Board Thread:Questions and Answers/@comment-10774412-20130617192915/@comment-9839593-20130619093744

Sank you very much. Zat's exactly what I was hoping for:

Now ze formulas imply following:

- it is not important as to how AS/ DS is distributed - it is only important how big AS+DS is!

You can see zis by answering the question how many full attacks can I do, given may AS or DS and SC: number of full attacks = (AS/ASRemoved - 1) = 2x(AS+DS)/SC - 1 = (DS/DSRemoved -1)

number of full attacks: If you want 2 full concquest attacks you need AS Required to do 1 full attack +  AS Removed to do another full attack. ASDouble = ASRequired + AS Removed = 3x ASRemoved (remember from above ASReq=2xASRemoved) -> ASDouble/ASRemoved=3= number of full attack +1

The formulas show that the ratio DS/DSRemoved and AS/ASRemoved depend only on AS+DS namely the sum.If you have another distribution the sum is still the same. You can still do the same number of attacks regardless of the AS/DS distribution.

This holds of course for the remainder - say if you can do only "1.5 attacks".

If you sum ASRequired +DS Required thn this too is dependend only on AS+DS.

THERE IS NO OPTIMAL RATIO IF YOU ARE CASH PLAYER.

If you dont use SW then ist is indeed better to allocate all your ability points to DS becasu then  DS/(AS+DS) -> 1 and your fast recovering DS have high impact (impact=1) on you fighting ability. If you had AS > 0 then DS/(AS+DS) < 1 and the recovery of DS would have not the full impact.

OPTIMAL RATIO FOR FREE PLAYERS is AS/DS ~ 0/Full ability Points (of course you have to have a minimum AS points like 15).