# Is 'using a smaller cylinder' really that easy??

Is the amount of force you get from a .75" cylinder proportional or differential when compared to a 1" cylinder??

I’m trying to make sure that, before we install smaller cylinders on our lifting mechanism, that i won’t have to play with the voltage and speed of our compressor to get the same push as the 1" cyl.

My belief is that the pressure buildup is differential with respect to the volume of a cylinder, not dirrectly proportional. Therefore assuming i convert the moles of air and that really freaky constant R (8.314LkPa/Kmol) and V = (4/3)pi.75^2, do i differentiate V or the equation:
P = nRT/V

Geez I’m confused. Any help would be great. Thanx!!!

The force of a cylinder is simply A*P, where A is the cross-sectional area of the cylinder and P is the gauge pressure of the system. For example:

.75" bore: (.75/2)^2pi60= ~26 pounds
1" bore: (1/2)^2pi60= ~47 pounds

Furthermore, if you check the Pneumatics Manual on FIRST’s website, you’ll find helpful charts that list the extended and retracted force of each diameter bore available to us at values between 20 and 60 psi.

Because the air is stored at 120psi, while the cylinders get a regulated suppy at up to 60, the input pressure to the cylinders will usually be constant while they’re extending, as long as the pressure of the air stored at 120 does not drop below 60. So you really don’t need to calculate rates of pressure change as the cylinder extends due to the changing volume, because the pressure won’t change. You do need to know, however, how many actuations you can get from the air at 120 before the stored air drops below 120. Calculate the volume of the tanks and cylinders you’ll use to store air, and separately the volume of the actuated cylinder. Use the ideal gas law to calculate the moles of air in the tanks at 120psi (8.16atm), and the moles used by the cylinder at 60psi (4.08atm). Then subtract the second molar quantity from the first to show the amount of air that was lost, and re-cacluate the pressure, again with the ideal gas law. All you’re interested in is how many actuations it will take for you to fall below 60psi. As far as re-charging with the compressor is concerned, I’d suggesting determining the re-charge time (from 60psi, or whatever) experimentally, because the rate of air output isn’t constant.

Thanx u all. You cleared up a huge headache for me.