You have to admit I'm consistent Peter. Even if I get the same response from you.

I'm still stuck on "Flow Makes it Go" and "Pressure is Resistance to Flow" since it explains many situations i an operating Fluid Power circuit.

Peter wrote:

quote:

I don't want to be a forum monitor where I must read everything you say about flow or pressure.

But you're so good at it Peter and I need all the help I can get in my waning years.

Gentlemen:
It's great to see such a lively discussion, but it seems to be diverging away from the spirit of our forums. If I may, I'd like to contribute from my perspective.

When I was first introduced to hydraulics, I learned that pumps convert mechanical power (torque and rotational speed) to fluid power (pressure and flow). I also learned that pumps only produce flow (theoretically, anyway), and pressure is created by a resistance to flow -- generally from a load or a restriction.

I later learned that pressure (∆P) is what moves the load. Here's why: When pressure on one side of a piston is great enough to apply a force on the piston that exceeds the load (acting on the other side), then the piston (load) will move. Once the load begins to move, the pump must produce enough flow to maintain sufficient pressure to keep the load moving.

I think the heated argument that has ensued results from mixing theoretical and reality. There might even be cross-mixing of the types of systems. In mobile hydraulics, power (flow and pressure)is regulated by a variable-displacement (variable-flow) pump. A load-sensing circuit ensures that the pump puts out only that much power (flow and pressure) to move the load at the required speed. So in this case, "flow makes it go" because we assume the pump always provides enough pressure to move the load.

An industrial servohydraulic circuit, though, generally requires a constant-pressure source. This is usually a pump and accumulators. We don't really care how much flow the pump puts out -- as long as it always provides constant pressure to our servo or proportional valve.

So I think this points out the fundamental differences. In mobile hydraulics, you assume you have adequate pressure to drive the load, so flow makes it go. In the industrial servo system, you assume you have adequate flow to maintain pressure, so pressure makes it go. Maybe I've oversimplified it, but that's how I see it.

I also remember leaning that pressure compensation and load sensing exist to save energy in mobile hydraulic systems. They don't have to respond in milliseconds, because they are controlled by a human.

With servohydraulics, however, speed is the name of the game. So if you have to waste a lot of energy to maintain a constant pressure source, so be it. Industrial machines must respond in milliseconds to maximize cycle rates -- so energy efficiency is sacrificed.

But that's not necessarily bad. I often compare it to driving. When gas prices are high, and I'm not in a hurry, I push gently on the accelerator to get going, and I look ahead and coast to red lights. But if I'm in a hurry, I may punch it as soon as the light changes to green and go as fast as I can, even if it means hitting the brakes frequently. It's the same driver and same car -- just different goals. Just like the differences between the two hydraulic systems.

Alan wrote:
"Here's why: When pressure on one side of a piston is great enough to apply a force on the piston that exceeds the load (acting on the other side), then the piston (load) will move."

Still sounds like Flow from a Pump met the Resistance of a Piston that was pushing a load and Pressure Increased until that Flow had a path to move through.

However if the Load was high enough the Flow would have increased Pressure to the Relief valve or Compensator setting and Flow would have been Diverted to Tank at Relief valve Pressure or would have decreased when Compensator Set Pressure was attained. In that case there would be Pressure but no movement even though the Pump was still turning at maximum Pressure?????

On the other hand, with a lighter load, the Pump Flow would have pushed the Cylinder Piston to the Load at Full Flow and Whatever Pressure it took to move the Cylinder Piston. At Load contact Pressure would have increased to Whatever it took to move the Load and it would move the Load at a speed set by Pump Flow filling the Volume of the moving Cylinder Piston.

There was ample Force to move the Load but Flow Made it actually move at a preset rate is the way I look at the process.

quote:

Originally posted by Jim K:
Pressure can only be changed by adding or subtracting oil (FLOW).

How about a cylinder mounted vertically Rod Up, at Mid-Stroke and full of Oil with 10 Sq.In. Area and a 1,000# weight setting on it? I believe pressure would read 100 PSI in that scenario.
Now, add another 1,000# weight and I believe Pressure would increase to 200 PSI. No OIl Flow, No Oil Added, just more Force added to Trapped Oil that has no way of Flowing. In other words the Oil is in a Full Resistance to Flow condition so the added Force could only increase pressure.

Peter;
When you have a Flow Port Plugged it has Resistance to ANY and ALL FLOW. Adding more Force only added more pressure since the oil trying to flow from the cylinder was blocked.

If more Weight (Force) is added to the rod the Blocked Flow Path will cause Pressure to incease due to the Resistance to the Flow that is blocked. Drilling an Orifice in the Plugged Flow Path which would allow 0.01 GPM to Flow will still see Pressure until the Cylinder reaches the end of stroke and all the pressurized oil has exited the cylinder.

Seems so simple to theis math challenged person.


Comtining to add Weight would cause pressure to increase until the cylinder failed from excess pressure.

Originally posted by Jim K:
Pressure can only be changed by adding or subtracting oil (FLOW).
I corrected this 2 weeks ago.

Peter, Iknow what Feed Forward is. I read your article on it; I thought it was excellent.

I found the discussion to be curious. I’m not seeing the mathematical significance in Peter’s argument so I’m thinking it’s mostly a matter of semantics. It is odd that Peter claims “flow makes it go” leads to errors but I didn’t notice that he offered proof. To reject an existing body of work he should offer clear proof as to how and why it leads to errors. However, the discussion seems to have been turned around to where people are trying to prove why “flow makes it go” is OK. I don’t know how many convolutions the arguments have been through before making it to this forum but Bud seems to ask a simple question,

BUD H&P OCTOBER 31-
I never got an answer I understood when I asked about a cylinder moving along 500 PSI contacted another load that raised the pressure to 1,500 PSI, How much faster would it go if flow remained the same?

Instead of answering the question Peter changes the question to say the flow would not remain the same and discusses the load pressure as if it were the supply pressure to a valve-controlled-actuator. Changing the load will change the load pressure so I assume Bud is referring to the load pressure being raised. Bud doesn’t say how the flow is being kept constant, adjusting the valve for a constant pressure system or if it’s a constant flow system, but if the flow is the same how can the speed change? Seems most of Akkamaan’s questions are from this confusion. I checked plctalk to see where the arguments were coming from and found,

Peter plctalk October 15 –
“Finally, there reason I started that post is that there is a section at the bottom of the hydraulic section that calculates speed of a hydraulic actuator incorrectly. It basically is the 'flow makes it go' equation where the flow is divided by the area of the cylinder to calculate a velocity. The flow makes it go equation is wrong and gets many into trouble. Those that use the 'flow makes it go' ,V=Q/A, equation don't take into account the force required to get to the desired speed. If there isn't enough force then the desired speed can't be reached. I/we get tired of explaining to customers that they or their hydraulic designer screw up the design after the blame our controller for not make the system go fast enough.”

There are many reasons a cylinder may run slow but I don’t see how V=Q/A would cause any of them. VCAs have a particular load vs rate curve for max valve opening based the characteristics of flow through a sharp edged orifice and a constant supply pressure. The feasible operating points lie between this curve and the origin. So of course points outside the max curve are not attainable. But to say there is not enough force or not enough flow is moot. So if the desired operating point were outside the feasible design region, how would Peter suggest solving it? Use a bigger cylinder to increase the available force? Changing the cylinder area will shift the curve to bring some points that were outside to inside the feasible design region. But it will also exclude some points that were inside the region. Using the same cylinder but with a higher flow valve will shift the curve outward at all points except the stall-load point. This solution takes in more operating points without excluding any. But I don’t see how V=Q/A causes the desired operating point to be outside the feasible design region. In fact it is needed to define the feasible region so it can’t be causing the problem. I know that Peter knows all this because he does a nice job of providing such a curve in the spread sheet “cylcalc”.

What else will make a cylinder run slow? Line losses? Yes, but using a bigger cylinder to provide more force will just increase the line losses at any cylinder speed. Also, V=Q/A is needed to estimate line losses and adjust the load vs rate curve so again it doesn’t seem to be the problem but rather a useful tool in preventing the problem. Again Peter is aware of this so it can’t be his problem with “flow makes it go”.

So if it is Peter’s goal to remind designers that the operating points must be inside the feasible region it will probably be better served if he drops his Quiotic quest to conquer the “flow makes it go” and Q=V/A equation. They certainly can’t be the monsters he’s imagines them to be. Isn’t “flow makes it go” the same as saying turning the input gear makes the output gear turn? If you’re driving a gear box with a torque converter with a maximum input speed and you want to operate at point outside the torque converters power curve do you blame the gear box designer for believing the input to output speeds are proportional to the gear ratio?



Pete plctalk November 17-
“If you let ΔVolume=Flow-Velocoty*Area then you can see that If Velocity=Flow/Area that ΔPressure must be 0. You can say that Velocity=Flow/Area only when ΔPressure needs to be 0 but that doesn't make the system go. There needs to be a net force to make it go. It may keep the system going under ideal conditions. It depends on other factors.”

Yes, compressibility adds some dynamics to the system, often represented as Q=A*V+(Volume(x(t))/Beta)*(dP/dt) in simulations and as a minimum spring rate in analysis. This can be important for the system dynamics and stability but does it invalidate V=Q/A anymore than compliance between the input shaft and output shaft of a gear box invalidates the gear ratio? No math model, with lumped sum, averaged parameters, is the real world. To be able to work with the model only the real world effects that have a significant impact on results are represented. If the purpose is to remind people to include the compliance of the fluid to determine transient behavior then again the goal will probably be better served by dropping the campaign against “flow makes it go”. On Peter’s spreadsheet “cylcalc” are some nice plots of load vs speed for an unequal area cylinder. The equations on the sheet labeled VCCM equations appear to be derived using a very standard valve flow equation which I am sure was developed for incompressible flow and V=Q/A equation even though the flow enters the valve at 2000psi and exits at 0psi. Operating at very high pressures might call for some adjustment of the valve flow equation but the expansion happens across the valve, not in the cylinder. Again, changes in load will cause changes in cylinder pressure which will cause some transient behavior due to the spring like behavior of the fluid column.

F=ma is also often cited as proof of a cause-and-effect relationship? As opposed to a=F/m? I don’t think there is a cause-and-effect relationship between the through variable (current, flow, displacement) and the across variable (voltage, pressure, force) in dynamic equations. Does a current flowing through a resistor cause a voltage drop or does a voltage drop across a resistor cause current to flow through the resistor? Does pressure drop across a piston accelerate the piston or does the fluid flowing into the cylinder at an increasing rate accelerate the piston and the acceleration of the piston create a reaction increasing the fluid pressure? Saying force causes acceleration is a matter of how we speak and not a point of mathematical relevance. With gas law, pv=RT, p&v don’t cause T.

Even though pressure has the same units as energy density, it is not an energy density. Equal volumes of two different fluids pressurized to the same pressure will have different amounts of stored energy if the average bulk moduli are different. So they obviously have different energy densities but same pressure so there is no way to relate pressure directly to energy. If the fluid was perfectly incompressible there would be no compression energy in the fluid no matter how high the pressure was.

Peter does provide a lot of useful information so people should make use of it, even if he may come across as bit arrogant at times. So I agree with Bud and “flow makes it go” doesn’t cause me the least trouble. Perhaps I’ve missed Peter’s point, as “trolls and ignorant people” are apt to do. So if Peter can demonstrate how “flow makes it go” leads to errors I’ll be glad to consider it but if not it might be best if he dropped his quest lest he run afoul of good sense.

(I see some of the symbols didn't translate)

This message has been edited. Last edited by: John Andderson,

There is a new blog out hydraulicanswers.com

Peter, often I find when someone disagrees with me that they are making no sense at all until I listen to what they are saying. However, no matter how I might try, I am not able to make much sense of what you write. Please don’t pretend the discussion has been about anything other than your criticism of an existing body of work that uses the equation V=Q/A that you describe as “flow makes it go.” You claim the equation is incorrect and that it leads to errors. You use Alaric’s question concerning what effect doubling the (supply) pressure to a VAC has on the velocity to criticize other work. You are unable to explain how or to what error it leads. You don’t even attempt to demonstrate any logical connection between the supposed screw ups you find and “flow makes it go.” A person may stand on a street corner screaming, every time there is a car accident, its because the stop signs are red. There may be a real connection in his mind but to every one else he’s the village idiot. I won’t say you’re the idiot but I do think you’re a Don Quioxte. You wish to create a monster that only you have the insight to recognize. I hope people reading these posts realize that much of what you post is nonsense.

“Can you predict the motion with Q=V*A? No.
Did Newton include flow in his laws of motion. No.
F=m*a determines the steady state speed.
So so designers screw up designs with flow makes it go, yes.”

What gibberish! If you know flow and area you know the motion within the limits of fluid compressibility and leakage. I thought you were trying to make a point concerning compressibility or cross-port leakage but you don’t. You don’t even address these effects except to muddy the discussion by throwing in compressibility in such a way that makes it sound as if the delta P across the piston compresses the fluid which is wrong. Can you predict the motion of a VAC without Q=V*A? Of course not. The steady state velocity is determined by the system, inputs to the system, and the loads on the system. F=m*a defines the relationship between force and momentum. An important equation but that is all it is and it is of no more importance than any other equation in the set of equations representing the system. How do you answer Alaric with F=m*a? You don’t. In fact you don’t even answer him except to say the speed won’t double because flow is proportional to the square root of pressure drop. How are you relating flow to speed? Why not just tell Alaric that V(no-load) will increase by sqrt(2) and V=V(no-load)*sqrt(1-F/F(stall))? There is no secret to these equations so why be such a drama queen? Just provide the equation so we might get some insight as to what you are making such a fuss about. And of course if Alaric had been operating near stall load you’d be wrong to say doubling Ps would not double the speed. It could more than double it.

I looked at http://forums.hydraulicspneumatics.com/eve/forums/a/tpc...1030121/m/6761057803. If your story is factual I expect there’s a system engineer that’s very upset with you. I don’t know if he failed to give the PLC guy adequate instructions or if PLC guy just can’t follow instructions, but he certainly doesn’t sound competent. If I read it right you really messed up. Seems you put the cart in front of the horse and you’re costing them a lot of time and energy. I’ll read it again and try to gather my thoughts on it and post there later. But I suspect you may lack adequate experience in utilizing hydraulic equipment to be making such inputs.

You may be right about the force usually being on the x-axis on cylcalc. That’s probably not a real big deal as long as the axes are labeled. But what you may want to do, and how they are usually shown, is normalize the values to extend stall force and extend no-load velocity so all values are between 0 and 1. You’re equations are a bit sloppy. They’re usually streamlined in the form
Extend
Fe = Ae*Ps – (Ve)^2/C or Fe/Fe(stall) = 1 – (Ve/Ve(no-load))^2

Retract
Fr = Ar*Ps – (Vr)^2/C or Fr/Fe(stall) = Ar/Ae – (Vr/Ve(no-load))^2

C = Kv^2 / (Ae^3 + Ar^3) (Kv is for a single orifice)

Fe(stall)=Ae*Ps and Ve(no-load)=sqrt(C*Ae*Ps)

And of course, these come from three basic equations
Q=Kv*sqrt(deltaP)
Q=V*A
F=A1*P1-A2*P2.

There is no F=m*a involved, so tell me Peter how do you answer Alaric with F=m*a and not Q=V*A? What F is used for is not specified. You know what load curves are so now you should learn to use them.

The real question, Peter, is do you ever get tired of wallowing in drivel. Your comments seem mostly pointless and useless. The best I can figure is that you are trying to point out that the flow across a sharp edged orifice is proportional to the sqrt of pressure drop across the orifice and that an unequal area actuator has different load curves in the extend and retract directions. I expect most people using hydraulics are aware of this but if plc and sawmill people are not it’s good to get them up to speed. This may lead to a few questions that can be concisely answered. But instead of making useful comments you make an inane comment that the “flow makes it go” equation V=Q/A leads to errors that you are always having to correct. This lead to 15-16 pages of drivel on plctalk. If your goal is to wallow in drivel you’re doing a great job. If not you may want to rethink your approach.

Taking a step off subject, I also find it disappointing that Peter is unable to follow logical connections among basic engineering principles. Somewhere in plctalk I think I read someone saying hydraulics needs a unified theory. I think that for practical engineering applications, system dynamics does allow the various, seemingly dissimilar effects, to be tied together quite nicely into a single set of equations. Electrical, magnetic, hydraulic and mechanical effects can all be analyzed as a single consistent system. I would expect all engineers are required to take at least one course in system dynamics. Cannon and others separate the variables into two general classes, through variables and across variables, whose product is always power. The elements can be divided into four basic classes; those that dissipate energy, those that store energy in proportion to the square of the through variable, those that store energy in proportion to the square of the across variable, and those that operate on the variables to change the relative ratios while keeping the power product constant. These elements are idealized mathematical models of real world effects. Components are modeled by combining various elements to account for the significant effects. Levers, gears and areas would fall into the last class. The piston and cylinder changes the flow and pressure into velocity and force. Just as a set of gears changes the input torque and velocity into output torque and velocity. If there is an error in one element of a consistent methodology for solving problems, then the same error should be seen in all elements in the same class. I submit that any supposed error Peter feels exists in a cylinder caused by V=Q/A can be modeled in a gear and have a similar effect on omega(out)=omega(in)/Kgear. The relative magnitudes may vary but the effects should be there. Peter’s disassociation of physical elements may be a general problem among US engineers. I cannot vouch for the factuality of it but I’ve heard it discussed that US manufacturers generally use more parts for similar product lines. This increases production costs by increasing inventory costs and decreasing economies of scale. It also dilutes the effect of money spent on product improvement because there are more parts to improve and optimize. This inability to see the commonality leads to more expensive and lower quality products.

To further broaden the discussion off subject with another wild step I’ll suggest the root cause of the current financial and economic problems is lack of quality engineering in the US. The US has been running large trade deficits since the 70s. Economists will argue that trade imbalances help flatten out economic cycles but I don’t know of any economic model that is stable under continuous deficits. I think the US chose to avoid a stagnant economy under the weight of our trade deficits by pursuing policies to encourage easy credit, low interest and large tax cuts. Such policies would cause rapid growth and inflation in a healthy economy but with the burdened economy they just provided modest growth. But it couldn’t be sustained and was bound to collapse. Certainly lack of regulation and oversight made the collapse more spectacular but I don’t think a deficit economy could be sustained forever. Better engineering would provide more competitive products that could improve the trade deficit. Bud is always saying we need to improve hydraulic education but in general we need to improve our technical base.

There’s been criticism people’s ability to communicate in English on this forum. In my opinion I’d rather read something sensible in poor English than nonsense in perfect English.

this is funny...reminds my about my day...a friend of mine came over to my house with his 8 months old Rottweiler female 80 lbs, Reba. I have a 4 months old german Shepherd male puppie 40 lbs, Kaiser. Reba and Kaiser are plying along next door neighbors fence. There is a little Yorkie female 3lbs, Coco across the fence.She is barking loud and she is safe there. Reba (Peter) is of course in power control due to weight. Kaiser (John A) is attcking agressively never giving up even if beat up. Coco (Maglub) standnig on the other side safe, takings side for some one that probobly would eat her alive at first chance.

Keep barking guys, I enjoy it....

quote:

Why would you say it is wrong(V=Q/A)?! That is the formula used when the actuator is in motion(constant speed), idiot! We can predict the the motion by using this equation. You are always implying the formula F= ma, what is the use of this formula? You mean to say that this formula alone can accelerate your load? Stupi

give it a break....

This is not addressed to mag, as I doubt it matters, but to other readers.
BOTH formulas are required. Q = V * A is valid only if you know V or Q. You cannot know the final velocity or flow without knowing acceleration and deceleration, which is a balance of forces. Q is NOT pump rated flow or servo rated flow or any known number. Nor is it defined by average distance / time. Peter harps on pressure only, which is only true if the supply is constant pressure and infinite flow. That is true in his motion control world. In that case, balance of forces is F = M A, that generates a motion profile of velocity and accel, which defines the REQUIRED flow to achieve that. If the system is flow limited to something less than the required flow, the pressures drop off at some velocity and the required profile is not obtained.

So if the system is PRESSURE limited, the forces and accel decel define it. If it is FLOW limited, the flow defines maximum speeds which may clip off the profile.

k

Woohoo! I made Peter’s s-list. I must be doing something right! But I am surprised you feel such affection for me Peter. You have a pet name for me. I’m touched. But I’m not sure whether it’s just a personality disorder or if you think you can bully people into acquiescing to your inanities? I believe technically it’s know as the empowered nerd syndrome.

When I say your equations are sloppy I’m not implying they are incorrect. They seem to be correct and the plots you’ve labeled “differential force without load” are important in hydraulic design. When I say sloppy I’m referring to your collection of terms. Such as max extend no-load velocity (Calculations!M4) you have

V=sqrt((Kv*Kv*Ae*Ps)/(Ae^3*(1+(1/(Ae/Ar)^3))))

I’m not sure why you have so many inversions but it’s harder to type, harder to debug and programmers usually tighten up the equations such as

V=Kv*sqrt((Ae*Ps)/(Ae^3+Ar^3))

Then it’s a geometric constant times a variable,Kv, which is important when running in real time. Not important in a spreadsheet but I’m accustomed to programmers habitually doing things that way.

For understanding the relationship between force and velocity it’s generally considered best to write the equations in terms of no-load velocity and stall force. So close to what Peter? There is nothing more to an actuators load curve. You have a way of trying to imply the discussions were about something other than what they were.



From Peter’s post:

“I think you should study Newton's laws of motion. I am not a drama queen. I have posted links to the VCCM equation many times. If the people on the forum don't know the equations then it is their fault not mine. The VCCM equation is similar to those equations you posted but combined into one equation that will predict steady state speed.”

“quote:

And of course if Alaric had been operating near stall load you’d be wrong to say doubling Ps would not double the speed. It could more than double it.

Yes, but the actuator wasn't moving near stall load. So how do I know? Because I had talked to Alaric about his application on the phone. If you look at post #3 you can see I qualify my response by saying it is for no-load. You are trying to misrepresent what I said as if I didn't know this. You are like a politician.”

Yes Peter, in the third post on plctalk you did say Alaric was talking about no-load condition. I assure you it was not an effort on my part to misrepresent what you said, it was simply an inability to absorb and retain the key points among 16 pages of drivel. I admit, I did not read 16 pages, inanity bores me. I glanced through it to try to determine the origins of the discussion Bud was referring to. The best that I could determine is that you were trying to make two remedial points about actuators. That the flow across a sharp edged orifice was proportional to the square root of pressure drop and that an actuator of unequal area has different power curves in extend and retract. But that’s my point. If you wallow in drivel it is hard to get anything out of it, but I suspect your intent is to generate drivel. Since you didn’t answer the question outright at the beginning by saying the speed would increase by sqrt(2) I assumed you were suggesting there could be a load. By mentioning there could be a load I was hedging my answer.

When did you post the equations? If you say the equations are available, why was Alaric asking what happens to no-load velocity when supply pressure is doubled? Are the simplest points lost in drivel? If so are you sure it’s always “their fault not mine.”?


I do have some experience with power systems but I’m not ashamed to admit I could use some help in selecting the optimal pressure. There’s so much to consider, cost, weight, volume, efficiency, how distributed the system is, reliability, etc. Figure of Merit or other procedures are applied but it always seems somewhat contrived because the weighting factors tend to be, to some extent, arbitrary. So if you have any sensible inputs I’d be glad to read them.

But I’m suspect Peter’s not asking about anything significant, probably another remedial point that he’s discovered. What do you mean Peter by “OK a hole. How do you derive the optimal operating pressure for power? The calculations use the equation used to generator the load curve. We all know the answer so lets see how you derive it.” Are you talking about the actuators peak power? Wow! Peter, do you really feel that takes some mathematical sophistication to multiply F*V and differntiate it relative to V(or F) and set it equal to zero! The point is Peter, yes, max power is important but do you need to pretend its difficult to calculate? Your tendency to talk down to people tends to distract.


Peter’s quote
“So close. You posted most of the VCCM equation in your post but can't put it all together.
I suggest you research the VCCM equation. Then you will see that speed determines flow not the other way around.”

I’ve misread you Peter. I thought you were serious but now I understand your doing a comedy routine. The output creates the input! The speed determines the flow, that’s real laugh. I’d like to see the difference equations for such a simulation. So lets see Peter, we open the valve causing the actuator to move forward and the actuator’s motion sucks fluid into the cylinder. You really have discovered a new principle that only Peter has the insight to see. So we have been doing it all wrong. I have been thinking that when the valve opens fluid flows into one side of the cylinder compressing the fluid, thus raising the pressure, and out the other side lower the pressure, and the change in pressure across the piston caused an acceleration that leads to velocity. But Peter says velocity determines the flow so opening the valve creates a force on the piston leading to a velocity. Then the velocity of the piston causes fluid to flow in and out of the cylinder. A strange understanding of cause and effect. That’s up there with your claim that pressure is an energy density since it has the same units as energy over volume. That the actuator velocity causes fluid to flow into the actuator has to be the apex of your routine so you’ll have to move on to some new material and give hydraulics a rest. I don’t think you’ll top that.

kevin j,
I assume we are talking about VCAs on constant pressure systems. Yes, when the net force on the piston is zero the acceleration will be zero. Lets assume we know the remedial stuff. But what determines the final speed, what determines when F goes to zero. If you have a 10,000lb mass on linear rollers so that there is no appreciable drag and a 100 pound mass on similar rollers and each is connected to an identical actuator, what are the final speeds? They’ll be the same. If you have a 10lb perforated plate in viscous oil and the same actuator the final speed will be less than the larger masses. Is saying the final velocity is reached when the net force is zero any different than saying the final velocity is reached when the velocity remains constant? Net force equal to zero is a condition of being at final velocity, not what determines final velocity. I don't get your point as to Q=V*A as being valid only if you know Q or V. The equation is valid whether they are known or not. Peter's VCCM equation, or power curve as it is sometimes reffered to, is composed of 5 equations and 6 unknowns, Pin, Qin,Pout, Qout, V and F. We want to know how fast the actuator can travel under various loads so we use the 5 equations to eliminate 4 variables, Qs and Ps, so that we are left with one equation and 2 unkowns. So the maximum speed the actuator can move under each load can be plotted. Ther is no cause and effect relationship implied by the unkown we choose to solve for. F=ma is not involved because it doesn't matter what the load is. This curve is important in sizing an actuator. What ever the load is, if all the load points are inside the actuator power curve they can be achieved with suitable command and the supply pressure being constant. If the hydraulic supply cannot maintain pressure the power curve is derated to the pressure expected. Ideally the actuator would be load insensitive but they're not. If you know the load, such as a mass, you can use the equation to compensate for load,as I think you were refering to. I appologize for being overly remedial but I'm afraid our differences are even more trivial than you may be thinking.

Peter’s quote
“John could have asked nicely why I think V=Q/A is poor for calculating actuator velocities. John didn't realize that he admitted it here.
quote:
V=V(no-load)*sqrt(1-F/F(stall))?

This is right, V=Q*A is wrong for calculating actuator velocities. They can't both be right. V=Q/A doesn't take into account the net force which is something I have been saying all along. I have no idea why he jumped on my ass.

During operation the pressure and load force will vary. I DON'T always assume that the supply pressure is constant. There is a load force term and a pressure term in the VCCM equation so I will simplify things by just saying
1. Vss=VCCM(Ps,Fl);
2. Q=V*A therefore;
3. Qss=Ape*VCCM(Ps,Fl);
Now we have the steady state flow. Do you agree so far? ”

Peter, I still don’t get your point. Q/A=V(no-load)*sqrt(1-F/F(stall)). They are the same V. Q=V*A was used in deriving the power curve equation. (Did someone type V=Q*A? Certainly you realize that is a typo.) It was used twice, for in flow and out flow. Qin and Qout were eliminated so as to have the equation in force and velocity. How is 3. any different from Q=A*V? It seems a circular argument. Is this why you state velocity determines the flow?

Now you’re trying to make me feel bad Peter. I didn’t know I was the one who jumped you out of the blue. I thought on plctalk you stated you liked to ruffle hydraulic guys feathers. Was that not your intent? I stated I had no problem with the “flow makes it go” statement and I didn’t see that it led to errors. To trash other work you need to clearly show why it is wrong. I thought I asked nice in my first post,"So if Peter can demonstrate how “flow makes it go” leads to errors I’ll be glad to consider it but if not it might be best if he dropped his quest lest he run afoul of good sense."


I don’t want you to be lonely Peter but I’m not going to be able to keep up with you on these posts. I’m glad you like to smash grapes, everyone should enjoy a hobby, but don’t drink so much of the juice. It tends to fog the thinking. Just joking.

quote:

Does force make it go?
Does flow make it go?

I can't spend much time due to death in the family and an upcoming wedding, but still want to wade in again.

I agree with both bud and peter, and disagree with both. Both sides are correct, IN CONTEXT, as far end special cases of the general system curve of the VCCM equations.

Flow and force are not mutually exclusive terms. Both are required and related. Newtons law is F=MA, which means force to keep moving, accel or decel. A porta power hand pump can create force imbalance on a cylinder. However, as soon as a piston moves more than the compression of the fluid, the pressure decays and goes to (what?) The pump must replace the fluid at a rate enough to maintain the pressure and force to maintain motion or acceleration. This means flow, which is pressure drop through lines and valve, which means piston pressures will be something less than pump pressure (ignoring pump and accumulator droop, which peter is correct on). The pump must be capable of moving this much flow, and prime mover must have enough torque to turn the shaft against the pressure loads. If any of these assumptions are not met, there are limits outside of the VCCM equations.

VCCM equations predict the entire system curve. Stall force (pressure limited, no flow) is one extreme. Slewing speed maximum velocity with no load is the other extreme. Speed at some force defines points along the curve. VCCM assumes enough flow and pressure to reach the steady state of the curve. Pump limited maximum flow is an upper limit that prevents achieving maximum speed. Mass and pressure/force may not achieve the acceleration and deceleration required.

Bud sees the flow limited condition, and speed is flow divided by area. If the system is only flow limited, then bigger pump = faster speed. Peter, flow makes it go, IN THIS CONTEXT. If the system was almost pressure limited by undersized open loop valves, or is on a servovalve control, then bigger pump goes across relief and ‘flow makes it go’ is wrong IN THIS CONTEXT.

Peter sees the pressure limited condition. I agree with Peter on the common mistakes. Speed is NOT valve or pump rating divided by cylinder size. I also see that mistake many times. Increasing pressure increases speed by the VCCM calcs, not directly. Assuming the pump is already ‘big enough’, then doubling the pump size has no effect and ‘flow makes it go’ is wrong.

So, ultimately, I will describe this as yes, flow makes it go. But ‘flow’ is from the VCCM calcs, not pump size, or servo rated size, or someone’s wish size. I agree with Peter here, do the math first, and find out which ‘flow’ we are talking about.

Peter: I looked at your site briefly. I think it would be helpful to add sketch of the lands and cylinder and loads, like johnson’s sketches. I am glad you are posting the material.

I struggled with these concepts a long time. They are not intuitive. Ultimately Johnson helped me work through the math til I understood it. Not higher math, simple algebra, just really ugly. VCCM is ultimately steady state for a small finite amount of time. May change an instant later, but it is simply balancing flows and pressures for this steady state instant. I think simple math.

In the Alice in Wonderland world in pressure limited conditions, yes, cylinders retract faster than extend. Most people see the logsplitter work and are convinced that extend = slow, retract = fast is some sort of universal hydraulic law. It is not, it is only a special case of the system curve where the system if FLOW limited by the gear pump. When PRESSURE limited, it is the other way around.

Peter’s math is the dynamics of the small time intervals and I struggle with that part. But it is essential if fluid power is to have any future.

- - - - -
I posted this last week in the thread about air cylinders. Maybe redundant, but easy to cut & paste. No time to edit.

BTW, this somewhat illustrates what I think are some errors in the ‘Flow/pressure makes it go argument’. Like the blind men describing an elephant based on what portion of the beast they are sensing, Bud and Peter are describing/arguing two different conditions on a system curve.

Bud describes open loop situations where the pressure drops in valves are small compared to load pressure. Speed is determined by flow because the system is FLOW limited, not pressure limited. Operating pressure is determined by load resistance (and pressure drops of course but they are small in comparison to the load.) Acceleration and deceleration are brief compared to the steady state part of the cycle, so total cycle time is close enough with just steady state calculations. Natural frequency and ringing are so small in time that they are ignored, sometimes in great error and with bad results as Peter often sees.

Peter is assuming (not always assuming, but usually seeing in motion control. I agree with peter’s comments. kcj edit 2/5/09) the system has infinite flow available and is PRESSURE limited, much like the electrical engineer assumes constant voltage available from the wall socket or electrical grid. In many of his posts he states this fact and I think forgets that this scenario is not true in many open loop or bang/bang systems. (It is of course essential in closed loop control to have constant supply source.) Pressure drops across the controlling devices are high, and the load pressure is smaller in comparison to the available pressures. Pressure/forces for acceleration and deceleration are of great importance in natural frequency and whether the cycle can be met, especially if it is a sine wave or repeating cycle.
I understand and agree with most of Peter’s writings, but I also think you see everything in terms of constant pressure supply and motion control and very brief time frames. Makes sense to me, as by definition of the Delta products, those are the type of applications you see.

Both worlds are part of fluid power. The open loop power world is one extreme. Still the bulk of applications in the real world, but in my opinion not where the future is. The closed loop dynamic motion control world is the other end, and I think the essential future for fluid power. Both are ruled by the same physics, just differ in which parts of the physics are small enough to ignore in which scenario.

And, yes, I think we need more training and technical knowledge, in I think two different focuses. In the last 24 hours I have dealt with field operators who need the training Bud pushes, and also dealt with engineers who don’t have the physics and energy and math understanding that Peter pushes. To me, it is like arguing if medical school should teach 2 year nursing skills or 10 year brain surgery skills. Both are needed for different students, and both types of students are needed for the market at hand.


kcj

Peter,
Centrifugal pumps are very seldom used in hydraulic systems and never in servo/proportional power units.

Positive displacement pumps have different characteristics (thin blue vertical line)as shown in the modified diagram downloaded from gouldspumps site.

Servo/proportional systems works better if the variation in flow doesn't produce pressure change so piston pump + accu should guarantee as much flow as the servo require without noticeable pressure variation, pressure variation are disturbs

Centrifugal pumps are pressure generators(thin red horizontal line), but real centrifugal pump because the slope of the characteristic curve, shown a remarkable reduction in pressure from zero to max flow.

Luciano