I have no idea why I did this. But I was thinking, probably because of my comments in my firearms-related Violent Resolution column.

But . . . I wondered to myself if there was a way to turn some sort of real-world number into D&D damage output.

I know, I know. Why would I ever do such a thing? I had noted (complained, really) that a 9mm was 2d6, and the mighty .50BMG was but 2d12.

So . . . I whipped out solver, and it turns out if you use the energy of the bullet, and only the energy of the bullet, if you use 4 * Log (Base 5) Energy you get a number that might just equate to the maximum damage you can roll on the dice. It compresses the scale even further than the usual result, but it’s not insane.

# Examples?

Cartridge Name |
D&D Damage? |
Roughly |

.22LR | 12 | 2d6 |

.380 ACP | 13 | 2d6+1 |

4.6x30mm PDW | 15 | 2d6+3 |

.45 ACP | 15 | 2d6+3 |

5.7x28mm | 16 | 2d8 |

.40S&W | 16 | 2d8 |

124gr 9x19mm | 16 | 2d8 |

.45GAP | 16 | 2d8 |

180gr 10mm Auto | 16 | 2d8 |

5.45x39mm | 18 | 2d8+2 |

240gr .44M | 18 | 2d8+2 |

.50 AE | 19 | 2d8+3 |

M855 5.56x45mm | 19 | 2d8+3 |

7.62x39mm | 19 | 2d8+3 |

6.8x43mm SPC | 19 | 2d8+3 |

12 Gauge Shotgun Slug | 20 | 2d10 |

150gr NATO 7.62x51mm | 20 | 2d10 |

.500 S&W | 20 | 2d10 |

.30-06 | 21 | 2d10+1 |

.300 Win Mag | 21 | 2d10+1 |

.338 Lapua Magnum | 22 | 2d10+2 |

.50 BMG | 24 | 2d12 |

14.5x114mm KPV | 26 | 2d12+2 |

120mm M829-A1 | 39 | 6d6+3 |

16″ Naval gun | 49 | 8d6+1 |

** Show the Work**

How did I do it?

I tried to make a .22LR 8 points (2d4), a 9mm 12 points (2d6), and a .50BMG 24 points. I used a formula to set a quantity of D = A * logB(Energy). I squared the difference and normalized it to the target squared . . . so (D-T)^2 / T^2. I also weighted the results, so the .22LR got 1000x the figured sum, the 9mm got 4000x, and the BMG got 9000x. That was to force Solver (in Excel) to give more weight to making the .50BMG 2d12 or 24 points. The energies I used were 130J for the .22LR, 585J for the 9mm, and 14,700 for the .50BMG, which assumes a man-portable 32″ barrel instead of the 43″ bbl on the machinegun (which is about 16,000J).

Solver gave an exact figure of A = 3.88 and B of 5.1. But setting A=4 and B= 5 is actually better at fitting the BMG, and puts the .22LR at the 2d6 value above. Given the energy involved, that’s probably as good as the d20 modern values.

When converting max damage to dice, I always use the largest dice I can, but don’t allow subtraction. So 19 points isn’t 2d10-1, but rather 2d8+3. That’s a quirk of mine. You can certainly convert any way you like, and 39 points could be 4d8+5, 4d10-1, or 3d12+3 easily enough. Heck, have at it and make it 9d4+3, and the 16″ Naval Gun 12d4+1 to keep the minimum damage high.

Note that the Naval Gun is just the kinetic energy. I haven’t yet figured out how to rate the explosion of 150 lbs. of high explosive inside about 2,000lbs of metal.

** Bah! The Damages are Too High!**

A comment on G+ noted that 3e humans only have 4 HP, which is a fair point. If you wanted purposefully lower numbers, then here are some nudges/hacks, as well as my line of thought.

I based them off of d20 Modern’s list, where a 9mm was 2d6 and a .50BMG was 2d12. The math forced the 9mm to 2d8 and put the .22LR, which I tried to make about 2d4, into 2d6.

In 5e, at least, a 1st level fighter is going to start with at least 10 HP, and you get a DEX bonus to the 1d6 or 1d8 base damage of a short or longbow, respectively. So from that perspective, 2d6 (ish) or 2d8 for a pistol is the equivalent, on the average of 1d6+3.5 and 1d8+4.5 for damage, neither of which is out of line for d20 Modern or 5e, at least.

If you lower the values to make them work for low level characters, you have the opposite problem – a high level fighter can shrug off a burst of .50BMG unless you invoke the harshest of harsh wounds rule where if you take more HP than your CON, you save or die (that’s a suggested threshold – the harshest one – from the d20 Modern SRD).

If you force the .22LR down closer to a shortbow, the formula becomes something like 2*log(base4) Energy. That makes a .22LR 1d6+1, a 9mm about 1d8+1, a 5.56mm 1d10+1, 7.62mmNATO 1d12, and a .50BMG 1d12+2.

This gives fewer categories of damage

- 1d6+1 for .22LR
- 1d8 for .380 ACP
- 1d8+1 for PDW rounds and all normal military pistols (.45 ACP, 9mm, 10mm, .40S&W)
- 1d10 for magnum pistols (.357M, .44M) and lower-powered assault rifles (4.73x33mmCLS, 5.45x39mm)
- 1d10+1 for standard military assault rifles (5.56, 6.8SPC, 7.62x39mm, 6.5 Grendel)
- 1d12 for battle rifles and sniper rifles from .308 to .338 Win Mag
- 1d12+1 for .338 Lapua or .416 Rigby
- 1d12+2 for .50BMG

Do you have the Weapons Locker splatbook for d20 Modern? Might be worth a perusal for comparison.

Also, what kind of Feats'll let me use the 16" cannon? Is that an Exotic weapon? 🙂

You are reminding me that I have d20M posts of my own to finish. Bah, so many words to slap onto the 'net, so little time…

I don't – my "data" was only from the SRD online, which has a decent selection of weapons.

I quite like your revised weapon damages and at the granularity level of D&D or Pathfinder , its got decent verisimilitude. I'm biased a bit since it kind of matches my own assumptions, small pistol 1d6 big pistol 1d8 with larger numbers for other guns

Testing his we'll take a L1 commoner. Commoners in pathfinder get d6 hit points an average of 3.5, round to 4 hit points for 1st level types so many people will survive being shot.

Take this commoner, say a poor guy getting robbed at a convenience store. He gets shot with a robbers .380 which gives him 5 points and drops him to -1 hit point.

He is dying and has a 45% chance to stabilize. Its a bit high but if you fudge it , eh not bad.

Say the bad guy uses a shotgun. That's not listed but I'll go with a d12. That's an average of 7 points, enough to drop t the commoner to -3 This drops the commoner to a 25% chance per round to stabilize.

Grainy but workable.

At higher levels its not as good but I don't play 3.5 or Pathfinder past L6 or 8 (low powered Supers really) at the most since the Epic 6 variant seems to balance realism and the D&D experience pretty well.

What I find amusing is a mid level fighter can stand in front of a 16" naval gun, take the hit and then continue fighting. 😛

A * logB(E) is a weird formula, since it can be rewritten as A/log(B) * log(E); since both A and B are arbitrary constants, you can rewrite the whole thing as K * log(E). A = 4, B = 5 is equivalent to 5.72 * log10(E); 130J is thus 12.09, 14,700 is 23.84. If you use A = 3.88, B = 5.1, the constant is instead 5.48.

Incidentally, there's a certain logic to using the Strength table — if 1J = 1 lb heavy load, 130J is equivalent to ST 12 (+1), 14,700J is equivalent to ST 49 (+19). That obviously fails to match d20 weapon damage, but it is consistent with a different part of the rules.

Interesting take on the problem of scaling firearm damage in a fantasy setting. Having some experience with naval guns, and having a good friend who was the Turret Two officer on the Iowa, you're damage is way too low. One does not catch a flying Volkswagen and walk away from it, even before the HE charge goes off.

I tried my hand at this last year using KE to scale the damage, http://alesmiter.blogspot.com/2014/03/firearms-in-fantasy-campaigns-iii-damage.html – it didn't satisfy me either.

This is what I call the "pinning" problem, in that my answer depends on what I think a few of my data points need to be. If I look at GURPS, which uses SQRT(KE) effectively to determine penetration, if our .22LR is 1, a 9mm is 2.1, an M16 is 3.7, a 7.62 is 5.1, our .50 BMG is 11.4, and the 16" Naval gun is 1658. From that perspective, you could put the big guns at 1000d6 or more and be happy, but that makes our rifles about 2d10 for the M16 and 3d10 for the battle rifle. That's going to turn low and high level characters into paste a lot.

I chose to take as correct the notion that 2d12 was the game-determined place where the .50 sat, and a pistol should be about the same as a longbow. That seriously compresses the upper end.

Interesting article!

I also had a go at this some years ago.

I was more interested in heavy weaponry so decided to work off armor penetration (and this also let me convert directly from GURPS (where dice x 1.27 = armor penetration in mm).

For d20 I determined average damage = 5mm x (cube root of penetration in mm of RHA)

7.62mm NATO = 8mm RHA penetration = 10 points = 2d8+1 damage

.50 BMG = 16mm of RHA penetration = 12.66 points = 2d12 damage

120mm tank gun = 600mm of RHA penetration = 12d6 damage.

With some help from various fellow Hellions ( Hans, Shawn, Kenneth) I statted up much of the NATO and Russian arsenal that way also using the same formula to assign Damage Reduction to vehicles.

It saw print as d20 Military Vehicles many years ago, and I'm still kind of fond of it.

For those of us without the calculator to do logs in base 5…am I correct in my conversion to (log(energy)/log (5))×4?

Yes, 4 x log5(KE) = 4/log10(5) * log10(KE).

Because it's all about that base, bout that base . . .

Ok so the next step…what modifications for automatic and burst fire? Making the damage be for one bullet seems a bit unreasonable. But giving a burst 3 weapon 3 times the damage seems a bit OP.