# Proportional valve control calculation. Part 1.

## Part 1. Proportional control of double acting hydraulic cylinder.

Fig.1. Double acting cylinder control.

The goal of this calculation is a definition of a cylinder velocity at any percent of the operating control signal at the valve’s solenoid (voltage or amperage).
For calculations we need to know: cylinder and valve parameters (take them from manufacturer catalog), values of forces at the cylinder rod (load), and pressure drop at the return line (can be assumed as zero if unknown).

Summary:
1.1. Hydraulic cylinder extension calculation.
1.1.1. Spool lands ratio.
1.1.2. Pressure drops at the spool lands
1.1.3. Balance of forces at the cylinder.
1.1.4. System pressure calculation.
1.1.5. Flow through spool lands: max. and at current voltage.
1.2. Hydraulic cylinder retraction calculation.
1.3. Full stroke time calculation.

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## 1.1. Hydraulic cylinder extension calculation.

### 1.1.1. Spool lands ratio.

For calculation we assume the pressure compensated pump with variable displacement is used as a source of pressure and flow. It means the constant pressure at port “P” of proportional valve and variable flow to keep the same pressure to compensate different contribution.

At the Fig.2. you can see simplified hydraulic schematic of proportional control the cylinder extension and its equivalent where proportional valve shown as two separate orifices:

Fig.2. System pressure description during cylinder extension.

Here:

$\triangle p_{PA}$ – pressure drop at the spool land P>A
$\triangle p_{BT}$ – pressure drop at the spool land B>T
$\triangle p_R$ – pressure drop at the return line
$p_A$ – pressure at the port A of proportional valve (at the cylinder blind side port)
$p_B$ – pressure at the port B of proportional valve (at the cylinder rod side port)
$p_T$ – pressure at the port T of proportional valve
$A_p$ – cylinder’s piston (bore ) area
$A_a$cylinder’s annular area

For future calculation let we assume $r_c$ as a cylinder area ratio:

$r_c=\frac{A_p}{A_p-A_r}=\frac{A_p}{A_a}$          (1)

The pressure drops at valve’s spool lands is different by two reasons:

1. There is different flow rate to cylinder blind port and from cylinder rod port trough cylinder work area difference;
2. The spool lands area could be different at proportional valve.

Let we a little bit discuss about valve’s spool lands area. Unlike a conventional solenoid valve, the proportional valve has a spool with notches cut into the edges:

Fig.3. Conventional and proportional valves spools

Depending upon the maximum flow to be controlled, different shape, size or number of spool notches can be applied. Because a flow to the cylinder blind port is bigger than flow from cylinder rod port, the spool with equal notches configuration will have unequal pressure drops across both lands. To prevent it manufacturers make different configurations of spool lands what helps to keep pressure drop on both lands fairly equal.

For example, to control the cylinder with area ratio 2:1 ($r_c=2$), a spool is machined with twice the number of notches on one side of the land as the other side, so spool lands ratio is $r_s=2$, or 2:1.

The most popular spool lands ratio for proportional valves are: 2:1, 1.6:1 and 1:1. The spool lands ratio shows how many times work area of the land from valve’s port A bigger than work area of the land from valve’s port B. This value you can select from the manufacturer catalog and add in valve order code (but sometimes it is a special option).

So, ideally, the spool lands work area ratio has to be the same like cylinder area ratio $r_c=r_s$ :

$r_s=\frac{A_{PA}}{A_{BT}}$          (2)

Here:
$A_{PA}$ – work area of the spool land P>A
$A_{BT}$ – work area of the spool land B>T

### 1.1.2. Pressure drops at the spool lands

All proportional valve spools can meter fluid in and out. Because of this orifice function, the equation for flow through an orifice applies:

$Q=c_d\cdot A_l\sqrt{\frac{2\cdot\triangle p}\rho}$

Here:
$c_d$ – orifice discharge coefficient
$A_l$ – spool land work area
$\triangle p$ – pressure drop at spool land
$\rho$ – fluid density

So, the flow flow at both spool lands will be:

$Q_{PA}=c_d\cdot A_{PA}\sqrt{\frac{2\cdot\triangle p_{PA}}\rho}$

$Q_{BT}=c_d\cdot A_{BT}\sqrt{\frac{2\cdot\triangle p_{BT}}\rho}$

The cylinder’s piston velocity is:

$v=\frac{Q_{PA}}{A_p}=\frac{Q_{BT}}{A_a}$

Therefore:

$r_c=\frac{A_p}{A_a}=\frac{Q_{PA}}{Q_{BT}}$

Now, using formulas (2) and (3) we can determine the relation between cylinder area ratio and spool lands work area ratio:

$r_c=\frac{A_p}{A_a}=\frac{Q_{PA}}{Q_{BT}}=\frac{c_d\cdot A_{PA}\sqrt{\frac{2\cdot\triangle p_{PA}}\rho}}{c_d\cdot A_{BT}\sqrt{\frac{2\cdot\triangle p_{BT}}\rho}}=\frac{A_{PA}\sqrt{\triangle p_{PA}}}{A_{BT}\sqrt{\triangle p_{BT}}}=r_s\frac{\sqrt{\triangle p_{PA}}}{\sqrt{\triangle p_{BT}}}$

$\Rightarrow r_c=r_s\frac{\sqrt{\triangle p_{PA}}}{\sqrt{\triangle p_{BT}}}$

$\Rightarrow \triangle p_{PA}=\triangle p_{BT}\left(\frac{r_c}{r_s}\right)^2$          (3)

Formula (3) will be used in the future calculations.

### 1.1.3. Balance of forces at the cylinder.

Let’s figure out all forces what are at the cylinder at the constant velocity, acceleration and deceleration of the piston.

At the left image the load pushes the rod, at the right image the load pulls the rod:

Fig.4. Forces at the cylinder during extension.

Here:

$F_{load}$ – Load force from mechanism or/and from tool weight.
$F_f$ – Friction force, what includes piston side breakaway force and friction in the mechanism.
$F_p$ – Force created by hydraulic pressure in the piston area (from port “A” of proportional valve): $F_p=p_A\cdot A_p$
$F_v$ – Force created by backpressure in the annular area (from port “B” of proportional valve, because it works as an orifice):$F_v=p_B\cdot A_a$

So, at the constant velocity, when external load pushes the rod, the force balance is:

$F_p=F_v+F_f+F_{load}$

and, when external load pulles the rod, the force balance is:

$F_p+F_{load}=F_v+F_f$

So, the generic formula for cylinder extension is:

$F_p=F_v+F_f\pm F_{load}$

where:
$F_{load}$ > 0 (positive value) when the load pushes the cylinder rod;
$F_{load}$ < 0 (negative value) when the load pulls the cylinder rod.

But, if the piston moves with acceleration or deceleration, in accordance with Newton’s Second Law:

$F_a=m\cdot a=F_p-(F_v+F_f\pm F_{load})$

or:

$F_p=F_v+F_f\pm F_{load}\pm m\cdot a$

where:
$F_a$ – is the acceleration/deceleration force;
$m$ – is cylinder’s pulled or pushed mass;
$a$ > 0 (positive value) – acceleration value;
$a$ < 0 (negative value) – deceleration value;
$a$ = 0 – for calculations of constant velocity.

To simplify future formulas I want to involve “Result external force” $F_{res}$:

$F_{res}=F_f\pm F_{load}\pm m\cdot a$

Now, just transfer this formula for future calculations:

$F_p=F_v+F_{res}$

$p_A\cdot A_p=p_B\cdot A_a+F_{res}$

$p_A=p_B\cdot \frac{A_a}{A_p}+\frac{F_{res}}{A_p}$

$p_A=\frac{p_B}{r_c}+\frac{F_{res}}{A_p}$

So, the generic force balance formula for cylinder extension is:

$p_A=\frac{p_B}{r_c}+\frac{F_{res}}{A_p}$          (4)

Formula (4) will be used in future calculations.

### 1.1.4. System pressure calculation.

With pressure compensated pump during the work operation the pressure equilibrium try to be established in the system all the time. The load creates a force at the cylinder, and the pressure in all points of the system depends from this force. If the force is not changed during lifting or lowering, the pressure in all points is not changed as well. We assume and accept it, because otherwise the calculations will be more complicated and should be considered specifically for each case separately.

Fig.5. System pressure during the cylinder extension.

Now we need to determine the pressure between all hydraulic components which connected in series circuit.
The pressure at the port “A” we found at formula (4), but in addition it can be determined by formula:

$p_A=p-\triangle p_{PA}$

The pressure at the port “B”:

$p_B=\triangle p_{BT}+p_T$

The pressure at the port “T”:

$p_T=\triangle p_R$

So, the next what we need to do is solving the system:

$\begin{cases} p_A=\frac{p_B}{r_c}+\frac{F_{res}}{A_p}\\ p_A=p-\triangle p_{PA}\\ p_B=\triangle p_{BT}+p_T\\ p_T=\triangle p_R \end{cases}$          (5)

Let’s do this:

$\frac{p_B}{r_c}+\frac{F_{res}}{A_p}=p-\triangle p_{PA}$

$\frac{\triangle p_{BT}+\triangle p_R}{r_c}+\frac{F_{res}}{A_p}=p-\triangle p_{PA}$

Next, use the formula (3) to replace the value of pressure drop at P>A land with the value of B>T land:

$\frac{\triangle p_{BT}}{r_c}+\frac{\triangle p_R}{r_c}+\frac{F_{res}}{A_p}=p-\triangle p_{BT}\left(\frac{r_c}{r_s}\right)^2$

$\triangle p_{BT}+\triangle p_R+r_c\cdot \frac{F_{res}}{A_p}=r_c\cdot p-r_c\cdot \triangle p_{BT}\left(\frac{r_c}{r_s}\right)^2$

$\triangle p_{BT}\cdot (1+r_c\cdot \left(\frac{r_c}{r_s}\right)^2)=r_c\cdot p-\triangle p_R-r_c\cdot \frac{F_{res}}{A_p}$

Next, use the formula (1) find pressure drop at the B>T land:

$\triangle p_{BT}\cdot (1+r_c\cdot \left(\frac{r_c}{r_s}\right)^2)=r_c\cdot p-\triangle p_R-r_c\cdot \frac{F_{res}}{A_a}$

$\Rightarrow\triangle p_{BT}=\frac{r_c\cdot (p-\frac{F_{res}}{A_a})-\triangle p_R}{1+r_c\cdot\left(\frac{r_c}{r_s}\right)^2}$          (6)

Put $\triangle p_{BT}$ to the formula (3) and find pressure drop at P>A land:

$\triangle p_{PA}=\left(\frac{r_c}{r_s}\right)^2\cdot\frac{r_c\cdot (p-\frac{F_{res}}{A_a})-\triangle p_R}{1+r_c\cdot\left(\frac{r_c}{r_s}\right)^2}$          (7)

Now we know values of pressure drop at both lands, so we can estimate both pressures at blind and rod ends of the cylinder from system (5):

$p_B=\frac{r_c\cdot (p-\frac{F_{res}}{A_a})-\triangle p_R}{1+r_c\cdot\left(\frac{r_c}{r_s}\right)^2}+\triangle p_R$

$p_A=p-\left(\frac{r_c}{r_s}\right)^2\cdot\frac{r_c\cdot (p-\frac{F_{res}}{A_a})-\triangle p_R}{1+r_c\cdot\left(\frac{r_c}{r_s}\right)^2}$

Good! So now we know pressure at all system points in all three modes: at acceleration, constant velocity and deceleration. The next step is definition of the flow through spool lands and the cylinder velocity.

### 1.1.5. Flow through spool lands: max. and at current voltage.

To help us to calculate the flow trough the valve spool lands (i.e. find the cylinder velocity) manufacturers in all proportional valves catalogs provide with two important values: rated flow $Q_N$ at rated pressure drop $\triangle p_R$. A 100% command signal (i.e. +10VDC = 100% valve opening) the valve flow at rated pressure drop $\triangle p_R$ per metering land is the rated flow $Q_N$. Please do not mix flow rate and pressure drop trough the spool and trough the land, when take values from the valve data sheet!

As we already know, the rated flow trough land is:

$Q_N=c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p_N}\rho}$

For other than rated pressure drop the valve flow changes at constant command signal according to the square root function for sharp edged orifices:

$Q=c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p}\rho}$

If the electrical signal is other than max voltage (10 VDC) we need to proportionally apply area correction:

$Q=c_d\cdot\frac y{y_{max}}\cdot A_{land}\sqrt{\frac{2\cdot\triangle p}\rho}$

here:

$y$ – operation voltage
$y_{max}$ – max. operation voltage (10 VDC)

The flow ratio can give us the relation between nominal and actual flow:

$\frac Q{Q_N}=\frac{c_d\cdot\frac y{y_{max}}\cdot A_{land}\sqrt{\frac{2\cdot\triangle p}\rho}}{c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p_N}\rho}}=\frac y{y_{max}}\sqrt{\frac{\triangle p}{\triangle p_N}}$

$\Rightarrow Q=Q_N\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p}{\triangle p_N}}$

Using this formula and value of pressure drop from formula (7), we can calculate the flow trough P>A land:

$Q_{PA}=Q_N\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p_{PA}}{\triangle p_N}}$

Now, the cylinder velocity is:

$v=\frac{Q_{PA}}{A_p}=\frac{Q_{BT}}{A_a}$          (8)

To find the flow trough B>T land, we need to correct spool area at these lands with the spool lands ratio:

$Q=c_d\cdot\frac y{y_{max}}\cdot \frac{A_{land}}{r_s}\cdot \sqrt{\frac{2\cdot\triangle p}\rho}$

$\frac Q{Q_N}=\frac{c_d\cdot\frac y{y_{max}}\cdot \frac{A_{land}}{r_s}\cdot\sqrt{\frac{2\cdot\triangle p}\rho}}{c_d\cdot A_{land}\sqrt{\frac{2\cdot\triangle p_N}\rho}}=\frac y{y_{max}}\cdot\frac 1{r_s}\cdot \sqrt{\frac{\triangle p}{\triangle p_N}}$

So, the flow trough B>T land could be find using formula (6):

$Q_{BT}=\frac{Q_N}{r_s}\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p_{BT}}{\triangle p_N}}$

As an option (or to double check calculations), you can find the flow trough B>T land by easy way:

$Q_{BT}=\frac{Q_{PA}}{r_c}$

The max flow trough spool lands be at $y=y_{max}$:

$Q_{PAmax}=Q_N\cdot\sqrt{\frac{\triangle p_{PA}}{\triangle p_N}}$

$Q_{BTmax}=\frac{Q_N}{r_s}\cdot\sqrt{\frac{\triangle p_{BT}}{\triangle p_N}}$

If you have an opposite task (find operation voltage for target flow), you can easy reconfigure the formula:

$y=Q_{PA}\cdot\frac{y_{max}}{Q_N}\cdot\sqrt{\frac{\triangle p_N}{\triangle p_{PA}}}=Q_{BT}\cdot\frac{y_{max}\cdot r_s}{Q_N}\cdot\sqrt{\frac{\triangle p_N}{\triangle p_{BT}}}$

## 1.2. Hydraulic cylinder retraction calculation.

The logic of  retraction calculations is the same like for cylinder extension calculations, so I will show it very quickly with minimum notes.

Balance of forces at the cylinder during retraction:

Fig.6. Forces at the cylinder during retraction.

Here:

$F_{load}$ – Load force from mechanism or/and from tool weight.
$F_f$ – Friction force, what includes piston side breakaway force and friction in the mechanism.
$F_p$ – Force created by hydraulic pressure in the annular area (from port “B” of proportional valve): $F_p=p_B\cdot A_a$
$F_v$ – Force created by backpressure in the piston area (from port “A” of proportional valve, because it works as an orifice):$F_v=p_A\cdot A_p$

The generic formula for cylinder retraction with constant velocity is:

$F_p=F_v+F_f\pm F_{load}$

where:
$F_{load}$ > 0 (positive value) when the load pulls the cylinder rod.
$F_{load}$ < 0 (negative value) when the load pushes the cylinder rod;

When the piston moves with acceleration or deceleration, in accordance with Newton’s Second Law:

$F_a=m\cdot a=F_p-(F_v+F_f\pm F_{load})$

or:

$F_p=F_v+F_f\pm F_{load}\pm m\cdot a$

where:
$F_a$ – is the acceleration/deceleration force;
$m$ – is cylinder’s pulled or pushed mass;
$a$ > 0 (positive value) – acceleration value;
$a$ < 0 (negative value) – deceleration value;
$a$ = 0 – for calculations of constant velocity.

To simplify future formulas I again involve “Result external force” $F_{res}$:

$F_{res}=F_f\pm F_{load}\pm m\cdot a$

Now, just transfer this formula for future calculations:

$F_p=F_v+F_{res}$

$p_B\cdot A_a=p_A\cdot A_p+F_{res}$

$p_B=p_A\cdot \frac{A_p}{A_a}+\frac{F_{res}}{A_a}$

$p_B=p_a\cdot r_c+\frac{F_{res}}{A_a}$

So, the generic force balance formula for cylinder retraction is:

$p_B=p_a\cdot r_c+\frac{F_{res}}{A_a}$          (9)

Next, the spool lands work area ratio:

$r_s=\frac{A_{AT}}{A_{PB}}$

Here:
$A_{AT}$ – work area of the spool land A>T
$A_{PB}$ – work area of the spool land P>B

The relation between cylinder area ratio and spool lands work area ratio:

$\triangle p_{AT}=\triangle p_{PB}\left(\frac{r_c}{r_s}\right)^2$

The simplified hydraulic schematic of proportional control the cylinder retraction and its equivalent where proportional valve shown as two separate orifices:

Fig.7. System pressure during the cylinder retraction.

Here:

$\triangle p_{PB}$ – pressure drop at the spool land P>B
$\triangle p_{AT}$ – pressure drop at the spool land A>T
$\triangle p_R$ – pressure drop at the return line
$p_A$ – pressure at the port A of proportional valve (at the cylinder blind side port)
$p_B$ – pressure at the port B of proportional valve (at the cylinder rod side port)
$p_T$ – pressure at the port T of proportional valve
$A_p$ – cylinder’s piston (bore) area
$A_a$cylinder’s annular area

The pressure in all system points, using formula (9):

$\begin{cases} p_B=p-\triangle p_{PB}\\ p_B=p_a\cdot r_c+\frac{F_{res}}{A_a}\\ p_A=\triangle p_{AT}+p_T\\ p_T=\triangle p_R \end{cases}$

Solving this system (the same way like at extension calculation) we get next values:

– pressure drop at P>B land:

$\triangle p_{PB}=\frac{p-\frac{F_{res}}{A_a}-r_c\cdot \triangle p_R}{r_c\cdot\left(\frac{r_c}{r_s}\right)^2+1}$

– pressure drop at A>T land:

$\triangle p_{AT}=\left(\frac{r_c}{r_s}\right)^2\cdot \frac{p-\frac{F_{res}}{A_a}-r_c\cdot \triangle p_R}{r_c\cdot\left(\frac{r_c}{r_s}\right)^2+1}$

– pressure the port “A”:

$p_A=\left(\frac{r_c}{r_s}\right)^2\cdot \frac{p-\frac{F_{res}}{A_a}-r_c\cdot \triangle p_R}{r_c\cdot\left(\frac{r_c}{r_s}\right)^2+1}+\triangle p_R$

– pressure the port “B”:

$p_B=p-\frac{p-\frac{F_{res}}{A_a}-r_c\cdot \triangle p_R}{r_c\cdot\left(\frac{r_c}{r_s}\right)^2+1}$

The next step – find flow through spool lands:

The flow trough P>B land:

$Q_{PB}=\frac{Q_N}{r_s}\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p_{PB}}{\triangle p_N}}$

The flow trough A>T land:

$Q_{AT}=Q_N\cdot\frac y{y_{max}}\cdot\sqrt{\frac{\triangle p_{AT}}{\triangle p_N}}$

The cylinder velocity is:

$v=\frac{Q_{AT}}{A_p}=\frac{Q_{PB}}{A_a}$          (10)

The max flow trough spool lands (at $y=y_{max}$):

$Q_{ATmax}=Q_N\cdot \sqrt{\frac{\triangle p_{AT}}{\triangle p_N}}$

$Q_{PBmax}=\frac{Q_N}{r_s}\sqrt{\frac{\triangle p_{PB}}{\triangle p_N}}$

The operation voltage for target flow:

$y=Q_{AT}\cdot\frac{y_{max}}{Q_N}\cdot\sqrt{\frac{\triangle p_N}{\triangle p_{AT}}}=Q_{PB}\cdot\frac{y_{max}\cdot r_s}{Q_N}\cdot\sqrt{\frac{\triangle p_N}{\triangle p_{PB}}}$

## 1.3. Full stroke time calculation.

Fig.8. Stroke time calculation

Time, what needs for cylinder piston acceleration from 0 to maximum velocity:

$t_{acc}=\frac{v_{max}}{a_{acc}}$

Time, what needs for cylinder piston deceleration from maximum velocity to 0:

$t_{dec}=\frac{v_{max}}{a_{dec}}$

Here $v_{max}$ can be defined using formulas (8) or (10) at $y=y_{max}$.

Cylinder’s stroke during the acceleration:

$s_{acc}=\frac{a_{acc}\cdot t_{acc}^2}{2}$

Cylinder’s stroke during the deceleration:

$s_{dec}=\frac{a_{dec}\cdot t_{dec}^2}{2}$

Cylinder’s stroke during the constant velocity:

$s_{const}=L_{cyl}-s_{acc}-s_{dec}$

here $L_{cyl}$ is full cylinder stroke.

Time, of cylinder piston during the constant velocity at extension:

$t_{const}=\frac{A_p\cdot s_{const}}{Q_{PA}}=\frac{A_a\cdot s_{const}}{Q_{BT}}$

Time, of cylinder piston during the constant velocity at retraction:

$t_{const}=\frac{A_a\cdot s_{const}}{Q_{PB}}=\frac{A_p\cdot s_{const}}{Q_{AT}}$

Full stroke time:

$t=t_{acc}+t_{const}+t_{dec}$

These calculations are only basis and can give you very approximate values but they are good enough for most applications.
If you have any questions or found a mistake, please let me know in the comments below.

1. shirdish patil

Hello ,
If we know the stroke and the Cycle time , how to find the max velocity ?? This is a more practical condition where in we know the stroke and time required to complete the stroke and then have to find out the acceleration ??
The average velocity can be defined as total distance traveled ( stroke ) divided by the time taken.

1. Dzyanis (Post author)

Hello Shirdish,
You are right about average velocity. But if the cylinder’s piston moves from 0 to max speed with acceleration, the max speed it can reach at the end of the stroke is: Vmax = 2*s/t where ‘s’ is stroke and ‘t’ is the time required to complete the stroke. (https://www.engineeringtoolbox.com/acceleration-velocity-d_1769.html)

1. shirdish patil

Hello Dzyanis,
You are right , but this applies only to a triangular profile and not the trapezoidal motion profile as shown in the fig 8 of your blog.
Moreover all our systems follow the profile you have explained in fig. 8 ( more correctly we want them to follow S type) ,where we have acceleration and then constant velocity and then deceleration.
In this case Vmax = S/(t-{(ta+td)/2}), where ta is acceleration time and td is deceleration time. I guess this is correct.
What is your opinion on this. Base on this Vmax then we can define the flow requirements.

1. Dzyanis (Post author)

Sorry Shirdish, I thought you asking just a generic question about cylinder’s piston acceleration.
If follow to Fig.8, the max velocity is definitely determined by max flow through the valve spool: Vmax=Qmax/A where Qmax is the max flow trough spool lands (taken from formulas above) and A is a cylinder annular or bore area (depends on cylinder retracting or extending). You can easily get max theoretical velocity using calculations above and, if you would like to proceed to my calculator by link: https://fluidpower.pro/cylinder-prop-control-calc/ you can find I calculated the value of Vmax.

2. Gee Kay

Mr. Patil,

Please study and note that a triangle and a rectangle are two extreme members of the trapezoid family. In the case of a motion profile. The triangular profile represents the only profile which can give the highest velocity for a given cycle while the rectangular profile represents the lowest velocity that can be achieved for the same cycle. Between these two extremes are infinite possibilities of trapezoidal profiles which is the only practical motion profile (along with its variants such as the sigmoid trapezoid) since for the case of the triangular profile there is zero dwell time (from acceleration to immediate deceleration) while for the rectangular profile there is infinite acceleration and deceleration. Both cases are physically unrealizable but important as they define the displacement as the area under the curves (which are equal for both profiles and whichever trapezoidal profile is selected as well). If you grasp this you have grasped the required reasoning behind constructing motion profiles. I must say Dzyanis has done a fantastic job deconstructing these formulae. Thanks a million!

1. HEH

still you have to check for maximum parameters like Vmax, acc ….. etc

2. HEH

if we still speak about trapezoidal profile :
Impose Tacc (=Tdec), calculate Aacc from forces then find Vmax
and then find Sacc ( = Sdec)….. then everything is solved

2. shirdish patil

Hello Dzyanis,
I have a feeling we should 1st calculate the vmax and then depending upon this calculate the Qmax = Vmax/A.
Once we know Qmax required for the application we can select the best valve.
The user will generally specify the Loads and the Cycle time, so going this way is obvious. What is your opinion.

1. Dzyanis (Post author)

Hello Shirdish,
Yes, you can easily reverse the calculations above, and start from the required cylinder velocity to estimate the flow. Next, select the valve and based on its parameters and re-calculate the system to get the real flow and real velocity. This way is correct from the theoretical aspect.
But my practice shows me the opposite way is more efficient. Usually, we select valves based on what our vendors can supply (brand, type, price, delivery time), and based on the cylinder and HPU parameters (for example, max flow can be provided by the pump and work pressure). So it is much easy to put cylinders and HPU parameters in the calculator and play with different valves to get the values of speed you to need.
And one more important thing. If you start to reverse calculations above to get the required valve parameters, at the end of the calculation you will need to determine many variables that will be required to select the valve. Problem is, each variable will affect other variables and in the end you can come to the valve with not realistic parameters that no one vendor can supply.
This is why much better preliminary select valves what vendors can supply and using existing HPU and cylinder data play with valve’s parameters from the manufacturer’s catalogue to select the best one.
Again, this just from my real practice. I share my ideas and my ways with everybody and it doesn’t mean only one way is correct.

3. Chris

Hi Dzyanis,

Thank you so much for sharing this information. I haven’t really been able to easily find it anywhere else.

I am pulling together my own spreadsheet for general cylinder calculations and used your formulas for blind and rod end pressures (pA and pB). For some reason, though, the numbers come out wrong. In this specific scenario:

p: 1000 psi
dpR: 100 psi
Bore: 3″
Rod: 1.5″ (Aa = 5.3in^2 and rc = 1.33)
Force: 3000 lbf compression
rs = 1

gives me:

extension blind end: 747 psi
extension rod end: 242 psi
retraction blind end: 719 psi
retraction rod end 575 psi

However, your online calculator appears to work fine. Could you double check the pressure equations listed on this page? Thanks a bunch!

1. Dzyanis (Post author)

Hello Chris,

Thanks for your notes. Unfortunately, you didn’t provide all the required data for calculation (for example, all required parameters of the proportional valve), so I can not check your numbers. Next, all formulas provided above I used to make my online calculator. Probably I could make a typo when added these formulas in the post, but a quick comparison with my paper notes didn’t find any mistakes… I used my online calculator in real applications and it shows the realistic values at the end. The Electrical Engineer in my team confirmed values of cylinder velocity vs voltage at the solenoid with a very small tolerance (theory vs practice). So, I’m pretty sure the formulas at least in the calculator are correct…

I will glad if somebody can go through the theory above and find the mistake or give me some advice on how to improve the article.

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