Difference between revisions of "Talk:Stacking penalties"

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m (Fix User: link)
 
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S(n) = 0.5^( (n-1)^2 / LN(140) )
 
S(n) = 0.5^( (n-1)^2 / LN(140) )
  
-- [[User|Naara elein]], 23:07, 4 May 2013 (UTC)
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-- [[User:Naara elein|Naara elein]], 23:07, 4 May 2013 (UTC)
  
  
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I also reset the penalty formula in terms of a zero-based index and changed/expanded the accompanying text accordingly. This eliminates a set of nested parentheses and reduces the visual complexity of the expression, which is a virtue when writing math in plain text.
 
I also reset the penalty formula in terms of a zero-based index and changed/expanded the accompanying text accordingly. This eliminates a set of nested parentheses and reduces the visual complexity of the expression, which is a virtue when writing math in plain text.
  
-- [[User:Rain kaessinde|Rain]] ([[User talk:Rain kaessinde|talk]]) 18:35, 25 November 2013 (UTC)
+
-- [[User:Rain kaessinde|Rain kaessinde]] ([[User talk:Rain kaessinde|talk]]) 18:35, 25 November 2013 (UTC)

Latest revision as of 16:57, 20 November 2021

2.22292081^2 is an approximation of LN(140)

The formula should be:

S(n) = 0.5^( (n-1)^2 / LN(140) )

-- Naara elein, 23:07, 4 May 2013 (UTC)


Changed the penalty formula to a base-e form, which matches in-game values to the full displayed precision of 10^-10. Note that the decay constant of 2.67 is an exact number, not a rounded decimal!

I also reset the penalty formula in terms of a zero-based index and changed/expanded the accompanying text accordingly. This eliminates a set of nested parentheses and reduces the visual complexity of the expression, which is a virtue when writing math in plain text.

-- Rain kaessinde (talk) 18:35, 25 November 2013 (UTC)