My apologies if this has been covered before, but a search of the DH forum alone yielded 15 pages which I read, and my eyes glazed over by the end.

How does Righteous Fury interact with automatic fire?

Example (unrealistic but needed to make it clear)

Heavy Bolter - dmg: 2d10, RoF: 10
Hit 3 times, and does max damage
i.e. (10+10) + (10+10) + (10+10) (we can safely ignore tearing here)

How many times do I roll for a chance of Righteous Fury? Once for the whole pattern, 3 times (once for each “hit”) or 6 times (once for each “10”).

Cheers,

Tam

Its one chance of righteous fury per individual bullet. It requires at least one 10 to achieve.

I swear I read somewhere on weapons that do 2d10 if you roll a 10 on both you automatically confirm righteous fury, but I can't recall where I found it.

As an aside, with weapons that do tearing and roll 2+ dice for damage, do you roll 3d10 (2d10 + the tearing die) and discard the lowest, or 4d10 (2d10 +2d10, discarding the lowest two)?

I'd say roll all the damage in separate chunks, since remember righteous fury can chain,

So with tearing on a 2d10 weapons, it'd be something like:

roll 2d10 +1 for tearing, get (10, 5, 7, use 10 and 7). 17

Roll for righteous fury, and hit

roll 2d10 +1 for tearing, get (10, 6, 4, use 10 and 6). 16

roll 2d10 +1 for tearing, get (9, 5, 7, use 9 and 7). 16

So in this case, a total of 49 damage (assuming just a 2d10 + 0 weapon with tearing)

We've been doing it like this:

Assume 2 hits, and since it's against Xenos, deathwatch auto confirms

1st hit: rolls 3d(lowest)+5 : Rolls 10, 7, 4. re-roll 10 and add (rolls 6), then lowest. = a 16, a 7, and the +5 = 28 damage

2nd hit: rolls 3d(lowest)+5 : Rolls 10, 10, 10. . re-rolls and add (roll 6, 3, 7), then lowest = a 16, a 17, and the +5 = 38 damage.

Not sure if that's correct, but it makes sense to us. Treating exploding dice this way comes natural to us after playing shadowrun back in the day.

Yeah, I forgot the context was in terms of autofire with my example. But yes, I'd say treat each individual shot in autofire as its own attack for the purposes of righteous fury. In theory, a heavy bolter could achieve 10 total righteous furies.

Larin said:

The method you are applying to the first hit is wrong. Assuming autoconfirm...

1st hit: 2d10 (+tearing, +5) : Rolls 10, 7, 4 = (two highest + 5) = 17 + 5 = 22 + (Righteous Fury triggered and confirmed) -> 2d10 (+tearing, +5) : Rolls 8, 6, 4 = (two highest + 5) = 14 + 5 = 19. Total for 1st hit = 22 + 19 = 41 damage

Acutally hit two is also wrong...

2nd hit: rolls 3d(lowest)+5 : Rolls 10, 10, 10. . = 10 + 10 + 5 = 25, re-rolls and add highest two (roll 6, 3, 7) = 25 + 6 + 7 + 5 = 43 Wounds.

According to the worked example on page 245 of RT.

Thanks for your time replying everyone.

Cheers,

Tam

I usually only let the player reroll the dice that came up 10 and add that to their total when they Fury.

SJE