| As air is applied to the piston, the rod extends with a force
proportional to the piston area and the pressure of the air source.
e.g. A 2" bore (1" radius) operating at 50
psig would generate over 157 lbs. of force
The area is π
r2
or 3.1416 x 12 or
3.1416 in2
at 50 psig the force is 50 x 3.1416 =
157.08 lbs |
