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I'm trying to calculate piston force pressures for double acting pistons for pneumatic and hydraulic pressures. I found the following formula for calculating both

$$\text{Force} = \frac{\text{Pressure} \times \pi \times (\phi_\text{Bore}^2 - \phi_\text{Rod}^2)}{4}$$

It seems to me that there should be some kind of conversion factor when going from pneumatic to hydraulic or vice versa. Is this formula valid for both pneumatic and hydraulic?

I got the formulas from the following sites:

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The formula is valid independant of the fluid.

The force in an hydraulic system is based on pressure and area - a height difference between pistons is usually small compared to the acting pressure from the pump - this is why is is usually ignored. Think of the pressure created by the difference in height of the pistons in tipper truck systems for example compared to the system pressure of 30 Bars or 700 Bars.

If you are looking at the pressure created by a column of fluid of 30m height then that will be affected by the different fluids (p = density * gravity * height).

Once you come to calculate the motor power necessary then the fluid becomes relevant.

Edit aded in response to question below:

The formula is valid independant of the fluid ie Pressure = Force / Area, so a fluid (air, water, hydraulic oil) at a given pressure acting on the surface of a piston creates a force proportional to the Area.

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  • $\begingroup$ Is the formula valid for calculating force in a pneumatic system as well? $\endgroup$ – Brian Kalski Aug 9 '17 at 13:51
  • $\begingroup$ The formula is valid independant of the fluid ie Pressure = Force / Area, so a fluid (air, water, hydraulic oil) at a given pressure acting on the surface of a piston creates a force proportional to the Area. $\endgroup$ – Solar Mike Aug 9 '17 at 14:54
  • $\begingroup$ The formula doesn't care what the fluid is, but the pressure will probably be dramatically different in the two cases. Compressed air pressures are typically one or two hundred psi, while hydraulic pressures are typically ten times higher. $\endgroup$ – Mark Aug 9 '17 at 22:16

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