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Bernoulli
Posted
Hello dear
I am new to this forum so hi to every one
I want to ask that have any body used series hydraulic circuit in order to sync. the movement of 02 hydraulic cylinders. If it is so then what is the mechanism of pressurising the line connecting the two cylinders.
 
Posts: 5 | Location: Pakistan | Registered: 21 August 2008Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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Attached is a Power Point presentation from a training book I use to teach and advanced Circuit Design Class. This is one of about eight ways to synchronize actuators with hydraulic components.

The book is Fluid Power Circuits Explained and can be seen at http://www.hydraulicspneumatics.com/200/eBooks/

It is only partially up so far. H&P is putting a Section up avery Month or so.

Also take a look at the Basic Book that is not tied to any mannufacturers equipment but shows all the different component designs I have ever come across.

I have a third training book to teach Trouble Shooting Hydraulic Circuits and I can send brochures with content and prices if you like.


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley


PowerpointSynSeriesFa.ppt (1,264 Kb, 90 downloads) Series Cylinders Synchronizing
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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I made the Power Point presentations as a help for teaching the class on circuit design. The book has a lotmore info on filling the blind area between the cylinders.

All the hydraulic ways of synchronizing actuators must have a way to re-synchronize when they get out of phase.


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
Pascal
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Slide 13 isn't quite right. Hint the problem is with the left cylinder.


Peter Nachtwey
Delta Computer Systems, Inc.
http://www.deltamotion.com
"Living is easy with eyes closed, misunderstanding all you see..." John Lennon
 
Posts: 325 | Location: Vancouver, WA | Registered: 09 August 2006Reply With QuoteEdit or Delete MessageReport This Post
Pascal
Picture of Nahum Goldenberg
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To Synchronize Cylindersin series use the followings:
NG-Pitagoras Equations.
-------------------------------------
1) Master Cylinder Dimeter..... = K x 21/2.
2) Master Cylinder Rod Diameter = K x 11/2
3) Slave Cylinder Diameter..... = K x 2
4) Slave Cylinder Rod Diameter. = Any < K x 2.
You can proceed similarly with Slave2 Cylinder
as a new master cylinder and so on.


Nahum Goldenberg
info@hydrocad.com
www.hydrocad.com
www.hydrocad.blogspot.com
 
Posts: 208 | Registered: 02 April 2006Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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quote:
Slide 13 isn't quite right. Hint the problem is with the left cylinder.


I give up Peter, What is wrong with the slide????


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
Pascal
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quote:
Originally posted by Bud T:
quote:
Slide 13 isn't quite right. Hint the problem is with the left cylinder.


I give up Peter, What is wrong with the slide????

The gage reads 0 psi. What about the load induced pressure?


Peter Nachtwey
Delta Computer Systems, Inc.
http://www.deltamotion.com
"Living is easy with eyes closed, misunderstanding all you see..." John Lennon
 
Posts: 325 | Location: Vancouver, WA | Registered: 09 August 2006Reply With QuoteEdit or Delete MessageReport This Post
Bernoulli
Posted Hide Post
Thank for the Slide
LEt me see these and then i will let you know the results
With regards
Muhammad Umair Nazar
 
Posts: 5 | Location: Pakistan | Registered: 21 August 2008Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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Peters answer to asking what the problem with Slide 13 Gauge PG1 reading 0 PSI.
quote:
The gage reads 0 psi. What about the load induced pressure?


Gauge "PG1" only has Line and Valve Flow Resistance Back Pressure and the only pressure it will see is "Whatever it Takes" to send oil through the Lines and Valves and on to Tank.

I know this is always some pressure but it also varies in any system with Resistance Level and Flow amount. In all my training sessions I refer to this Back Pressure as "WE" (Whatever it Takes) since it may be an amount that can be figured but usually not something easily modified.

If this Flow Resistance was not addressed by the Circuit Designer, as it should have been, then there needs to be more training for that person.

In my way of thinking it is not "Load Induced Pressure."

The Flow Resistance in this line should be under 75 PSI. I usually size Return Lines for 25-50 PSI Back Pressure since all the Back Pressure here is resistance to Pump Flow and cannot do any WORK so it is all WASTED ENERGY/HEAT.


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
Pascal
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'What ever' is not an engineering term.

Lets assume the area of the piston is 10 in[sup]2[/sup]
The load can be calculated by using the pressures and multiplying by the areas.
The left side has a load of
(1250-450)*10=8000 lb
the right side as a load of
(450-0)*10=4500 lb
This makes sense as the load is mostly on the left side. There is no movement or flow so these calculations are easy.

When going down the calculations are more difficult but no much. If friction is 100 lb then the pressure on the bottom side of the cylinder is the total force pushing down divided by the area. The total force is

8000+450*10-100=12400

If total force is divided by the area

12400/10=1240 psi.

We know the load induced pressure must be this high so the sum of forces acting on the left side is 0. If not then the system will never be at a steady speed and the system will bind. Newton strikes again.

Since PG1 is at 0 PSI there must be a lot of pressure dropped across the counter balance valve. Smiler

If you were right about the pressure being at 0 then the left side of the platen would try to fall with an acceleration rate of 12400/8000 or 1.55g which is faster than what the load on top the platen would fall. That would be interesting to see. I am sure the 'guided' platen would bind first and the resistance of twisted metal would slow down the left actuator. BTW, 1 lb force ( lbf ) applied to 1 lb mass ( lbm ) will accelerate at 1g. ( It would be nice to use metric units. )

There is no indication what the pressure really is on the top right cylinder but lets just do some math.
The force pushing down is

Pb*10+4500-100

where Pb is the b port pressure to the top of the right cylinder
and the force pushing up from the bottom in your diagram is still 4500 lb. If you do the math we have

Pb*10+4500-100=4500 so Pb=10 psi. That means there must be a lot of pressure drop across that land of your Bang-Bang valve. If there is not pressure drop across the valve then the next force would be

2000*10+4500-100=24490 lbf

If I divide that my the load on the right side I get an acceleration of 24490/4500=5.4422g which is MUCH faster than the load on top of the platen will fall. You can see that the right side will fall much faster than the left side but again I am sure the resistance to motion due to twisting metal will slow things down.

Actually, the accelerations will be worse than I predict because after the system starts accelerating down the load on top of the platen will not be in contact with the platen so the hydraulic force will be applied to the platen only.

The point is that both of these cylinders must go up and down at the same speed and will accelerate at the same rate ( ignoring oil compression ). This means the force relative to the masses must be the same while accelerating and the the net forces for both actuators must be 0 at a constant velocity. I suspect you are counting on a lot of structural stiffness in the guided platen to 'balance' the forces.


Peter Nachtwey
Delta Computer Systems, Inc.
http://www.deltamotion.com
"Living is easy with eyes closed, misunderstanding all you see..." John Lennon
 
Posts: 325 | Location: Vancouver, WA | Registered: 09 August 2006Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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See answers at your input:

quote:
'What ever' is not an engineering term.

I knew that, But the untrained seem to undestand the term.

Since PG1 is at 0 PSI there must be a lot of pressure dropped across the counter balance valve.

You can read about Counter Balance Valves and their function in Chapter 14 Pages 3-6. Essentially they are a Relief Valve used in a diferent circuit location to control flow in relation to pressure. They are set 100-150 PSI higher than the Load Induced Pressue and must be opened by pump flow to the retract side of the cylinders. This keeps the cylinders from moving until the circuit wants them to regardless of the Load force. Been around for years and can be Internally Piloted as shown, Externally Piloted and Internally/Externally Piloted to achieve different end results. That means the Runnibg Away Load will not move any faster than the pump flow is forcing it to. Stop pump flow and the cylinders stop.

If you were right about the pressure being at 0 then the left side of the platen

Pressure is not at ZERO on the left side of the Platen it's at ZERO at PG1 on the Right Side (Outlet) of the Counter Balance Valve.

Pressure on the Left Side of the Coubter Balance Valve is Whatever Load Induced Pressure is while the platen is stopped and Whatever the Counter Balance Valve Pressure Setting is while the Cylinder is being pushed down as fast as the pump supplies oil to its given area on the retract side of the piston. That means the Load will not fall butonly move as fast as the pump pushes it. The Counter Balance Valve causes an Overrunning Load to act like a resistive load and the resistance can be set as desired.

would try to fall with an acceleration rate of 12400/8000 or 1.55g which is faster than what the load on top the platen would fall. That would be interesting to see. I am sure the 'guided' platen would bind first and the resistance of twisted metal would slow down the left actuator. BTW, 1 lb force ( lbf ) applied to 1 lb mass ( lbm ) will accelerate at 1g. ( It would be nice to use metric units. )

There is no indication what the pressure really is on the top right cylinder but lets just do some math.
The force pushing down is

Pb*10+4500-100

where Pb is the b port pressure to the top of the right cylinder
and the force pushing up from the bottom in your diagram is still 4500 lb. If you do the math we have

Pb*10+4500-100=4500 so Pb=10 psi. That means there must be a lot of pressure drop across that land of your Bang-Bang valve. If there is not pressure drop across the valve then the next force would be

2000*10+4500-100=24490 lbf

If I divide that my the load on the right side I get an acceleration of 24490/4500=5.4422g which is MUCH faster than the load on top of the platen will fall. You can see that the right side will fall much faster than the left side but again I am sure the resistance to motion due to twisting metal will slow things down.

Actually, the accelerations will be worse than I predict because after the system starts accelerating down the load on top of the platen will not be in contact with the platen so the hydraulic force will be applied to the platen only.

Acceleration will only be as fast as the pump moves the cylinder since the Counter Balance Valve must be forced open to allow flow after its pressure seting is reached. Anytime Pressure at the Counter Balance Inlet drops below its setting it will start closing and throttle flow at its setpressure.

The point is that both of these cylinders must go up and down at the same speed and will accelerate at the same rate ( ignoring oil compression ). This means the force relative to the masses must be the same while accelerating and the the net forces for both actuators must be 0 at a constant velocity. I suspect you are counting on a lot of structural stiffness in the guided platen to 'balance' the forces.


I am counting on the Load to resist Pump Flow while Extending and the Counter Balance Valve to RESIST THE LOAD WHILE THE CYLINDERS ARE RETRACTING SO THEY ONLY MOVE AS FAST AS THE PUMP FILLS THEM DURING THEIR DECENT.

It's in the Basic Book.


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
Darcy
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I'm just now entering this conversation, but figured I might be able to supplement this conversation with some additional calculations.

At the moment I'm a little pressed for time so I won't list the equations, but will touch back on that with a follow up post, but both sides have a valid point.

Peter: your calculations, on face value are correct, from what i've seen. There are some Force balance problems with the circuit, which i agree with you on. The only thing that caught my eye was the values you used for the "g force" number, being that lbf/lbm, whereby you produce an acceleration value, shouldn't your lbm be:

lbm = 8000 lbf / 32.8 ft/s^2

since you already defined 8000 to be lbf when calculating the F = P * A. Although this may correct the equation, it increases the "g force" even higher then what you are saying!!!

Bud: From what I understand, this model was used as an educational purpose and although the numbers may not be exactly on, it displays the required task of conveying the type of system logic required to preform series cylinder operation.

I will post my equations later, but it will need to include such assumptions as flow rate, counterbalance settings and resistivity values for valves and cylinder efficiency. When we start looking at calculations, there is a lot to be considered...
 
Posts: 11 | Registered: 04 August 2008Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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Stroupwrote:
quote:
Bud: From what I understand, this model was used as an educational purpose and although the numbers may not be exactly on, it displays the required task of conveying the type of system logic required to preform series cylinder operation.

Thanks for your feedback. Yes it is Section 22 on "Synchronizing Circuits" in the Fluid Power Circuits Explained book you have.

These circuits have never been critiqued by any of the persons I have sent them to. I know how much trouble it is to do a critique and have passed on several request's to do so for others.

During some of the Circuit Design classes the circuits were used I and the students found mistakes which were corrected before I made these Power Point presentations. Actually I have never used but a few of these presentations in a classroom setting. There never was time and often nothing but an Overhead Projector which required colored slides for viewing.

This book was used as a followup class for anyone who had been through a Basic class and wanted to learn more about Circuit Design. Since most of the circuits had been shown in the basic book I made an assignment each week for the students to design circuits for two or three sections of the book. At the next class I would collect their offerigs and pass out the sections to be covered and go over each different aspect of each design problem. I would grade the students circuits with notes so they could learn more.

Surprisingly I found some unique ideas from the untrained such as the Motor Flow Divider Regeneration Circuit from an Illinois Farmer/Maintenance person at one of my customers. Would you believe I told him I didnt thik it would work and he did'nt have the expertise to argue the point.

After sayig all that, Gauge PG2 reading 450 PSI on the Retract Stroke could be correct if someone had set the Counter Balance Valve 450 PSI above Load Balance. However, this is around 300 PSI higher than I would recommend in a typical circuit using Counter Balance to resist a Running Away load. That high a setting would be unnecessary and create extra heat.

BTW, did you notice the response by Nahum. I know I'm Math illiterate but I don't follow his response at all.


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
Darcy
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I did notice his equation listing and it does look foreign to me as well. I'm going to do some research on the equation name he listed before i post my results and reccomendations on the circuit at a later date. I'll also try to review the chapter you mentioned in the book.
 
Posts: 11 | Registered: 04 August 2008Reply With QuoteEdit or Delete MessageReport This Post
Bourdon
Picture of Bud T
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Could it be the Pythagorean theorem he is referrin to?


Bud Trinkel
FP Consultant Retired
"Fame is a vapor, popularity an accident, riches take wings. Only one thing
endures, and that is character." -- Horace Greeley
 
Posts: 1271 | Location: Newburgh, Indiana | Registered: 07 October 2005Reply With QuoteEdit or Delete MessageReport This Post
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