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mjt902
01-28-2003, 08:38 PM
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!!!

rbayer
01-28-2003, 09:14 PM
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)^2*pi*60= ~26 pounds
1" bore: (1/2)^2*pi*60= ~47 pounds