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Old 03-23-2007, 08:13 PM   #13
Kromm
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Join Date: Jul 2004
Location: Montréal, Québec
Default Re: Collision Damage, Momentum and Energy

To elaborate on what David said . . .

My lazy statements that we used "momentum" weren't strictly true. Damage is complicated. For collisions -- be they people with people or bullets with people -- energy is kinetic energy (T), and a function of mass (m) and speed (v):
T = 1/2mv^2
If you assume that area (A) and volume (V) are simply powers of some hidden linear scale (L):
A = L^2
V = L^3
then you get:
A = V^2/3
And since mass depends on volume and density (ρ) as follows:
m = ρV
you can solve to get:
A = (m/ρ)^2/3
Penetration is tricky, because in addition to needing to know energy per unit area, you have to know how tough the penetrator is, to estimate the extent to which it will break up. We assumed a toughness factor equal to hit points (H), which are a dimensionless function of mass and some constant (c):
H = cm^1/3
As p. B558 suggests, c = 2 for flesh, 4 for structural materials with hollows inside, and 8 for solid structural materials. Since we wanted to work entirely within game units, we solved for m as a function of H:
m = (H/c)^3
and wrote:
T = 1/2[(H/c)^3]v^2
A = [(H/c)^2]/ρ^2/3
The final damage formula calculated a completely abstract concept -- damage (D) -- which we took to be the square root of the product of energy per unit area (T/A) with the toughness factor (H). We used the square root because this worked well for anti-tank weapons, frankly. Cranking through all the substitutions, this ended up making damage:
D = (HT/A)^1/2 = kHv
Where:
k = (ρ^1/3)/(2c)^1/2
Now k isn't constant, but it doesn't vary much for ordinary objects. A lump of iron with specific gravity 7.85 that's also inanimate and gets c = 8 has k = 0.50. A lump of mostly watery meat with specific gravity 1.1 or so, and that's animate (c = 2), has k = 0.52. Even crazy things like depleted uranium give k = 0.66 or thereabouts (so maybe DU ammo should get more damage, and indeed it will). Hollow machines end up with k = 0.5 to 0.6. We therefore took k to be approximately constant, with no more error than any of our other assumptions.

Thus, damage ended up being proportional to the product of hit points and speed, to within a fudge factor. We then chose convenient scales for speed and damage to get the formulae in the book. My earlier, lazier statements about "damage proportional to momentum" were a reference to the dependency on v rather that v^2, and incorrect.
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