| Pages: [1] :: one page |
| Author |
Thread Statistics | Show CCP posts - 0 post(s) |
FITTIEEE
   |
Posted - 2009.08.08 21:52:00 -
[1]
Guys,
Trying to do my invention math and somethings just occured to me. I've been using the 9 run decryptor for frigs for a while now. But im looking at all possibilitys of decryptors for cruisers etc and i dont understand something.
http://www.eve-guides.com/invention/decryptors.php
and
http://games.chruker.dk/eve_online/invention_chance.php?base_item=none&base_chance=25&decryptor=modifier2&skill_encryption=5&skill_datacore_1=5&skill_datacore_2=5&precalculated_chance=absolute
Both list a "+9" run mod, for caldari "alignment chart"
When i use this decryptor, i get a 9 run BPC upon sucess.
If this is the case, i dont understand the fact that they list BOTH a +1 run modifier decryptor (id expect to give a 1 run copy in the above logic) and they also list a decryprtor with no run modifier at all (which, from the above logic, id expect to give a 0 run BPC, wtf?)
I dont get it? Can someone shed any light? The reason why i ask is i want to ensure my maths re: runs on the "+4" decryptor are 100% correct before financially commiting.
|
FITTIEEE
|
Posted - 2009.08.08 21:55:00 -
[2]
P.s, im using 1 run blue print copies. I only use max run copies for inventing stuff like drones and ballistic controls. Is this where something is going amiss?
Thanks
|
Gavin Miner
|
Posted - 2009.08.08 22:02:00 -
[3]
using a less then max run ship BPC reduces the invented BPCs total runs by one.
so yes, if you want a 10 run BPC use a max run frigate BPC.
|
Horchan
Gallente
|
Posted - 2009.08.08 22:05:00 -
[4]
Ok, to start with, directly from the first link you gave:
Invention Output Runs = min(max( round down ( (Input BPC Runs / T1 Production Limit) * (T2 Production Limit / 10) + Decryptor Runs ), 1), T2 Production Limit)
Let's plug in some numbers into the meat of this:
(1/15) * (10/10) + 9) = 9.06667
This is then rounded down to 9. If you're doing ship invention and you want the extra run from a decryptor, you MUST use a max-run T1 BPC.
---
DesuSigs |
Construxz
|
Posted - 2009.08.09 01:58:00 -
[5]
Edited by: Construxz on 09/08/2009 02:02:05
Edited by: Construxz on 09/08/2009 02:00:06
gah, nm. Deleted my response. apparently I can't read brackets.
But the rationale used to invent ships is generally: use single run T1 copies UNLESS you're using a run-modifying decryptor. In that case, use a max run T1 BPC always.
|
Horchan
Gallente
|
Posted - 2009.08.09 02:05:00 -
[6]
 Originally by: Horchan
(1/15) * (10/10) + 9 = 9.06667
There. I put an extra close parentheses. The "+ 9" is done last. And the answer is still 9.0667. If you look at the original equation, each of those parts is separate:
( (Input BPC Runs / T1 Production Limit) * (T2 Production Limit / 10) + Decryptor Runs )
---
DesuSigs |
Gavin DeVries
|
Posted - 2009.08.09 02:49:00 -
[7]
You always have a minimum 1 run on a T2 BPC from invention. So for ships, if you use a 1 run BPC to invent from and no decryptor, even though the result is a fraction of a run you still get 1 run with it. If you use a decryptor that adds runs, those add in before rounding takes place, and now you wind up losing a run.
______________________________________________________
Isn't it enough to know that I ruined a pony making a gift for you? |
| |
|
| Pages: [1] :: one page |
| First page | Previous page | Next page | Last page |