Author 
Topic 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 01:52:00  [ 1]
Edited by: Jim Hsu on 09/12/2005 18:39:56Edited by: Jim Hsu on 09/12/2005 18:39:27Edited by: Jim Hsu on 07/12/2005 18:30:30Edited by: Jim Hsu on 07/12/2005 00:11:57Edited by: Jim Hsu on 06/12/2005 21:54:30Edited by: Jim Hsu on 04/12/2005 00:42:26Edited by: Jim Hsu on 03/12/2005 01:52:32UPDATE: Mr.M has his calculator on. It's superior to mine. check www.evegeek.com. UPDATE: Check this. http://www.evefiles.com/media/12/stackingreal.zip I'm currently looking at the stacking equation for Sisi. I have made considerable progress, but there are errors. I will use this guy's data, for instance. Blame him if the data's wrong. Originally by: "Harry Voyager"
Raw data for Sisi Module stacking penalty
Fitting: Tempest with D650 Base RoF: 2.87 sec Base Damage Mod: 2.8875x
Fitting: Gyro I's: DMx1.07, RoFx0.925
1x RoF: 2.66 DM: 3.089265
2x RoF: 2.48 DM: 3.27759283745
3x RoF: 2.38 DM: 3.40850258315
4x RoF: 2.33 DM: 3.47601431929
5x RoF: 2.31 DM: 3.50180455707
6x RoF: 2.30 DM: 3.50915618132
I took the damage mod into account. Let f(x) equal a function that expresses the relative amount that performance is degraded on a single mod considering the number of modules stacked. For example, an f(2)=0.85 means that the 2nd mod has a modifier of 85% of the mod itself (If the modifier was 1.07x, it would be (1.071)*0.85 + 1 = 1.0595x.) f(1)=1 as we all know. I postulated that the total modded value is equal to value*(mod*f(1))*(mod*f(2))*(mod*f(3))...(mod*f(n)). Where n is the number of modules fitted. Doing this, and using the above data, I obtained the following values: f(1) = 1 f(2) = 0.8708860 f(3) = 0.5705831 f(4) = 0.2829552 f(5) = 0.1059927 f(6) = 0.0299912 I obtained the values first by figuring out the damage mod for the additional mod (by dividing the total value for (N+1) mods by the total value for (N) mods [e.g. 3.408502583/3.277592837). That gives the amount that the (N+1)th mod will change it by. Then, I divided that value by the modifier (in this case, 1.07x), and got a decimal from 0 to 1, posted above. Doing this, I saw a definite gaussian/logistic pattern to the points. I tried fitting it, and obtained: 1.08115690  + 0.0015763657 (1 + 0.0231600863 * E^(1.207799629x) This is what f(x) is supposed to be equal to. Obviously, this is weird and messy. The R value is 99.96%, which indicates that this is a very close, but not exact match. Now, questions. 1) Why is it not exact, but very, very close? 2) Are there any other values being modified? Do I have to multiply the mod by a constant, for example? 3) Does this apply for other examples? I think I'll rest for now. 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 02:21:00  [ 2]
Edited by: Jim Hsu on 03/12/2005 02:21:26This is what the graph of number of mods vs. f(x) looks like. http://www.evefiles.com/media/12/test.GIFWithout getting technical, basically this is saying that you get quite severe penalties for fitting after 3 mods. Technically, the inflection point of the graph is at x=2.928, and that the derivative of this function looks like a classic bell curve. 
Altrex Stoppel Sniggerdly Pandemic Legion 
Posted  2005.12.03 03:30:00  [ 3]
Hmm wow that drop is quite drastic for stacking...should really change the game good job CCP 
hylleX Herrow Inc

Posted  2005.12.03 04:08:00  [ 4]
bleh 
Vishnej Demonic Retribution Pure. 
Posted  2005.12.03 04:29:00  [ 5]
A different way of looking at it:
Right now, the formula is (mod1*mod2*mod3...*modn)^((1/n)^0.25) Or, to shorten it, mod1*mod2*mod3..*modn ^ X I'll remove the mod1*mod2 etc and just focus on X. Currently, X = ((1/n)^0.25) The new stacking penalty is not of the exact same form with altered constants, that much I know from playing around with them.
But the sum total of the X, the second part of the equation, whatever has taken its place, in this case ends up being: 1 mod: 0.9982189240.975609756 (actual value 1, indicates rounding error somewhat) 2 mods: 0.9344193090.938986964 3 mods: 0.8121611590.806621411 4 mods: 0.6780841980.677019285 5 mods: 0.5618748820.566454012 6 mods: 0.4718747060.483039455
These numbers are: ((sample/basestat)^(1/n)1)/(basebonus1) The first figures are for Dmod, the second for RoF  the second have more rounding and are therefore less accurate.
So the values above are for X: if you want to compute what 6 of the same mods gives you, take (modifier^6)^(0.471874706)
Next step, if anyone wants to take it: Take new tests with varying quality modules (a tech 2 mod and a tech 1, for example), and figure out a way to compute whether the exponent X is the same as here. 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 04:36:00  [ 6]
Edited by: Jim Hsu on 03/12/2005 04:36:33The following is data from shield hardening a scorp with kinetic hardeners. Hardeners used: Tech 1 (50% active, 1% passive) Base kin resist: 40% http://www.evefiles.com/media/12/8hardenersisi.zipTo sum it up, here are my results, based on number of hardeners activated: Calculated (actual) deviation 0= 41.7036% 41.70161420% 0.00198935% 1= 70.7023% 70.55637081% 0.14597168% 2= 83.1570% 83.20542905% 0.04839982% 3= 87.9755% 87.92789721% 0.04760049% 4= 89.6213% 89.60641989% 0.01491866% 5= 90.1430% 90.14679777% 0.00377118% 6= 90.3065% 90.29164064% 0.01489737% 7= 90.3281% 90.32213645% 0.00596034% 8= 90.3496% 90.32704425% 0.02256338% 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 04:42:00  [ 7]
Originally by: Vishnej A different way of looking at it:
Right now, the formula is (mod1*mod2*mod3...*modn)^((1/n)^0.25) Or, to shorten it, mod1*mod2*mod3..*modn ^ X I'll remove the mod1*mod2 etc and just focus on X. Currently, X = ((1/n)^0.25) The new stacking penalty is not of the exact same form with altered constants, that much I know from playing around with them.
But the sum total of the X, the second part of the equation, whatever has taken its place, in this case ends up being: 1 mod: 0.9982189240.975609756 (actual value 1, indicates rounding error somewhat) 2 mods: 0.9344193090.938986964 3 mods: 0.8121611590.806621411 4 mods: 0.6780841980.677019285 5 mods: 0.5618748820.566454012 6 mods: 0.4718747060.483039455
These numbers are: ((sample/basestat)^(1/n)1)/(basebonus1) The first figures are for Dmod, the second for RoF  the second have more rounding and are therefore less accurate.
So the values above are for X: if you want to compute what 6 of the same mods gives you, take (modifier^6)^(0.471874706)
Next step, if anyone wants to take it: Take new tests with varying quality modules (a tech 2 mod and a tech 1, for example), and figure out a way to compute whether the exponent X is the same as here.
Interesting, that also looks strikingly similar to a logistic graph, and it also "almost fits but not quite". Perhaps there is another factor involved? Perhaps it is simply client error? 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 05:01:00  [ 8]
Potential problem spotted.
In my formula, switching the order of stacking around (with varying attribute mods) changes the resist. This should not happen.
Now what? 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 05:10:00  [ 9]
I shall now answer my own question. :)
Dmg mods stack with the largest one first (e.g. for a setup with 2x 25% invuln fields and one 50% hardener), the 50% hardener is applied first, then the invuln fields. EVEN IF the invln fields are activated first.
So, no there is no "activation order". 
Vishnej Demonic Retribution Pure. 
Posted  2005.12.03 05:37:00  [ 10]
Edited by: Vishnej on 03/12/2005 05:41:36 Edited by: Vishnej on 03/12/2005 05:38:10 False.
There is no order like it highest% to lowest or first put on the ship.
Here's the stacking equation right now:
TotalModifier = (Modifier1*modifier2*modifier3...*modifierN) ^ ((1/n)^0.25)
You'll note there's no need to put them in order.
I think it's likely that the first part of the equation (mod1*mod2...) holds true, and also likely that the second part of the equation is a power of that. I've just noted that whatever is used to create that exponent, X, is purely a function of # of mods, rather than being based on modifier  the Dmod and RoF figures match up closely enough to explain the difference with rounding errors.
Most likely, find me that function X(n) to a reasonable degree of accuracy, and we'll have the stacking equation  all that will be left will be matching it up to something made with simple constants.

Vishnej Demonic Retribution Pure. 
Posted  2005.12.03 05:43:00  [ 11]
And yes, there is "client error" involved  the client rounds the numbers, or simply cuts off the rest of the digits. 
ELECTR0FREAK Eye of God 
Posted  2005.12.03 05:51:00  [ 12]
Gah, I went through all this with missiles... good luck folks. 
xOm3gAx Caldari Stain of Mind

Posted  2005.12.03 06:12:00  [ 13]
Originally by: ELECTR0FREAK Gah, I went through all this with missiles... good luck folks.
i remember =( was truly horrible (great job on your part though) 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 06:17:00  [ 14]
Edited by: Jim Hsu on 03/12/2005 06:18:50The following equation for X(n) has an R^2 value of 0.9996. About the best I can do. X(n) = 0.695384707 / (1 + E^(2.89517392+0.795065316*n) ) + 0.381401531 Technical data: Originally by: NLREG
Test Number of observations = 6 Maximum allowed number of iterations = 500 Convergence tolerance factor = 1.000000E010 Stopped due to: Relative function convergence. Number of iterations performed = 21 Final sum of squared deviations = 8.1591434E005 Final sum of deviations = 2.9228286E012 Standard error of estimate = 0.00638715 Average deviation = 0.00339333 Maximum deviation for any observation = 0.00597077 Proportion of variance explained (R^2) = 0.9996 (99.96%) Adjusted coefficient of multiple determination (Ra^2) = 0.9991 (99.91%) DurbinWatson test for autocorrelation = 3.028 Analysis completed 3Dec2005 00:12. Runtime = 0.01 seconds.
 Descriptive Statistics for Variables 
Variable Minimum value Maximum value Mean value Standard dev.      Mods 1 6 3.5 1.870829 X 0.4718747 0.9982189 0.7427722 0.20837
 Calculated Parameter Values 
Parameter Initial guess Final estimate Standard error t Prob(t)       K 0.7 0.695384707 0.04850193 14.34 0.00483 a 1 2.89517392 0.2982305 9.71 0.01045 b 0.8 0.795065316 0.08336475 9.54 0.01082 c 0.1 0.381401531 0.02777393 13.73 0.00526
 Analysis of Variance 
Source DF Sum of Squares Mean Square F value Prob(F)       Regression 3 0.2170087 0.07233624 1773.13 0.00056 Error 2 8.159143E005 4.079572E005 Total 5 0.2170903
Could you clean this up? :) PS Thanks a lot for this alternate view. It helps to simplify the excel programming sufficiently. Now ... how do hardeners work... 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 06:38:00  [ 15]
Originally by: Vishnej A different way of looking at it:
Right now, the formula is (mod1*mod2*mod3...*modn)^((1/n)^0.25) Or, to shorten it, mod1*mod2*mod3..*modn ^ X I'll remove the mod1*mod2 etc and just focus on X. Currently, X = ((1/n)^0.25) The new stacking penalty is not of the exact same form with altered constants, that much I know from playing around with them.
But the sum total of the X, the second part of the equation, whatever has taken its place, in this case ends up being: 1 mod: 0.9982189240.975609756 (actual value 1, indicates rounding error somewhat) 2 mods: 0.9344193090.938986964 3 mods: 0.8121611590.806621411 4 mods: 0.6780841980.677019285 5 mods: 0.5618748820.566454012 6 mods: 0.4718747060.483039455
These numbers are: ((sample/basestat)^(1/n)1)/(basebonus1) The first figures are for Dmod, the second for RoF  the second have more rounding and are therefore less accurate.
So the values above are for X: if you want to compute what 6 of the same mods gives you, take (modifier^6)^(0.471874706)
Next step, if anyone wants to take it: Take new tests with varying quality modules (a tech 2 mod and a tech 1, for example), and figure out a way to compute whether the exponent X is the same as here.
Um.. are you sure that is correct. Look at this: Assume base: 100, modifier = 1.07x New Value Change Fitting 1 mod 107.006728 Fitting 2 mods 113.3868397 5.96% Fitting 3 mods 118.0105419 4.08% Fitting 4 mods 120.2009964 1.86% Fitting 5 mods 120.7623671 0.47% Fitting 6 mods 121.210488 0.37% Fitting 7 mods 122.3799773 0.96% Fitting 8 mods 124.3401723 1.60% Why does it start increasing FASTER when we go down to 7 or 8? I think there is something wrong. 
Vishnej Demonic Retribution Pure. 
Posted  2005.12.03 07:18:00  [ 16]
It should end in a horizontal asymptote, at a certain % benefit... IE, on the 10th mod, you have to use an absurd number of decimal places to show its benefit over having 9 mods on.

Vishnej Demonic Retribution Pure. 
Posted  2005.12.03 07:20:00  [ 17]
Edited by: Vishnej on 03/12/2005 07:21:26 Hardeners work with the same formula, however, you have to think a bit differently. Think in terms of a "vulnerability" stat rather than a "resistance" stat. A 20% resistance is an 80% vulnerability.
So right now, three 20% hardeners = (0.8*0.8*0.8) ^ ((1/3)^0.25)
BTW, when I asked you to find the formula, I was implying that I didn't know it it was a logarithm, hyperbola, exponent, what. Tweaking either constant that's there now (1 or 0.25) does not produce the right shaped result, so I presume it's in a different form. 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 18:45:00  [ 18]
Edited by: Jim Hsu on 03/12/2005 18:46:54OMG! R = 1.0000! The formula is ABSURDLY complicated though. Behold: Definitions: Let parameters p1p5 be the following: P1 = asymptotic minimum P2 = asymptotic maximum P3 = slope parameter P4 = value at inflection point p5 = slope parameter And let the following hold: c = 2*p3*p5/abs(p3+p5); f = 1/(1+exp(c*(p4n))); g = exp(p3*(p4n)); h = exp(p5*(p4n)); Function modvalue = p1+(p2/(1+f*g+(1f)*h)); Where p1p5 are the above, n is the number of mods used, and modvalue is the relative performance increase (from the base value). I let p5 be a constant (1.07). And used the data: 0,1 1,1.069875325 2,1.135097087 3,1.180433795 4,1.203814483 5,1.212746167 6,1.215292184 (caculated by dividing the modded value by the base value *2.8875)) And obtained the following: Originally by: nlreg
 Final Results 
NLREG version 6.3 Copyright (c) 19922005 Phillip H. Sherrod.
Mod value vs. number Number of observations = 7 Maximum allowed number of iterations = 500 Convergence tolerance factor = 1.000000E010 Stopped due to: Both parameter and relative function convergence. Number of iterations performed = 6 Final sum of squared deviations = 1.7340339E006 Final sum of deviations = 6.6613381E016 Standard error of estimate = 0.000760271 Average deviation = 0.000440172 Maximum deviation for any observation = 0.000729685 Proportion of variance explained (R^2) = 1.0000 (100.00%) Adjusted coefficient of multiple determination (Ra^2) = 0.9999 (99.99%) DurbinWatson test for autocorrelation = 2.169 Analysis completed 3Dec2005 12:42. Runtime = 0.01 seconds.
 Descriptive Statistics for Variables 
Variable Minimum value Maximum value Mean value Standard dev.      n 0 6 3 2.160247 modvalue 1 1.215292 1.145323 0.08267536
 Calculated Parameter Values 
Parameter Initial guess Final estimate Standard error t Prob(t)       p1 0 0.848843439 0.007575417 112.05 0.00001 p2 1 0.369053789 0.008004306 46.11 0.00002 p3 0.5 0.447261067 0.02644867 16.91 0.00045 p4 0.5 0.495751353 0.05486836 9.04 0.00286
 Analysis of Variance 
Source DF Sum of Squares Mean Square F value Prob(F)       Regression 3 0.04100955 0.01366985 23649.80 0.00001 Error 3 1.734034E006 5.780113E007 Total 6 0.04101129
Observe that p5 is not in there (I made it a constant). Basically, the function looks like a modified logistic / asymtopic. Observe that there are none of the weird problems with 78 mods increasing anymore Updates coming... 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 18:49:00  [ 19]
For reference, if anyone wishes to verify the results, download NLREG at www.nlreg.com and use the following script: Quote:
Title "Mod value vs. number"; Variable n,modvalue; //Parameter K = 1; //Parameter a = 2; //Parameter b = 1; //Parameter c = 0.8;
//Function modvalue = ( a + b * exp(c*n+K) );
Parameters p1=0, p2=1, p3=0.5, p4=0.5, p5=0.3; Double c, f, g, h; c = 2*p3*1.07/abs(p3+1.07); f = 1/(1+exp(c*(p4n))); g = exp(p3*(p4n)); h = exp(1.07*(p4n)); Function modvalue = p1+(p2/(1+f*g+(1f)*h));
Plot Domain = 0,10, ylabel = "Efficiency"; Data; 0,1 1,1.069875325 2,1.135097087 3,1.180433795 4,1.203814483 5,1.212746167 6,1.215292184
My impression of the program is that it's pretty powerful, but the user interface is somewhat lacking. 
Grimpak Gallente Midnight Elites Echelon Rising 
Posted  2005.12.03 19:23:00  [ 20]
Originally by: Jim Hsu Edited by: Jim Hsu on 03/12/2005 18:46:54 OMG! R = 1.0000!
The formula is ABSURDLY complicated though. Behold:
Definitions:
Let parameters p1p5 be the following:
P1 = asymptotic minimum P2 = asymptotic maximum P3 = slope parameter P4 = value at inflection point p5 = slope parameter
And let the following hold:
c = 2*p3*p5/abs(p3+p5); f = 1/(1+exp(c*(p4n))); g = exp(p3*(p4n)); h = exp(p5*(p4n)); Function modvalue = p1+(p2/(1+f*g+(1f)*h));
Where p1p5 are the above, n is the number of mods used, and modvalue is the relative performance increase (from the base value).
I let p5 be a constant (1.07). And used the data:
0,1 1,1.069875325 2,1.135097087 3,1.180433795 4,1.203814483 5,1.212746167 6,1.215292184
(caculated by dividing the modded value by the base value *2.8875))
And obtained the following:
Originally by: nlreg
 Final Results 
NLREG version 6.3 Copyright (c) 19922005 Phillip H. Sherrod.
Mod value vs. number Number of observations = 7 Maximum allowed number of iterations = 500 Convergence tolerance factor = 1.000000E010 Stopped due to: Both parameter and relative function convergence. Number of iterations performed = 6 Final sum of squared deviations = 1.7340339E006 Final sum of deviations = 6.6613381E016 Standard error of estimate = 0.000760271 Average deviation = 0.000440172 Maximum deviation for any observation = 0.000729685 Proportion of variance explained (R^2) = 1.0000 (100.00%) Adjusted coefficient of multiple determination (Ra^2) = 0.9999 (99.99%) DurbinWatson test for autocorrelation = 2.169 Analysis completed 3Dec2005 12:42. Runtime = 0.01 seconds.
 Descriptive Statistics for Variables 
Variable Minimum value Maximum value Mean value Standard dev.      n 0 6 3 2.160247 modvalue 1 1.215292 1.145323 0.08267536
 Calculated Parameter Values 
Parameter Initial guess Final estimate Standard error t Prob(t)       p1 0 0.848843439 0.007575417 112.05 0.00001 p2 1 0.369053789 0.008004306 46.11 0.00002 p3 0.5 0.447261067 0.02644867 16.91 0.00045 p4 0.5 0.495751353 0.05486836 9.04 0.00286
 Analysis of Variance 
Source DF Sum of Squares Mean Square F value Prob(F)       Regression 3 0.04100955 0.01366985 23649.80 0.00001 Error 3 1.734034E006 5.780113E007 Total 6 0.04101129
Observe that p5 is not in there (I made it a constant).
Basically, the function looks like a modified logistic / asymtopic. Observe that there are none of the weird problems with 78 mods increasing anymore
Updates coming...
I just had a brain fart by looking at the first 5 lines. most defenitely, numbers are not my thing 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 20:44:00  [ 21]
Edited by: Jim Hsu on 03/12/2005 20:47:06 The formula works perfectly when all the mods are the same, but what if they are different?
Um.. let multipler = (modvalue1*modvalue2 ... modvaluen) ^ (1/n). Will that work? In effect, if the modvalues are the same, this will be (modvalue^n)^(1/n), which is what we wanted. If they are different, it will take an "average" of the values.
I will try to verify this in game, if it works or not. 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 20:55:00  [ 22]
Alternatively, I can try to do some multiple regression (ick) with the data if it really doesn't work.
E.g.
0,1.07,1 1,1.07,1.069875325 2,1.07,1.135097087 3,1.07,1.180433795 4,1.07,1.203814483 5,1.07,1.212746167 6,1.07,1.215292184 0,1.085,1 1,1.085,? 2,1.085,? 3,1.085,? 4,1.085,? 5,1.085,? 6,1.085,? 0,1.035,1 1,1.035,? 2,1.035,? 3,1.035,? 4,1.035,? 5,1.035,? 6,1.035,?

Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 22:57:00  [ 23]
Edited by: Jim Hsu on 03/12/2005 23:25:59Edited by: Jim Hsu on 03/12/2005 23:17:52Upon further consideration, the previus formula was WAY too complicated. The following is a logistic function of four parameters: Originally by: nlreg
NLREG version 6.3 Copyright (c) 19922005 Phillip H. Sherrod.
Mod value vs. number Number of observations = 7 Maximum allowed number of iterations = 500 Convergence tolerance factor = 1.000000E010 Stopped due to: Relative function convergence. Number of iterations performed = 7 Final sum of squared deviations = 6.1563129E006 Final sum of deviations = 2.9433123E011 Standard error of estimate = 0.00143252 Average deviation = 0.000798255 Maximum deviation for any observation = 0.00134958 Proportion of variance explained (R^2) = 0.9998 (99.98%) Adjusted coefficient of multiple determination (Ra^2) = 0.9997 (99.97%) DurbinWatson test for autocorrelation = 2.202 Analysis completed 3Dec2005 16:52. Runtime = 0.00 seconds.
 Descriptive Statistics for Variables 
Variable Minimum value Maximum value Mean value Standard dev.      n 0 6 3 2.160247 modvalue 1 1.215292 1.145323 0.08267536
 Calculated Parameter Values 
Parameter Initial guess Final estimate Standard error t Prob(t)       K 1 1.2194265 0.001462092 834.03 0.00001 a 2 0.334306464 0.01655995 20.19 0.00027 b 1 0.964036977 0.03971412 24.27 0.00015 c 0.8 0.667902481 0.1019214 6.55 0.00723
 Analysis of Variance 
Source DF Sum of Squares Mean Square F value Prob(F)       Regression 3 0.04100513 0.01366838 6660.66 0.00001 Error 3 6.156313E006 2.052104E006 Total 6 0.04101129
Current working formula: modvalue = 1.2194265 + (0.334306464 * (mod1*mod2*mod3...modn)/(1+exp(0.964036977*(n0.667902481)))); PS This is also broken. A big problem I'm having is that the models all fit for mod=1.07, but nothing else. I have to reconsider.. 
keepiru Omega Fleet Enterprises Executive Outcomes 
Posted  2005.12.03 22:58:00  [ 24]
IF that's the actual formula forumfu maths on dmg mods is gonna be pretty tough, heh. 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.03 23:57:00  [ 25]
Edited by: Jim Hsu on 04/12/2005 00:08:15 Fit this:
0,1.07,1 1,1.07,1.07 2,1.07,1.135097087 3,1.07,1.180433795 4,1.07,1.203814483 5,1.07,1.212746167 6,1.07,1.215292184 0,1.077,1 1,1.077,1.077 2,1.077,1.149075251 3,1.077,1.199559759 4,1.077,1.225695224 0,1.098,1 1,1.098,1.098 2,1.098,1.191520786
Um.. yea. A 3d logistic decay thing in both directions. (huh??!!)
If this does not fit (I'm not very confident...), then I'll have to go back to my original idea... 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.04 00:42:00  [ 26]
Working but unelegant version of stacking: http://www.evefiles.com/media/12/stackworkbeta1.zipThings to note: 1. Edit values in bordered areas, don't edit any other values. 2. Put in the mods in descending order (highest dmg mod in 1st) 3. The values are inexact, but very close. 
Jim Hsu Deep Core Mining Inc.

Posted  2005.12.04 03:58:00  [ 27]
Generally:
3 mods is the "sweet spot".
Example statistics for a Gyro II:
ROF mod
Base: 100
Fitting 1 mod 89.5 10.50% Fitting 2 mods 81.41714826 18.58% Fitting 3 mods 76.45498308 23.55% Fitting 4 mods 74.21879518 25.78% Fitting 5 mods 73.42019139 26.58% Fitting 6 mods 73.15929612 26.84% Fitting 7 mods 73.0715301 26.93% Fitting 8 mods 73.03670702 26.96%
Damage mod
Base: 100
Fitting 1 mod 110 10.00% Fitting 2 mods 119.4611726 19.46% Fitting 3 mods 126.3953157 26.40% Fitting 4 mods 129.9161386 29.92% Fitting 5 mods 131.2474856 31.25% Fitting 6 mods 131.6916589 31.69% Fitting 7 mods 131.8421205 31.84% Fitting 8 mods 131.9019595 31.90%

Jim Hsu Deep Core Mining Inc.

Posted  2005.12.04 20:06:00  [ 28]
Edited by: Jim Hsu on 04/12/2005 20:09:33Edited by: Jim Hsu on 04/12/2005 20:06:09Vishnej: After thinking some more, I doubt you can take a simple function and multiply and/or exponentiate the (mod1*mod2*...modn). It would have to be automatically sorted, and likewise applied to each mod individually. (As per my first example). Consider the following: I have 8 50% hardeners in a scorp. Hardeners have an active resist of 50% and a passive resist of 1%. Let's say I activate one hardener. By your formula, I would get (1.5*1.01^7)=1.6082. Taking that to be "approximately" equal to a multiplier of 1.5, I would get (1.5*1.01^7)^x(8) = 1.5, so x(8) has to be 0.853403 (by using logs). But then if I activate all of my hardeners, that would be (1.5^8)^0.853403 = 15.929960. (n=8 for both examples of course.) There is NO WAY that that can ever be correct. Multiplying by (K) or (K^n) or (K^(n+c)) or (K^(n*c)) before the mod stacking doesn't help, I've verified. The reason is that n does not change, so you're just multiplying by a constant in all cases. Conversely, If we let the (1.5^8)^x = (whatever it is), we would get an absurdly small value for x(8) when we apply it to (1.5*1.07^7). Recall: Originally by: blog
Also the dreaded problem of stacking super and lamer modifiers giving zero and negative values has been fixed (like the crazy tracking computer modified by cormack himself and a basic tracking enhancer could do if you got unlucky when dragging the modules to your ship).

Denrace Amarr PURE Legion Pure. 
Posted  2005.12.04 20:13:00  [ 29]
This game was much better back before people knew math. Oh wait, haven't I heard that line before? Anywho, good job on the equation chaps! 
Vishnej Demonic Retribution Pure. 
Posted  2005.12.05 01:47:00  [ 30]
Edited by: Vishnej on 05/12/2005 01:50:50 Originally by: Jim Hsu Edited by: Jim Hsu on 04/12/2005 20:09:33 Edited by: Jim Hsu on 04/12/2005 20:06:09 Vishnej:
After thinking some more, I doubt you can take a simple function and multiply and/or exponentiate the (mod1*mod2*...modn). It would have to be automatically sorted, and likewise applied to each mod individually. (As per my first example).
Consider the following:
I have 8 50% hardeners in a scorp. Hardeners have an active resist of 50% and a passive resist of 1%.
Let's say I activate one hardener.
By your formula, I would get (1.5*1.01^7)=1.6082. Taking that to be "approximately" equal to a multiplier of 1.5, I would get (1.5*1.01^7)^x(8) = 1.5, so x(8) has to be 0.853403 (by using logs). But then if I activate all of my hardeners, that would be (1.5^8)^0.853403 = 15.929960. (n=8 for both examples of course.) There is NO WAY that that can ever be correct. Multiplying by (K) or (K^n) or (K^(n+c)) or (K^(n*c)) before the mod stacking doesn't help, I've verified. The reason is that n does not change, so you're just multiplying by a constant in all cases.
Conversely, If we let the (1.5^8)^x = (whatever it is), we would get an absurdly small value for x(8) when we apply it to (1.5*1.07^7).
Recall:
Originally by: blog
Also the dreaded problem of stacking super and lamer modifiers giving zero and negative values has been fixed (like the crazy tracking computer modified by cormack himself and a basic tracking enhancer could do if you got unlucky when dragging the modules to your ship).
You need to get down basic, prestacking penalty resistance stacking before you even try something like this with a fancy math program. 50% hardener + 50% hardener = 75% hardener [1((10.5)*(10.5)=0.25) = 0.75] NOT 1.5*1.5 = 225% It's best to think of it as tolerance rather than a resistance. A 20% hardener tolerates 80% of that damagetype to go through. So for two hardeners, 0.8*0.8 = 0.64 tolerence for that damagetype. It fits into the same stacking penalty as everything else, and it's multiplicative, meaning there is NO ORDER TO IT WHATSOEVER. Take a 55%, a 20%, and a 10% hardener. 0.45*0.8*0.9 = 0.324 = 0.8*0.45*0.9 It's called a commutative property, and it's one thing that WILL be part of whatever stacking penalty is tossed on. Here's how the stacking penalty CURRENTLY works with resistances: Take the same 55%, 20%, and 10% resistance hardeners. They equal 0.45, 0.8, and 0.9 "tolerence" hardeners. So (0.45 * 0.8 * 0.9) ^ ((1/3)^0.25) = 0.288 tolerence = 71.2% resistance. This is how the game works right now, verified by all manner of testing. The stacking penalty is the same equation for almost all manner of positive and negative mods  it's proven itself very flexible. There IS an F(n), whether it's the type of equation your program is searching for or not, which will describe the new stacking penalty thusly: (mod1*mod2*mod3...*modn) ^ F(n) 
