How does BAB stack, exactly?

Hi folks.

Call me stupid, but I still don't understand how BAB stacks when a character has more than one class. The PHB gives an example for a low level character, which is obvious anyway, but could someone please give me a more precise explanation, or perhaps post a few examples for a high level character. What BAB does a level 4 fighter / level 16 rogue have, for instance?
 

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Look at the tables for each of the classes in the PHB. Notice how one column is labeled "Base Attack Bonus"? Go to your character's class level and take the first number in that column. That's the BAB from that class. Do this for every class your character has, and add them up. That's your total BAB.

When you take the full attack action, you get an extra attack when your total BAB hits +6, +11, and +16. Each of these is at a cumulative -5 penalty from your total BAB.

A ftr4/rog16 would have a BAB of 4 (fighter) + 12 (rogue) = +16.
 

hong said:

When you take the full attack action, you get an extra attack when your total BAB hits +6, +11, and +16. Each of these is at a cumulative -5 penalty from your total BAB.

A ftr4/rog16 would have a BAB of 4 (fighter) + 12 (rogue) = +16.

Thanks.
It was multiple attacks (when a character takes a full attack action) that confused me. So a fighter 4/rogue 16 would have the following bonuses for a full attack action... +16/+11/+6/+1 ...right?
 


Rary the Traitor said:
It was multiple attacks (when a character takes a full attack action) that confused me.

The important rule is where on PH p. 55 it directs you to "Find the base attack value on Table 3-1: Base Save and Base Attack Bonuses to see how many additional attacks the character gets and at what bonuses."

Hence in your example, we add the base attack bonus of Ftr4 (+4) with that of Rog16 (+12) to get +16. Looking that up on Table 3-1 (PH p. 22), we see that the only results that start off with "+16" look like "+16/+11/+6/+1".

The end result is that, yes, attack progressions always fall off by factors of 5.
 

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