I have an application I am working on where the customer would like to control the force of two hydraulic cylinders at various force settings ranging from 60 lbs-f each up to 2200 lbs-f each. These cylinders would be pressing down on two ends of a 4"-8" diameter roller. The roller would then be driven at 500 RPM through direct contact with a different “drive” roller (drive roller speed varies to ensure 500 RPM on test roller across all diameters). The test roller can vary in roundness up to ~0.04", which would result in an load scenario that varies and repeats every 0.125 ms based on the 500 rpm speed of the roller. The customer is planning on using individual load cells for each cylinder (0-3,000 lbs; +/- 0.1% accuracy).
My question is to what accuracy can I reasonably tell the customer a force control system can maintain? Would the smallest flow on a valve available be ideal or is there a flow range I should look for (test runs for days; speed of cylinder during setup is not an issue). Do the load cells need to be more accurate? Can I achieve both the low end forces and high end forces using the same system (valve,feedback device, cylinders)?
1.25ms per rev is impossible but at 500 rpm = 8. 333 rps or 0.12 seconds per rev which is possible. To figure worst case assume there is no controller, valve and the oil is trapped in the cylinder. Then the force change roughly proportional to the pressure change. The pressure change is ΔP=-βΔV/V where β is the bulk modulus of oil. If the valve is mounted on the cylinder then the volumes can be factored out so ΔP=-βΔx/x where x is the distance from the cap end of the cylinder and Δx is positive when extending. Since the pressure changes on both sides of the cylinder and the force is proportional to the area so ΔF = -(ArβΔx)/(l-x) +(AcβΔx)/x or ΔF= -βΔx(Ar/(l-x))+ (Ac/x)). Where l is the length of the cylinder and Ar is the area of the rod end and Ac is the area of the cap end. This question can’t be answered without knowing the Ac, Ar and L and it assume the valve if mounted on the cylinder so the volume of piping between the valve and cylinder can be ignored. This should be the worst case. Δx would be 0.04 inches.
A motion controller can add or subtract oil at 8.333 Hz to get the force relatively constant but the key to success will be anticipating how the shaft is bent. I would be good to have an encoder on the spinning shaft so the motion controller can know or learn the sinusoidal motion of the bent shaft. Then the motion controller can proactively anticipate the increase and decrease in force instead of just relying on the PID to react to the changes in the force.
An example calculation. If a 2 inch diameter cylinder with a length of 2 feet is 10 inches from the cap end and the position is pushed backwards 0.001 the pressure on the cap side will go up 200,000001/10 = 20 psi. If the cylinder is a 2 inch diameter cylinder the force on the cap side of the piston will go up 62.8 psi. The pressure on the rod side will drop 2000000.001/14 or 14 psi. The force on the rod side will drop 23.65 psi so pushing the rod back 0.001 will result in the opposing force to go up 62.8-23.65 psi = about 39 pounds.
So if the motion controller can anticipate the wobble within 2 thousandths instead of reacting to a 0.04 inch wobble you should be able to keep the force constant within 100 lbf.
0.125 ms was not the correct units. I meant 0.125 s or as you calculated 0.12 s. In regards to predicting the roundness, do you envision a slow rotate of one revolution to teach the controller each test part, or is it something that the controller would adapt to automatically after n number of revolutions at full speed? Would the +/- 100 lbs accuracy be the same at 65 lbs. target as it is at 2200 lbs target? Or does the small target force hamper our control in some way?
No, the wobble or deflection as a function of angle must be recorded. There are a couple ways of doing this. One could simply fit a sine to the wobble and use that to offset the radius or the radius could be recorded using one of our curves or splines. See the curve tool.
We have done something like this for the FAA. We had to map the bumps in the runway and try to maintain a constant force even though the wheels were rolling up and down over bumps or ruts. controleng.com/single-articl … b3882.html
Roughly. The formula above only cares about the change in the pressure as the volume is reduced by some percentage. The initial pressure isn’t part of the equation BUT that assumes the bulk modulus of oil doesn’t doesn’t change as a function of pressure. It does but not much.