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# Higher spell damage through math

Discussion in 'Sorceress' started by scwizard, Aug 5, 2010.

1. ### scwizardDiabloii.Net Member

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Higher spell damage through math

First let it be known that this doesn't apply to the cold tree, because its mastery contributes -res instead of a damage multiplier.

We know that the formula for the total damage of a spell is:
spell base damage * (1 + %bonus from synergy) * (1 + %bonus from mastery + %bonus from items)

Now we break the individual bits down further:
%bonus from synergy = synergy base + [additional] hard points in synergy * synergy bonus
%bonus from mastery = mastery base + mastery bonus * ([additional] hard points in mastery + item points in mastery)

So you're respecing and you have a pool of unallocated skill points to put into a certain spell. You've already put 20 points into the spell itself. Now the question is, with a minimum of one in mastery and each of the skill's prerequisites how many skill points should be put into synergy and how many into mastery if I want to maximize the damage multiplier for that particular spell?
This question reveals another formula:
[additional] hard points in synergy + [additional] hard points in mastery = total unallocated points

To answer this question, first we need to algebratize it.
x = (additional) hard points in synergy
y = (additional) hard points in mastery
For constants, I will use capital letters:
S = synergy bonus
M = mastery bonus
I = item points in mastery
B = mastery base = level 1 mastery bonus
P = synergy base = the amount of synergy you're required to give the skill due to preqs
T = %bonus from items (Temper bonus)
U = unallocated points

Now we restate the problem:
(1+P+Sx)(1+B+T+M(y+I)) = Multiplier
U = x + y

y = -x + U
(1+P+Sx)(1+B+T+M(-x+U+I))

As you learned in school, (a+b)(c+d) = ac + ad + bc + bd
(1+P)(1+B+T) + (1+P)(M(-x+U+I)) + (Sx)(1+B+T) + (Sx)(M(-x+U+I))

And a(b + c) = ab + ac
1+B+T+P+PB+PT + -Mx+MU+MI-PMx+PMU+PMI + Sx+SBx+STx + Sx(M(-x+U+I))

The fourth term needs some additional factoring out:
M(-x + U + I) = -Mx + MU + MI
Sx(-Mx + MU + MI) = -SMxx + SMUx + SMIx

Now we have:
1+B+T+P+PB+PT + -Mx+MU+MI-PMx+PMU+PMI + Sx+SBx+STx + -SM(x^2)+SMUx+SMIx

This is a second degree polynomial. You could graph it if you so wished. To find the maximum value of a second degree polynomial with respect to x, we examine the following values:
The earliest value in x's domain. 0 hard points beyond the prereqs in this case.
The latest value in x's domain. U hard points beyond the prereqs in this case.
The value where x "peaks."

One quality of x's "peak" is that a line drawn tangent to that point will be flat. Its slope will be zero. Here is a picture of what I'm talking about.
To find the slope of a tangent line of a function, we take that function's derivative. Here's the derivative, we want to know when it's equal to 0:
-M + -P + S+SB+ST + -2SMx+SMU+SMI = 0
So we do algebra.
-M-P+S+SB+ST+SMU+SMI = 2SMx
(-M-P+S+SB+ST+SMU+SMI)/2SM = x
That is the value of x that we are interested in, in terms of known constants. To be safe we'll want to examine both the floor and the ceiling of that value.

Example coming later, I need a break from math right now.

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3. ### scwizardDiabloii.Net Member

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Re: Higher spell damage through math

DOH >_>

Whatever, it was something to do while I was on the train yesturday.

The equation I ended up with: (-M-P+S+SB+ST+SMU+SMI)/2SM = x

Is slightly more general though, taking things like "+5% to lightning skill damage" into account.