"Direct driven hydraulic drive for new powertrain topologies for non-road mobile machinery"
To be clear, I'm not an expert on hydraulics (far from it, in fact-- haven't seen em since college), so a lot of the questions you're asking I'd have to research in order to answer. Plus, I'm pretty sure the "why is it inefficient?" question would be worth a published paper or fifty....
If I were to "shoot from the hip" to answer (which means I'm 99% wrong), I'd say the different efficiency cases comes down to the properties of the fluid (or gas) that you're moving, especially viscosity and compressibility-- imagine trying to pump honey with a "propeller", vs. a screw pump (which I imagine would work quite well)
Intuitively (so now I'm 99.9% wrong), if we think of a piston pump, the only real "wasted motion" is retracting the piston and refilling the void it's about to pump. Valves would cut the piston of from the system any time it's not contributing energy, and thermal losses from momentary compression aren't a big issue for stuff with such low compressibility.
That's just my 2c tho, and it's worth noting that the paper I referenced is specifically for small-scale, direct-driven pumps with commercial hydraulic fluid.
> (re: driveshafts)
Thanks, man. I’ll need to come up with a mechanism that allows a keyed gear to slide along the keyshaft’s axis, while not allowing rotation. Thankfully, a bit of slop (backlash) is tolerable, but if you have any ideas to minimize it, let me know. (rel >>24744
> (re: hydraulic volume change under pressure)
I think we had a bit of a miscommunication there, I was referring to NoidoDev's idea of a passive, compressed storage system for active operation. The compressed-gas hydraulic bladders you mentioned will work, of course, but in that case, it's not the water "storing" that pressure, just transmitting it. Bottom line with "incompressible" stuff is you'll never get more volume out of it than the tank displaced in the first case, no matter how much pressure the tank has (both by-definition and purely theoretical, I know)
As I understand it, the easiest way to understand how much energy there is "in" a compressed system is to think of the ratio between compressed and uncompressed densities. (multiplied by the force exerted at each particular pressure, of course)
For example, let's take two pistons exerting (the equivalent of) 1000lbs on a 1in plunger (say 1000psi inside), one full of compressed water, the other with compressed air, and say they're each at "100% volume" (not yet extended).
Now, as they extend, the volume the compressed medium fills will increase, so the pressure exerted by the piston will decrease. Once the piston full of compressed air extends to 101,103, 105...% volume, the force it exerts (and internal pressure) will have decreased a little, but there's still plenty of Work (physics-def.) it can do.
The piston full of water, on the other hand, no longer exerts any force once it reaches 100.2% volume (since water's compressibility is ~0.2% per 1000 psi), and the internal pressure (ignore gravity for a bit) is now ambient, so "zero".
In this example, the density ratio of water was 1.02 @ 1000psi, but air has a density ratio over 69.1 at the same pressure (see https://www.engineeringtoolbox.com/air-temperature-pressure-density-d_771.html
-- I was incorrect with the 5% factor, that was the compression ratio at the bottom of the ocean lol >>24990
(Formal explanation of it: https://www.sciencedirect.com/topics/earth-and-planetary-sciences/compressibility)