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[Arrkhal]

The first thing that comes to mind while wading through the raws is that having 6 different yield strengths is unnecessary for metals.  In general, for metals only, there's only one "yield strength," that's valid for any direction of force.  The values will differ because properties are given for a particular shape of metal, like a bar or rod.  A rod is obviously weaker in compression, bending, etc., than it is in tensile strength.  Ultimate strengths also tend to depend somewhat on the size and shape of what's being tested.

Impact strength should be in units of energy, not pressure.
http://en.wikipedia.org/wiki/Charpy_impact_test

MAX_EDGE is too abstract, and you'd be better served by using the vickers or brinell hardness, as hardness really does tell you how sharp something can be made, in practical terms.  Hardness is also probably more useful than yield and ultimate strengths, in telling you what will put a hole in what.

The other units, while nice, probably aren't necessary.  Yield strength, impact strength (brittleness), and hardness will probably result in a system that's both simpler and more realistic.

Other materials will need a more complex modeling system, though.

[G-Flex]

Keep in mind that the material system is in place for far more than just metals. It would be silly to have metals following a completely different system from other things, and probably wouldn't work, at any rate.

Whatever system is in place has to work for metals, rocks, plant fiber, skin, and basically everything else.

[Squirrelloid]

I don't know about the similarity of metal response to different types of stresses.  I'd need to see the experimental results.  It would not surprise me, however.

But Yield (Strength), Fracture, and Elasticity would seem to be the relevant variables, although it doesn't give you enough information to know what should happen after Yield is exceeded but before Fracture is reached.

Yield Strength is in units of Force, like it should be.  Its defined as the stress (force) at which a material stops deforming elastically. 

F is obviously in kgm/s^2.  I can't remember what the generic units for E are offhand, but based on E=mc^2, it must be kgm^2/s^2, so the energy of the system would be Stress*Strain?  (Because the only distance that seems relevant to the energy is the deform distance).

E = Stress*Strain = Stress^2*Elasticity (unit check: kg^2m^2/s^4 * s^2/kg = kgm^2/s^2)

Using Yield Energy might let you combine Yield Strength and Elasticity into a single variable, but that depends on how involved the combat calculations are.  Knowing what elasticity is could be important if the combat model is sufficiently good.

[Arrkhal]

Keep in mind that the material system is in place for far more than just metals. It would be silly to have metals following a completely different system from other things, and probably wouldn't work, at any rate.

Wouldn't be completely different.  It's just that metals tend to have one single yield strength that's applicable no matter how and where force is applied, except when the shape of the object changes it (which will happen with stuff made of anything).

Which means that the yield strengths across the board for metals should usually be the same, while most rocks will have a much higher compressive strength than tensile, etc.

-----

But Yield (Strength), Fracture, and Elasticity would seem to be the relevant variables, although it doesn't give you enough information to know what should happen after Yield is exceeded but before Fracture is reached.

Elasticity tells you that.  And the difference between yield and fracture strength allows fairly good guesstimation of the elongation % at break.

And yes, stress * strain is one way to arrive at energy.  The main other ones are force * distance and mass * velocity^2.

Really, I think elasticity should mainly come into play for durability, not so much for effectiveness.  If your weapon is made of very soft material, it doesn't matter whether it bounces back or stays dented, it's not going to be very effective.  Rubber hammer vs. wax hammer.  The rubber one lasts longer, but neither hits as hard as a steel one.  Elasticity would mainly matter for armor.  Inelastic armor would dent and stay dented, which could exaberate a wound.  On the other hand, that could be a subset of item condition; wearing a dented piece of armor aggravates injuries under it, whether the dent and the injury happened at the same time or not.

But anyway, yield strength (of the relevant type if applicable), brittleness, and hardness should tell you everything relevant about a material, for the purposes of what gives first, and how it gives, when things are whacked together.  Add in elasticity to determine durability, too.  And I guess brittleness could be determined either by comparing yield to fracture strength (brittle materials fracture when they yield, or very shortly after), instead of an impact test.  But still, impact yield is generally expressed in energy, because force alone tells you nothing about how rapidly the force is applied.

[Andeerz]

:3  Me likey lots.

[Osmosis Jones]

Isn't strain unitless? Specifically, it's (change in length)/(length) = m/m = 1.

Thus stress * strain has units of stress, which =/= units of energy.

[Squirrelloid]

Isn't strain unitless? Specifically, it's (change in length)/(length) = m/m = 1.

Thus stress * strain has units of stress, which =/= units of energy.

You know, you may be right.  I thought strain was in units of distance for some reason, but I'm remembering like 5 or so years after I actually did anything real with this stuff.

[cooky173]

Yield Strength is best given as a stress, not a force. Simply put, a bar with twice the cross sectional area will need twice the force, but the same stress to cause yield, as stress is Force/Area.

Stored elastic potential energy can be considered as applied force x displacement, assuming no plastic deformation.

For a rod in tension or compression

Force is stress x area.
displacement is strain x undeformed length

Elastic potential energy is stress x area x strain x undeformed length
EPE = stress x strain x area x undeformed length
EPE = stress x strain x volume.

The interesting result then, is that the energy stored per volume is then the stress x strain

Now, for a material under a bending load, I think you use this fact and perform an area integral.

Hope this aids the discussion

[BlckKnght]

Before you go too far in simplifying the material values, you may want to think about the fact that many materials can be manufactured with non-uniform properties. Hammering on most metals hardens them, for example.

A skilled craftsdwarf is likely to use the full range of material properties an material is capable of to give each part of an item the best qualities possible. A case hardened steel axe head, for example, would have a hard cutting edge but an inner core that is softer and resists shattering when the axe hits something.

[Arrkhal]

Before you go too far in simplifying the material values, you may want to think about the fact that many materials can be manufactured with non-uniform properties. Hammering on most metals hardens them, for example.

That would probably have to be done by either treating hardened and unhardened material as seperate materials, or having ranges.  Heat-treated steel would have a higher yield strength and hardness compared to annealed, but (roughly) proportionally higher brittleness, too.

Case-hardening, differential tempering, or selective work-hardening could be simulated pretty closely by using the high hardness of the hardened area, low brittleness of the soft area, and a yield strength intermediate between the two.  It'd definitely be true enough to life for a fantasy universe simulator.

[Arrkhal]

Not to necro and double-post, but we seriously need hardness modeled.  Using realistic shear strengths for wood makes wooden weapons unable to pierce skin.  The strongest woods have shear strengths of about 2600 to 3000 PSI.  Human epidermis has a shear strength of 3000 to 4500 PSI.

[Andeerz]

I agree.  And the ability for things to break should as well.  But are you saying that wooden weapons should not be able to pierce skin?  What about wooden spears and the like?   

[Fetus4188]

The only way to feasibly include materials properties in DF is with a completely abstracted system.  I think the current one has a bunch of values that aren't even being used, so those should probably be cut out unless something is going to be done with in the future.

You're not going to be able to anything remotely close to realistic behavior with DF, it would be a huge program in itself.  Metals gain their strength from having a ordered crystalline structure which is interspersed with impurities and shifts in the crystalline pattern, it's these breaks from uniformity that give it strength, there's no way you're going to model that into a game.

[Arrkhal]

But are you saying that wooden weapons should not be able to pierce skin?  What about wooden spears and the like?

Inability to pierce skin is the current behavior.  It needs to be fixed.  Wood has a lower shear strength, but the higher hardness should allow it to pierce skin anyway.

There's also the fact that shear strength doesn't have much to do with actual cutting, too.  Shear strength just means stresses applied this way:

>=>
>=>
<=<
<=<

It tells you how easy something is to cut with scissors or tin-snips, not with a knife.

[G-Flex]

The values in the raws must be wrong, then, which is odd since Toady states he's using real-life values.

Wood:

Code: [Select]

[SHEAR_YIELD:40000] used pine
[SHEAR_FRACTURE:40000]
[SHEAR_ELASTICITY:1000]
Skin:

Code: [Select]

[SHEAR_YIELD:20000] used data for human skin
[SHEAR_FRACTURE:20000]
[SHEAR_ELASTICITY:50000]

You make some good points, though. I didn't really understand how shear stress was supposed to be involved for weapon penetration either.


Any comments on this, though? I mean, I guess it does make sense that something impacting something else gets subject to some level of shear stress (since it wants to get "squashed") and not just compressive, but I don't know much about this, and it seems like different types of strain would affect each other in a case like that.