.
The effect of potential EQ does not change the outcome because %INT, %ATT and %DMG all equally affect the equation ASSUMING you're just adding MA or INT. If you're deciding between a %INT and a %MA eq, then it depends on what you are currently wearing because +% equipment adds additively, i.e.
(1+%D)D = (1+%A)A[4(1+%I)I + (1+%L)L]
So if you already have MW or some %INT eq and you are considering between a %INT or a %MA eq, then %MA will be better. If you don't have anything of those and are buying them new, then for the same %, they make no difference.
It doesn't take anymore complicated math to show why stacking %INT is not as good as distributing %MA and %INT. Let's try this: if you could add .1 to %A or %I and you have a current %I only of .15, then which would be larger?
If I add .1 to the already existing .15 from MW, then
D'/D = A[4(1+.15+.1)I] / A[4(1+1.15)I] = 1.25/1.15
Whereas if I add .1 to MA, then
D''/D = (1+.1)A[4(1+.15)I] / A[4(1+1.15)I] = 1.1
And if you punch your calculator, 1.25/1.15 = 1.087 < 1.1
This will ALWAYS be true because (1+%I) and (1+%A) are all greater than 1, and if you didn't know, squares have a larger area per unit surface area than rectangles. Don't believe? Try differentiating it yourself. The perimeter is how much % you can give, and the area is the amount of damage you can deal. Same goes for %D.
CONCLUSION: Each MA is worth TotalINT/TotalMA amount of INT, and this effect is independent of %potentialgear for the case of direct addition of INT and/or MA.
AND: Spread out your %MA, %INT and %DMG. Currently not sure if %stat adds directly to %INT or is a secondary external multiplier.
*edit* I found the old thread where I properly did everything nicely short and sweet. Do try it on WolframAlpha. Don't know what it is? Go Google it.
http://forums.asiasoftsea.net/showthread.php?t=903623
Hadriel