Random Question

[ShowNoMercy]

I was wondering what kind of force would be placed on a sear holding a 1 inch piston back from say 80 psi? I know I should know how to figure this out but I am drawing a blank here. So any help would be appreciated.

[jackssmirkingrevenge]

surface area of piston x pressure = (0.5" x 0.5" x 22/7) x 80 = 62.84 lb

[ShowNoMercy]

Thanks a lot, regular cold rolled steel should have no problem with that riight?

[jackssmirkingrevenge]

it should be ok, if in doubt you could always hang 62.84lb of weights from it to simulate the strain caused by the pressure.

[ShowNoMercy]

I think I should be good, Im planning on making something similar to the HEAR valve.

[Hawkeye]

You can always anneal a section of a metal file and shape the sear from it and then re-temper it. Very easy to do and you don't have to get it perfect. It will wear better as well.

[VH_man]

shouldnt the equation for the force be

(Pi X Diamter of the piston) X pressure of the cylinder?

that would be 3.14in2 X 100 PSI, wich would mean : approximately 314 pounds of force.

314 is alot of force, but ive seen pneumatic cylinders with a 1 inch bore that have bent 1/4 inch steel.......

[jackssmirkingrevenge]

shouldnt the equation for the force be

(Pi X Diamter of the piston) X pressure of the cylinder?

No.

Force = area of piston x pressure acting on piston = pi x radius squared x pressure

[VH_man]

ah. that makes sense now......

that helps me alot... wow now i have to re-examine the actual strentgh of my robotics project.

pneumatic rams are powerufl, but apparently ive overestimated their strenght.... dang nabit......