Force of a pneumatic piston
Force of a pneumatic piston
(OP)
How much force would a 2.5 inch diameter 4 lb. piston traveling 6 inches pressurized by 150 psi. be hitting a stop? I know the force of a 2.5 inch piston with 150 psi is 736 lbs.
Thanks
Jay
Thanks
Jay





RE: Force of a pneumatic piston
Use impulse = momentum to calculate force acting over time to stop. The time to stop is the tricky variable to guess or want by design.
Ted
RE: Force of a pneumatic piston
Thanks
Jay
RE: Force of a pneumatic piston
W=integral(F*ds)
W=736.2lb*(6/12)ft=368.1 lb-ft
Ke=0.5*m*v^2
368.1lb-ft=(0.5)*(4/32.2)*v^2 => v=76.98 ft/s
736.2/(4/32.2)=5926.41 ft/s^2
s=0.5*a*t^2 => t=1.687*(10^-4) sec.
momentum. assuming linear force not varying with time
m*v=integral(F*dt)
(4/32.2)*(76.98)/(time to stop)=F
pick a time to stop as Ted said. Ex. pick half the time it takes to go from rest to 6in in distance give about 113.4 kips.
Note: this assumes a completely constant pressure behind the piston with absolutely no losses.
Also, the assumption about the constant linear force not varying with time is not really correct but the analysis cannot be concluded without making something of the force.
This is not the only way to do this, there are better more extensive methods that are not worth the time in a lot of cases, its just the only way that takes 3min. and can give you a feel for the problem you have.
ttyl,
Fe
RE: Force of a pneumatic piston
Keep this and how you might use this in the above simple analysis in mind. (to give you a hint, you can play with the dt by playing with the k)
Fe
RE: Force of a pneumatic piston
Not all impact energy is absorbed, some is returned in rebound.
Ted
RE: Force of a pneumatic piston
Thanks Again
Jay
RE: Force of a pneumatic piston
Fe