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Linear with Piston

A place to ask general spud cannon related questions.
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Is this a good idea?

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Linear with Piston

Unread postAuthor: -T0OL- » Sun Oct 08, 2006 7:58 pm

First off im going to say i have no idea if anyone has tried this before or not or thought of it... This idea came to me as I was thinking about piston cannons (coaxial, barrel sealing, ... ) and while reading a whole bunch of posts saying "dont do that... all the turns slow down the air..." so here is my idea:

pictures first words later.

--click to make larger--
Image

my idea was to make a piston cannon linear so that the air has less corners to go around. All the piston cannons i have seen so far need the air to do at a 180* turn (two 90* turns).

now the picture is color coded with numbers for the colors :D so i will explain what each is.

1. This is a wall in that blocks the pilot area. It also holds the air input and output.

2. tihs is a tube that goes along the inside of the chamber. Also used for the piston to slide on. This pipe will also block off the equilizer holes when the piston slides.

3. This is the air in the chamber.

4. This is where the air comes in

5. This is where the air goes out (trigger)

6. This is the pilot area.

7. This is the piston, it goes on the outside of #2 and slides along it... it also goes into the reducer to block off air flow when pressurizing the chamber. It also has the equilizing holes in it.

8. This is the casing for the chamber, this is also the reducer.

9. This is the barrel. (there is also something in it to stop ammo from rolling into chamber (JB weld, hot glue, screw, ... )

10. This is the ammo ( be creative )

11. These two lines are rubber bands***

12. hooks for the rubber band to connect to.

13. equilizing holes (NOTE: the holes are such that the piston closes off the barrel before they are exposed)

*** Rubber Bands: Im not too sure if these are needed... when the pilot area is depressurized im not too sure if that will be enough to move the piston all the way back... i think it should be but not too sure. If not these bands would only be strong enough to help pull the piston back.

Here is a .gif i made to illustrate it.( what does everyone here use for the gif/flash animations?)

--click to make larger--
Image

so there is my idea. Any ideas suggestions? if you think it sucks or something like that and you feel and urge to say so, say something constructive to improve the design also.
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Unread postAuthor: boilingleadbath » Mon Oct 09, 2006 2:13 pm

It's not workable as it is, as the piston has no area to push it rearward, and some to push it forwards. This will require rather stout rubber bands.

However, if one where to make the piston fit <i>inside</i> the piston-tube, and have this piston tube be a larger diameter than the barrel, you have a workable design... no spring needed, even.

However, using 1.7" pipe (1.5" SDR 21) as the piston tube and 1.59" pipe (1.5" sch 40) as the barrel, the pressure in the pilot area needs to drop to about 5-6 psi for the valve to trigger, when the chamber is filled to 100 PSI. (not happening if you don't have a checkvalve... or an insanely large vent)

Using 2" sch 40 and 1.5" sch 40, it'll trigger at about 50 psi, given the same fill pressure... so that's my recomendation if you acctualy go to bulid the thing.
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Unread postAuthor: Velocity » Mon Oct 09, 2006 3:01 pm

Yeah, I think boiling lead bath is right. You would need the guide tube to be on the outside of the piston.

And anyway, why are you firing during an earthquake :P ? JK, thats just what the shaking reminded me of...
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Unread postAuthor: -T0OL- » Mon Oct 09, 2006 5:02 pm

how do you find out stuff like how much PSI needs to be released from the pilot for the piston to move and such info?
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Unread postAuthor: boilingleadbath » Mon Oct 09, 2006 9:27 pm

Free body diagrams, foo!
(add 3-5 pounds friction, per O-ring)

In this specific case?
Force = (3.14/4) * ((1.70^2 * pilot pressure) - ((1.70^2 - 1.59^2) * 100. psi))

Set force = -10 pounds (O-ring friction)

-10 = (3.14/4) * ((1.7^2 * P) - 36.2)
-10 = (3.15/4) * (2.89*P - 36.2)
-12.7 = 2.89P - 36.2
23.5 = 2.89P
Pressure = 8.13 PSI

(My initial estimate wasn't far off, considering it's mostly-mental nature...)
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Unread postAuthor: -T0OL- » Tue Oct 10, 2006 11:19 am

ok so

a= Piston Diameter
b= Pilot Pressure
c= Barrel Diameter
d= Psi

F= (PI/4) x ((a^2 x b) - ((a^2 - c^2) * d))

or does this simplify more? (and what are the proper variables). Oh and does this formula have a name?
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