Here is another example problem...if people like to practice...

A carriage with a gear ratio of 38/17 is to move a load weighing 8100 lbs to a linear velocity of Vmax=3.28ft/sec and then stop within 1.5 ft. Therefore the required acceleration is 3.56 ft/sec^2. The other parameters of the system are as follows:

hyd motor displacement= 6.7 in^3/rev
desired motor speed= 272 rpm
tube inside diameter= 1/2"
tube length= 32ft
Bulk Modulus= 200,000psi

Will the required acceleration is obtainable? If not how can we correct it?


To those interested...goodluck!!!

This message has been edited. Last edited by: maglub,

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Originally posted by maglub:
Here is another example problem...if people like to practice...

I have that problem in my book too. The problem is that natural frequency does not make things go so except for the natural frequency calculation the rest is bogus. The system can be ramped down in 0.4 seconds WITHOUT shortening the tubing. This easily which is 3 times faster and saves 0.8 seconds which is significant to the production people.

Peter, if the proportional valve is used(not servo), by ramping too fast the system might become bang-bang?! Otherwise you need to use a bigger valve and the resolution is not good because the maximum spool stroke is not fully utilized.

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Originally posted by maglub:
Peter, if the proportional valve is used(not servo), by ramping too fast the system might become bang-bang?!

Yes, in fact that is true whether you are using proportional valve or a servo valve.

The purpose of the exercise is to calculate the time one should use to ramp the servo or proportional valve to the desired speed. If one simply opens the valve instantly to full open then the system will oscillate but it you ramp up the control signal the right way you can get the load to move without oscillating.

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Otherwise you need to use a bigger valve and the resolution is not good because the maximum spool stroke is not fully utilized.

It is obvious that the whole problem is bogus if you don't pick a valve that is big enough to go at the desired speed. One can big a valve that is just big enough so the valve must be opened 100% or nearly so. You can also get a valve that us twice as big and open it only 50% to get the same result.

The resolution is a function of the digital to analog converter. A good motion controller has 16 bit resolution or maps -100% to 100%, -10v to +10v or -20ma to +20ma to -32767 to 32767 digital to analog counts.

It should be obvious that the rise time from 0 in/sec to 10 in/sec is going to be much shorter than the rise time from 0 in/sec to 20 in/sec. This is not a function of a time constant. It is a function of the force available due to the system pressure and the valve.

The acceleration and deceleration of hydraulic motors and cylinders refers to the change of flow per unit time. The positive or negative change in flow takes place via the proportional valve. The time, within which this change in flow and therefore the change in position of the control spool is to take place is preset at the electronic control for the proportional valve's solenoid. The electrical component that is used to accelerate and decelerate is called the ramp generator. To set the ramp up or down, it is being adjusted to the point before the instability occurs(approximately 6 time constant).

Regarding the selection of the valve size, IMO, the criteria is based on the adjustability range of the selected valve size. Peter pointed out that any valve can be selected as long as the movement can be controlled. Not a problem at all as long as it can be easily adjusted. If we select a big valve and utilizing 50% of spool travel for maximum flow, the resolution of the valve(flow wrt percentage of current applied) is lesser than choosing a smaller size that can utilize almost the maximum spool travel.

Input power must also be taken into consideration when selecting acceleration. We need to calculate the force due to acceleration and consequently the pressure that is needed to accelerate(force make it go? Maybe not...). This pressure is the resistance to flow. The flow has to be resisted to have a pressure build up...by load( in this case). And at a constant load, the maximum pressure during acceleration(due to F=ma) is Fa/A, where A is the area to which the pressure is being applied.

If a flow rate(at constant load) necessary to accelerate to Vmax is determined(using Q=Vmax *A) and by choosing a valve(for example only) to open it at say 90 percent, it is still possible to increase the velocity to 100% at rated valve pressure drop by increasing the solenoid current thereby openning the spool to maximum and increasing the flow. In this case the pump pressure is constant while the pressure drop across the valve(delta P1 +delta P2) is in the minimum pressure drop requirement(10 bars for proportional and 70 bars for servo). We can also decelerate by reducing the flow by decreasing the input current thereby reducing the spool openning. How and why?!

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Originally posted by maglub:
The acceleration and deceleration of hydraulic motors and cylinders refers to the change of flow per unit time.

Actually, you are close. Remember that Q=v*A ONLY when the rate in pressure change is 0. In this case the differential pressures across the motor only needs to be about 25 psid, I have done the calculations, so you are close, but it is really the torque and inertia that determine the angular acceleration.

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To set the ramp up or down, it is being adjusted to the point before the instability occurs(approximately 6 time constant).

I repeat, the motor can be smoothly accelerated in 0.4 seconds instead of 1.2 seconds. That is 3 times faster.

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Input power must also be taken into consideration when selecting acceleration. We need to calculate the force due to acceleration and consequently the pressure that is needed to accelerate(force make it go? Maybe not...)

I didn't bother to calculate the required torque. I just plugged the values in the book into my simulator and adjusted the ramp to be much faster. I tweaked on the valve size to get the approximately the same speed as that in the book. However, my simulator does calculate torque for every millisecond or 100 microseconds when doing the simulation.

There should be an accumulator to store energy so the supply pressure is kept constant.

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This pressure is the resistance to flow.

I was hoping that this saying would go away. Pressure is force per area.

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The flow has to be resisted to have a pressure build up...by load( in this case).

More importantly there needs to be a pump to increase pressure. You can have all the 'resistance to flow' in the world but if nothing compresses the oil there will be no pressure. It takes work to compress oil and compresses oil has potential energy.

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And at a constant load, the maximum pressure during acceleration(due to F=ma) is Fa/A, where A is the area to which the pressure is being applied.

Not quite F=m*a -> P*A=m*a -> P=m*a/A

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If a flow rate(at constant load) necessary to accelerate to Vmax is determined(using Q=Vmax *A) and by choosing a valve(for example only) to open it at say 90 percent, it is still possible to increase the velocity to 100% at rated valve pressure drop by increasing the solenoid current thereby openning the spool to maximum and increasing the flow. In this case the pump pressure is constant while the pressure drop across the valve(delta P1 +delta P2) is in the minimum pressure drop requirement(10 bars for proportional and 70 bars for servo). We can also decelerate by reducing the flow by decreasing the input current thereby reducing the spool openning. How and why?!

That is pretty much what happens. Opening the valve reduces the pressure drop across the valve and allows for more pressure drop across the load. The pressure drop across the load is multiplied by the area to get force or in the case of a motor the pressure drop is multiplied by its displacement/(2π) to get torque. Flow happens because pressure will try to equalize. Closing the valve increases the pressure drop across the valve and reduces flow into and out of the motor. Pressure on the meter out side builds up and so does opposing torque and this provides braking.

You must remember that Q>v*A on the pushing side while accelerating because it takes some oil to compress the oil that is there and build up pressure to build up force or torque. Q<v*A while decelerating because pressure must be reduced. This problem is about accelerating and decelerating a hydraulic motor. During that time Q<>v*A but in this particular problem the error would not be much because so little torque is required.

The ramp generator on you valve controller generates linear ramps. This doesn't take into account that it does take time to compress and decompress oil to build up the necessary torque or force required to accelerate the load.

Read this
http://intechweb.org/downloadp...kgaf3flaaog56bdr51r5
On Algorithms for Planning S-curve Motion Profiles
On the first page it says that trapezoidal motion like your ramp generator uses tend to cause overshoot and excite residual vibrations.
If acceleration ramps are use as on page 7 then one can smoothly accelerate 3 times faster. I am sure the manufacturers of these ramp generators know this but they keep on selling this linear ramp generator garbage and lose jobs to the servo motor people. My objection to the whole hydraulic motor problem is that it is obsolete and became obsolete over 15 years ago. There are better solutions to the problem now. Use better ramp generators that can generate s curves. The book I have with the hydraulic motor problem was copyrighted in 1985. Back then the books solution was OK because computers where not so common. Now this solution is a loser to new technology.

One more thing. We were using the techniques described in the pdf back in 1993. The first sentence is in the abstract is wrong. The motion control people keep their algorithms for smooth motion secret.