Suppose you want to make springs out of tool steel, which may not be the best possible choice but is close to feasible at least. (Normal people use music wire, apparently.)
http://www.matweb.com/search/datasheet.aspx?matguid=5abd35ce6bd64254b6db980b83683f27&ckck=1 says that AISI type A2 tool steel (which can be hardened by air-cooling after heating) has a modulus of elasticity of 203 GPa, a shear modulus of 78 GPa estimated from the elastic modulus (presumably with E = 2G(1 + ν), where E is Young’s modulus, G is the shear modulus, and ν is Poisson’s ratio), and a density of 7.86 g/cc. Unfortunately this doesn’t supply the strength data that we need to figure out when it will break.
http://www.matweb.com/search/datasheet.aspx?matguid=8188603e5f954e53b479aaa4b07cdffb is a page about particular brand of A2 tool steel, giving its yield stress as 1580 MPa, its ultimate tensile strength as 2050 MPa, and its modulus of elasticity as 203.4 GPa, with a 1% elongation at break. (This makes sense because 2050 MPa is about 1% of 203 GPa.)
https://en.wikipedia.org/wiki/Shear_strength says that in steels, the shear yield stress is about 0.58 of the tensile yield strength (this is called the von Mises yield criterion, and apparently applies to ductile materials in general?)
This suggests that the shear yield stress of A2 should be about (* .58 1580) = 916 MPa, at which point it’s distorted at a slope of (/ .916 78) = 0.0117, or 1.17%.
Let’s consider the case of twisting a cylindrical torsion bar made of this material, say 1 m long, 200 mm diameter, and 1 mm thick, thin enough that we can mostly ignore the difference in radii between inner and outer walls, but thick enough that it won’t collapse as we twist it. Twisting it by 1.17% means a twist of 11.7 mm over that meter length, which is 0.117 radians of twist. The cross-sectional area there is about 1 mm · 200 mm · π ≈ 628 mm². At the yield stress of 916 MPa, this cross-sectional area generates (* 916 628) = 575248 N of force, which is almost sixty tons. (We could convert this to a torque by multiplying by the 100 mm radius, but we don’t need to.) Building linearly to that force over 11.7 mm of travel distance gives us (* 575248 .5 .0117) = 3365 J of energy stored in the bar.
What’s the energy density of that? The metal occupies 0.628 ℓ of volume, so that’s (/ 3365 .628) = 5358 J/ℓ. At 7.86 g/cc or kg/ℓ, that’s (/ 5258 7.86) = 669 J/kg. This is close to, but larger than, the 300 J/kg cited in https://en.wikipedia.org/wiki/Energy_density_Extended_Reference_Table.
If we were somehow able to stress the metal in tension instead of torsion, we’d get to 1580 MPa at an elongation strain of (/ 1.58 203) = 0.0078 or 0.78%, or 7.8 mm, at a force of (* 1580 628) = 992240 N, and an energy of (* 992240 .5 .0078) = 3870 J, which is better, but only by 15%. (/ 3870 .628) = 6162 J/ℓ; (/ 6162 7.86) = 780 J/kg.
Nested torsion tubes in series offer the opportunity to exploit this entire 669 J/kg energy capacity; normal coil springs only manage about two thirds of it, and springs such as garage door torsion springs that are stressed in bending rather than tension or torsion only get to half of the tension number. You need some space in between the tubes, and some mass in the coupling between the tubes at the ends, but those can be very small numbers. So 5kJ/ℓ should be a totally reachable energy density in practice.
Unfortunately, these specific energies are substantially lower than I want for the compact application I have in mind, in which I would like to hold several hundred to several thousand joules in a spring weighing under 200g, then release it in submillisecond timescales. 200g · 669 J/kg = 134 J, barely acceptable.
I should investigate whether spring steels, beryllium copper, or nitinol can provide larger energy capacities. Apparently ASTM A228 music wire is a common spring material, as are SAE 1074 and 1075 steels, while AISI 1095 steel (ASTM A684) is used for more demanding applications.
The speed of sound in a material is √(K/ρ), where K is the relevant modulus of elasticity and ρ is the density of the material. In this case, for transmission of shear, sqrt(78 GPa/7.86 (g/cc)) comes out to 3150 m/s. This means that it takes 317 μs for a movement to travel a meter through the spring; the suggested meter-long torsion bar won’t be able to respond faster than that. You can, in effect, fold up the spring and nest it inside itself into a series of nested torsion tubes, which in theory won’t affect either the response time of the spring or its rate. You can decrease the spring rate (i.e. increase the compliance) by using thinner-walled tubes, up to a point where the spring buckles and collapses; but, if you do that while keeping the energy capacity and mass constant, you have to increase the response time proportionally.
If your torsion spring is a single tube of constant diameter, it is useful to give it a constant thickness as well; otherwise, the thinner part will fail before the thicker part is fully charged. This is because the torque is constant along the entire length of the tube, but that torque translates into different stresses at different thicknesses. Once you start nesting the tube inside of itself in series, you still have the constant torque, but now you have different radii in different parts of the spring, which translates to different tangential forces inversely proportional to the radii; this means that the wall thicknesses also need to be inversely proportional to the radii in order to keep the stress constant through the entire spring.
This has the problem that once the radius is small enough, the inner and outer radii start to differ significantly, which means that the stress on the inside radius of the tube is significantly lower than the yield stress. For example, the 1mm-thick 20mm-wide tube in the example above, in which the stress is below optimal by only 10% on the inner wall, could be put in series with a 10mm-wide 5mm-thick tube — which is no tube at all, but merely a rod! It has no torsional stress at all at its center.
The upshot of all of this is that you can get arbitrarily fast reaction times only at the cost of arbitrarily high forces or arbitrarily low energy capacities per spring.
What about Dyneema? If it has an ultimate tensile strength (and also yield stress!) of 2.5 GPa and a Young’s modulus of 100 GPa, and we somehow stress it in tension instead of torsion, then that same 628 mm² cross-section rope of it would hold up to 1.57 meganewtons, about 160 tons; if it were 1 m long and stretched by 2.5%, or 25 mm, then we’d have 19.6 kJ stored, 31.25 kJ/ℓ. And it would weigh only 609 g, so it’s 32.2 kJ/kg.
In energy density per kg, that’s 48 times better than the steel spring. Per liter, it’s only about six times better. And that’s without getting into weird effects like rubber’s hyperelasticity or nitinol’s pseudoelasticity which I worry might convert their theoretically higher energy capacities into delusions at these time scales. According to ARL’s tests http://www.dtic.mil/get-tr-doc/pdf?AD=ADA606636 Dyneema is capable at least of absorbing energy at these moduli at a kilostrain per second.
http://www.sciencedirect.com/science/article/pii/S1877705811005649
This suggests that maybe nylon would be an even better place to look for spring energy density! It’s not as strong as Dyneema, but it has a dramatically lower modulus, and I think it also avoids the weird pseudoelasticity thing where the “elastic” energy gets lost as heat. http://www.engineeringtoolbox.com/engineering-materials-properties-d_1225.html claims that nylon 6/6’s density is 1.15 g/cc, its tensile modulus is 2 to 3.6 GPa, its tensile strength is 0.082 GPa. Dividing, that gives it 24% (or more) elongation at break; 24% .082 GPa/2 = 9.8 kJ/ℓ. That’s not as good as Dyneema, but it’s still volumetrically better than most steels, and a lot better per mass, at 8.6 kJ/kg. Nylon springs would easily meet the 200g limit I’m trying for: 200 g 8.6 kJ/kg = 1.7 kJ. In fact, you could probably get by with 100g or 50g. And nylon is a lot cheaper than Dyneema still.
The √(K/ρ) speed of tensile sound in nylon, by these numbers, should be 1318 m/s, only four times its speed in air, and low enough that you can only get 1300 mm of nylon spring to respond in a millisecond. However, this should be plenty of time. I really only need a millisecond or so; 100μs response time is more than adequate, and I have in mind for the entire spring to be under 100mm long.
Disappointingly, it seems that nylon is indeed hyperelastic and behaves much more stiffly at high strain rates; http://scholarbank.nus.edu.sg/bitstream/handle/10635/37891/PhD%20Thesis,%20Habib%20Pouriayevali%20(%20Mechanical%20Dep)%20HT081385J.pdf?sequence=1 and http://www.sciencedirect.com/science/article/pii/S1877705811005649 show the results of H. Pouriayevali with respect to the issue, showing that his nylon sample compressed 30% under a 50 MPa strain under quasi-static conditions, but at strain rate of -3203 per second, it had only compressed 2% at that same strain (which I suppose means he reached it in 6 μs). At a more moderate strain rate of -980 per second, it had compressed some 8% at that strain, in, I suppose, about 80 μs. The dissertation also, alarmingly, says that nylon “is notably rate-dependent and exhibits a temperature increase under high rate deformation”, but it turns out that he’s talking about like five kelvins for fairly large strains like 21%.
Pouriayevali’s 6-6 nylon samples failed in quasi-static tension at strains which appear to have been around 1.0, 1.2, and 1.4, with stresses around 80 or 100 MPa, which seems improbably rubbery, but fairly plausible strength. He reports a quasi-static elastic modulus of 958 MPa at small strains, but as the graphs show clearly, this drops precipitously at higher stresses.
This stiffness varying with strain rates presumably means you have high hysteresis losses in nylon, more usually known as “vibration dampening ability”.
However, I’d probably be fine with >50% losses, as long as I can get the rest of the energy out in well under a millisecond. 12% strain in a millisecond is a strain rate of 120 strains per second, which may be low enough that nylon will have low losses.
Pouriayevali did do lower-rate experiments under tension. His lowest-rate dynamic tension experiment involved an impact of 150 strains per second, reaching some 30 MPa and 5% strain; the quasi-static condition had only 20 MPa at that same 5% strain. That means that if you stretched your nylon by 5% at 150 strains per second, you’d’ve put in 30 MPa * 5% / 2 = 750 J/ℓ, and then if you unloaded it slowly, you’d only get out 20 MPa * 5% / 2 = 500 J/ℓ. In this case I propose to do essentially the reverse: load it slowly, then unload it quickly. It looks like this will probably be in the neighborhood of that 50% efficiency.
(Actually the stress-strain curves aren’t very linear, and they’re actually maybe a bit closer together than that makes it sound.)
In chapter 5 of his dissertation, Pouriayevali fits numerical models to the properties of the nylon which seem to suggest something like 10% or 20% energy losses at 150 strains per second when extended to larger strains, with actually much lower losses at higher strains, up to 0.6. Also, the stresses at higher strains are more nearly constant, which should ease design substantially.