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Desmont McCallock
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Posted - 2008.08.27 17:45:00 -
[31]
Edited by: Desmont McCallock on 27/08/2008 18:16:37 Edited by: Desmont McCallock on 27/08/2008 17:58:42 Edited by: Desmont McCallock on 27/08/2008 17:52:43 Lexander, you are the man too!
That's exactly the formula I had in mind. Due to my limited knowledge in maths, I wasn't able to combine my thoughts into one formula. Thanks alot for the help!
You all must admit though, that this one looks a lot more like CCP style, doesn't it?
I never said that the formula would be linear. In fact I suspect that little CCP's formula's are linear. Take a look here and here.
What if F% could be expressed as function of Fstd with some kind of math type, so that F% varies according to Fstd from -10 to 10? And C% the same? Just a thought.
P.S. If you use the formula in Excel, using the "rounddown" function you get the same 2digit accuracy as in EVE.
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Lexander Morinex
Caldari LDD Investments
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Posted - 2008.08.27 18:18:00 -
[32]
Edited by: Lexander Morinex on 27/08/2008 18:20:43
Originally by: Desmont McCallock Edited by: Desmont McCallock on 27/08/2008 17:58:42 Edited by: Desmont McCallock on 27/08/2008 17:52:43 Lexander, you are the man too!
That's exactly the formula I had in mind. Due to my limited knowledge in maths, I wasn't able to combine my thoughts into one formula. Thanks alot for the help!
You all must admit though, that this one looks a lot more like CCP style, doesn't it?
I never said that the formula would be linear. In fact I suspect that little CCP's formula's are linear. Take a look here and here.
What if F% could be expressed as function of Fstd with some kind of math type, so that F% varies according to Fstd from -10 to 10? And C% the same? Just a thought.
Glad to be of help. One thing I should mention from my own experience.
I have seen the formula for (1 - a) before, as the cumulative probability function of an exponential random variable, with a = e^x. It is a natural byproduct of repeatedly multiplying over and over, so it fits your idea.
- Lexander Morinex
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Namco
Balls of Steel
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Posted - 2008.08.27 18:22:00 -
[33]
Very nice work there, guys. :)
If someone is wanting to tackle the missions related stuff, I'd be willing to put in any assistance I can. I'm not overly concerned with it all, but it'd be kinda neat to see if that formula could be derived from the information that IS available.
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Vallintinus
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Posted - 2008.08.27 19:27:00 -
[34]
http://au.youtube.com/watch?v=HY-03vYYAjA
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Treelox
Amarr Market Jihadist Revolutionary Party
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Posted - 2008.08.27 19:30:00 -
[35]
Originally by: Vallintinus http://au.youtube.com/watch?v=HY-03vYYAjA
lol, an epic linkage if ever.
yes math can be "mind blowing" for some, tis true. --
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Danari
Syncore
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Posted - 2008.08.27 20:26:00 -
[36]
And this is why math majors don't get paid squat: Always two others willing to do the work for free because it's 'fun'.
But I admit it's like she-male dwarf ****, strangely arousing to watch.
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Lexander Morinex
Caldari LDD Investments
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Posted - 2008.08.27 20:32:00 -
[37]
Originally by: Danari And this is why math majors don't get paid squat: Always two others willing to do the work for free because it's 'fun'.
But I admit it's like she-male dwarf ****, strangely arousing to watch.
Not all math majors get paid badly. I had a choice, before I went to graduate school. I could become a pure mathematician, do the usual courses in real and complex analysis, topology, etc. Or I could pick an applied math for the money. Professional statisticians (not the kind who accumulate numbers, but the kind who analyze data) make pretty good money, sometimes great money. I am still looking for the job I want when I graduate, but I hardly intend to work for peanuts.
As for solving the problem here, to me I didn't think I could easily get paid anything besides ISK for this analysis, and I thought that any credibility it gained me would far exceed any trivial fee I might charge. Sometimes just showing people you can solve one problem is helpful in getting paid to solve another.
- Lexander Morinex
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Desmont McCallock
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Posted - 2008.08.28 13:04:00 -
[38]
Ok. Ones again I need your help, guys. I'm very close in completing the formula, but... my limited maths knowledge won't let me fininsh. I need someone to come up with an equation that provides results for F% from 13.5 to 6.5 according to Fstd from -10 to 10. The curve looks like a wave and the axis crosses at 0.
```````````F% --------------13.5 ...o............... .......o........... ..........o........ ............o...... ..............o.... ...............o... ................o.. -10..............0.............10 FStd ....................o .....................o ......................o ........................o ..........................o .............................o ................................o ---------------6.5
just to give you an idea.
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Shadarle
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Posted - 2008.08.28 17:02:00 -
[39]
Originally by: Lexander Morinex and I thought that any credibility it gained me would far exceed any trivial fee I might charge. Sometimes just showing people you can solve one problem is helpful in getting paid to solve another.
If people chipped in and showed they were very capable and smart and willing to help out before they launched IPO's their success rate would be exponentially higher.
So I completely agree that doing work for nothing can sometimes be far more profitable than getting payed for it.
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Desmont McCallock
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Posted - 2008.08.28 21:12:00 -
[40]
I found the equation I was looking for. It's y = x^(1/3).
So F% = 10-(Fstd*(x)^(1/3)) and C% = 5-(Cstd*(x)^(1/3)).
Still trying to define x for F% and C%.
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