Thot

A bleeding rule is easily added, though. Well, at least it seems agreed that there is room for improvement.

So, here is a list of the weapons I have looked up and damage as computed per the old formula and as per a new one, which reads: =MAX(((LOG(Joules;1,225))-26);1). Name Dmg J Avrg Dam J/AvrgDmg Old formula New Formula 5.56×45mm NATO (e.g. G36, SA80, FAMAS, M16, AR15) 2D6 1.796 11,0 163,3 8,4 10,9 7.62×51mm_NATO (e.g. G3, L1A1 SLR, M17, M14) 2D6+4 3.506 11,0 318,7 11,7 14,2 9×19mm_Parabellum (e.g. Walther P1, Uzi, HK MP7) 1D8 678 4,5 150,7 3,6 6,1 7.62×39mm (e.g. AK-47) 2D6+2 2.179 9,0 242,1 9,4 11,9 Pistol, Derringer (.41 Short) 1D6 71 3,5 20,3 -7,5 1,0 Pistol, Flintlock 1D6+1 434 4,5 96,5 1,4 3,9 .45 ACP 1D10+2 835 7,5 111,3 4,6 7,1 .38 Short Colt (light pistols and revolvers) 1D6 245 3,5 70,0 -1,4 1,1 Revolver, Heavy (.50 Action Express) 1D10+2 2.200 7,5 293,3 9,4 11,9 Rifle, Bolt-action (7.62×63mm) 2D6+4 4.042 11,0 367,5 12,4 14,9 Rifle, Elephant (130g, 430 m/s) 3D6+4 12.019 15,5 775,4 17,8 20,3 1861 Springfield Rifle-Musket, 33 gram bullet, 290 m/s v 1D10+4 1.367 9,5 143,9 7,1 9,6 Rifle, Sniper (12.7×99mm NATO) 2D10+4 20.195 15,0 1.346,3 20,3 22,8 Rheinmetall Rh 202 (Damage derived from new formula) 5D10+2 81.070 29,5 2.748,1 27,2 29,7 Bofors 40 mm L/70 (Damage derived from new formula) 7D10 461.492 38,5 11.986,8 35,7 38,3 76mm Bofors naval ship gun (Damage derived from new formula) 7D12+1 2.637.259 46,5 526,6 44,3 46,9 120 mm Tank Gun 15D6 13.000.000 52,5 3.505,4 52,2 54,7 Looks fine for me. The lighter weapons come out way smaller, but then again, they should be.

Which is good, because we want the PC's to be able to make a difference. Again, this is extremely dependent on the country and area they attack first. Hm, If such magic is part of the invaders' arsenal, I'd want the player characters to be able to be immune from it. Little is more frustrating than the GM taking over a PC. BRP's standard magic doesn't allow for such complete control. Of course, there may be people who actually want to join the Evil Empire for selfish reasons, or out of fear for loved ones. Well, that is always true for all campaigns, isn't it. But discussing the ideas you guys have helps me not overlooking things that the players would possibly see and maybe even find annoying. Has anyone figured out how to format these remnants of deleted quoted text?

Or throw a dart!

Don't worry, rigorous critique is the best way to improve. Force is ma, not mv. But in the end, the two are equivalent. You have to find some way of translating either to damage, and while those equations will look different, in the end they will be equivalent, too (in a sense). Well, I am collecting the data for various rounds from all around the net, but Wikipedia is a very encompassing source, yes. You can get more detailed info from each individual cartridge's wiki page, though, which sometimes includes measured steel penetration. I am not at all done with the spreadsheet, but the fact that a Derringer's .41 Short round has a muzzle energy of a mere 71 Joules is worth mentioning. For comparison: A punch with a fist will reportedly be around 100 Joules for a normal human... granted, that energy is but into effect on a much larger surface, so penetration will not be as deep, but the Derringer seems to have been a particularly unimpressive weapon, unless you hit the eye. I mean, seriously, use a knife instead!

Interesting. Thanks! .22 Short is 110-120 Joules, which is below the range the present formula can cover. Damage is given 1D6 or 3.5 average. .45 Automatic is 500 to 835 Joules, depending on exact cartridge. Damage is given as 1D10+2, or 7.5 average. AK-47 is 7.62×39mm, 2000 to 2200 Joules. Damage is given ans 2D6+2. .30-06 is 3800 to 4000 Joules. Damage is given as 2D6+4. I'll make a more complete spreadsheet, fiddle with the numbers a bit more and report back.

Hm. A smartphone video or ten or a hundred, uploaded to the Net during the initial attack might draw some attention from the country's authorities rather quickly if contact to the local authorities is lost. If we assume the government in question not to be the US, the availability of firearms will be much more limited. But the biggest issue for the invaders, I believe, would be to wrap their minds around all the things they see, from 50-story buildings, cars, smartphones, and airplanes to tanks that crush them from three kilometers distance, and let us not forget howitzers and self-propelled guns... a PzH 2000 has a range of 50-60 km, after all. So the battles will begin without the invaders even seeing any opponents. Even some people who grew up in today's first world have trouble wrapping their minds around that kind of situation. That is just something not even seasoned enemy commanders will be used to, and when they get used to it, it may be too late for the initial invasion force. Some survivors will, of course, escape and report back... but the next wave will have to be much more careful, maybe even resort to infiltration.

After our other discussion, I would probably place the light revolver more in the .32 or .38 range, so definitely less than 1000, more like 250 J. For the medium, I used one of the heavier .45 Colt rounds: https://en.wikipedia.org/wiki/.45_Colt But of course, there's always a range here, so it is not so easy to assign which is which.

The problem is that most weapons in the BGB table cannot be easily identified (or at all) with any certainty. So we could probably both find weapons that fit or don't fit with the formula, but you are right that the formula fails for the really small guns like a Derringer or other .38 calibre weapons.

Of course they didn't. The formula is just an approximation, of course. It does work for the weapons I used for playing with the numbers, and seems like a good approximation for me. I tested it with the assault rifle, various handguns and the tank gun, liked the result, and reported back here. A range from a light pistol to a 120mm tank gun is a wide enough range for me. The force and the energy are equivalent in amount, so I do not see any relevant difference here? Well, nothing is going to be completely the same as the gut-feeling stats. Not even another person's gut feelings.

There was no "standard musket", and especially not a standard load. The musket shown in he BGB may be of a different type, or simply loaded differently when they tested it. Ah, didn't catch that. Thanks for the hint. But... there is no .58 Springfield rifle in the BGB?

From where do you get the notion that those weaker weapons were more powerful than 9mm parabellum? I'd be very interested in reports about the actual penetration of those bullets that you deem more powerful than the energy amount would indicate. I would be inclined to subsume this under http://tvtropes.org/pmwiki/pmwiki.php/Main/KatanasAreJustBetter , if you get my meaning. That said, yup, the formula doesn't work here. The .45 ACP comes out as too light, and for the https://en.wikipedia.org/wiki/.500_S%26W_Magnum which one could also understand as a "heavy revolver" round, it gives too high a value. Hm. Personally, when in doubt, I'd prefer something deriving the stats from the energy over some cinema-induced gut feeling. But yes, some firearms in the table apparently don't follow the formula.

Hm. According to this: http://historum.com/war-military-history/37754-kinetic-energy-ancient-modern-weapons.html "Matchlock Full Musket - 1,943" That is above the level of 5.56mm NATO, and should produce an average of 9 points of average damage. Works fine for me. An elephant gun that shoots a 130 gram bullet at a speed of 430 m/s, which are numbers I get from: https://en.wikipedia.org/wiki/Elephant_gun will do an average of 18 points of damage (damage given by the BGB is 3D6+4, or an average of 14.5, but that may be a smaller gun than I used in my computation now - there were many types of 'elephant guns', apparently). Head hit location hitpoints are one third of total hit points, so an elephant with no more than 54 hitpoints should be a somewhat safe kill - for stubborn elephants, history seems to indicate that many needed more than one shot.

https://en.wikipedia.org/wiki/9×19mm_Parabellum 570 to 680 Joules. (3 to 4 points of average damage according to the formula) https://en.wikipedia.org/wiki/.45_ACP Depending on manufacturer and subtype (of which there are many), 483 to 835 Joules (2 to 5 points of average damage according to the formula) I would reckon that the discrepancy you see stems from comparing different manufacturers' rounds.

Well, for such variances, we do have the damage dice, which allow for a wide range of actual results. The formula in the OP just gives you the average damage. Of course, you could still argue with the details, such as energy vs. shape of the bullet. But do we need it that precise? What we need is a plausible number to distinguish a 5.56 mm from a 7.62 mm cartridge, and a way to figure out what the 40mm ship gun can do.

It does work for all weapons that I looked at, and that includes the "tank gun" given in the BGB. Any one that it doesn't cover?

The BGB sais you should halve any pre-modern armor vs firarms. And while 1D8 isn't much against plate armor (which will protect with 4 against it), a medium pistol isn't a particularly deadly weapon and could conceivably be stopped by full plate half of the time. An assault rifle could not really avoid to penetrate it, though (except under very rare circumstances). You could do that, of course. But that is a level of detail that, at least in my opinion, just isn't worth it in a game.

There basically is one for firearms. See the OP. It does not work for energy weapons, at least not in a way I have found yet, but for firearms, yes, it does. The formula in the OP works for a Derringer, a Colt or a tank gun alike. And that is practical and useful. Yes, of course RPG's are simulators, in that they are models of the way the game world works. They don't need to be very precise in their simulation (so for many weapons and damage types, a guess for damage will do), but they do deliver consistent results that allow for planning and are generally thought to be somewhat mirroring the way things work in the real world. No, the reason is because it is a complicated and difficult undertaking, and each type of damage would require different formulas. It is just easier to go by gut feeling in many cases, while getting just-as-good results. But for firearms, it is more practical to have a formula.

Well, that of course will very much be decisive, won't it. Opening a gate in the middle of the Namib isn't going to do anything, much like gates that open over some Earth ocean. Thinking about that, they will probably have some kind of "best practice" for attacking a new world - opening gates all across the globe, quickly abandoning those that aren't worth it, colonizing uninhabited but fertile areas, attacking any settlements to conquer them. A medieval world of Earth's size (which I would assume all the available worlds out there to be) should have a population of no more than 100 million, and by the usual 2%-rule-of-thumb, the whole world would then not have more than about 2 million warriors, scattered across the continents. Most worlds out there will have some kind of magic (including travel magic of various kinds), so we can safely assume that the invaders will have to face much of those 2 million warriors pretty quick, and will be used to it. The evil empire has the element of surprise, so they wouldn't need 3 times as many soldiers for an attack, but just to be on the safe side, an initial invasion force would need to be about 3 million strong, I would say. Unbeatable for a medieval world, but a big 21st century country like Russia or the US could take on that number of attackers on their own even at equal levels of technology. The attackers would probably try to take the first thing they perceive as a major population center (so any town with more than 5000 inhabitants...) with all their might, to be quick and fortify their positions for holding their new bridgehead and garrison it with a fraction of their force, while the vast majority of their army marches on to conquer and occupy more. Native troops will be force-recruited along the way, either by extortion or magic (if anything fails, as undead). In the case of Earth, there are many possible scenarios how this plays out. They could start at a midwestern US town, a small city in the middle of Africa, or a Chinese metropolis, or anything in between. Their first contact with defenders of Earth will not be with military, but with police in riot gear, supported by water gun police cars, equipped with transparent shields, night sticks and possibly handguns. And one-on-one, those would probably already be more than a match, at least until the dragons, mages and undead move in, but the police will of course be hopelessly outnumbered and will probably have to retreat quickly. In some countries, the military will be available immediately, in others, further attempts to get rid of this plague with police forces will occur, costing valuable time during which the invaders try to figure out any captured firearms. Regular citizens will probably flee, but should some of them decide o fight... Honestly, that I don't believe. An angry man with a car will be an "effective military" to them, worth 10 mounted knights at least. Yeah, that would be boring, so it won't happen. To use it, I would guess pretty quickly, within hours. To build an maintain it might prove more difficult, unless they can capture enough natives that can be forced to do it for them... with all the negative side effects such an approach would have (sabotage, intentional neglect, escaping mechanics who are supposed to repair a car, but steal it instead...). A steady supply of ammunition would not be easy to come by, so any captured firearms will only have limited uses. They might even mistake firearms for magic item and wonder why they stop working after a few dozen shots.

Hm. Energy weapons would work completely differently, I would reckon. We don't have any real-world examples to compare the BGB's sample energy weapons with, so we'd have to use fire as a reference. A small candle fire is supposed to do 1 point of damage per turn (which is 12 seconds). Edit: In the following I try something that in the end doesn't work. So don't spend too much effort on trying to work with this. According to http://www.thenakedscientists.com/HTML/questions/question/1857/ , a candlelight has a power of around 80 watts (joules per second), so 1 point of damage across 12 seconds apparently equals 960 Joules. We can simplify this to 1000 joules, but as joules to damage isn't a linear conversion, that doesn't help us much. A torch supposedly does 1D6 (3.5 average) of heat damage per turn, a large bonfire 1D6+2 (5.5 average), molten lava or a rocket 3D6 (10.5 average). A torch might be worth, well, about 30 times the volume of a candlelight. That's a very rough estimate, but that would mean increasing energy by a factor of 30 increases damage by a factor of about 3.5. Adding one point of damage then should then require an increase in energy by about a factor of (very roughly) 5.5. (Lots of very, very, very rough estimates here!) Thus, the spreadsheet formula would be (average damage) = 1+ LOG((Energy/1000); 5.5) A pool of lava would would then inflict just short of 4.6 billion joules on a human body over a time of 12 seconds. Lava has a temperature of about 700 to 1200 degrees Celsius. Inflicting that kind of temperature on a (very muscular or very fat, but in either case more easily computable) human body of 100 kg at 37 degrees Celsius would require (energy required to heat a gram by 1 degree) x (number of degrees) x (number of grams). Now, the amount of energy needed to heat given piece of matter by one degree increases with the temperature of that piece of matter, but we'll just ignore that for simplicity. Just over 4 Joules are needed to heat one gram of water by 1 degree at human temperature ranges, so the above formula at the lower end of the range for lava temperature would be: 4x700x100.000=280.000.000 or 280 million joules per second, times 12 for a full combat round is 3,36 billion Joules. But a ton of TNT contains about 4.1 gigajoules. So the simple logarithmic formula doesn't work here, I would say.

Looking at the various weapons in the BGB, I wondered what the formula behind damage is. Obviously, it is some sort of logarithmic scale, but how can it possibly be computed for new weapons? After having played around with the available numbers a bit, the following seems to work: Starting at 400 J, which I put equivalent to 1 point of average damage, each additional point of damage requires a multiplication of the muzzle energy by 1.225. So 400 for the first point of damage, a total of 490 Joules for 2 points of damage, 600 J for three, and so on. (Those numbers 1.225 and 400 are sort of arbitrarily chosen via trial-and-error, but should be close enough.) As a formula, that is (average damage)=1+ Logarithm [to the base of 1.225] of (400-number of joules). Or, in a LibreOffice spreadsheet: 1+(LOG(A1/400;1,225)) ... where "A1" is the field in which you put the projectile's muzzle energy. That formula will give you average damage - maximum damage will probably be 30-50% more than that. Translating this average damage number into damage dice is probably not done via a formula, but I guess selecting a die type (d4, d6, d8, d10 or d12), dividing by the expectation value for that die and keeping any rest as a bonus to the damage dice should do the trick. Now, projectile muzzle energies you can find all over the Internet for a given type of ammo, but if you don't find it or want to use a fictional weapon, you could simply do the E=0.5mxv², equation, which, if you set m as the mass in kilograms and v as the velocity in m/s, gives you the correct joule value.

Just for the fun of it: The German WW2 standard rifle ammunition was 7,92 × 57 mm with a muzzle energy of 3600 to 4100 joules. Using the above assault rifle as a reference, that would give it a max damage of about 28 (!) and a damage of about 18, or in other words, something like 4D6+4. Obviously, at some point, the linear joules-to-damage equation breaks. Or the Germans won WW2. Your choice. A tank gun (using the 120mm smooth bore cannon from Rheinmetall which is used in the German Leopard MBT as a reference) has a muzzle energy of around 13,000,000 joules (13MJ). As that equates to 15 dice of damage, and using the above weapons as a reference: 1,800 J -> 2D6+2 (average 9, max 14) 13,880 J -> 2D10+4 (average 15, max 24) 13,000,000 J -> 15D6 (average 52.5, max 90) I am sure the designers just handwaved the numbers a bit to get approximately plausible values, but I wonder if we could find a progression that works...

Wikipedia, actually. I took some standard calibres and looked the muzzle energy up. I had to make guesses for the calibre, of course: 5.56mm NATO for the assault rifle, 14mm for the sniper rifle, .44 for the Heavy pistol, etc. (Some had just muzzle velocity and bullet mass, but that's then just (energy in joules)=1/2*(mass in kg)*(muzzle velocity in m/s)² ) With that and the above, one can, for example, conclude that the Soviet AK-47 with its 7.62×39mm cartridge and a muzzle energy of roughly 2100J (depending on cartridge type) should have a single-shot max damage of around 16 and an average damage of around 10.5, so something like 2D6+4 should do it for that weapon/cartridge.

Hm, I think I'll give it a try with just magic and sorcery, and should the need arise, I will have to add other magic systems on top of that.

Well, with such a vast multidimensional empire, there will of course be smart and not so smart leaders. I mean, we want the PC's to have to do something meaningful, after all. Well, thing is: In the setup, the evil empire's base assumption when opening the gates (which I imagine as mobile, summonable-with-a-simple-spell portals) will be that this is just another world to conquer easily. Which means they initially don't plan to spend all of their empire's military might for such a conquest. And that means they will have other conquests running elsewhere which bind troops, at least in the beginning, so we have an escalation chain for the military involvement of the evil empire on Earth: Standard invasion force led by young and inexperienced leader gets shredded to pieces, then a bigger force led by a veteran commander comes, who might actually be willing to report back home how and why his army might not be able to pull this off, and only then we'd see the massive masses that the empire could theoretically field. Somewhere during step 2 (which may have several substages) at the latest, I would expect some players to learn magic... to open portals, become invisible, teleport, etc. But that assumes that the players (and any powers on Earth who support them!) quickly understand the evil empire's power structure, overall goals and vastness. A big element in military SF is to learn who your enemy actually is, what they want, and why this war is happening in the first place.