Worms in hydraulic systems

Tom Jenkins

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Tom and Alaric are both wrong.
Well pardner, I can't speak for Alaric, but I is often wrong. Not this time, though. (Actually, I was kinda feelin' ornery and did my post partly jist to git
yer goat, Peter.)

Tom and Alaric are both wrong.
An actuator will accelerate until the sum of forces acting on it are 0. At that point there will be no more acceleration and maximum velocity is achieved.

Well, yes 'n no. Ya gots to remember, Peter, thet not all hydraulic systems are highly engineered precision motion control packages like the ones you deal
with. (And I might add you is undoubtedly a world class expert in thet field.)

Some sysetms is jist brute force producers. Dynamic considerations is usually negligible. A good example is the old manure bucket I used as a kid back on th'
farm. It was an open loop system, with a single acting lift cylinder and a gear pump direct driven off the engine and a relie valve that dumped plain ole 10W
engine oil direct to the hydraulic reservoir.


Now, when I shifted the spool in the valve they was obviously accelleration, but the time frame was imperceptible to the human eye. The cylinder hit max
velocity durn near instantly, seein' as how max velocity was very slow. If I wanted more flow I'd open the throttle on the engine and the bucket would go
from very slow to jist slow. More FLOW meant more GO, on accont of cuz the pressure capability of the pump was way more than the load induced by the bucket
of @*$&^! (Let's jist call it a fragrant dairy by product).

So, You is right, Peter, that the load would accellerate when flow was given a path to the cylinder. The force and rate of acceleration was determined by the
pressure setting of the relief valve and the area of the cylinder, up to the point that the terminal velocity of the cylinder was reached. This hyar teminmal
velocity was determined by the flow rate of the pump. When terminal velocity was reached, then the pressure x area inside the cylinder equalled the load of
dairy by product so accelleration stopped. Flow made the cylinder go, and more flow made the cylinder go more faster!

Now if I added load to the bucket while lifting (say, by driving the tractor into the pile of dairy by product while raising the bucket) my cylinder wouldn't
slow at all. Thet's cuz the load still weren't enough to exceed the pressure x area of the cylinder. The gear pump kept pumpin' at a constant rate and the
cylinder kept movin' at the same speed.

On the other hand, if'n I hooked the bucket under the corner of the barn, the load exceeded the force capability of the cylinder at the max relief pressure
setting. I had force, of course, but not enough to cause accelleration or movement. The net force wern't zero, but there weren't no accelleration cause the
net force was in the wrong direction for cylinder movement.

A net force causes acceleration

Yep - no doubt about it. And when the net force is zero - like my bucket at steady state lift velocity - acceleration is zero too, but velocity ain't.

Acceleration is integrated to get velocity.

Well, thet is jist a typo on your part, Peter. a = dV/dt ....which ain't integration. No change in velcotiy means no accelleration. Wen my bucket was
climbing towards the sky with full pump flow to the cylinder velcity was constant so accelleration were zero.

Flow is caused by a difference in pressure.

Well, actsually more often the change in pressure is caused by flow. The pressure drop and movement of oil through them hoses and tubes caused a frictional
pressure drop between the pump and the cylinder. If I hooked a hose on a tree branch and tore it loose at the pump, I had all kinds of flow but the pressure
drop was negligible. Thets cause the pump kept pumping, but hte system had no restriction so the discharge pressure was zero. Thet is also why the spool
vents direct back to the reservoir in center position - full flow, but small restriction so small pressure drop and less wasted power.

It works the other way, too, of course. With thet ole single acting cylinder the wieght of the bucket and product dropped the bucket when I moved the spool
to zero. The pump flow was diverted direct to reservoir. The weight in the bucket created pressure in the cylinder, which created a pressure difference,
which created flow. I could add more restriction to flow by feathering the valve. This would decrease the downward velocity, since the bucket was a constant
pressure source. Forces would balance because the pressure drop through the valve had to equal the pressure in the cylinder. Pressure drop through the valve
would vary as the square of the flow. So pressure difference was creating flow, at the same time flow was creating pressure differnce. Cute!
 
Barn yard hydraulics vs Newton's laws of motion

Some sysetms is jist brute force producers.
All systems are subject to the same laws of physics. Even the brute force ones.

Now, when I shifted the spool in the valve they was obviously accelleration, but the time frame was imperceptible to the human eye. The cylinder hit max
velocity durn near instantly, seein' as how max velocity was very slow. If I wanted more flow I'd open the throttle on the engine and the bucket would go
from very slow to jist slow. More FLOW meant more GO, on accont of cuz the pressure capability of the pump was way more than the load induced by the bucket
Until the end of travel. Then what happens to your flow? It stops because the pressure equalizes. The bucket stops because the opposing force at the end of travel is equal to the hydraulic force. No net force so no motion. Oil flows because there is motion which causes the pressure to drop which allows more oil to flow to fill the space.

Flow made the cylinder go, and more flow made the cylinder go more faster!
Then why didn't the actuator go beyond the end of travel? You still had plenty of flow didn't your?

Now if I added load to the bucket while lifting (say, by driving the tractor into the pile of dairy by product while raising the bucket) my cylinder wouldn't
slow at all. Thet's cuz the load still weren't enough to exceed the pressure x area of the cylinder. The gear pump kept pumpin' at a constant rate and the
cylinder kept movin' at the same speed.
If you have a lot of power relative the load you won't see much of a slow down but that is a specific case. Most systems are designed to have the required power and no more because extra power cost money. Part of what I have to do is make calculations about how changing loads affect the speed and therefore the tuning.

On the other hand, if'n I hooked the bucket under the corner of the barn, the load exceeded the force capability of the cylinder at the max relief pressure
setting. I had force, of course, but not enough to cause accelleration or movement. The net force wern't zero, but there weren't no accelleration cause the
net force was in the wrong direction for cylinder movement.
You are forgetting about Newton's third law about equal and opposite reaction. The barn puts the same force on the bucket as the bucket puts on the barn. The net force is 0 and there is no motion. Newton's laws are very consistent.

Yep - no doubt about it. And when the net force is zero - like my bucket at steady state lift velocity - acceleration is zero too, but velocity ain't.
That is good because otherwise you would be violating Newton's first law.

Well, thet is jist a typo on your part, Peter. a = dV/dt ....which ain't integration. No change in velcotiy means no accelleration.
No error. You must remember that I do this all the time.
You had better review your calculus again.
a = dV/dt or
dv =a dt now integrate both sides.
v = ∫ a dt
Usually the acceleration isn't constant but a function of time so
v = ∫ a(t) dt
This basic stuff can be found in these links.
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity.html
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

Stop confuuzzin' the youngin's heads with your barn yard hydraulics.
 
I think we should all get together and go fishing.
747119_can-o-worms.gif


Hunting would probably be a bad idea.
129185867237633005.gif



🍺
 
All systems are subject to the same laws of physics. Even the brute force ones.


Until the end of travel. Then what happens to your flow? It stops because the pressure equalizes. The bucket stops because the opposing force at the end of travel is equal to the hydraulic force. No net force so no motion. Oil flows because there is motion which causes the pressure to drop which allows more oil to flow to fill the space.

Flow into the cylinder stops when you reach end of stroke. The force of the cylinder is at max then, of course. No flow, no go, and the no flow into the cylinder is caused by a mechanical constraint of the system and has nothing to do with the pressure increas. The pressure increase is in fact induced by the mechanical restraint preventing movement. Flow from the pump continues but dumps over relief.

Then why didn't the actuator go beyond the end of travel? You still had plenty of flow didn't your?

No flow into the cylinder with no travel!


If you have a lot of power relative the load you won't see much of a slow down but that is a specific case. Most systems are designed to have the required power and no more because extra power cost money. Part of what I have to do is make calculations about how changing loads affect the speed and therefore the tuning.

Most systems are designed to have the right power at rated load, plus a little safety factor. Most systems operate most of the time at much less than rated load. Those of us who have ridden alone in a hotel elevator can verify that!


You are forgetting about Newton's third law about equal and opposite reaction. The barn puts the same force on the bucket as the bucket puts on the barn. The net force is 0 and there is no motion. Newton's laws are very consistent.


That is good because otherwise you would be violating Newton's first law.

I didn't state it clearly - you have me there. The net force of the total system is indeed zero! The force the bucket produced was less than the total weight (force) of the barn, and the difference was made up by the resisting force of the dirt under the foundation.

No error. You must remember that I do this all the time.
You had better review your calculus again.
a = dV/dt or
dv =a dt now integrate both sides.
v = ∫ a dt
Usually the acceleration isn't constant but a function of time so
v = ∫ a(t) dt
This basic stuff can be found in these links.
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/applications/velocity.html
http://hyperphysics.phy-astr.gsu.edu/hbase/acons.html

Mathematically you are correct, Peter, but for the large class of problems where V = a constant and a = 0 the integration isn't much use. It is more common, and certainly more instructional, to look at acceleration as change of velocity over time. The integration adds no clarity to the discussion.

Stop confuuzzin' the youngin's heads with your barn yard hydraulics.

Actually, the barnyard hydraulics served two purposes. First of all, it described a simple system, easy to visualize, that includes all the elements needed for instruction. Besides, I hoped a little humor would ease the acrimony.

The youngin's is the point. Too many discussions in this forum have tried to ignore either pressure or flow. Neither extreme is correct. Proper analysis requires consideration of mechanical configuration and restraints, characteristics of the components, and effects of these on flow AND pressure. You can't say flow doesn't count, because in some systems flow is the dominant factor. In other systems, pressure is dominant. But in all systems both must be considered!
 
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No flow into the cylinder with no travel!
Exactly, it is the travel that allows flow. Not the other way around.

Mathematically you are correct, Peter, but for the large class of problems where V = a constant and a = 0 the integration isn't much use.
What? Acceleration is never instant. Instant acceleration requires infinite force and power.

It is more common, and certainly more instructional, to look at acceleration as change of velocity over time. The integration adds no clarity to the discussion.
Well there is the problem. What time? Did you see the motion profile at the bottom of this?
http://www.deltamotion.com/peter/Maxima/VCCM difeq.html

Actually, the barnyard hydraulics served two purposes. First of all, it described a simple system, easy to visualize, that includes all the elements needed for instruction. Besides, I hoped a little humor would ease the acrimony.
I object! You don't see all the mis designed systems that I do. You don't see all the frustrated control people that work long and wasted hours trying to get poorly designed hydraulic systems to go. Many think it is their problem when in actuality it is a hydraulic design problem. If the control guy knew hydraulics he would know that the system is poorly designed.

The youngin's is the point. Too many discussions in this forum have tried to ignore either pressure or flow.
Oil is just another way of getting power from here to there like gears or a drive shaft. Still everything starts with knowing the force or torque necessary to achieve the desired motion.

You can't say flow doesn't count, because in some systems flow is the dominant factor. In other systems, pressure is dominant. But in all systems both must be considered!
Flow is how the energy is transfered but this energy does no work until it applies a force and moves something. Flow just equalizes pressure or energy. Pressure can be thought of as kind of an energy density.

I attached a very simple hydraulic motor simulation. It assumes the valve is instantly fully open. You can see the motor takes time to accelerate. There are basically 3 differential equations that are implemented as very crude first order difference equations. Two for pressure and one for the angular velocity. The angular velocity is calculated by integrating the angular acceleration. You can't simply take the rated flow and divide by the cubic inches per revolution of the motor to compute speed.
 
Love the can of worms, alaric!

Peter, the calculations are very elegant. They raise some interesting questions.

They also serve very nicely to prove my point.

First off, on page 1 my good buddy flow shows up immediately! The pump dispacement (Cubic Inches per Revolution) and system flow both show up in your formulas. This is exactly the point I''m trying to make. You can't talk about a system's speed without including flow. (Flow makes it go to coin a phrase!) Of course, we know that flow can't make anything go without adequate pressure to provide the needed torque or force.

The second point is that there is a universe of applications outside of the sophisticated motion control systems that you are involved in. If I'm interpreting the time scale correctly, your system reaches terminal velocity in less than a second. A short accelertion time isn't a usually goal with a winch or swing circuit on a back hoe. In fact, they want to make sure the acceleration is slow - that's why they sometimes use valves with feathering notches in the spool!

In a winch circuit they create a pressure drop by throttling the flow, limiting pressure and torque and acceleration at the motor. The excess flow goes over the relief valve, since with a positive displacement pump total flow is a constant. To keep the speed of the winch motor down you need to keep the flow to the winch motor down. To keep the torque down you need to keep the pressure down by loosing some of it through the valve spool. You can't say flow doesn't matter - you need to consider both!

At any given point in time the speed of the motor is always flow divided by displacement! Once the load is accelerated and all the flow goes through the motor you ceratinly can take the rated flow divided by displacement to get velocity. If part of the flow goes to the relief, or if it is a pressure compensated pump, the velocity is determined by actual flow through the motor. A speed lower than "rated" speed means flow through the motor is less than rated flow. The reduction in flow is accomplished by either dumping some flow over the relief valve or de-stroking a variable displacement pump.

And that brings us to another point your calculations help illustrate. Implicit in your assumptions is that the power source is a constant pressure device with flow capacity greater than or equal to that needed by the motor at terminal velocity. But there are lots of systems out there, like my good ole manure bucket, that use a constant displacement gear pump of low flow capacity. At 100 rad/sec (15.9 rev/sec) your motor needs 80 cu in/sec flow. What happens in your system if the pump is only good for 40 cu in/sec? Your velocity won't get over 8 rev/sec because there isn't enough flow to make it go 15.9!

And, finally, you show the pressure at the inlet and discharge ports of the motor as equal, which implies no external load. Many applications are interested in moving a load and acceleration and other motion control niceties are insignificant. My question is, am I correct in assuming you included no external load to simplify the formulas?

These are the points I want the youngun's to understand. You can't eliminate flow and mechanical constraints from your analysis, any more than you can eliminate pressure from the analysis. They are all interelated and they all work together in the system. You must consider them all in the analysis.

I know you won't change your mind, Peter, and that doesn't bother me. You are an expert in a specialty, and you know how to make your systems work. A lot of guys here need to know the other things to consider in more general systems.
 
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Once the load is accelerated and all the flow goes through the motor you ceratinly can take the rated flow divided by displacement to get velocity.
It is that once the load is accelerated part that I have problems with. You have no idea what the final speed will be until you know the load that is being over come. You are also looking at it backwards. It is the fact that the torque has accelerated the motor that you get flow. At the beginning of the ramp up there is little flow until the motor starts moving. Then you can see the pressures across the motor drop and the flow will then increase to equalize the pressure.

I never said pressure and flow aren't important but look at the last line of the iteration
block. It is torque that causes the angular acceleration. In the first two pressure equations you can see that the flow is very small because the pressure across the motor is large. Only when the motor gets moving does flow increase.

This is a very crude starter simulation. Notice that the vale is open fully so there is no attempted to control the speed. If I add a torque load at 1 second the system will slow down. I can change Kv to be anything I want or increase the mass/inertia to get any rise time I want. The point is that you can't calculate the speed with V=Q*A. Speed is calculated by integrating acceleration.

I have versions 1 through 8 that get increasingly more complicated. The more complicated versions have a target generator and PID with feed forwards. There is a pump in the model that doesn doesn't supply full flow until it is at the bottom of its proportional band. I assume the pump is a pressure compensate pump. I have an accumulator. I can compute the oil into and out of the accumulator and calcuate the pressure in the accumulator. I simulate valve spool response and leakage. Leakage is very important.

You are trying to bring up fuzzy factors. I can model all you suggest and more.
The point is that a net torque or net force makes the motor go. The equation for computing the veocity doesn't need to change for any of the things you suggest.
The equation in the bottom line gets two more terms in the advanced models. One is a drag or friction proportional to the angular velocity and the other is a torque load function that allows be to change the load at any particular point in the motion.

Force/torque make it go.
Flow just attempts to equalize pressure.

You are looking at things backwards.
I think I have done my home work.
 
Reading this discussion yet again caused the following image:

Two guys standing on opposite sides of a pump violently disagreeing about which direction the motor is turning: clockwise or counterclockwise. However, they both agree that it is turning the correct way and the pump is operating normally.
 
Not quite, we are arguing which end of the egg to break and it makes a difference!
Actually there is a subtle but big difference in what we are saying. This is the problem I have with many hydraulic designers that use their barn yard methods to design servo systems. I don't care if you can get by with V=Q/A in a barn yard. It may work most of the time but it will not work all of the time. In servo systems V=Q/A doesn't work except for when the pressure in the cylinder is not changing but that only happens when some steady state condition is reached. V=Q/A does explain how that velocity is reached or if it ever can be. Pressure is always changing in a dynamic system.
 
I do get it Peter.

I'm with you on this one.

The special case where there is no acceleration can be handled with a basically wrong approach that still generates the right numbers. The fact that the method is used successfully by a lot of people doesn't make it correct.

The general solution illustrates what is wrong with that approach and handles all situations. That is why it is correct. Period.


People were able to navigate at sea for many years believing the world was flat, and for a limited area it works ok. That doesn't make it correct. Understanding the true shape of the world has little effect on how you sail short distances. That doesn't make it irrelevant.
 
I know the last thing you guys need is another farm boy sticking his nose in and asking questions.... but I'm always looking to learn more about hydraulics.

I may misunderstand the dispute, but I see it like this: Sufficient pressure must be present to cause the load to move, but since hydraulic fluid doesn't expand or contract, the amount of "movement" must be compensated for by REPLACING that volume of fluid.

The act of replacing that volume of fluid is "flow".

To perform the amount "work" you wish to perform requires "pressure" at "flow" or "flow" at "pressure". These TWO factors BOTH "define" or are design parameters of the system.

If this isn't true, I need to understand why.

Thanks

Stationmaster
 
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Dang! Intresting discussion! Shoulda made some popcorn!

If Tom was looking to get Peters goat, he couldnt have picked a better example to illustrate with, a manure bucket.

Its one thing to have someone refute your highly scientific thesis with ordinary examples, but for someone to call you out with a big pile of cow **** is another!
No way thats gonna slide by!

At any rate I really need to brush up on hydraulics myself, because I only "kind of get" the concepts discussed.
 
I may misunderstand the dispute, but I see it like this: Sufficient pressure must be present to cause the load to move,
Close but not quite. There must be suffucient force. Alaric had a problem a year or so back where he had normal system pressure but the piston backed up to the port so instead of having a whole piston area to push on there was only the port area so the piston didn't move.

but since hydraulic fluid doesn't expand or contract, the amount of "movement" must be compensated for by REPLACING that volume of fluid.
Hydraulic fluid does expand and compress but very little. Even water expand and compresses. The bulk modulus of oil, β , is a measure of how much force it takes to compress the oil.

The act of replacing that volume of fluid is "flow".
Yes, oil equalizes pressure or energy. I should redo that pdf to show flow.

To perform the amount "work" you wish to perform requires "pressure" at "flow" or "flow" at "pressure". These TWO factors BOTH "define" or are design parameters of the system.
Work=ForceXDistance or PressureXAreaXDistance.

Force makes things move. I love to p!ss off the hydraulic guys by constantly reminding them that Newton didn't include oil, flow or pressure in his three laws of motion.

I can calcuate velocity as a function of time using my force and and torque cause acceleration method. Tom can't with his V=Q/A.

May the Force be with you.

carwashblues said:
No way thats gonna slide by!
I can't let it slide by because then there will be people that will think Tom is right and design poor systems. This is bad for our business because usually we end up having to re-educated these people after they come to see that their servo system doesn't work as desired. This is usually very expensive.

I have pretty sophisticated simulators where I can get an pretty good idea of what will work and what won't. The simulation can never tell you a system is really going to work but it will tell you that a system will not. If the system seems to be OK then the simulator will point out where the week spots are going to be.
 
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Work=ForceXDistance or PressureXAreaXDistance.

I can calcuate velocity as a function of time using my force and and torque cause acceleration method. Tom can't with his V=Q/A.

Still, to MAINTAIN velocity, you must maintain "force". And to maintain force you must maintain "pressure". So you have to select a pump that can continue move the replacement fluid (flow) into the cylinder AT pressure if you want the 'calculated velocity' to continue. Otherwise your pump is expressed as having theoretically limitless capacity for flow. Real pumps have limits.
 
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Yes, the pump or the accumulator most supply the required amount of oil but notice this. Even though my motor has an endless oil supply at at constant pressure there is still a speed limit based in where the ∑ forces = 0.
Getting the oil supply right is easy. Usually people oversize the pump and don't realize that they can use big accumulator so the pump can run a constant velocity not ramp up and down supply with the motion.
 

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