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| SPG |
 Posted: 01/02/2008 9:58 AM Post subject: Hookes Law and Pressure - I'm confused, can you help. |
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2nd Lieutenant
Joined: 09 Feb 2006
Posts: 236
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The force in the compression spring is found from Hooke's Law,
F=k * (Lfree - Ldef)
where
k is spring constant
Lfree is the uncompressed length of the spring
Ldef is the compressed length of the spring
The force on a piston is found from
F = P * A
where
F is force
P is pressure
A is area
So combining these two we should be able to work out suitable springs for the whole series of JSR and SPG spring related autos. Only thing is that's as far as my physics gets me, I can't work it out on practical level.
Lets say I have a blow-forwards design. I need to have 1/2 inch of piston travel, and I want that to happen at 110psi. What spring should I pick?
And next problem, I know that coil springs aren't constant rate, but Ideally I want the force on the spring when uncompressed to be sufficient to shut a valve. and not have the whole thing hang half open in a JSR like way.
I'm thinking that it might be better to pre-load a spring, but again my physics lets me down.
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| jackssmirkingrevenge |
| Posted: 01/02/2008 10:20 AM Post subject: Re: Hookes Law and Pressure - I'm confused, can you help. |
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space monkey
Joined: 15 Mar 2007
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| SPG wrote: | | and not have the whole thing hang half open in a JSR like way. |
I resent that remark 
In my opinion from what I've done so far, no amount of playing with spring strength and travel is going to make those bolt mechanisms work with constant flow of air.
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| SPG |
| Posted: 01/02/2008 11:13 AM Post subject: |
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2nd Lieutenant
Joined: 09 Feb 2006
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No I'm sure it must work, what we're essentially doing here is building a modified pressure release or pop off valve. If you think about how they work they have very very limited travel over the safe pressure range, followed by long travel above this range to release.
And they use springs so someone's done the work already once, it's just replicating it by understanding the physics.
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| jackssmirkingrevenge |
| Posted: 01/02/2008 11:15 AM Post subject: |
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space monkey
Joined: 15 Mar 2007
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| I don't think pop-off valves (as opposed to simple pressure relief valves) are as simple as that, maybe some kind sould can take one apart? My search for diagrams online has been largely inconclusive.
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| Skywalker |
| Posted: 01/02/2008 12:21 PM Post subject: |
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Sergeant First Class
Joined: 05 Aug 2007
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I think you are right JSR, they have some kind of setup so that they only close after a certain amount of pressure drop has occurred. I think Hawkeye may have suggested this, but it seems like you'd need a stepped piston and a stepped tube to go with it, so that the exhaust comes out thru a hole in the larger diameter tube. That way, the thing will only pop open when the pressure pushes the small plug out, but the large area of the big plug will keep the valve open till the pressure drops a good bit.
What we need is the difference between a pop-off and a regulator. Without some kind of step like that, we just have a regulator.
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| POLAND_SPUD |
| Posted: 01/02/2008 14:15 PM Post subject: |
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Brigadier General
Joined: 13 Oct 2007
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yo might just buy the valve and use it in your design JSR.... but I assume you want to build it youreslf becase otherwise it would be too easy right.... 
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| SpudMonster |
| Posted: 01/02/2008 21:25 PM Post subject: |
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| Pilgrimman |
| Posted: 01/02/2008 23:55 PM Post subject: |
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Major
Joined: 28 Jun 2007
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If I read your post right, here's my 2 cents.
If you can find the force, which you already know, you just multiply that by 2 to get the spring constant, "k". Then get a spring with a slightly lower "k" than what you calculated, and have the spring compress the piston enough to close the valve, but still have a force difference between the air and spring compression that is high.
Example:
If you have a piston with 1" of surface area for the air to push on (to open), and you want 1/2" of piston travel at 110 psi, your spring constant, "k", will be 220 lb./in. Equations: (110 lbs./in^2 * 1 in^2. = 110 lbs.) Then you take the force, 110 lbs., and divide by the change in distance, in this case, 1/2 in. , and you get a spring constant of 220 lb./in. (110 lbs. / 1/2 in. = 220 lb./in.)
So then you find a spring with slightly less than that constant, probably on the order of 210 - 215 lb./in. That way, you can put the spring in your gun in such a way that when the system is unpressurized, there is 5 - 10 lbs. of force acting on the piston to close it. When you've got 110 lbs. of force due to the air trying to open the piston at firing time, and only 5 - 10 lbs. of spring force acting the opposite way, you still get a force that will push the piston very close to 1/2 in. If you factor in the mass of the piston, which should ideally be low, you probably get around a 1/2 in. of travel.
It should be noted that mass and friction were assumed to be negligible.
Give us your piston's diameter, and the outer diameter of your barrel, and one of us could crunch the numbers for you! 
I'm going to assume that your piston is 2 in. in diameter, if you are using the barrel / 4 piston travel principle.
Am I right?
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| jackssmirkingrevenge |
| Posted: 01/03/2008 0:10 AM Post subject: |
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space monkey
Joined: 15 Mar 2007
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| SpudMonster wrote: | http://www.spiraxsarco.com/images/resources/steam-engineering-tutoria ls/9/1/fig9_1_8.gif
There is a cutaway of an adjustable safety valve. The adjustable ring controls the blow-down pressure, depending on how far it is screwed out. |
The critical element seems to be the greater piston diameter (in blue) that is exposed to the air after the piston starts to move, because that gives the air much more space to act that the previous red area - basically as suggested by Skywalker:
| Quote: | | they have some kind of setup so that they only close after a certain amount of pressure drop has occurred. I think Hawkeye may have suggested this, but it seems like you'd need a stepped piston and a stepped tube to go with it, so that the exhaust comes out thru a hole in the larger diameter tube. That way, the thing will only pop open when the pressure pushes the small plug out, but the large area of the big plug will keep the valve open till the pressure drops a good bit. |
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| SPG |
| Posted: 01/03/2008 6:25 AM Post subject: |
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2nd Lieutenant
Joined: 09 Feb 2006
Posts: 236
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I think I might have solved the dilemma it's on the auto v4.0 thread.
And piston size? about 1/4 of an inch square tops.
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| Pilgrimman |
| Posted: 01/03/2008 15:14 PM Post subject: |
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Major
Joined: 28 Jun 2007
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If the piston has 1/4 " of it being acted upon by air, you want a spring with a "k" of about 54 lb./in.
1/4 in.^2 * 110 ~= 27 lbs.
27 lbs. / 1/2" = 54 lb./in.
Hope I helped!
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