I just typed this all out in barnest's excellent little PbP down in the forums, but I wanted to share it all with you.
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Rules question for your most esteemed GM:
I've always thought that Barrage rules were a good idea, but poorly explained. As long as one accepts that the barrage rules are a major abstraction and does some very strict reading of the rules, I think they actually can work pretty well.
The part that has always confused me, though, is how individual weapon damage gets taken into account. Under some of the examples, it doesn't. But there is one sentence that DOES explain what you are supposed to do and I'm wondering what you think about it.
On page 74, the third sentence under the boldfaced heading "Barrages & Defenses" says, "Add up the protection offered by the defenses, and then subtract it from the individual weapon damage score to determine final DM."
This sentence is key, IMO, to making the barrage attack rules work right.
Unfortunately, in the example that actually ends on page 75, they don't follow their own rule and it has created vast confusion on these forums.
In that example, the Bucephalus fires a 200-Nuclear Missile-Long-2 barrage at the Victory. The Bucephalus rolls a 7 on the 2d6 +0 (range) +3 (skill) + 4 (fire control) -2 (dodging enemy) for a total of 7+5 = 12. In the next little example, the Victory calculates her defense DM of 7, and so the example gives a net barrage score of 5, which results in 50% damage. I believe this is incorrect. I believe the final barrage score should be 7, doing 100% damage.
Here is why:
The sentence I quoted above instructs the player to "subtract it (the protection offered by defenses) from the individual weapon damage (in this case it's 2) to determine the final DM." So, subtracting the IWD, which is 2, from the total defenses DM, which is 7, gets us a -5 DM.
Now, you might be saying, "well that's odd, if I subtract a -5 DM from the attack roll, won't I be ADDING +5?" Yes, but that's not what the rules say. It just says "defenses such as armor or sand provide a DM to the attack roll." It doesn't say, "add up the defenses and subtract that number from the attack roll." (And yes, I know I'm lawyering the crap out of these rules, but they DO work if you read them super strictly!)
So, we simply apply the "final DM" of defenses to the attack roll of 12 and we get 7, not 5.
Why does this matter? When I did the math, it makes a HUGE difference when using powerful weapons, especially when armor is around in large quantities.
For example, let's use the simulation in your game that didn't quite finish due to the radiation leak.
You were rolling attack and subtracting 8 for armor and 4 for range regardless of the weapons being fired. The range DM is standard. However, if you had used the above interpretation of the rules, the DM from defenses (only armor, in this case) would have differed between the heavy particle beam bay and the particle beam barbettes. In fact, by a lot!
Take the barrages from
http://www.mongoosepublishing.com/phpBB2/viewtopic.php?p=657555#657555 that post. One heavy bay got an 11, one barbette got a 9.
Armor on those ships is 8, and heavy particle bays do 9 barrage damage and that is also their individual weapon damage. Defenses subtracted from IWD is 9-8=+1 DM. So, instead of 11-8 (armor)-4 (range) = -1, miss, you'd have 11+1 (armor) -4 (range) = 8 = 125% damage!
But for the particle barbette, which does only 4d6, defense DM becomes 4-8=-4, so 9-4 (defense)-4 (range) = 1 = 0% and thus no damage. If we give the particle barbettes the same roll (11) for sake of argument, you end up with a 3, which does 10% damage.
So, final damage (assuming that both attacks were 11s) would be as follows:
4 heavy particle beam bays at 9d6 and 125% damage = 45 damage split between two sections.
16 particle barbettes at 4d6 and 10% damage = 6 damage split between two sections.
And that is why that one little sentence is so important and (IMO) makes or breaks the barrage damage rules.
(takes breath)
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Rules question for your most esteemed GM:
I've always thought that Barrage rules were a good idea, but poorly explained. As long as one accepts that the barrage rules are a major abstraction and does some very strict reading of the rules, I think they actually can work pretty well.
The part that has always confused me, though, is how individual weapon damage gets taken into account. Under some of the examples, it doesn't. But there is one sentence that DOES explain what you are supposed to do and I'm wondering what you think about it.
On page 74, the third sentence under the boldfaced heading "Barrages & Defenses" says, "Add up the protection offered by the defenses, and then subtract it from the individual weapon damage score to determine final DM."
This sentence is key, IMO, to making the barrage attack rules work right.
Unfortunately, in the example that actually ends on page 75, they don't follow their own rule and it has created vast confusion on these forums.
In that example, the Bucephalus fires a 200-Nuclear Missile-Long-2 barrage at the Victory. The Bucephalus rolls a 7 on the 2d6 +0 (range) +3 (skill) + 4 (fire control) -2 (dodging enemy) for a total of 7+5 = 12. In the next little example, the Victory calculates her defense DM of 7, and so the example gives a net barrage score of 5, which results in 50% damage. I believe this is incorrect. I believe the final barrage score should be 7, doing 100% damage.
Here is why:
The sentence I quoted above instructs the player to "subtract it (the protection offered by defenses) from the individual weapon damage (in this case it's 2) to determine the final DM." So, subtracting the IWD, which is 2, from the total defenses DM, which is 7, gets us a -5 DM.
Now, you might be saying, "well that's odd, if I subtract a -5 DM from the attack roll, won't I be ADDING +5?" Yes, but that's not what the rules say. It just says "defenses such as armor or sand provide a DM to the attack roll." It doesn't say, "add up the defenses and subtract that number from the attack roll." (And yes, I know I'm lawyering the crap out of these rules, but they DO work if you read them super strictly!)
So, we simply apply the "final DM" of defenses to the attack roll of 12 and we get 7, not 5.
Why does this matter? When I did the math, it makes a HUGE difference when using powerful weapons, especially when armor is around in large quantities.
For example, let's use the simulation in your game that didn't quite finish due to the radiation leak.
You were rolling attack and subtracting 8 for armor and 4 for range regardless of the weapons being fired. The range DM is standard. However, if you had used the above interpretation of the rules, the DM from defenses (only armor, in this case) would have differed between the heavy particle beam bay and the particle beam barbettes. In fact, by a lot!
Take the barrages from
http://www.mongoosepublishing.com/phpBB2/viewtopic.php?p=657555#657555 that post. One heavy bay got an 11, one barbette got a 9.
Armor on those ships is 8, and heavy particle bays do 9 barrage damage and that is also their individual weapon damage. Defenses subtracted from IWD is 9-8=+1 DM. So, instead of 11-8 (armor)-4 (range) = -1, miss, you'd have 11+1 (armor) -4 (range) = 8 = 125% damage!
But for the particle barbette, which does only 4d6, defense DM becomes 4-8=-4, so 9-4 (defense)-4 (range) = 1 = 0% and thus no damage. If we give the particle barbettes the same roll (11) for sake of argument, you end up with a 3, which does 10% damage.
So, final damage (assuming that both attacks were 11s) would be as follows:
4 heavy particle beam bays at 9d6 and 125% damage = 45 damage split between two sections.
16 particle barbettes at 4d6 and 10% damage = 6 damage split between two sections.
And that is why that one little sentence is so important and (IMO) makes or breaks the barrage damage rules.
(takes breath)
