Mathcad - Hydraulic Cylinder

Hydraulic Cylinder Simulation

Simulating a hydraulic cylinder starts with the basic F=m*a and then is enhanced by doing a series of substitutions. The net force is equal to the force applied on the piston minus the opposing force required to push oil out the other side and the opposing force due to friction. Fnet = Fa − Fb − Fl − Ff Substitute for the for the force on both sides of the piston and the frictional and load forces. Assume the friction is proportional to the velocity. Ignore static friction for now and just model the kinetic friction

a=

Pa⋅ Aa − Pb ⋅ Ab − Fl − Ff

The acceleration that must be integrated to provide velocity and position.

mass

Since a is the same as

d v dt

Pa⋅ Aa − Pb⋅ Ab d v= mass dt

The acceleration or rate of change in velocity

β⋅Q d P= V dt

Expressed as a simple ODE

Since the moving piston changes the volume or ∆V on both sides of the piston the velocity of the piston must also be taken into account β d P = ( Q + v⋅ A) V dt

Calculate how pressure changes as a function of net flow Va and Vb are the volumes of oil between the valve or pump and the cylinder piston. The subroutines Qa(Pa) and Qb(Pb) are needed to calculate the net flow.

β d Pa = ( Q( Pa) − v⋅ Aa) Va dt

Calculate the rate of change in pressure as a function of flow and velocity.

β d Pb = ( Q( Pb) + v⋅ Ab) Vb dt

(c) 2010 Delta Computer Systems, Inc.

1/27

1/16/2013 10:10 AM


Simulate the Hydraulic Actuator Acceleration Pt := 14.7

Tank pressure psia

Ps0 := 2000

Initial system pressure, psia

CylDia := 2 RodDia := 1.375 Aa := Ab :=

π 4 π 4

2

⋅ CylDia

(

Aa = 3.142 2

)

2

⋅ CylDia − RodDia

Ab = 1.657

CylLen := 24 W := 600 mass :=

Weight in lbf W

mass = 1.554

386

Fl := W⋅ sin( 0 )

Slinch. A slinch is 12 slugs. 1 lb force accelerates a slinch at 1 in/s^2

Fl = 0 lbf ⋅ sec

Kf := 5

frictional velocity constant

in

Vamin := 5

Cubic inches of oil between the valve and the pump Try setting the volumes to 1000 and then 10000.

Vbmin := 5 + 0.25⋅ CylLen β := 160000

The bulk modulus of oil

Calculate the natural frequency of the cylinder and system ωn :=

2 2  Aa Ab  + mass  CylLen CylLen  Vbmin + Ab⋅  Vamin + Aa⋅ 2  2

β

⋅

ωn = 181.497

Natural frequency in radians per second

18

Calculate the acceleration time using the natural frequency method.

ωn

= 0.099


2/27

1/16/2013 10:10 AM


Pa0 :=

Pb0 :=

0.5⋅ Ps0⋅ Ab − Fl

Pa0 = 345.269

Aa + Ab 0.5⋅ Ps0⋅ Aa + Fl

The initial pressure on the A and B port should be 1/2 the supply pressure if the valve leakage from P->A and A->T is identical. The same applies to the B port.

Pb0 = 654.731

Aa + Ab

− 13

Pa0 ⋅ Aa − Pb0⋅ Ab = 2.274 × 10

Calculate the valve flow coefficient GPM := 27

Rated valve flow at rated pressure drop.

∆P := 1015

Rated pressure

GPM⋅ Kv :=

231 60

∆P

Calculate the flow constant for the powered land. Yields flow in cubic inches per second

Kv = 4.614

2


3/27

1/16/2013 10:10 AM


Valve Spool Response The valve spool response can be roughly simulated by a natural frequency and a damping factor. For the purposes of this model the spool's natural frequency and damping factor are assumed to be fixed but in reality the natural frequency drops as the amplitude increases. This may be due to the fact that is takes significantly more power to move at twice the frequency. ωs := 20⋅ 2 ⋅ π

ωs = 125.664

A 20 Hz valve

ζ s := 0.7

Spool damping factor. The closer to 1 the better. ωs

Ts( s) :=

2

Assume the spool is a second order system and possibly under damped. The Bode plots show this to be the case

2

2

s + 2 ⋅ ζ s⋅ ωs⋅ s + ωs n := 0 .. 320 n

hz := 10

Iterate the frequency from .01 HZ to 1000 HZ using 64 steps per decade for a total of 256 iterations.

−2

64

n

 Mn  := r ← Tsj ⋅ ( 2 ⋅ π ) ⋅ hz  n   φn    a←

Calculate magnitude and phase for T(s)

arg( r) deg

a ← a − 360 if n > 0 ∧ a − φ n− 1 > 100

 20⋅ log( r )    a  


4/27

1/16/2013 10:10 AM


Valve Spool Bode Plot 20

200

150

20 100 40 50

60

80 0.01

Phase deg

Magnitude db

0

0.1

1

10

100

03 1 .10

Frequency Hz Magnitude Phase The Bode plot looks something like the plots provided by the valve manufacturers. It must be acknowledge that this approximation is better than nothing but there is much more to modeling just the valve. For instance the valve has on board electronics and closed loop control which alter the response. The controller may even have a PID control but I best most have just a simple proportional controller. Still the magnitude is down -3db at 20 Hz. This means that at 20 Hz one must provide a 20% signal to get a 10% response from the valve. This works at low amplitudes but the controller can at most provide a 100% signal and get 50% amplitude from the valve 20Hz. There are two difference equations for the spool because it is a second order system. There is one for the velocity of the spool and one for the position of the spool d xs = vs dt

Calculate the spool position

(

)

2 d vs = −2 ⋅ ζ s⋅ ωs⋅ vs + ωs ⋅ u ( t ) − xs dt


Calculate the spool velocity

5/27

1/16/2013 10:10 AM


Verifying the valve spool response y :=

0   0

Initial spool position and velocity are 0

u ( t ) := 100

The control signal to the valve is a simple step signal of 90%

Time := 0.1

Simulation time

 xs  ←y  vs 

D( t , y) :=

vs     ω 2⋅ u ( t) − x − 2 ⋅ ζ ⋅ ω ⋅ v  s) s s s  s ( S := rkfixed( y , 0 , Time , 100 , D)

Spool Response to a Step Input 6000

4000 100

2000

50 0

0

0

0.02

0.04

0.06

0.08

Spool Velocity, percent per second

Spool Position, percent

150

2000 0.1

Time Spool Position Spool Velocity One can see that the model is close to 100% open in 25 milliseconds. The valve is actually open 101% at 25 milliseconds which is close to the valve documentation. The over shoot is caused because the spool is under damped as indicated by the damping factor being less than 1. In reality the spool will probably hit an end of travel stop instead of going over 100%.


6/27

1/16/2013 10:10 AM


Target Generator for Position, Velocity and Acceleration Generation motion profile using a cosine ramp target generator. This target generator will calculate a position, velocity and acceleration at time t. The motion profile assumes the ramp up, constant velocity and ramp down sections each take 1/3 of the total move time. All that is required is the move time and the distance to move.

Cosine Ramps MoveTime := 2 MaxVel :=

3 MoveDist ⋅ 2 MoveTime

MoveDist := 16

x0 := 4

MaxVel = 12

AvgAcc :=

9 2

⋅

MoveDist MoveTime

2

AvgAcc = 18

   π   sin 3 ⋅ ⋅ t    3 ∆x  1  ∆t   ⋅ ⋅  t − ⋅ ∆t⋅   + x0 π   4 ∆t  3   if 0 ≤ t < ∆t 3 ∆x  π (r t , ∆t , ∆x) :=   3 ⋅ ⋅ t  ⋅ ⋅ 1 − cos       3 4 ∆t   ∆t     9 ∆x    π  ⋅ ⋅ sin 3 ⋅ ⋅ t ⋅ π   4 2 ∆t   ∆t    3 ⋅ ∆x ⋅  t − 1 ⋅ ∆t + 1 ⋅ ∆x + x0  2 ∆t  3  4      ∆t ∆t ≤ t < 2⋅ 3 ∆x   if ⋅ 3 3   2 ∆t   0   2    t − ⋅ ∆t   1 ∆x   3  1  ∆x 3  4 ⋅ π ⋅ sin 3 ⋅ π ⋅ ∆t  + 4 ⋅  t + 3 ⋅ ∆t ⋅ ∆t + x0       2       t − ⋅ ∆t      if 2 ⋅ ∆t ≤ t < ∆t 3 ∆x  3 ⋅ ⋅ 1 + cos 3 ⋅ π ⋅       3 4 ∆t  ∆t      2    t − ⋅ ∆t     −9 ∆x 3 ⋅ ⋅ sin 3 ⋅ π ⋅  ⋅π   4 2 ∆t   ∆t    ∆x + x0   0  if ∆t ≤ t    0 


7/27

1/16/2013 10:10 AM


Linear Ramps  9 ⋅ ∆x ⋅ t2 + x0    4 2  ∆t   9  ∆x (r t , ∆t , ∆x) :=  2 ⋅ 2 ⋅ t  if 0 ≤ t < ∆t   3 ∆t   9 ∆x   ⋅ 2  2  ∆t    3 ⋅ ∆x ⋅  t − 1 ⋅ ∆t + 1 ⋅ ∆x + x0  3  4 2    ∆t   ∆t ∆t ≤ t < 2⋅ 3 ∆x   if ⋅ 3 3   2 ∆t   0    −5 9 t 9 t2    + ⋅ − ⋅  ⋅ ∆x + x0 2 ∆t 4 2  4  ∆t    9 ∆x   if 2 ⋅ ∆t ≤ t < ∆t ⋅ ⋅ ( ∆t − t ) 2  2  3 ∆t   9 ∆x   − ⋅   2 2 ∆t    ∆x + x0   0  if ∆t ≤ t    0  Step, No Ramps  ∆x  4  r( t , ∆t , ∆x) :=   0   0   


8/27

1/16/2013 10:10 AM


Define the valve control signal a function of time, u (t) The control signal can be ramped shut to simulate quick stops and check for pressure increases. ∆t := 0.001

t 0 := 0 t 1 := t 2 :=

Start ramping up

MoveTime

End of ramp up

3 2 ⋅ MoveTime

Start of ramp down

3 End of ramp down. Make the difference between t3

t 3 := MoveTime N :=

and t2 bigger to get a ramp that doesn't oscillate.

MoveTime + 1

N = 3000

∆t

Ending time and number of iterations

Open loop Control

u ( t ) :=







t − t0

   if t0 ≤ t < t1  t1 − t0  

0.5⋅  1 − cos π ⋅

The control signal as a function of time. The control signal has a range of -1 to 1 where -1 will move the spool to 100% P->B and A->T, 0 is leakage only and 1 will move the spool to 100% P->A and B->T.

1 if t 1 ≤ t < t 2







t − t2

   if t2 ≤ t < t3  t3 − t2  

0.5⋅  1 + cos π ⋅ 0 if t 3 ≤ t


9/27

1/16/2013 10:10 AM


Estimate Valve Leakage Calculate the leakage flow constant K l from the leakage specification The total leakage flow is the sum of the flow from P->A and P->B. Since the orifice is assumed to be symmetrical we can assume the flow from P->A is equal to the flow from P->B therefore the total leakage flow Ql is equal to 2 times the leakage from P->A, Q pa. 231 Ql := 0.25⋅ 60

Ql = 0.963

The specification for valve leakage for my NG10 is about .25 gpm. Convert to cubic inches per second.

The total valve leakage, Ql, is equal to the leakage from P->A plus the leakage from P->B. Assuming the leakage is symmetrical then the total leakage is 2 times the leakage from P->A.

Ql = 2 ⋅ Qpa = 2 ⋅ Kl ⋅

Ps

The leakage flow is 2 times the flow from A->B

2

Ql = Kl ⋅ 2 ⋅ Ps Kl :=

Combine some constants

Ql

Divide both sides by the square root term

2 ⋅ Ps0

Kl = 0.015

Leakage flow constant. This is very small relative to K v.

Select valve over/under lap type. The valve overlap is determined by x l. The spool position, x s, is in percent of full spool travel. x l, should be a small number in per cent overlap or under lap. A value of 0 signifies a 0 lapped spool whereas negative value specify under lapped spools and positive values signify overlapped spools. x .lt allows one to specify a different leakage coefficient A->T and B->T that differs from P->A and P->B. Spool over/under lap variable in percent. Set negative of under lap and positive for over lap spools. For the pressure port

xl := 0 xlt := xl


Spool over/under lap for the tank port. Normally this should be the same as the pressure port over/under lap.

10/27

1/16/2013 10:10 AM


Calculate the flow coefficients as a function of spool position The flow around each edge of the spool is modeled using a simple function. Basically there are two states. Either the edge is in a normal flowing condition or it is blocked and in a leakage state. This results in two linear segments that will determine the flow coefficient as a function of the position as it goes from -100% to +100%.

( )

Kpa xs :=

xs − xl

(Kv − Kl)⋅ 100 − x

l

(

)

+ Kl if xs − xl ≥ 0

 100 + xs  Kl ⋅   otherwise  100 + xl 

 −xs − xlt   + Kl if ( −xs − xlt) ≥ 0 lt   Calculate the flow coefficient from the A to T ports as a function of the spool  100 − xs  Kl ⋅  otherwise position   100 + xlt 

( )

( Kv − Kl) ⋅  100 − x

( )

( Kv − Kl) ⋅ 100 − x

Kat xs :=

Kpb xs :=

Calculate the flow coefficient from the P to A ports as a function of the spool position

−xs − xl l

+ Kl if xs + xl ≤ 0

Calculate the flow coefficient from the P to B ports as a function of the spool position

 100 − xs  Kl ⋅   otherwise  100 + xl 

( )

Kbt xs :=

 xs − xlt  Calculate the flow coefficient from the  + Kl if −xs + xlt ≤ 0 B to T ports as a function of the spool lt   position  100 + xs  Kl ⋅   otherwise  100 + xlt 

(Kv − Kl)⋅  100 − x

Note, the spools are assume to be cut so that the flow changes linearly with spool position. This is not always the case. There are 'curvilinear' and 'nick' spools. Modeling either one of these spools would require modifications to the four flow coefficient functions above. For instance, and 'nick' spool can be simulated by adding a third linear segment with a higher gain when the spool reaches 20% or 40% depending on the spool.


11/27

1/16/2013 10:10 AM


Sanity check the flow constant formula by graphing the flow coefficients as a function of the control signal. Note that the control signal goes from -1 to 1. x := −100 , −099.9 .. 100

Flow Coefficient P to A

Flow Coefficient P to B 6

Flow Coefficent

Flow Coefficient

6

4

2

0 100

50

0

50

2

0 100

100

50

0

50

Control Signal %

Control Signal %

Flow Coefficient A to T

Flow Coefficient B to T

4

2

0 100

100

6

Flow Coefficient

6

Flow Coefficient

4

50

0

50

100

Control Signal %


4

2

0 100

50

0

50

100

Control Signal %

12/27

1/16/2013 10:10 AM


Simulating Pump and Accumulators A real hydraulic system does not have a constant supply pressure. The supply pressure drops when a lot of flow is required. To offset this accumulators are used to supply energy. This allows the designer to use a pump that supplies just a little more than the average flow. During slow or dwell times the oil/energy is stored in accumulators and during times when the flow exceeds the pump capacity the accumulator can make up the difference. To start lets assume the pump is a fixed displacement pump of 60 gpm and that any excess oil flows over a relief valve at 2000 psi. Also, assume the accumulator capacity is 10 gal accumulator.

GPM := 5

GPM⋅

231 60

= 19.25

Pump capacity

PPB := 200

Pump proportional band

τp := 0.050

Pump time constant

AccumSize := 5

Accumulator size in gallons

Va := AccumSize⋅ 231

Va = 1155

Accumulator size in cubic inches

γ := 1.4

Ratio of specific heat Cp/Cv

Calculate the Supply Pressure and Accumulator Volume As A Function of Time. P0 := 0.9⋅ Ps0 K := P0 ⋅ Va

P0 = 1800

γ

Pre charge pressure Calculate K, nrT, using pre charge values calculated above for pressure calculations below.

7

K = 3.49 × 10

K   Ps 

ln

solve , Vg →e explicit

γ

K = Ps⋅ Vg

γ

 K    Ps0 

ln

Vg0 := e

Ps( Vg) :=

γ

K Vg

γ

Vg0 = 1071.268

Initial gas volume in cubic inches

Ps( Vg0) = 2000

Supply pressure as a function of the accumulator volume.


13/27

1/16/2013 10:10 AM


Total Flow Equations Qa and Qb are the net flow to either side of the cylinder piston. The total flow equations are important since they determine the rate of change in pressure,∆P=β*∆V/V or ∆P=β*Q*∆t. The current flow equation has four parts. The top equation calculates the flow from the supply to the A or B port and the second equation calculates the flow from the A or B port to the tank.The leakage terms are very important. Without the leakage terms the system will tend to oscillate for an unrealistically long time since there is little if any energy loss when the valve is shut. When accelerating there is plenty of energy loss due to the pressure drop across the valve. You can see there isn't any hint of oscillating because the damping factor is very high when the spools is wide open.

(

)

(

)

Qpa Pa , xs , Vg := Kpa xs⋅ 100 ⋅ sign ( Ps( Vg) − Pa) ⋅

(

)

(

)

Qat Pa , xs := −Kat xs⋅ 100 ⋅ sign ( Pa − Pt) ⋅

(

)

(

Pa − Pt

)

Qpb Pb , xs , Vg := Kpb xs⋅ 100 ⋅ sign ( Ps( Vg) − Pb) ⋅

(

)

(

)

Qbt Pb , xs := −Kbt xs⋅ 100 ⋅ sign( Pb − Pt) ⋅

(

)

(

)

(

)

(

)

(

Ps( Vg) − Pa

Ps( Vg) − Pb

Pb − Pt

)

Qa Pa , xs , Vg := Qpa Pa , xs , Vg + Qat Pa , xs

(

)

Qb Pb , xs , Vg := Qpb Pb , xs , Vg + Qbt Pb , xs

Force load as a function of time and distance The force can be changed as a function of time or distance. Fl( t , x) :=

2000⋅ 

t − 0.40⋅ MoveTime 

 if 0.40⋅ MoveTime ≤ t < 0.43⋅ MoveTime  0.03⋅ MoveTime 

2000 if 0.43⋅ MoveTime ≤ t < 0.47⋅ MoveTime 2000⋅ 

0.5⋅ MoveTime − t 

 if 0.47⋅ MoveTime ≤ t < 0.5⋅ MoveTime  0.03⋅ MoveTime 

0 ( x − 19) if x ≥ 19 0 otherwise


14/27

1/16/2013 10:10 AM


When extending against a load ρv := 1

The spool is symetrical

Ape := Aa

Ape = 3.142

FL := 0

FL = 0

ρc :=

Aa

ρc = 1.896

Ab

Ratio of the powered end area the exhausting end area

Compute maximum steady state velocity in each direction to estimate velocity feed forwards

vss_ext := Kv⋅

Ps0⋅ Ape − FL



Ape ⋅  1 + 3

 

2

ρv ρc

vss_ext = 61.342



3

Calculate the maximum velocity when the control output is one using the VCCM equation



Ape := Ab

FL := 0

ρc :=

No load when retracting

Ab

ρc = 0.527

Aa

vss_ret := Kv⋅

Ps0⋅ Ape − FL 2  ρv  3 Ape ⋅  1 +  3 ρc  


Ratio of the powered end area the exhausting end area

vss_ret = 44.545

15/27

1/16/2013 10:10 AM


Closed Loop Control Define the valve control signal a function of error between the target and actual position, velocity and acceleration. ∆t := 0.0001

Update period. The update period must be smaller as the system response get faster.

Controller gains. Ki := 0.5

Integrator gain

Kp := 0.1

Proportional gain

Kd := 0.001 Kv_ext :=

Kv_ret :=

Derivative gain 1

vss_ext 1 vss_ret

Kv_ext = 0.016

Velocity feed forward

Kv_ret = 0.022

Ka := 0.00018

Acceleration feed forward

Control output goes from -1 to 1 PID( u , xt , vt , at , x , v , a) :=

Kv ← Kv_ext if vt > 0 Kv ← Kv_ret if vt < 0 Kv ← 0 if vt = 0

 xt − x  u' ← ( Ki Kp Kd ) ⋅  vt − v     at − a  u ← u + u'⋅ ∆t + Kv⋅ vt + Ka⋅ at if u > 1 u←1 u' ←

1 − u − Kv⋅ vt − Ka⋅ at ∆t

if u < −1 u ← −1 u' ←

( −1 ) − u − Kv⋅ vt − Ka⋅ at ∆t

u    u' 


16/27

1/16/2013 10:10 AM


The Simulation Using Runge-Kutta. Execute the equations simultaneously by putting the equations and results in a matrix. There are now eight differential equations that are computed simultaneously. ActAcc( t , Pa , Pb , x , v , xt , vt) :=

 0   0   Pa0   Pb0    y :=  0   x0     0   Qp ← 0   Vg0 

( Pa⋅ Aa − Pb⋅ Ab) − Kf ⋅ v − Fl( t , x) mass

Calculate the actual acceleration

Initial spool position initial spool velocity Initial A port pressure Initial B port pressure Initial velocity Initial position Initial control output Initial accumulator gas volume


17/27

1/16/2013 10:10 AM


 xt  D( t , y) :=  vt  ← r( t , MoveTime , MoveDist)    at   xs     vs   Pa     Pb   v ←y    x u   Qp     Vg  a ← ActAcc( t , Pa , Pb , x , v , xt , vt)

 u  ← PID( u , xt , vt , at , x, v, a)    u'  vs     2 ωs ⋅ ( u − xs) − 2 ⋅ ζ s⋅ ωs⋅ vs     β ⋅ ( Qa( Pa , xs , Vg) − v⋅ Aa)   Vamin + x⋅ Aa     β ⋅ ( Qb( Pb , xs , Vg) + v⋅ Ab)    Vbmin + ( CylLen − x) ⋅ Ab    a   v     u' 1  Ps0 − Ps( Vg)   231  ⋅  max min , 1  , 0  ⋅ GPM⋅ − Qp  PPB 60     τp      Qpa( Pa , xs , Vg) + Qpb( Pb , xs , Vg) − Qp   SimTime := MoveTime + 1 N :=

SimTime ∆t

N = 30000

Simulation time Number of iterations

S := rkfixed( y , 0 , SimTime , N , D)


18/27

1/16/2013 10:10 AM


〈0〉 t := S

〈1〉 xs := S

〈3〉 Vg := S

〈3〉 Pa := S

〈4〉 Pb := S

〈5〉 v := S

〈6〉 x := S

〈7〉 u := S

〈8〉 Qp := S

〈9〉 Vg := S

n := 0 .. N

 xtn   vt  := r( n⋅ ∆t , MoveTime , MoveDist)  n  atn   

(n

a := ActAcc t , Pa , Pb , x , v , xt , vt n

n

n

n n

n

(

n

)

)

CO := PID u , xt , vt , at , x , v , a 0 n n n n n n n n


19/27

1/16/2013 10:10 AM


t

xs 0

0 1 2 3 4 5 6 7 8 9 10 11 12

S=

13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

vs 1

Pa 2

Pb

v

x

3

4

345.269 345.58

654.731 654.794

0 0

345.892

654.857

0

0 1.625·10-9 0 3.821·10-9

0

346.203

654.92

0

4-2.496·10-7 2.66·10-7

0

346.513

654.983

0

0.001 7.403·10-9 0.001 1.269·10-8

0

346.822

655.046

0.001

4-4.449·10-7 4.726·10-7 4 -6.97·10-7 7.379·10-7

0

347.131

655.109

0.001

0.001 1.998·10-8 0.001 2.958·10-8

0

347.438

655.173

0.001

0

347.744

655.236

0.002

0.001 4.176·10-8 0.001 5.681·10-8

0

348.049

655.3

0.002

0

348.352

655.364

0.003

0.001 7.498·10-8 0.001 9.652·10-8

0

348.653

655.429

0.003

0

348.952

655.494

0.004

0.001 1.217·10-7 0.001 1.507·10-7

0

349.25

655.559

0.005

0

349.545

655.625

0.005

0.002 1.837·10-7 0.002 2.211·10-7

0

349.837

655.691

0.006

0

350.128

655.757

0.007

0.002 2.629·10-7 0.002 3.094·10-7

0

350.415

655.824

0.008

0

350.7

655.892

0.009

0.002 3.607·10-7 0.002 4.17·10-7

0.001

350.982

655.96

0.01

4

-0

0

0.001

351.261

656.029

0.011

4

-0

0

0.002 4.785·10-7 0.002 5.453·10-7

0.001

351.536

656.098

0.012

4

-0

0

0.001

351.809

656.168

0.013

4

-0

0

0.002 6.176·10-7 0.002 6.955·10-7

0.001

352.078

656.239

0.014

4

-0

0

0.001

352.343

656.311

0.016

4

-0

0

0.003 7.792·10-7 0.003 8.686·10-7

0.001

352.604

656.383

0.017

4

-0

0

0.001

352.862

656.456

0.018

4

-0

0

0.003 9.64·10-7 0.003 1.066·10-6

0.001

353.116

656.53

0.02

4

-0

0

0.001

353.365

656.604

0.021

4

-0

0

20/27

6

Qp

0 0 0 06.118·10-11 1.83·10-6 04.855·10-10 7.243·10-6


5

u 7

8

4 0 0 4-2.758·10-8 2.96·10-8 4-1.106·10-7 1.183·10-7

4-1.006·10-6 1.062·10-6 4-1.372·10-6 1.444·10-6 4-1.796·10-6 1.885·10-6 4-2.278·10-6 2.384·10-6 4-2.818·10-6 2.941·10-6 4-3.416·10-6 3.557·10-6 4-4.072·10-6 4.23·10-6 4-4.786·10-6 4.961·10-6 4-5.559·10-6 5.749·10-6 4 -6.39·10-6 6.595·10-6 4-7.279·10-6 7.499·10-6 4-8.227·10-6 8.46·10-6 4-9.232·10-6 9.478·10-6

1/16/2013 10:10 AM


The first Spool Response graph shows the control signal, valve spool position and as a function of time.

Spool Response to Control Signal 15

0.2 10 0.15

5

0.1

0.05

Velocity in in/s

Conrol Signal and Spool Positon -1 to +1

0.25

0 0

0.05

0

0.5

1

1.5

2

2.5

3

5

Time Control Signal Spool Position Velocity


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1/16/2013 10:10 AM


Hydraulic Cylinder Response to Spool Position 25

0.25

Position in Inches

0.15 15 0.1 10 0.05 5

0

Spool Position -1 to 1

0.2

20

0

0

0.5

1

1.5

2

2.5

3

0.05

Time Actuator Position Spool Position


22/27

1/16/2013 10:10 AM


Velocity and Pressure Response 2500

2000

Velocity in in/sec

10 1500 5 1000 0 500

5

0

0.5

1

1.5

2

2.5

A, B and Supply Port Pressure, psi

15

0 3

Time Velocity Pressure Port A Pressure Port B Supply Pressure The load can induce a large pressure difference across the piston. The supply pressure is plotted to compare with the A and B port pressure. When the valve shut quickly pressure spikes are induced and these pressure spikes can cause the oil to flow back into the accumulator. This is why the code for flow must use sign ( Ps( Vg) − Pa) ⋅ Ps( Vg) − Pa when computing the flow. This avoids errors caused by taking the square root of a negative number.


23/27

1/16/2013 10:10 AM


Force Response 3000

Force, lbf

2000

1000

0

1000

0

0.5

1

1.5

2

2.5

3

Time Cylinder Force Friction Force Load Force Net Force

Accumulator Gas Volume and Accumulator Pressure

4.8 2000 4.75 1950 4.7 1900 4.65

4.6

0

0.5

1

1.5

2

2.5

Accumulator Gas Pressure, psi

Accumulator Gas Volume, gal

2050

1850 3

Time Gas Volume Presssure


24/27

1/16/2013 10:10 AM


A and B Port and Pump Flow and Pressures 2500

50

40

20

1500

10 1000

0

Pressures, psi

Cubic Inches per Second

2000 30

10 500 20

30

0

0.5

1

1.5

2

2.5

0 3

Time A Port Flow B Port Flow Pump Flow A Port Pressure B Port Pressure Supply Pressure The pump flow is not as great as the flow into the cylinders A port. This can happen because the extra oil comes from the accumulator.


25/27

1/16/2013 10:10 AM


Target and Actual Positions and Velocities 15

20 10 15 5 10 0 5

0

0

0.5

1

1.5

2

2.5

3

Target and Actual Velocity in in/sec

Target and Actual Position in inches

25

5

Time Target Position Actual Position Target Velocity Actual Velocity Control Output -10 to 10 Ki = 0.5

Integrator gain

Kp = 0.1

Proportional gain

Kd = 0.001

Derivative gain

Kv_ext = 0.016

Velocity feed forward extend

Kv_ret = 0.022

Velocity

Ka = 0.00018

Acceleration feed forward

∑

(r(n⋅∆t , MoveTime , MoveDist)0 − xn) N

2

= 0.002674

Normalized sum of squared errors. Adjust gains to minimize

n


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Mathcad - Hydraulic Cylinder

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