Structural HP and AC

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Esa the Wanderer
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Joined: 2015-01-28 11:15
Structural HP and AC

I've been trying to build some conversion of sea vessels from GURPS to ACKS in addition to playing around with GURPS (3rd ed.): Vehicles and the construction rules in GURPS (4th ed.): Low Tech Companion 3. What I noticed is that the SHP of vessels in ACKS seem to be roughly linear with the length of the vessel, but the SHP of fortifications is closer to the volume of the structure. (To complicate things, GURPS 3rd ed. HP scaled with the surface area!) In addition the raft SHP scale with their area.

Which one should be used as a general rule, if for example creating magical flying buildings?

With the possibility of calculating siege weapon damage with the formula on Guns of War it's possible to compare penetration and damage values between systems (and to any real world data) and benchmark these.

Note: Vessels have around 1 SHP per foot of length (and about ½ of GURPS 4th ed. HP) and stone fortifications have around 1 SHP per 12 cubic feet of wall (or ton of mass).

Also AC of both types is based on the material (and possible angling/rounding bonus), and not modified by the size of the construct (unlike monsters). The increased thickness of ships of war and the English naval ship rating may have been the original source of the Gygax/Arneson decreasing AC, but in ACKS this is not the case. Here SHP are the measure of thickness (linear scaling) and thicker walled ships should have more SHP.

The different treatment of SHP could be thought to represent two different things: Enough damage to breach the structure (to sink a ship or allow egress into a fortress) vs. enough damage to completely demolish the structure. Reading some historical anecdotes (no real data or statistics) I was surprised how many hits ships could survive without sinking.

To have an idea on the the amount of SHP for ships of the line, I'll have to calculate some real world gun penetration vs. hull thickness data and compare to ACKS gun damage. I'm suspecting that getting the ships' endurance in line with the size of the guns carried (and required to breach them) will lead to either increasing their SHP, requiring more breaches to sink larger vessels or taking hull thickness into account by increasing AC. Going backwards into increasing ACs for ships would incidentally eliminate the need for a structural saving throw for them. I'll have to think about that when I look into streamlining firing broadsides of tens of guns in naval battles.

As an example "Sovereign of the Seas", a 1st rate ship of the line is 170' long (ca. 170 SHP) has 20 "cannon of 7" (36-pdr cannon), 28 culverin (18-pdr gun) and 54 demi-culverin (8-pdr gun). A broadside will do on average around 1800 SHP damage if all shots hit, 800 SHP for a point blank shot with 0-level gunnery crews (assuming ship profile is larger than scatter and either AC vs. to-hit or Structural save). So two similar ships firing point-blank broadsides at each other will either sink or breach each other more than 4 times. To give chance at a 1-round sinking, but having the ships survive more than one broadside on average to keep things playable (and more interesting) would require one of these:

  1. Multiply SHP by 10 (basing them on volume/mass may give similar result, as the ships tonnage [not actual structural mass!] is 1500)
  2. Divide damage to SHP by 10 like mechanical artillery
  3. Increase AC to 9 vs. to-hit
  4. Increase AC to 18 vs. structural save

As a 1st rater it's estimated hull thickness is 28" (possibly lower as the rating/thickness table is for a later century), compared to 2"-5" of boats, civilian ships and viking vessels.

Anyways, something needs to be made more clear or changed, but more data is needed to know how much. What to change is then another question.

Alex
The Autarch
Joined: 2011-06-30 18:10

Hello, Esa!

First off, let me say that the SHP rules of ACKS are not as coherent as other aspects of the rules. I focused very heavily on making the economics coherent, but I didn't make as much of an effort in making hit points, structural hit points, and damage entirely coherent - something I regretted heavily as I delved into D@W and then Guns at War.

With that caveat, let's address some of your questions:

  • Stone structures typically have around 1 shp per 12 cubic feet, as you surmised. 
  • Wood structures typically have around 1 shp per 30 cubic feet (it ranges from 29 to 32, with one gross exception).
  • Ships are modeled as triangular prisms of three rectangular side, each with a length, height, and thickness. Volume of each side is computed, then they are summed to give a total volume.
    • Large Gallery - 150' length x 20' height' x 0.33' thickness; volume 990 cf/side; total volume 2,970 cf; 95 SHP; 1 shp per 31.3 cf
    • Small Galley - 100' length x 15' height x 0.33' thickness; volume 495 cf/side; total volume 1,485 cf;  75 SHP; 1 shp per 19.8 cf
    • Large Ship - 150' length x 30' height x 0.33' thickness; volume 1,485 cf/side; total volume 4,455 cf; 150 SHP; 1 shp per 29.7 cf
    • Small Ship - 80' length x 30' height x 0.33' thickness; volume 792 cf/side; total volume 2,376 cf 75 SHP; 1 shp per 31.7 cf
    • Large Warship - 150' length x 30' height x 0.4' thickness; volume 1,800 cf/side; total volume 5,400 cf; 175 SHP; 1 shp per 30.9 cf
    • Small Warship - 80' length x 30' height x 0.4' thickness; volume 960 cf/side; total volume 2,880 cf; 90 SHP; 1 shp per 32 cf
  • Wooden rampart is modeled as 100' long x 10' high x 0.083' thickness, or 83 cubic feet; 1 shp per 30 cf yields 2.8 shp, rounds to 3.
  • Gunpowder artillery deals average shp damage equal to the square root of the weapon's kinetic energy in Joules / 10.
    • A cannon with 250,000 joules of muzzle-energy would deal (250,000 ^ 0.5) / 10 = 50 shp of damage, or about 10d10. 
    • Gunpowder artillery which does *not* hit a structure has a linear area of effect based on the size of the cannon, while gunpowder which does hit a structure has a small area of effect of fragments.
  • Ancient artillery is, uh, a little more complex. Find the weapon's kinetic energy in Joules. It will deal approximately 1 hit point of damage to a single "target area" of 19 square feet per 100 joules. But damage is semi-arbitrarily spread across multiple target areas, to give an area of effect. (A single human target occupies a 19 sf target area.)
    • A light catapult capable of fling a 2.7kg (6lb) stone to 300 meters has 4,000 joules of kinetic energy. Therefore it will deal 40 hit points of damage to a single target area of 19 square feet. We semi-arbitrarily decide to spread this across 4 target areas, each of which will suffer 10 hit points of damage. An average of 10 hit points of damage translates to 3d6 points of damage (averaging 10). 4 target areas of 19 sf yields 76 sf. The catapult's area of effect is therefore set at a 5' radius (5 x 5 x 3.4 = 78.5 feet).
    • Ancient artillery then deals structural hit points equal to its damage dice, disregarding area of effect, and dividing by 10 for stone structures.
  • The  underlying assumption is that gunpowder artillery directs most of its kinetic energy into to the structure, while ancient artillery wastes much of its kinetic energy pulvering its own stone ball against the structure and causing area of effect damage around it. 
    • To model the fact that cannon balls would sometimes bounce off or ricochet from structures, the structural saving throw rules were added. In their absence, structures fell too quickly to gunpowder weapons, and ships fell MUCH too quickly. 

Hope that helps!

 

The Dark
Sinister Stone of Sakkara BackerLairs And Encounters Backer
Joined: 2013-07-05 19:55

Esa,

Sovereign of the Seas is actually a bit larger than stated. 170' was the length of her gundeck, but length overall was 232'.

My house rules (which I really need to finish at some point) use tonnage in Builders' Old Measurement as shp. For Sovereign, with a length of 232', beam of 48' and draught of 23.5', BOM is (232-(0.6*48)*48*23.5)/94, or 2438 shp. Barring some sort of critical hit, this ship would survive a few broadsides.

Magus7a
Joined: 2017-01-20 08:13

Interesting houserule, though I'd note that if you do this with the ships in the core book, they become 3.5-4.5x as hard to kill

The Dark
Sinister Stone of Sakkara BackerLairs And Encounters Backer
Joined: 2013-07-05 19:55

Interesting houserule, though I'd note that if you do this with the ships in the core book, they become 3.5-4.5x as hard to kill


-Magus7a
The differences become more extreme - a small galley actually drops to 40 shp while a large galley goes up to 147 and a large warship up to 421. Honestly, though, part of the reason I'm good with that is that until the age of cannon, artillery very rarely sank ships.

Esa the Wanderer
Patreon Supporter
Joined: 2015-01-28 11:15

Thanks for the info!

I'll probably set wood SHP to 1/30 cf and calculate the structural volume of the vessels to see how it looks with more examples.

The formula for kinetic damage is also useful, I'll compare it to other formula (game and real world). I already reverse engineered the black-powder formulas for caliber and shot mass. I'll play around with BTRC 3G and some historical examples to see if I can come up with some adjustments for early (medieval) and late (18th century) gunnery.

I'll also see if I can benchmark the penetration vs. general destruction to kinetic energy on various structures.

Esa the Wanderer
Patreon Supporter
Joined: 2015-01-28 11:15

Esa,

Sovereign of the Seas is actually a bit larger than stated. 170' was the length of her gundeck, but length overall was 232'.

My house rules (which I really need to finish at some point) use tonnage in Builders' Old Measurement as shp. For Sovereign, with a length of 232', beam of 48' and draught of 23.5', BOM is (232-(0.6*48)*48*23.5)/94, or 2438 shp. Barring some sort of critical hit, this ship would survive a few broadsides.


-The Dark

Hi!

The measurements of the ship vary somewhat depending on the source and which measurments were given. threedecks.org & Wikipedia give them as gundeck 168, beam 46.5 (or breadth 48.3) and depth of hold 19.3, which give a minimum for the displacement which in total is bigger. Burthen tons are 1522 or 1683 (bm tons 1650). With these minimums a prism area is around 11 000 sf, and with the bigger measurements 15 000 sf. With 4" thickness that would be 120/170 SHP. If it was 28" thick all around (highly unlikely), the values would have been 870/1170 SHP. These would be enough to resist the point-blank broadside, at least for a round.I need to find out the total volume of scantlings to estimate the average thickness (and have a ready value for the structural volume) and add that to the plank thickness.

Alex
The Autarch
Joined: 2011-06-30 18:10

I did a deep dive into ballistic weapon damage while researching Guns of War, as it was tangential to some other projects I was working on for ACKS post-apocalypse and ACKS cyberpunk.

It turned out to be both robustly studied and yet also impossibly complex to model in a tabletop game. As far as I have been able to gather, there are four basic types of penetration, and each of them has to be modeled differently if one wants to be scientific:

  • Fracture due to inertial stress wave Compressive waves propagate into the plate upon impact. If the stress magnitude of this wave exceeds the dynamic yield strength of the target, failure may occur in an unconfined region of the target plate. For a plate target, failure or fracture would occur near the rear target surface. The probability of this type of failure decreases with an increase in target density, hardness or compressive yield or ultimate strength
  • Radial fracture behind initial wave front Tensile radial stresses are built up as the compressive wave propagates away from the impact sight. If the target material behavior is tensile and the magnitude of the built up stresses are higher than the ultimate dynamic tensile strength, radial and/or circumferential cracks may occur. The hoop or circumferential stresses will be tensile because of the Poisson’s effect, as the compressive wave propagates outward. Radial cracks are caused by this circumferential tensile stress.
  • Spallation The compressive waves reflect from the rear surface as tensile waves. First, the tensile wave cancels the compressive wave. As the compressive wave propagates to the back of the plate, the amplitude of the compressive wave decays, than the net tensile stress may exceed the ultimate dynamic tensile strength of the target material. In this case, a tensile fracture will occur.
  • Plugging This type of failure occurs when the projectile pushes a plate plug through the rear surface of the plate. This plug has approximately an equal radius to that of the deformed projectile. As the hardness of the plate is increased (related to the yield strength or the hardness number), the tendency for plugging increases. The reason is that, it becomes harder for the plate material to be pushed radially outward by the projectile. Thus a narrow shear zone builds up in front of the projectile in the periphery region and the plastic flow is confined to this region. Other parameters that effect the formation of plugging are the relative plate thickness and projectile nose shape. Plugging occurs more easily in thinner plates such that even softer plates may fail in this in case that the impact velocity is not sufficiently close to the ballistic limit so that radial momentum transfer causes severe plate bending.

I decided to use "plugging" as the model I'd work from, as it is the most likely outcome for both thick,hard plates and for thin, soft materials.  

In plugging, the material in front of the projectile from the face to the back fails IN ONE PIECE by being force out of the projectile's path out the plate back, either as a cork-like plug or by being torn in the middle and folded back to the sides as thick petals, all as one solid piece of material the total plate thickness deep. This can be approximated by a flat-nosed projectile slicing out a disk of armor like a cookie cutter, but it also occurs in moderately-thick plates hit by pointed projectiles when the resistance is primarily by tearing open in the center over the tip of the point and wedging the hole open by petal-formation, that is, by pushing the plate material forward as it splits open in the center and then having it bend to the sides, since it is still attached to the armor ringing the impact site.

  • For the plate thickness T, the plug will not move at all (it is assumed to be incompressible) until the applied force is greater than Fr, the resistance force, and thus allows the entire plug to move as one piece.
  • Fr, the resistance force is friction resistance along the sides of the plug, like a cork in the bottle. 
  • Assume that, like a cork in the bottle, the friction resistance along sides of the plug to be pushed out is constant everywhere (reasonable enough), so that the resistance force Fr = KrT for some constant Kr. (The thicker the plug, the  more resistant force it exerts.)
  • The Kinetic Energy required to push the plug is the amount of distance moved (its thickness T) against a resistance force (KrT), so Work = KrTT = KrT2.
  • The KE available is the projectile's motion energy KE = (0.5)(Projectile Mass M)V2
    • To barely penetrate the plate, they are equal: (0.5)MV2 = KrT2
  • Since damage is effectively depth of penetration T, then it is proportional to the square root of Kinetic Energy. 

Another similar analysis can be found here - http://forums.sjgames.com/showthread.php?p=135955. I quote:

"It depends on what sort of material is being penetrated. This gets complicated, but the short answer is that

  • For materials that respond elastically, the depth of penetration is roughly proportional to the square root of the energy.
  • For non-elastic materials that can simply be "crushed" out of the way, penetration is proportional to the energy.
  • For materials that primarily interact with a projectile though viscosity, penetration is proportional to the momentum
  • For materials that are non-viscous fluids penetration is formally infinite, but the energy and velocity of the projectile decrease exponentially with distance until you reach a velocity regime where non-hydrodynamic forces dominate.

Things used for armor tend to be elastic solids. This means when exposed to a force, they will deform to distribute the strain of the force over a significant part of the volume of the solid. The thicker it is, the more the "layers" behind the first "layer" can buttress the first layer, helping it to resist deformation and failure (there are not really layers, since the material is uniform).

For this reason, the GURPS formulas for damage where damage is roughly proportional to the [square root of the] kinetic energy and DR is roughly proportional to armor thickness tends to reflect real life trends.

Now, things get more complicated when the armor gets thicker than the typical radius of elastic response, or when the projectile is travelling supersonically in the armor medium, so it simply does not have time to respond elastically (in either of these cases, penetration is going to be roughly linear with energy).

Things also get complicated when the projectile itself can deform - in an extreme case you would treat it as a fluid plume interacting hydrodynamically with the armor medium (shaped charge explosive jets fall into this regime)."

 

Esa the Wanderer
Patreon Supporter
Joined: 2015-01-28 11:15

Hi!

Thanks for the elaboration of the model behind the numbers. The relation of the penetration dept to the square root of kinetic energy seems to come up in different sources and data set, so it seems pretty usable.

How did you come up with the kinetic energies for the black powder weapons? The amount of powder per shot mass in Guns of War seems high. The data for Trafalgar era naval guns I came across gives powder mass as 1/3 of the shot mass and it seems that was cut down to 1/6 for close range shots. The velocities (and energies) calculated from the damage dice feels low too. Is that because of the difference in technology? (late medieval to 17th Century guns vs. late 18the Century to early 19th Century naval guns)

8d10 gun (18-pounder), 44 SHP on average, 193.6 kJ KE, velocity 207 m/s (690 fps)

From "Ship structures under sail and under gunfire" by Fernández-González I found the "table average ideal penetration in oak in centimeter with 1/3 charge" from Le vaisseau de 74 canons. 4 vols." by Jean Boudriot as 120 cm (48") and velocity at 100 m (333') as 400 m/s.

BTRC's Gun's, Gun's, Gun's gives a maximum safe charge of 33% at TL5 and 38% with energies of 1,7/2.7 MJ, muzzle velocities of 470/618 m/s and penetrations in wood as 117/154 (TL5 being 1400 and TL6 1700).

1/3 weight charge of black powder for the 18-pounder is 6 lbs and has ca. 9 MJ (and a muzzle velocity of 1000 m/s). With 10% total efficiency it has a KE of 0.9 MJ, velocity of 316 m/s. 3G would give it's penetration in wood as 123 cm.

I'm not trying to be pedantic, I'm just looking for ACKS benchmarks to compare to other data and models to figure out things.

Alex
The Autarch
Joined: 2011-06-30 18:10

No one ever needs to apologize for being pedantic on the Autarch forums!

The amount of powder used was taken straight from the sources. The era I worked from was 100-200 years earlier than the data you are working from and the firearms, the cannon, and the powder were all more primitive.

The kinetic energies were calculated by reverse-engineering from cannon's range and ball weight.

 

 

The Dark
Sinister Stone of Sakkara BackerLairs And Encounters Backer
Joined: 2013-07-05 19:55

Powder charges changed fairly rapidly. Fifteenth-century guns using mealed powder used a charge of about 15% shot weight (and had barrels only 5 times the caliber of the shot). In the sixteenth century, with corned powder slowing the burn rate and improved powder quality, barrel lengths increased to 14-50 times shot caliber, and powder charges were 50-100% of shot weight, with some small cannon having charges of 117% shot weight. Later gunners found this was wasteful - using 1/3 the powder would provide 2/3 the muzzle velocity, and it didn't drop off as quickly, so the penetrating power was even closer. Rifling dropped powder charges even further by eliminating windage (the gap between a cannonball and the cannon) and allowing all the energy to be transferred.

Esa the Wanderer
Patreon Supporter
Joined: 2015-01-28 11:15

Oh! So basically nothing was constant black powder guns. :( I'll have to make sure I do comparisons only within the same era/TL. And the baseline for ACKS guns is 16th Century?

Alex
The Autarch
Joined: 2011-06-30 18:10

I would say late 16th/early 17th century.