I’d previously made some notes somewhere on using mechanical vibration to counter stiction when driving or removing screws, thus preventing screwdriver camout; this would be less annoying if it was above the threshold of human hearing, although that might not be practical, because the energy of each vibrational pulse needs to exceed the stiction energy. It occurs to me that it would also be useful for driving nails, where stiction is pretty much the name of the game.
The ideal graph of force against time of such a sonic-screwdriver system would be the derivative of a sawtooth: a Dirac delta in the desired direction large enough to make progress on the fastener, which necessarily creates momentum within the sonic screwdriver in the opposite direction, followed by a recovery time during which a constant small force is applied to arrest and eventually reverse that momentum, all the while maintaining contact with the fastener. The recovery force is applied against, say, the hand of the person using the device.
A control system can pause if contact is lost, and detect whether progress is being made and increase the impact energy (lengthening the recovery time and lowering the frequency) if not, or decreasing the energy if so, moving the frequency into the ultrasonic; and some degree of randomness in the interval would seem to be desirable to spread acoustic annoyance across the spectrum, also avoiding the danger of dumping a damaging amount of energy into a single resonance mode of the system.
(For screw installation, it would be very useful to be able to set a torque.)
This is all very similar, actually, to the action a hammer drill or rotary hammer uses for drilling holes in masonry. Bosch’s 5.7 kg 1500-watt “Titan 5 kg SDS Plus” hammer drill advertises 5 J maximum impact energy and is rated for drilling 32-mm holes in masonry, and other hammer drills from the same “Farmers Weekly” seller advertise impact energy in the 1.7 to 5 J range. Makita advertises their “23 lb. AVTDemolition Hammer” HM1213C rotary hammer as “delivering 18.8 ft.lbs. of impact energy, which is 25 J; it runs on 14 amps, which is presumably 1.7 kW. One shill site advertises the DeWalt DCH773Y2 rotary hammer (US$1099, 21 pounds) at 19.4 J of impact 1000-2000 times a second and marvels at how it “just pulverizes concrete”.
For no particularly good reason, there doesn’t seem to be a tool that applies the same sawtooth force profile to pulling an actual flat saw blade or wire saw through stone or concrete.
Some of these hammer drills and rotary hammers have a new anti-vibration feature which creates two simultaneous impacts with each hammer blow: one on the tool bit to make the hole, and the other on an internal counterweight, so that what rebounds from the impact is only the counterweight rather than the tool you’re struggling to hold in your hands. A different way to accomplish the same thing would be to just impact the counterweight directly against the toolbit: apply the tool’s motor to gradually accelerating the counterweight away from the toolbit against the counterweight’s spring, then toward the toolbit once the spring wins and the counterweight starts accelerating it back toward the toolbit. Either version of this same feature can be applied to the helical motion of an impact screwdriver (like those I’m discussing here) just as well as to the linear motion of a rotary hammer.
How hard is it to drive screws? Well, how hard can I twist a screwdriver? In a simple test just now with a water bottle and a metal pipe, I was able to rotate my wrist in a T-handle sort of configuration hard enough to lift 5.5 kg with a lever arm of 330 mm (the water bottle) and 430 g at 500 mm (the metal pipe itself), which is a torque of about 20 N m. This is probably about as hard as I can twist a T-handle screwdriver to unscrew a screw, but smaller amounts of torque usually suffice. About a tenth of a turn is pretty much always enough for a screw to make progress rather than springing back to its old position, and that would work out to about 13 J at that torque. (This explains the quasi-unit-compatibility of torque and energy: 20 N m of torque is really 20 N m of energy per radian!)
Normally, of course, screws don’t have to turn nearly that far to not spring back, and I don’t need a T-handle screwdriver to remove them.
Engineering Toolbox gives withdrawal forces for some nails in spruce ranging from 17.6 pounds to 348 pounds (79-1550 N), with a common 16-penny nail of length 3½” (89 mm) requiring 141 pounds (630 N). Driving such a nail 1 mm would probably be far enough that it wouldn’t spring back out, and that would be 0.6 J. Framing carpenters drive them all the way in in three hammer blows, which is about 20 J per blow, disregarding the mass of the nail and the vibrations of the structure, as you should.
So a simple sonic screwdriver would probably need to be able to ramp up to on the order of 1-10 J per impact to make progress in difficult cases, and then it would be able not only to remove screws and drive framing nails, but also drill and saw concrete, wood, and stone, and engrave or center-punch steel. But most of the time 0.1 J would be enough. If it were running at 1000 W, the average impulse frequency would be in the range 100 Hz to 10 kHz, unfortunately all well within the audible range. 1000 W would also drive one of those 60 J nails in 60 ms.
You could power it off Li-ion batteries in the now-conventional way, but it really only needs to contain on the order of 128-256 J at any given time, so it might often be more convenient to charge it with the necessary energy just before use. Even a clockwork spring sort of arrangement might be adequate with a pullstring; I can do two pushups to about 600 mm, lifting half my 110-kg weight, which is about 320 J. So normal people should be able to repeatedly pull a pullstring out to about a meter under a tension of about 50 N, like starting a lawnmower. Easy jobs might need a single pull occasionally, while hard jobs might need five or six pulls before every fastener or whatever, although at some point you just want to plug the thing in.
The pullstring need not be visible in normal operation; you could split the tool into two parts joined by the pullstring when charging it, then reassemble them once charged.
Clockwork springs have the advantage of having almost arbitrarily high power density, unlike batteries, both for charging and for discharging. Rechargeable lithium batteries typically have a “fast charge rate” of “1 C”, meaning 1/1 hour, or less, perhaps 0.5 C, meaning 1/.5 = 2 hours. If you were going to charge them up with a pullstring, you would have to pull the string continuously over the course of that hour or two.
The modulus of resilience of a material is the amount of energy it can store per unit volume as elastic deformation. For tensile elastic deformation of ductile linear materials, the relevant figures are Young’s modulus E and the yield stress σy, and the integral of deformation from 0 to the yield strain σy/E gives us ½σy²/E. Most types of steel have the same Young’s modulus regardless of their hardness, about 200 GPa. Soft steels have yield stresses as low as 300 MPa, but normally we make springs from music wire, which is more like 2800 MPa. This gives us a tensile modulus of resilience for music wire of some 20 MJ/m³, or 20 J/cc. 256 J then would require 13 cc of spring steel, or 100 g.
Conventional clock mainsprings do in fact deform in tension and compression, but the mainspring is only stressed to its limit at its inner and outer surfaces; on its neutral axis it isn’t strained at all. So the situation is actually even worse: you only get ¼ of the possible tensile energy storage that way, and you’d need 400 g of mainspring. This is getting to be a rather heavy screwdriver!
Coil springs instead deform the spring material in torsion, which is to say, in shear, and much more of the material is closer to the maximum shear strain. For shear deformation we’re interested in the shear modulus G, about 79 GPa for steels, and the yield shear strength τy, which for steels is about [0.58 of σy][10], or say 1600 MPa; the factor 0.58 comes from 3-½. So if we just calculate the shear modulus of resilience as ½τy/G, which I’m not sure is the right thing to do, we get 16 MJ/m³ or 16 J/cc, about 15% lower than the tensile modulus. I guess I should do the integral to see how much the distribution of the shear strain in the circular coil spring cross-section affects the situation, but it seems clear that using shear rather than tension (and compression) doesn’t make a huge difference.
By twisting a tube rather than a solid bar, the way a lot of torsion bars in car suspensions do nowadays, you can get the full shear modulus of resilience of your metal, but that doesn’t get you more energy per volume, just more per mass.
So what about carbon fiber? I hear truck suspensions nowadays are starting to use carbon-fiber-reinforced plastic rather than steel.
sawtooth components