Post/Author/DateTime	Post
#1 deedsy Jul 23, 2008 16:49:27	i would like to know where you can find the rules about the addition of critical hit multiplier exemple: i scrore a crital hit with a lance (crit x3) during a spirit charge that already triple my damage.... would id be (1d8 x3) x3 or it become 1d8x4 or x5... you see what i mean so can someone explain me how it works and an other question is if you got pounce so you can make like 2 or 3 attack in a charge and u got the feat headlong rush that double your damge on a charge... does it double all your 3 attack or only the first??
#2 iconoplast Jul 23, 2008 17:11:06	would id be (1d8 x3) x3 or it become 1d8x4 or x5... "D&D Math" works as follows: each "doubling" is an addition of the original damage. SO, 1d8 * 3 * 3 = 1d8 * 5 (Since "tripling" is, in this sense, adding twice your original damage)
#3 arcane_dup Jul 23, 2008 22:15:27	What iconoplast says, but for clarification on the threat range (because that's also wonky): a doubling of 20 is 19-20. Therefore, a doubling of [x-20] is {[20 - (20 - x + 1) (2) -1] to 20.} For example, 18-20 is: [20 - (20 - 18 + 1)] = 3. 3 * 2 = 6. 6 - 1 = 5. 20 - 5 = 15. 15-20 is the result. A bit complicated, yes, but it keeps things more balanced. Otherwise people could just stack multipliers and end up with x9 or more.
#4 inarai Jul 23, 2008 22:41:57	There is a much simpler way to say that... If you crit on a natural 20 only, the threat range is one, as in, one number. On 19-20, it's 2, as in 2 numbers. Double 19-20 would be 17-20, a four number range. As for iconoplast's math, I don't understand that explanation...
#5 arcane_dup Jul 23, 2008 23:06:12	iconoplast's math works in a "step" system, as D&D does. x1 is zero steps. x2 is one step. so on. x3 is two steps. x3 + x3 = x5 because two steps up from x3 is x5.
#6 inarai Jul 23, 2008 23:09:25	So, literally, it's (m-1)+(n-1), where m is the first multiplier and n is the second?
#7 deedsy Jul 23, 2008 23:14:16	thx guy for the reply but its not the threat range taht i got problem with .. maybe i wasent very clear sry.. if im a orc with the feats headlong rush that mean double damage on a charge.. if i use a falchion... the critique range is 18-20 x2... so if i hit the guy and do a critique on my charque ... will it be 2d4 x3 or 2d4 x2 then i multiply by 2 the total for the feat.... (2d4x2) x2 ??? or what...
#8 inarai Jul 24, 2008 0:06:04	Multiplier stacking is additive, as iconoplast said and Arcane clarified. Bascially, subtract 1 from each multiplier, and add it all together, a la the formula that I posted to simplify it.
#9 arcane_dup Jul 24, 2008 0:20:10	I know you weren't asking about it, but quite a few people have issues with the threat range. Inarai and iconoplast have the multiplier parts correct. If you have a x2 multiplier plus another x2 multiplier, it's x3.
#10 deedsy Jul 24, 2008 0:58:59	ty very much now everything is clear
#11 robbypants Jul 24, 2008 9:13:45	So, literally, it's (m-1)+(n-1), where m is the first multiplier and n is the second? Close. It would be (m)+(n-1). Also, note that the "-1" applies to each multiplier after the first. So it would really be something more like: (m)+(n1-1)+(n2-1)+...+(nk-1) So a few examples: