So, I have 72 hydraulic McKibben actuators with 30cm in length and 14cm in width under 400 kPa (or 4 bar) that lifts a 5 ton weight that I want to actuate in a third of a second (continuously, actuating and un-actuating).
The explanation below is about McKibben actuators, if you have no clue about how these work, just imagine that these are hydraulic cylinders that work at 4 bar, have 30cm in length and 40cm of diameter and just move by 6cm.
McKibben muscles normally contract 20% of its length and increase 30-40% of its diameter.
So this means that a single one of these muscles have an uncontracted volume of 4.618 liters inside of it.
Taking into consideration the change of length in 20% and diameter in 40%, the contracted McKibben muscle would have a volume of 6.244 liters inside of it.
The difference between these two would be 1.626 liters, which means that if I wanted to actuate all of the muscles in a third of a second, I would need 578.88 liters per second, so 34732,8 liters per minute.
Most commercial hydraulic pumps have a certain limit of RPM, liters per minute and pressure. I don't think that even if the torque of the pump motor was super low, it would be able to pump as much fluid as this.
So, regardless of making a custom pump or not, I believe that this is a really inefficient way of actuating these.
One would need a lot of energy to pump 34000 liters of hydraulic oil, and it would also take a lot of energy to actuate cylinders that work around 12000 PSI, but need a few ml.
So, the question:
Is there a way of achieving some kind of ideal balance between working pressure to fluid flow? Or I will need to test/calculate option to option until I find a certain balance?