Help figuring out hammer valve calculations

[JDP12]

I'm tryin to design a hammer valve as BTB drew here...
http://www.spudfiles.com/forums/files/r ... ed_277.png

but im havin problems with the spring calculations. If the part of the red hammer that the spring pushes against is .5" diameter, and I have 800psi in the chamber, that means the air pushes against that with 157 pound-force..

and the spring would have to exceed that in order to keep it shut... which is a ridiculous spring. I don't want its relaxed length to be more than 1" ideally, so how can I do it? Are my calculations correct?

I was thinkin about making that part stepped, but it'll have to be insane difference..

Suggestions?

[jackssmirkingrevenge]

[JDP12]

while funny... not sure how momentum comes into play here, as the red part is at rest... I'm worried about it staying sealed while at rest.

What am I missing?

[ramses]

Air pressure serves to force the valve closed; it pushes the piston against the stop/seal, much like in a normal piston valve. The volume of high pressures air at the far right exerts force on a slightly smaller area, such that the force required to open the valve is relatively small (say, 20 lbs). The spring actually helps force the valve closed when there is no pressure in the valve( to make sure it seals at the start)

If the two areas of high pressure act on the same area, the only force required to open the valve is that required to overcome the spring and piston. In this configuration, the valve will actually become harder to hold open momentarily after the valve is opened, since the pressure at the far right would then be greater than the pressure immediately next to the outlet of the valve. The spring would also then help ensure the valve reseals between shots.

[jackssmirkingrevenge]

Whoops, sorry, thought you were talking about hammer springs.

Ramses gave a more useful answer, if the red parts are of equal diameter, then the spring only needs to overcome the friction of the piston movement as the net pressure force is zero.

[JDP12]

so say the two red parts are equal diameter... wouldnt the force on the far right still be greater since the far left is a cone, so the force isn't acting directly straight? Trying to remember stuff from classes last semester....

EDIT: Although I suppose the far right piece will be a smaller diameter if the barrel vent hole and the far right hole are equal diameter, since then the far left piece is going to be bigger than the hole.

Makes sense.

[jackssmirkingrevenge]

so say the two red parts are equal diameter... wouldnt the force on the far right still be greater since the far left is a cone, so the force isn't acting directly straight? Trying to remember stuff from classes last semester...

Noooooooooooooo!

Vectors JDP, vectors!!!!

[JDP12]

oh shit. right. *facepalm*.... im an idiot... i was thinking of the x and y components, but there's still the total force acting on it.

I'm an idiot. My brain is mush from so much homework these days.... Oh aeronautical engineering.

[jackssmirkingrevenge]

Oh aeronautical engineering.

Made me feel a little less safe about flying there

[JDP12]

Feel even less safe... Im in the Air Force ROTC program.. when I graduate hopefully I'll be flying fighter jets

For you foreigners- ROTC is a program college kids can be in in which you receive military training while at college, when you graduate you commission as an officer.

[warhead052]

Feel even less safe... Im in the Air Force ROTC program.. when I graduate hopefully I'll be flying fighter jets

For you foreigners- ROTC is a program college kids can be in in which you receive military training while at college, when you graduate you commission as an officer.

Good luck! You'll need it.

But as for the valve, it should work that way. As JSR and Ramses said, you only need to over come the opening force of the valve.